Matrix rep of Lorentz transformation in Dirac spinor space

[iuval]

I'd like to have the explicit form of the 4x4 matrix representing a
Lorentz transformation of a Dirac Spinor, with parameters: theta1,
theta2, theta3, phi1, phi2, phi3, where the thetas are rotation angles
about x, y, or z axes, and the phis are rapidities in those
directions. I know how to get this for the standard 4-vector (1/2,1/2)
rep: Find the generators of the Lorentz algebra by differentiating the
standard matrices corresponding to a rotation (by angle theta) or a
boost (by rapidity phi)at 0 rotation or rapidity (J1, J2, J3, K1, K2,
K3). Then exponentiate: L(theta1,...,phi3)=Exp[theta1 J1+...phi3 K3].
So the same thing should work for the Dirac spinor rep. But what do
the generators look like in that rep? According to my understanding of
wikipedia they are just (J_i+I K_i)/2 and (J_i-I K_i)/2 with J_i being
I/2*Pauli matrices, but I don't know what the K_is are. And what are
the corresponding parameters to the above group element that must
multiply these generators in the exponential? Do the (Ji+I Ki)/2 apply
to the "top" part of the bispinor, and the (J_i-I K_i)/2 to the bottom
(with each block a 2x2 matrix) so that the matrix looks like (with
implied summation over i):
Exp[(theta_i-I phi_i) (J_i+I K_i)/2] 0
0 Exp[(theta_i+I
phi_i) (J_i-I K_i)/2] or is this only correct in the Weyl
representation of the spinor?
I think this is wrong. Since each (J_i+I K_i)/2 and (J_i-I K_i)/2 is
now a generator of SU(2).

Should the matrix then look like:
Exp[(theta_i-I phi_i) (J_i) 0
0 Exp[(theta_i+I phi_i) (J_i)

And again, is this only true in the Weyl rep?

I am trying to understand:
1. if Dirac spinors can be generalized to more than one particle/
antiparticle pair
2. if the spin connection which arises by making the Dirac Lagrangian
locally Lorentz invariant has a physical significance--is there a
Lagrangian just for the free "gauge" field and what are its Euler-
Lagrange equations? Perhaps this is a way to derive the Einstein field
equations?
3. Is there a way to understand the Dirac spinor in purely geometric
(classical) form? The fact that it satisfies a classical equation
obtained from minimizing a classical action suggests so. Is it just a
mathematical coincidence that this can be done?

[Norbert Dragon]

* iuval writes:

> I'd like to have the explicit form of the 4x4 matrix representing a
> Lorentz transformation of a Dirac Spinor, with parameters: theta1,
> theta2, theta3, phi1, phi2, phi3, where the thetas are rotation angles
> about x, y, or z axes, and the phis are rapidities in those
> directions.

Eq. 6.50 and 6.68

http://www.itp.uni-hannover.de/~dragon/rel_e.pdf

give the explicit 2x2 matrices of a Weyl representation in case of either
a rotation or a boost. But one can easily use the arguments given there,
to sum the exponential series for a general Lorentz transformation.

> I know how to get this for the standard 4-vector (1/2,1/2)
> rep: Find the generators of the Lorentz algebra by differentiating the
> standard matrices corresponding to a rotation (by angle theta) or a
> boost (by rapidity phi)at 0 rotation or rapidity (J1, J2, J3, K1, K2,
> K3). Then exponentiate: L(theta1,...,phi3)=Exp[theta1 J1+...phi3 K3].

In case of spin 1/2 this exponentiation can be summed explicitly.

> So the same thing should work for the Dirac spinor rep.

A Dirac representation is a sum of a Weyl representation and its
complex conjugate.

> But what do the generators look like in that rep?

The Lie algebra is spanned by complex linear combinations of the
Pauli matrices sigma_1, sigma_2 and sigma_3. They constitute a real six
dimensional Lie algebra.

> According to my understanding of
> wikipedia they are just (J_i+I K_i)/2 and (J_i-I K_i)/2 with J_i being
> I/2*Pauli matrices, but I don't know what the K_is are. And what are
> the corresponding parameters to the above group element that must
> multiply these generators in the exponential? Do the (Ji+I Ki)/2 apply
> to the "top" part of the bispinor, and the (J_i-I K_i)/2 to the bottom
> (with each block a 2x2 matrix) so that the matrix looks like (with
> implied summation over i):
> Exp[(theta_i-I phi_i) (J_i+I K_i)/2] 0
> 0 Exp[(theta_i+I
> phi_i) (J_i-I K_i)/2] or is this only correct in the Weyl
> representation of the spinor?
> I think this is wrong. Since each (J_i+I K_i)/2 and (J_i-I K_i)/2 is
> now a generator of SU(2).

Nothing wrong with this.

> Should the matrix then look like:

> Exp[(theta_i-I phi_i) (J_i) 0
> 0 Exp[(theta_i+I phi_i) (J_i)

It looks like this with J_i = sigma_i / 2.

Note that a Dirac field is a map from spacetime to the space of spinors.
It does not transform only by multiplication with a representation
matrix S(Lambda) but also its space time argument x transforms

psi' = S psi Lambda^(-1) or in other words

psi'(Lambda x) = S psi(x) .

> And again, is this only true in the Weyl rep?

Each Dirac representation looks like this in a suitable basis.

> I am trying to understand:
> 1. if Dirac spinors can be generalized to more than one particle/
> antiparticle pair

This just means that you deal with several Dirac spinors. Usually one
calls the additional labels which enumerate them flavor or isospin.

> 2. if the spin connection which arises by making the Dirac Lagrangian
> locally Lorentz invariant has a physical significance

It is the gravitational field, the Vierbein or the metric.

> --is there a Lagrangian just for the free "gauge" field

Einstein gave the equations of motion 1915, well before Dirac spinors
became known.

> and what are its Euler- Lagrange equations?
> Perhaps this is a way to derive the Einstein field equations?

Your suggestion is late. If you try to introduce a gauge field for
Lorentz transformations, Einstein gravity is essentially unique
(if the physical degrees of freedom are required to have positive
energy density). "Essentially" means: one is free to allow for a
cosmological constant or for Brans-Dicke like scalars.

> 3. Is there a way to understand the Dirac spinor in purely geometric
> (classical) form?

It is essential, that a spin 1/2 field it is not a bosonic
field but anticommutes rather than to commute. Up to now I have not
seen a mathematical definition of such a Gra?mann field leave alone a
geometric interpretation.

The quantized free Dirac field is much simpler: it creates and
annihilates particles in a Fock space. But we need Gra?mann fields
as arguments of time ordered products (the fermionic arguments
of time ordered products anticommute).

But where do Gra?mann fields take their values? Does this target
space have a norm, such that differentiability can be defined?
As far. as I see, Gra?mann fields take values in something like a
universal graded commutative algebra. Differentiation is defined
formally and all Gra?mann fields which occur in Lagrangians and in
time ordered products are smoothly differentiable by assumption though
nothing allows to check it.

--
Superstition brings bad luck.

www.itp.uni-hannover.de/~dragon

[iuval]

On Tuesday, April 3, 2012 12:58:00 PM UTC-4, Norbert Dragon wrote:
> * iuval writes:
>
> > I'd like to have the explicit form of the 4x4 matrix representing a
> > Lorentz transformation of a Dirac Spinor, with parameters: theta1,
> > theta2, theta3, phi1, phi2, phi3, where the thetas are rotation angles
> > about x, y, or z axes, and the phis are rapidities in those
> > directions.
>
> Eq. 6.50 and 6.68
>
> http://www.itp.uni-hannover.de/~dragon/rel_e.pdf
>
> give the explicit 2x2 matrices of a Weyl representation in case of either
> a rotation or a boost. But one can easily use the arguments given there,
> to sum the exponential series for a general Lorentz transformation.

As I said, I need the 4x4 matrices, not 2x2.

>
> > I know how to get this for the standard 4-vector (1/2,1/2)
> > rep: Find the generators of the Lorentz algebra by differentiating the
> > standard matrices corresponding to a rotation (by angle theta) or a
> > boost (by rapidity phi)at 0 rotation or rapidity (J1, J2, J3, K1, K2,
> > K3). Then exponentiate: L(theta1,...,phi3)=Exp[theta1 J1+...phi3 K3].
>
> In case of spin 1/2 this exponentiation can be summed explicitly.

Sure, but this doesn't really help. Mathematica can exponentiate matrices with no problem (probably by diagonalizing, exponentiating the eigenvalues, and transforming back to original basis).

>
> > So the same thing should work for the Dirac spinor rep.
>
> A Dirac representation is a sum of a Weyl representation and its
> complex conjugate.

Not exactly, since the Weyl rep is 2 dimensional and the Dirac is 4 dimensional.

>
> > But what do the generators look like in that rep?
>
> The Lie algebra is spanned by complex linear combinations of the
> Pauli matrices sigma_1, sigma_2 and sigma_3. They constitute a real six
> dimensional Lie algebra.

That statement may be close, but it isn't helpful. I need 4x4 matrices. The pauli matrices are 2x2.

>
> > According to my understanding of
> > wikipedia they are just (J_i+I K_i)/2 and (J_i-I K_i)/2 with J_i being
> > I/2*Pauli matrices, but I don't know what the K_is are. And what are
> > the corresponding parameters to the above group element that must
> > multiply these generators in the exponential? Do the (Ji+I Ki)/2 apply
> > to the "top" part of the bispinor, and the (J_i-I K_i)/2 to the bottom
> > (with each block a 2x2 matrix) so that the matrix looks like (with
> > implied summation over i):
> > Exp[(theta_i-I phi_i) (J_i+I K_i)/2] 0
> > 0 Exp[(theta_i+I
> > phi_i) (J_i-I K_i)/2] or is this only correct in the Weyl
> > representation of the spinor?
> > I think this is wrong. Since each (J_i+I K_i)/2 and (J_i-I K_i)/2 is
> > now a generator of SU(2).
>
> Nothing wrong with this.
>
> > Should the matrix then look like:
>
> > Exp[(theta_i-I phi_i) (J_i) 0
> > 0 Exp[(theta_i+I phi_i) (J_i)
>
> It looks like this with J_i = sigma_i / 2.

I think it is it (I)* sigma_i/2.

>
> Note that a Dirac field is a map from spacetime to the space of spinors.
> It does not transform only by multiplication with a representation
> matrix S(Lambda) but also its space time argument x transforms
>
> psi' = S psi Lambda^(-1) or in other words
>
> psi'(Lambda x) = S psi(x) .

Actually, since you brought it up, Lambda is a passive transformation here, a change of coordinates. The argument of psi is the same spacetime point on both sides of the equation, just expressed in different coordinates. Do you agree?

>
> > And again, is this only true in the Weyl rep?
>
> Each Dirac representation looks like this in a suitable basis.

The basis matters because I am trying to verify the identity in wikipedia, which ensures that Psi^bar. Psi is a Lorentz invariant:

gamma_u=Lambda_{uv} Rep.gamma_v.Inverse[Rep], where the gammas are dirac matrices, and Rep is the sought after 4x4 representation of a Lorentz transformation Lambda. I guess the basis for Lambda has to be consistent with the basis for the dirac matrices and Rep.
I couldn't verify this in Mathematica, I posted the code, maybe you can see that post if the moderator accepts it. I also tested:
gamma_u=Lambda_{uv} ConjugateTranspose[Rep].gamma_v.Rep, which is what would be required (in disagreement with wikipedia) for psi_bar.psi to be an invariant (for the case u=0, the time coordinate, which is u=1 in the code).

>
> > I am trying to understand:
> > 1. if Dirac spinors can be generalized to more than one particle/
> > antiparticle pair
>
> This just means that you deal with several Dirac spinors. Usually one
> calls the additional labels which enumerate them flavor or isospin.

Flavor or isospin is not relevant to the question at hand. I am looking for an analog of a many particle Schrodinger wavefunction, which is not usually the product of the individual wavefunctions if there is entanglement.

>
> > 2. if the spin connection which arises by making the Dirac Lagrangian
> > locally Lorentz invariant has a physical significance
>
> It is the gravitational field, the Vierbein or the metric.

Not exactly. The metric does not figure in it, but a combination of the Levi-Civita connection and the Vierbein does.

>
> > --is there a Lagrangian just for the free "gauge" field
>
> Einstein gave the equations of motion 1915, well before Dirac spinors
> became known.

But I am not sure if what results is the Hilbert Lagrangian (from which the Einstein equation results as an Euler Lagrange equation).

>
> > and what are its Euler- Lagrange equations?
> > Perhaps this is a way to derive the Einstein field equations?
>
> Your suggestion is late. If you try to introduce a gauge field for
> Lorentz transformations, Einstein gravity is essentially unique
> (if the physical degrees of freedom are required to have positive
> energy density). "Essentially" means: one is free to allow for a
> cosmological constant or for Brans-Dicke like scalars.

This I was not aware of. I suppose I should find the free spin connection Lagrangian and see what it looks like.

>
> > 3. Is there a way to understand the Dirac spinor in purely geometric
> > (classical) form?
>
> It is essential, that a spin 1/2 field it is not a bosonic
> field but anticommutes rather than to commute. Up to now I have not
> seen a mathematical definition of such a Gra?mann field leave alone a
> geometric interpretation.

This I don't understand. The field operators have to anticommute, but the dirac spinor is just a classical field, since ultimately it gives probabilities when squared (for the 4 different spin and particle states, which can be summed to a total probability at each spacetime point, or integrated over a region, just like a Schrodinger field). Actually, even the electromagnetic classical field can be treated like a wavefunction, as in the Jones formalism. Both the Dirac and EM field can be further quantised, one with anticommutators and the other with commutators.

>
> The quantized free Dirac field is much simpler: it creates and
> annihilates particles in a Fock space. But we need Gra?mann fields
> as arguments of time ordered products (the fermionic arguments
> of time ordered products anticommute).

I suppose one can look at it this way if one does path integral quantization, but is it necessary? Canonical field quantization of a classical (non-grassman) dirac field does not require grassman variables, does it?

>
> But where do Gra?mann fields take their values? Does this target
> space have a norm, such that differentiability can be defined?
> As far. as I see, Gra?mann fields take values in something like a
> universal graded commutative algebra. Differentiation is defined
> formally and all Gra?mann fields which occur in Lagrangians and in
> time ordered products are smoothly differentiable by assumption though
> nothing allows to check it.

Sorry, I don't know anything about that.

[Hendrik van Hees]

On 01/04/12 12:50, iuval wrote:
> I'd like to have the explicit form of the 4x4 matrix representing a
> Lorentz transformation of a Dirac Spinor, with parameters: theta1,
> theta2, theta3, phi1, phi2, phi3, where the thetas are rotation angles
> about x, y, or z axes, and the phis are rapidities in those
> directions.

Although Norbert Dragon has answered your question completely, I'd like
to add the explicit treatment with Dirac spinors. Unfortunately my QFT
manuscript is in German, but the formulae are all there. Look at

http://theory.gsi.de/~vanhees/faq/qft/node22.html

I use the chiral representation that makes the logical structure of the
Dirac spinors most explicit since this is the representation which
represents it in a 2x2 block structure in terms of the direct sum of two
Weyl spinors (the upper components in this representation are the left
handed the lower the right-handed part of the Dirac spinor), i.e., as
the representation (1/2,0) \oplus (0,1/2) of the proper orthochronous
Lorentz group, for which this representation is explicitly reducible.
For the full orthocrhonous Lorentz group, which includes space
reflections (paritly transformations) its irreducible since a space
reflection interchanges the left- and right-handed parts.

On the above quoted web page you find the explicit expression for a
Lorentz boost (already "exponentiated") in Eq. (1.5.34), where eta
denotes the rapidity and \vec{n} the boost direction.

The rotations are simply expressed as its Spin-1/2 representation acting
on the left- (upper two) and right-handed (lower two) components.

--
Hendrik van Hees
Frankfurt Institute of Advanced Studies
D-60438 Frankfurt am Main
http://fias.uni-frankfurt.de/~hees/

[Norbert Dragon]

* iuval writes:

>* Norbert Dragon wrote:

>> Eq. 6.50 and 6.68

>> http://www.itp.uni-hannover.de/~dragon/rel_e.pdf

>> give the explicit 2x2 matrices of a Weyl representation in case of either
>> a rotation or a boost. But one can easily use the arguments given there,
>> to sum the exponential series for a general Lorentz transformation.

> As I said, I need the 4x4 matrices, not 2x2.

Simply use the 2x2 matrices and their complex conjugate as two blocks
on the diagonal of a 4x4 matrix.

You can, if you want to, dress these representation matrices with A
and A^(-1) from the left and right, to get each Dirca representation
which you like -- not that the gamma matrices, from which the
Dirac representation is constructed, are also unique only up to
the adjoint transformation A gamma A^(-1).

>> In case of spin 1/2 this exponentiation can be summed explicitly.

> Sure, but this doesn't really help. Mathematica can exponentiate
> matrices with no problem (probably by diagonalizing,
> exponentiating the eigenvalues, and transforming back to original
> basis).

If Mathematica would solve your problem, you would not ask here.

>> A Dirac representation is a sum of a Weyl representation and its
>> complex conjugate.

> Not exactly, since the Weyl rep is 2 dimensional and the Dirac is 4 dimensional.

You have not understood the word "sum" of representations A and B.
It does not mean the sum (A+B) of linear maps of the same vector space,
but matrices which act on the direct sum of vectors,

( A 0 )
( 0 B )

>>> Should the matrix then look like:
>>
>>> Exp[(theta_i-I phi_i) (J_i) 0
>>> 0 Exp[(theta_i+I phi_i) (J_i)
>>
>> It looks like this with J_i = sigma_i / 2.

> I think it is it (I)* sigma_i/2.

No. If you restrict yourself to purely imaginary linear combinations
of the Pauli matrices, you generate only SU(2), not the (cover of the)
Lorentz group SL(2,C)

You should read

http://www.itp.uni-hannover.de/~dragon/rel_e.pdf

really.

>> Note that a Dirac field is a map from spacetime to the space of spinors.
>> It does not transform only by multiplication with a representation
>> matrix S(Lambda) but also its space time argument x transforms

>> psi' = S psi Lambda^(-1) or in other words

>> psi'(Lambda x) = S psi(x) .

> Actually, since you brought it up, Lambda is a passive transformation
> here, a change of coordinates. The argument of psi is the same
> spacetime point on both sides of the equation, just expressed in
> different coordinates. Do you agree?

If you consider passive transformations, then the Dirac field does
not change at all, only the coordinates of spacetime point and the
spinorbasis in which you represent the field.

Normally one is interested in the change of the components as functions
of the coordinates of the spacetime points. These functions change by
passive transformations by a change of the argument x and by
multiplication with the matrix representation.

The active transformation differs from the passive one just by the
reflection g |--> g^(-1) .

I only stressed that Lorentz-transformations change the spacetime
argument, because normally the generators J and K are also considered
to generate the infinitesimal changes of these spacetime arguments.

Then J and K are differential operators which are not easily
exponentiated by Mathematica.

>> Each Dirac representation looks like this in a suitable basis.


> The basis matters because I am trying to verify the identity
> in wikipedia, which ensures that Psi^bar. Psi is a Lorentz invariant

To check the invariance, if is sufficient, to investigate infinitesimal
transformations.

But also the invariance under finite SL(2,C) transformations is
simple, knowing, that the four component Dirac spinor

Psi = (psi, chi^*) where psi an chi^* are two-spinors,

transforms into

Psi' = (M psi, M^* chi^*)

Psi'^bar Psi' is

(M chi)^T epsilon, (M psi)^T epsilon)(M psi, M^* chi^*)

where M^T epsilon M = det M = 1 is the determinant of M, so

Psi'^bar Psi' = Psi^bar Psi

> I couldn't verify this in Mathematica, I posted the code,
> maybe you can see that post if the moderator accepts it.

One verified the invariance of psi^bar psi not by experimenting
with advanced versions of Mathematica, but by mathematics, which
one can comprehend. Mathematic and Physics are sciences, not magic.

>>> I am trying to understand:
>>> 1. if Dirac spinors can be generalized to more than one
>>> particle/antiparticle pair

Mathematica is not the suitable tool to address this question.

If you have a matrix representation M_g, then one does not need
the computer to realize that the matrices

(M_g, 0 )
( 0 , M_g)

are also representations.

>> This just means that you deal with several Dirac spinors. Usually one
>> calls the additional labels which enumerate them flavor or isospin.

> Flavor or isospin is not relevant to the question at hand.

As long as you do not understand flavor or isospin you cannot judge
whether they are the answer to your question: they are.

> I am looking for an analog of a many particle Schrodinger wavefunction,
> which is not usually the product of the individual wavefunctions
> if there is entanglement.

Even if 2 + 2 = 4 = 2 * 2 one has to distinguish between the sum of
representations and their product. Entanglement has nothing to do with
sums of representations.

> Not exactly. The metric does not figure in it, but a combination of
> the Levi-Civita connection and the Vierbein does.

The metric and the vierbein are related intimitly such that physically
one does not need to distinguish them painstakingly. One defines the
other (up to gauge transformations), each of them generates the
gravitational interactions.

> But I am not sure if what results is the Hilbert Lagrangian
> (from which the Einstein equation results as an Euler Lagrange equation).

The Einstein Hilbert action is unique -- up to the cosmological
constant and up to Brans-Dicke type scalars -- if the energy
of the gravitaional fields is required to be positve.

>>> and what are its Euler- Lagrange equations?

The Einstein-equations, multiplied with an invertible matrix, the
Vierbein.

>>> Perhaps this is a way to derive the Einstein field equations?

>> Your suggestion is late. If you try to introduce a gauge field for
>> Lorentz transformations, Einstein gravity is essentially unique
>> (if the physical degrees of freedom are required to have positive
>> energy density). "Essentially" means: one is free to allow for a
>> cosmological constant or for Brans-Dicke like scalars.

> This I was not aware of. I suppose I should find the free spin
> connection Lagrangian and see what it looks like.

The spin connection omega_{ab}^c contains Spin-2 and 3 spin 1
and one spin 0 part. Concentrate on the spin-2 part and answer
the question, whether its energy density is positive. In
Lorentz invariant theories this cannot happen -- parts of the
spin-2 have to be gauge degrees of freedom, only the helicity 2
part can be physical. The results of these
considerations tell you, that spin-2 has to be accompanied by
invariance under general coordinate transformations (and that
there can be only one interacting spin-2 particle). As a result
of the invariance under general coordinate transformations the
spin-2 particle can couple only to the energy momentum tensor --
it is the graviton.

> The field operators have to anticommute,
> but the dirac spinor is just a classical field,

Not at all: if Psi were to commute, then

Psi^bar Psi = - Psi^bar Psi

(because the matrix epsilon is antisymmetric)

> since ultimately it gives probabilities when squared

Proof by denomiation, because you call the Dirac field Psi?

Theorem: The position wave function of a Spin-1/2 field does
not transform locally, i.e. not as

Psi'(x') = S Psi(x)

therefore, even if the position wavefunction fulfills the Dirac
equation, it is not the Dirac field which you consider.

That the position wavefunction transforms nonlocally is already true
for scalar particles.

The local transformation applies to operator fields, they create and
annihilate particles as follows readily from their transformation law.

> (for the 4 different spin and particle states,
> which can be summed to a total probability at each spacetime

> point, or integrated over a region, just like a Schr?dinger field).

Position wavefunctions are the Fourier transform of the momentum
wave functions.

If the scalar product of momentum wavefunctions is (stick for
simplicity with scalar particles)

< chi | phi > = int d^3 k 1/sqrt(m^2 + k^2) chi^*(k) phi(k)

then their Lorentz transformed function is simply

chi'(Lambda k) = chi(k)

But given the above scalar product, |phi(k)|^2 cannot be the
probability density to find the particle, which is in the state phi,
in a momentum range d^3 k around k.

With the above scalar product, phi/sqrt(sqrt(m^2 + k^2)) is the
momentum wave function. It transforms nonlocally because of the
convolution theorem, because its fouriertransform is the product
of a locally transforming factor with the function
sqrt(sqrt(m^2 + k^2)).

The argument is true though nearly all text books neglect it --
many of them even neglect, that Psi^bar Psi = - Psi^bar Psi
for commuting Dirac spinors.

[iuval]

I have corresponded with Hendrik and now I am more confused than before.
He seems to be saying that sometimes the gamma matrices transform a
certain way under a (passive) Lorentz transformation, and sometimes they
do not. He kindly wrote out by hand and scanned a derivation of the
invariance of psibar .psi, which also assumed the invariance (not
covariance) of gamma0. That is, gamma0'=gamma0, and
psi'^bar=ConjugateTranspose[psi'].gamma0, instead of
psi'^bar=ConjugateTranspose[psi'].gamma0' (I omit the spacetime
arguments for simplicity, but they also get transformed passively) He
refers to the Dirac matrices as transforming as in
http://theory.gsi.de/~vanhees/faq/qft/node22.html eqn 1.5.13, but gamma0
apparently is exempt from this transformation law. Perhaps eqn 1.5.13 is
not a transformation law at all (as there is no prime on the RHS), but
an identity. I haven't been able to verify that identity yet, but
perhaps that is due to needing to change basis in the defining
representation matrix that appears there.

It seems like the gamma matrices in the derivative part of the Dirac
equation on or Lagrangian had better transform as contravariant 4
vectors do in order for the equation and the Lagrangian to be invariant,
but that gamma0 in psibar in the Lagrangian is invariant and does not
transform at all between different observers. I may post my
correspondence with Hendrik here after I reformat it.

[Norbert Dragon]

* iuval writes:

> now I am more confused than before.

> He seems to be saying that sometimes the gamma matrices transform a
> certain way under a (passive) Lorentz transformation, and sometimes they
> do not.

Your confusion results from confused ideas about transformations.

In field theory the objects which transform are the fields and
their derivatives. As a result polynomials of these fields transform.
Lagrangians are such polynomials and are often restricted by the
requirement to transform as a scalar field.

Matrices, which appear as coefficients in polynomials of
(derivatives of) the fields do _not_ transform, neither do derivatives.

In psi^bar gamma^m psi, it is the field psi, which transforms under
Lorentz transformations into

(T_Lambda psi)(Lambda x) = chi(Lambda x) = S(Lambda) psi(x)

The gamma-matrices satisfy

S^(-1)(Lambda) gamma^m S(Lambda) = Lambda^m_n gamma^n

and the bar-operation is constructed such that

chi^bar(Lambda x) = psi^bar(x) S^(-1)(Lambda)

which is why

psi^bar(x) gamma^m psi(x)

transforms like a vector field.

Also devivatives do not transform

T_Lambda d_m = d_m T_Lambda ,

which is why psi^bar d_m psi transforms as a vector field and

psi^bar gamma^m d_m psi transforms as a scalar field.

In sloppy physical language, the transformation is attributed to
the gamma matrices and the derivatives, while actually it results
from the transformation of the Dirac field.

Not the gamma-matrices neither the gamma^0 matrix in the definition
of psi^bar transform.

The literature, which you used as introduction to field theory, does
not seem to be optimal.

--
Superstition brings bad luck

www.itp.uni-hannover.de/~dragon

[Hendrik van Hees]

On 13/04/12 00:03, iuval wrote:

> I have corresponded with Hendrik and now I am more confused than before.
> He seems to be saying that sometimes the gamma matrices transform a
> certain way under a (passive) Lorentz transformation, and sometimes they
> do not. He kindly wrote out by hand and scanned a derivation of the
> invariance of psibar .psi, which also assumed the invariance (not
> covariance) of gamma0. That is, gamma0'=gamma0, and
> psi'^bar=ConjugateTranspose[psi'].gamma0, instead of
> psi'^bar=ConjugateTranspose[psi'].gamma0' (I omit the spacetime
> arguments for simplicity, but they also get transformed passively) He
> refers to the Dirac matrices as transforming as in
> http://theory.gsi.de/~vanhees/faq/qft/node22.html eqn 1.5.13, but gamma0
> apparently is exempt from this transformation law. Perhaps eqn 1.5.13 is
> not a transformation law at all (as there is no prime on the RHS), but
> an identity. I haven't been able to verify that identity yet, but
> perhaps that is due to needing to change basis in the defining
> representation matrix that appears there.
>
> It seems like the gamma matrices in the derivative part of the Dirac
> equation on or Lagrangian had better transform as contravariant 4
> vectors do in order for the equation and the Lagrangian to be invariant,
> but that gamma0 in psibar in the Lagrangian is invariant and does not
> transform at all between different observers. I may post my
> correspondence with Hendrik here after I reformat it.

For the other readers' information, here's the mentioned scan of my
little proof for the fact that \bar{psi}(x) \psi(x) is a scalar field,
using the explicit representation of boosts and rotations in the Dirac
representation of the orthochronous Lorentz group O(1,3)\uparrow.

http://fias.uni-frankfurt.de/~hees/tmp/lorentz-trafos-diracology.pdf

[iuval]

I tried to respond to the above, but for some reason the moderator did not =
allow it. Suffice it to say that Norbert's words were not very enlightening=
for me. But I would like to respond to the following:

>
> Theorem: The position wave function of a Spin-1/2 field does
> not transform locally, i.e. not as
>
> Psi'(x') = S Psi(x)

But you say in this same post that this equation IS true:

"psi'(Lambda x) = S psi(x) . "

So do the psis have different meanings here? One with capitals P and one wi=
th small p? I am only for the moment interested in wavefunctions, not in cr=
eation and annihilation operators. is your psi (with small p) corresponding=
to an annihilation operator?

>
> therefore, even if the position wavefunction fulfills the Dirac
> equation, it is not the Dirac field which you consider.
>
> That the position wavefunction transforms nonlocally is already true
> for scalar particles.
>
> The local transformation applies to operator fields, they create and
> annihilate particles as follows readily from their transformation law.
>

But the Dirac equation, and Hendrik's manuscript (and yours too) apply to w=
avefunctions (Classical Dirac Field), not operators.

>
> > (for the 4 different spin and particle states,
> > which can be summed to a total probability at each spacetime
> > point, or integrated over a region, just like a Schr?dinger field).
>
> Position wavefunctions are the Fourier transform of the momentum
> wave functions.
>
> If the scalar product of momentum wavefunctions is (stick for
> simplicity with scalar particles)
>
> < chi | phi > = int d^3 k 1/sqrt(m^2 + k^2) chi^*(k) phi(k)
>
> then their Lorentz transformed function is simply
>
> chi'(Lambda k) = chi(k)
>
> But given the above scalar product, |phi(k)|^2 cannot be the
> probability density to find the particle, which is in the state phi,
> in a momentum range d^3 k around k.

Right, but phi(x) is defined as the probability to find the particle at pos=
ition x (within d^4 x).

>
> With the above scalar product, phi/sqrt(sqrt(m^2 + k^2)) is the
> momentum wave function. It transforms nonlocally because of the
> convolution theorem, because its fouriertransform is the product
> of a locally transforming factor with the function
> sqrt(sqrt(m^2 + k^2)).

Could you write this out explicitly? I almost understand, but not
quite. Yo= u mean that the FT of phi(k)/sqrt(sqrt(m^2 + k^2)) is not
phi(x)? What does= that have to do with the Lorentz transformation
properties of phi(x)? The = Fourier transform has nothing to do with the
Lorentz transform, except in f= inding invariant integration measures in
momentum space that have the mass = shell constraint.

>
> The argument is true though nearly all text books neglect it --
> many of them even neglect, that Psi^bar Psi = - Psi^bar Psi
> for commuting Dirac spinors.

Sorry, but that last paragraph sounds like gibberish. Do you mean for
anti-commuting Dirac Field Operators?

[Hendrik van Hees]

I don't understand, where all this confusion comes from. By construction
of a local Lorentz covariant particle's wave function with spin 1/2,
admitting spatial reflections, you end up with Dirac spinors. Of course,
they are to interpreted as fermion-field operators, but concerning their
tranformation properties you can as well argue with the Dirac equation
of a classical (c-number) Dirac field. It transforms under orthochronous
Lorentz transformations as a local field, i.e., via

Psi'(x')=S(Lambda) Psi(x),

where S(Lambda) is a (non-unitary!) representation of the orthochronous
Lorentz group.

Restricted to the proper orthochronous Lorentz group this representation
is reducible into two irreducible representations, which act on two Weyl
spinors, out of which thus the Dirac spinor is constructed as a direct
sum. The two reprsentations can be characterized by the chirality
(handedness), i.e., chosen as eigenvectors of gamma^5 with eigenvalues
\pm 1. Each eigenspace is two-dimensional. These are the two 2x2-matrix
representations Norbert has given. It's also all written in my
manuscript. If you don't understand German, use my QFT manuscript in
English:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

Now concerning the transformation behavior of the Dirac equation itself.
By construction, the above decribed Dirac representation of the
orthochronous Lorentz group, leaves the Dirac equation invariant, i.e.,
if psi(x) obeys the Dirac equation

(i \fslash{\partial}-m) \psi(x)=0,

the same holds true for \psi'(x'). In the Feynman slash the Dirac
matrices for both functions are the same. The invariance of the Dirac
equation then leads to the identity

S(Lambda) gamma^{mu} S^{-1}(Lambda)={\Lambda^{mu}}_{nu} gamma^{nu},

and it is said in the usual physicist's slang that the Dirac matrices
transform as a four-vector under Lorentz transformations.

Further, since in proving the transformation properties of the standard
sesqui-linear forms of the Dirac spinor, you never have to interchange
Dirac-spinor components (be it as a c-number field or a field operator),
in both cases they have the very same behavior as scalar, pseudoscalar,
vector, and tensor fields as it should be.

The representation of the Lorentz transformation for the momentum-space
modes is different since this provides a unitary representation of the
(orthochronous) Poincare group. You can read about this in my above
linked manuscript.

The by far best textbook on this subject is

S. Weinberg, Quantum Theory of Fields, Vol. I, Cambridge University Press

It is also worth to read the original paper by Wigner,

E. P. Wigner, On Unitary Representations of the Inhomgeneous Lorentz
Group. Annals of Mathematics 40 (1939), 149.
http://dx.doi.org/10.1016/0920-5632(89)90402-7

On 15/04/12 09:50, iuval wrote:
> I tried to respond to the above, but for some reason the moderator did not

> allow it. Suffice it to say that Norbert's words were not very enlightening

> for me. But I would like to respond to the following:
>>
>> Theorem: The position wave function of a Spin-1/2 field does
>> not transform locally, i.e. not as
>>
>> Psi'(x') = S Psi(x)
>
> But you say in this same post that this equation IS true:
> "psi'(Lambda x) = S psi(x) ."
> So do the psis have different meanings here? One with capitals P and one

> with small p? I am only for the moment interested in wavefunctions, not
> in creation and annihilation operators. is your psi (with small p)
> corresponding to an annihilation operator?

>>
>> therefore, even if the position wavefunction fulfills the Dirac
>> equation, it is not the Dirac field which you consider.
>>
>> That the position wavefunction transforms nonlocally is already true
>> for scalar particles.
>>
>> The local transformation applies to operator fields, they create and
>> annihilate particles as follows readily from their transformation law.
>>
>
> But the Dirac equation, and Hendrik's manuscript (and yours too) apply to

> wavefunctions (Classical Dirac Field), not operators.

>>
>>> (for the 4 different spin and particle states,
>>> which can be summed to a total probability at each spacetime
>>> point, or integrated over a region, just like a Schr?dinger field).
>>
>> Position wavefunctions are the Fourier transform of the momentum
>> wave functions.
>>
>> If the scalar product of momentum wavefunctions is (stick for
>> simplicity with scalar particles)
>>
>> < chi | phi> = int d^3 k 1/sqrt(m^2 + k^2) chi^*(k) phi(k)
>>
>> then their Lorentz transformed function is simply
>>
>> chi'(Lambda k) = chi(k)
>>
>> But given the above scalar product, |phi(k)|^2 cannot be the
>> probability density to find the particle, which is in the state phi,
>> in a momentum range d^3 k around k.
>
> Right, but phi(x) is defined as the probability to find the particle at

> position x (within d^4 x).

>
>>
>> With the above scalar product, phi/sqrt(sqrt(m^2 + k^2)) is the
>> momentum wave function. It transforms nonlocally because of the
>> convolution theorem, because its fouriertransform is the product
>> of a locally transforming factor with the function
>> sqrt(sqrt(m^2 + k^2)).
>
> Could you write this out explicitly? I almost understand, but not
> quite. Yo= u mean that the FT of phi(k)/sqrt(sqrt(m^2 + k^2)) is not
> phi(x)? What does= that have to do with the Lorentz transformation
> properties of phi(x)? The = Fourier transform has nothing to do with the
> Lorentz transform, except in f= inding invariant integration measures in
> momentum space that have the mass = shell constraint.
>
>>
>> The argument is true though nearly all text books neglect it --
>> many of them even neglect, that Psi^bar Psi = - Psi^bar Psi
>> for commuting Dirac spinors.
>
> Sorry, but that last paragraph sounds like gibberish. Do you mean for
> anti-commuting Dirac Field Operators?
>

[Norbert Dragon]

* Hendrik van Hees writes:

> I don't understand, where all this confusion comes from. By construction
> of a local Lorentz covariant particle's wave function with spin 1/2,
> admitting spatial reflections, you end up with Dirac spinors.

Not at all.

The state of a spin-1/2 particle is characterized by _two_ position
wave functions,

(Psi_up(x), Psi_down(x))

which give the probabilities

|Psi_up(x)| d^3 x and |Psi_down(x)| d^3 x

to find the particle with spin up or down around x in a domain of size
d^3 x.

This holds throughout Quantum mechanics, irrespective whether it is
relativistic or not.

To construct a Dirac field out of the two component position wave
function, one needs an additional argument.

If the Hilbert space of states carries a unitary representation of
Poincar? transformations then one easily confirms that it acts on the
momentum wave functions by

psi'~(Lambda k) = sqrt(k^0 / Lambda^0_n k^n) U(W(Lambda,k)) psi(k)

The Fourier transform of this transformation shows that the position
wave function does not transform as a local field, but as a convolution
of a local field with the Fourier transform of

sqrt(k^0 / Lambda^0_n k^n) U(W(Lambda,k))

> Of course,
> they are to interpreted as fermion-field operators, but concerning their
> tranformation properties you can as well argue with the Dirac equation
> of a classical (c-number) Dirac field.

No. Inspecting your arguments, you will realize that your sole
justification to call psi a position wave function is its denomination
psi -- which is cheating. If you inspect the property of a position
wave function, to be the Fourier transform of the momentum wave
function, then you cannot avoid the conclusion that position wave
functions have _two_ components and transform nonlocally under boosts.

> It transforms under orthochronous
> Lorentz transformations as a local field, i.e., via

> Psi'(x')=S(Lambda) Psi(x),

> where S(Lambda) is a (non-unitary!) representation of the orthochronous
> Lorentz group.

This is not the transformation of the position wave function and it is
not the unitary representation of Lorentztransformations in some
Hilbert space (of square integrable functions).

> Restricted to the proper orthochronous Lorentz group this representation
> is reducible into two irreducible representations, which act on two Weyl
> spinors, out of which thus the Dirac spinor is constructed as a direct
> sum. The two reprsentations can be characterized by the chirality
> (handedness), i.e., chosen as eigenvectors of gamma^5 with eigenvalues
> \pm 1. Each eigenspace is two-dimensional. These are the two 2x2-matrix
> representations Norbert has given.

Which is the Hilbert space in which the map Psi |--> Psi' is unitary?

Note well: the adjoint transformation of the space of operators is
_not_ unitary. You should care to distinguish between the Hilbert space
and the operators which act on it.

> Now concerning the transformation behavior of the Dirac equation itself.
> By construction, the above decribed Dirac representation of the
> orthochronous Lorentz group, leaves the Dirac equation invariant, i.e.,
> if psi(x) obeys the Dirac equation

> (i \slash{\partial}-m) psi(x)=0,

> the same holds true for psi'(x'). In the Feynman slash the Dirac
> matrices for both functions are the same. The invariance of the Dirac
> equation then leads to the identity

> S(Lambda) gamma^{mu} S^{-1}(Lambda)={\Lambda^{mu}}_{nu} gamma^{nu},

> and it is said in the usual physicist's slang that the Dirac matrices
> transform as a four-vector under Lorentz transformations.

I disagree concerning the exponent:

S^(-1)(Lambda) gamma^{mu} S(Lambda) = {Lambda^{mu}}_{nu} gamma^{nu},

S_2(-1) S_1(-1) gamma^m S_1 S_2 = S_2(-1) L_1^m_n gamma^n S_2

= L_1^m_n S_2(-1) gamma^n S_2

= L_1^m_n L_2^n_k gamma^k

= (L_1 L_2)^m_n gamma^n

> Further, since in proving the transformation properties of the standard
> sesqui-linear forms of the Dirac spinor, you never have to interchange
> Dirac-spinor components (be it as a c-number field or a field operator),
> in both cases they have the very same behavior as scalar, pseudoscalar,
> vector, and tensor fields as it should be.

You have to check that the object which you are dealing with does not
vanish. In case of a _commuting_ Majorana field

psi = (Psi, Chi) , Chi = epsilon (Psi^*)

the Dirac scalar psi^bar psi vanishes and psi^bar gamma^m d_m psi is

a complete derivative. As you can write each Dirca field as a sum

psi = M_1 + i M_2

of two Majorana spinor this means, that the Dirac Lagrangian and
consequently the energy momentum tensor for commuting spinors is off
diagonal of the type

M_1^bar M_2 + M_2^bar M_1

therefore the energy density of commuting spinor fields cannot be
bounded from below.

> The representation of the Lorentz transformation for the momentum-space
> modes is different since this provides a unitary representation of the
> (orthochronous) Poincare group.

Take the Fourier transform of the momentum wave functions to determine
the transformation of the position wave functions. Of course, the
transformation is also unitary, but it is non-local.

[Norbert Dragon]

* iuval wrote:

>> Theorem: The position wave function of a Spin-1/2 field does
>> not transform locally, i.e. not as
>>
>> Psi'(x') = S Psi(x)

> But you say in this same post that this equation IS true:
> "psi'(Lambda x) = S psi(x) . "

> So do the psis have different meanings here? One with capitals P and one

> with small p?

Exactly. I claim that the Dirac equation applies to several different
objects and that the set of solutions of the Dirac equation allows for
different realizations of Poincaré transformations. Position wave
functions of spin-1/2-particles have two components: the wave function
for spin up and for spin down.

Psi_i(x) , i in {up, down}

They satisfy the Klein-Gordon-equation.

(Box + m^2) Psi_i(x) = 0

Therefore, their derivatives sigma^n d_n Psi / m = Chi

define a second two component spinor, which together with Psi constitute

a Dirac spinor psi = (Psi, Chi) which satisfies the Dirac equation.

( -m , sigma^bar d ) (Psi)
( ) ( ) = 0
( sigma d , -m ) (Chi)

> I am only for the moment interested in wavefunctions, not in creation
> and annihilation operators.

Wave functions do not transform locally under Lorentz boosts, because
their Fourier transformation, the momentum wave function, transforms as

psi~'(Lambda k) = sqrt( k^0 / Lambda^0_n k^n ) U(W(Lambda, k)) psi~(k)

with a nontrivial normalization factor and the spin-1/2 representation
of the Wigner rotation W. Because the prefactors and the Wigner rotation
depend on k, therefore the transformed position wave function is the
convolution of the original wave function with the Fourier transform
of the prefactor and the representation of the Wigner rotation.

> is your psi (with small p) corresponding to an annihilation operator?

psi denotes a solution of the Dirac equation. If it transforms locally
as a tensor field

psi'(Lambda x)) = S(Lambda) psi(x)

it creates and annihilates spin-1/2 particles. If psi consists of the
two position wave functions of a spin-1/2 particle and their
derivatives, then it cannot transform locally under Lorentz boosts.

> But the Dirac equation, and Hendrik's manuscript (and yours too) apply to

> wavefunctions (Classical Dirac Field), not operators.

See above. A spin-1/2 particle has two, not four position wave
functions. From them one can construct a Dirac spinor, but it
transforms different to what is claimed in many books.

> Right, but phi(x) is defined as the probability to find the particle

> at position x (within d^4 x).

Not at all. Position wave funtions yield probability desities

|psi_i(x)|^2 , i in {up, down}

to find the particle with spin up or spin down at x in a volume d^3 x
if one measures at time x^0.

There is no time operator and no probability distribution of times in
quantum mechanics.

I shall answer your remaining questions separately.

[Moderator's note: The relevant RFC specifies a space after the -- to
denote the start of .sig; this is used by some software to recognize
them and thus display them or not depending on some setting.]

--
Aberglaube bringt Unglück

www.itp.uni-hannover.de/~dragon

[Hendrik van Hees]

I think this adds to the confusion, although I don't think that we
disagree on the basic principles.

As posted already yesterday, one has to distinguish between the proper
orthochronous Lorentz group SO(1,3)^, which in its fundamental
representation, acting on space-time four-vectors etc., consists of all
Lorentz matrices with determinant +1 and {Lambda^0}_0>=1, and the
orthochronous Lorentz group O(1,3)^ which in its fundamental
representation consists of all Lorentz matrices with {Lambda^0}_0>=1.
SO(1,3)^ is generated by all rotations and boosts and forms a
6-dimensional connected (but not simply connected) Lie group. Any
O(1,3)^ transformation is either an SO(1,3)^ transformation or an
SO(1,3)^ transformation followed by a space reflection (parity
operation), having determinant -1.

If you look for irreducible finite-dimensional representations of
SO(1,3)^ you find two inequivalent representations of its covering group
which reduce to SU(2) transformations (representing spin 1/2 particles)
for rotations. The covering group is SL(2,C) with its fundamental
representation and the conjugate complex transformation.

The corresponding spinors are called Weyl spinors and denoted by psi_L
and psi_R. One can show that, up to phases uniquely, one can extend the
SO(1,3)^ representations to O(1,3)^ representations by considering the
direct sum of the two inequivalent SL(2,C) representations. This
representation is thus reducible if restricted to SO(1,3)^
transformations by construction, but spacial reflections (parity
transformations) flip the left- and right-handed components.

Then you can construct local field equations for these Dirac spinors,
leading to a Dirac equation. Of course, these field equations cannot be
interpreted simply as wave functions in the sense of non-relativistic
quantum theory (at least not for interacting fields), but one has to
quantize these fields. In order to obtain a local representation of the
orthochronous Poincare group, one has to use bispinors and thus obtains
two spin-1/2 particles, one called "particle" and the other one called
the corresponding "anti-particle". At the same time, to have also a
Hamiltonian bounded from below, you have to quantize the Dirac field as
fermions, which is a special case of the general spin-statistics theorem.

You can find the very general and quite lengthy proof for particles of
any spin (and also for the somewhat different case of massless particles) in

S. Weinberg, Quantum Theory of Fields, Vol. 1, Cambridge University press.

[Jos Bergervoet]

On Apr 16, 10:06 pm, Norbert Dragon <dra...@itp.uni-hannover.de>
wrote:
...

> Wave functions do not transform locally under Lorentz boosts, because
> their Fourier transformation, the momentum wave function, transforms as

...

and later you write:

> |psi_i(x)|^2 , i in {up, down}
>
> to find the particle with spin up or spin down at x in a volume d^3 x
> if one measures at time x^0.

Do you mean that a particle inside a box may be observed
outside the box by a moving observer? (And if not, wouldn't
that restrict the wave function transform to be purely local?)

--
Jos

[Norbert Dragon]

* Jos Bergervoet writes:

>* Norbert Dragon wrote:

>> Wave functions do not transform locally under Lorentz boosts, because
>> their Fourier transformation, the momentum wave function, transforms as
> ...
> and later you write:

>> |psi_i(x)|^2 , i in {up, down}
>>
>> to find the particle with spin up or spin down at x in a volume d^3 x
>> if one measures at time x^0.

> Do you mean that a particle inside a box may be observed
> outside the box by a moving observer?

> And if not, wouldn't that restrict the wave function transform
> to be purely local?

You take it for granted that one can localize a particle strictly
inside some region. This assumption is wrong.

If at time t=0 the wave function vanishes outside some region, then
its time derivative d_t psi does not vanish in any open region, because
of the Schroedinger equation

i d_t psi = sqrt(m^2 + p^2) psi

So strict localization occurs only at exceptional times and only for
exceptional observers for whom the corresponding events are
simultaneous.

[Jos Bergervoet]

On Apr 18, 6:38 pm, Norbert Dragon <dra...@itp.uni-hannover.de> wrote:
> * Jos Bergervoet writes:
..

> > Do you mean that a particle inside a box may be observed
> > outside the box by a moving observer?

> > And if not, wouldn't that restrict the wave function transform

> > to be purely local?

> You take it for granted that one can localize a particle strictly
> inside some region. This assumption is wrong.

Actually I was just referring to properties of the
wavefunction as a solution of the equations: A
wavefunction vanishing outside some region
would be seen by a moving observer as nonzero
outside this region.

> If at time t=0 the wave function vanishes outside some region, then
> its time derivative d_t psi does not vanish in any open region, because
> of the Schroedinger equation
>
> i d_t psi = sqrt(m^2 + p^2) psi
>
> So strict localization occurs only at exceptional times and only for
> exceptional observers for whom the corresponding events are
> simultaneous.

*If* you have solutions of the Schrodinger equation.
And directly after those exceptional points in time
the wavefunction will pervade all of space, because
it propagates infinitely fast! This is all true for
solutions of the Schrodinger equation.

With the Dirac equation all these strange things
do not happen, so one can be sure that a particle
inside some region is seen by all moving
observers to be inside this region. Or not?

So can we say then that Dirac solutions transform
locally and Schrodinger solutions do not? (Beside
having other odd properties..)

--
Jos

[Norbert Dragon]

* Jos Bergervoet writes:

> *If* you have solutions of the Schroedinger equation.

> And directly after those exceptional points in time
> the wavefunction will pervade all of space, because
> it propagates infinitely fast! This is all true for
> solutions of the Schrodinger equation.

Side comment:

Your considerations are "antiquantum". A particle in
quantum mechanics is not found at some place _because_ is was
in a definite place before. This is disproved by double slits.

The change of probabilities of positions in quantum mechanics
is not caused by motion of particles but by propagation of
wave functions.

Moreover, in relativistic theories, there is no local current

j^m(Psi, d Psi)

which completes the position probability density

j^0 = |Psi|^2

to a conserved current. There is a electric conserved current
for a Dirac field psi, but up to its name it has little in common
with a position wave functions.

> With the Dirac equation all these strange things
> do not happen,

The Dirac equation has to be shown to be of relevance for the position
wave function. For this it is not sufficient to denote the Dirac spinor
by psi. Given the position wave functions Psi of a spin-1/2 particle,
then the spinor

psi = (Psi, sigma^n d_n Psi / m)

satisfies the Dirac equation, but it cannot vanish in any open region
due to the Schroedinger equation.

> so one can be sure that a particle inside some region is seen
> by all moving observers to be inside this region.

If you consider solutions of the Dirac equation which vanish outside
some bounded domain you can be sure that they are not the position
wave functions of a spin-1/2 particle.

You cannot avoid the Schroedinger equation for a position wave
function (if the time evolution is linear). It just states that the
total probability to find the particle somewhere is conserved for
all states.

i d_t Psi = sqrt(m^2+p^2) Psi

is unavoidable for each free, relativistic particle of mass m. It
disallows position wave functions which are restricted to a
bounded carrier.

> So can we say then that Dirac solutions transform

> locally and Schroedinger solutions do not?

Position wave functions do not transform locally, whether they are
part (two components) of a Dirac spinor or not. This follows from
the k-dependence of the factor

sqrt(k^0 / Lambda^0_n k^n) U(W(Lambda, k))

in the transformation

ps^~'(Lambda k) = sqrt(k^0 / Lambda^0_n k^n) U(W(Lambda, k)) psi~(k)

of the momentum wave function. Do you agree on this transformation?

To the fact that position wave functions transform nonlocally there
corresponds the fact that the three position operators X are not three
operators of a finite dimensional multiplet which transforms
among itself under Lorentz boosts.

To answer an anticipated counter argument:

No, there is no time operator t in quantum mechanics or in quantum
field theory, which completes X to a four vector (t,X).

String theory is different, however it is worse.
There is a Heisenberg pair of operators

[t,H] = i .

But this is why string theory is physically empty.
It does not have any physical state because the physical states have
to satify constraints, among others

(P^2 - m^2) Psi_physical = 0 (where m^2 is from some discrete set).

The operators t and H act on square integrable momentum wave functions
of R^26 or R^10 which because of the constraint vanish everywhere
apart from a set of measure zero.

Such wave functions are 0 in Hilbert space.

If you neglect this problem of string theory and work with wave
functions which are defined on the mass shell and have a scalar
product in a Hilbert space concerning 25 or 9 momenta only (as in field
theory), then there does not exist the Heisenberg pair [t,H]
and all the locality and completeness relations of string theory
which make use of these operators are violated.

[Jos Bergervoet]

On Apr 22, 10:17 am, Norbert Dragon <dra...@itp.uni-hannover.de>

wrote:
> * Jos Bergervoet writes:
..

>> *If* you have solutions of the Schroedinger equation.
>> And directly after those exceptional points in time
>> the wavefunction will pervade all of space, because
>> it propagates infinitely fast! This is all true for
>> solutions of the Schrodinger equation.
>
> Side comment:
>
> Your considerations are "antiquantum". A particle in
> quantum mechanics is not found at some place _because_ is was
> in a definite place before. This is disproved by double slits.

We were merely discussing the properties of
the solutions of the wave equations. But still,
Quantum field theory and relativistic wave
equations should both have a finite speed of
propagation, as far as I'm concerned..

...

>> With the Dirac equation all these strange things
>> do not happen,
>
> The Dirac equation has to be shown to be of relevance for the position
> wave function.

I would also be satisfied with a Klein-Gordon
equation, or the Maxwell equations. So let's
take the latter:

Take an EM wave packet restricted to a finite
region in space (zero fields outside). Now,
will a moving observer see fields *outside*
this region? Or even everywhere in space?

I would prefer to think one would see a Lorentz-
contracted, blue/red-shifted, wave packet that
is still restricted to the (transformed) finite
region. Where does this fail?

..

> If you consider solutions of the Dirac equation which vanish outside
> some bounded domain you can be sure that they are not the position
> wave functions of a spin-1/2 particle.

Why? (Of course if we cannot consider those
solutions then the paradox is gone..)

> You cannot avoid the Schroedinger equation for a position wave
> function (if the time evolution is linear). It just states that the
> total probability to find the particle somewhere is conserved for
> all states.
>
> i d_t Psi = sqrt(m^2+p^2) Psi
>
> is unavoidable for each free, relativistic particle of mass m. It
> disallows position wave functions which are restricted to a
> bounded carrier.

Is there a proof that one cannot avoid the
Schroedinger equation? Is probability
conservation not possible in another way?

..

> Position wave functions do not transform locally, whether they are
> part (two components) of a Dirac spinor or not. This follows from
> the k-dependence of the factor
>
> sqrt(k^0 / Lambda^0_n k^n) U(W(Lambda, k))
>
> in the transformation
>
> ps^~'(Lambda k) = sqrt(k^0 / Lambda^0_n k^n) U(W(Lambda, k)) psi~(k)
>
> of the momentum wave function. Do you agree on this transformation?

Where is the necessity coming from to have
this sqrt( k^0 / k^0' )?

--
Jos

[Norbert Dragon]

* Jos Bergervoet writes:

>* Norbert Dragon wrote:

> We were merely discussing the properties of
> the solutions of the wave equations. But still,
> Quantum field theory and relativistic wave
> equations should both have a finite speed of
> propagation, as far as I'm concerned..

page 85 and page 91

http://www.itp.uni-hannover.de/~dragon/rel_e.pdf

should convince you that I know how the solutions of wave equations
propagate.

The problem is, that the position wave function Psi and its time
derivative

i d_t Psi = sqrt(m^2 +p^2) Psi

cannot both be initially restricted to a bounded domain. The
Schroedinger equation is an uncertainty relation for both.

> I would also be satisfied with a Klein-Gordon
> equation, or the Maxwell equations. So let's
> take the latter:

The solutions of both the Klein-Gordon equation and the Maxwell
equations depend in each event E only on the initial conditions
and the inhomogeneities in the backward lightcone of E. But one
cannot localize both the position wave function and its time
derivative.

> Take an EM wave packet restricted to a finite
> region in space (zero fields outside).

A wave packet of electromagnetic fields is not the position
wavefunction of photons.

For photons it is not even possible to define a two component
position wave function which transforms locally under rotations
because one cannot define a real basis of transverse polarizations
for all directions of the momentum k.

> Now, will a moving observer see fields *outside*
> this region? Or even everywhere in space?

A localized wave packet appears localized to all inertial observers.
But no position wave function can be localized together with its
time derivative.

> Is there a proof that one cannot avoid the
> Schroedinger equation? Is probability
> conservation not possible in another way?

In my lectures on quantum mechanics I derive the Schroedinger equation
from the assumed linearity of the time evolution

Psi(t) = U(t) Psi(0)

the requirement that the evolution is differentiable and the
requirement that it preserves the overall probability

< Psi(t) | Psi(t) > = < Psi(0) | Psi(0) > for all Psi(0)

Consequently U(t) is unitary and d_t U antihermitian. This yields

i d_t Psi = H Psi

with some hermitian H. The Hamitonian H, which generates the time
evolution is by definition the energy in terms of momenta
(which generate spatial translations) and position. So

i d_t Psi = sqrt(m^2 + p^2) Psi

is inevitable in relativistic quantum mechanics.

>> psi^~'(Lambda k) = sqrt(k^0 / Lambda^0_n k^n) U(W(Lambda, k)) psi~(k)

> Where is the necessity coming from to have
> this sqrt( k^0 / k^0' )?

In text books the scalar product of "momentum wave functions"
(for scalar particles) is for convenience taken to be

< psi | chi > = integral d^3k / k^0 psi^*(k) chi(k)

where k^0 = sqrt(m^2 + k^2)

Then the transformation

(U psi)(k) = psi(Lambda^(-1) k)

is unitary (see e.g. page 97 rel_e.pdf). But psi and chi cannot be
exactly the momentum wave functions Psi and Chi, because the
probability density is |Psi(k)|^2 without a factor 1/sqrt(k^0)

So Psi(k) = psi(k) / sqrt(k^0) and the transformation of psi leads to

(U Psi)(Lambda k) = sqrt(k^0 / Lambda^0_n k^n) Psi(k)

--
Supersition brings bad luck

www.itp.uni-hannover.de/~dragon

[Jos Bergervoet]

On Apr 27, 12:39 am, Norbert Dragon <dra...@itp.uni-hannover.de>
wrote:
..

> The problem is, that the position wave function Psi and its time
> derivative
>
> i d_t Psi = sqrt(m^2 +p^2) Psi
>
> cannot both be initially restricted to a bounded domain. The
> Schroedinger equation is an uncertainty relation for both.

Only if you use the Schroedinger equation (which
has this time derivative). But this thread was
about the Dirac equation, which has another time
derivative.

>> I would also be satisfied with a Klein-Gordon
>> equation, or the Maxwell equations. So let's
>> take the latter:
>
> The solutions of both the Klein-Gordon equation and the Maxwell
> equations depend in each event E only on the initial conditions
> and the inhomogeneities in the backward lightcone of E. But one
> cannot localize both the position wave function and its time
> derivative.

This is not clear.. You apparently do not want
to call the solution to these equations position
wave functions? In that case, how can they lead
to any conclusion about the position wave function
at all?

>> Take an EM wave packet restricted to a finite
>> region in space (zero fields outside).
>
> A wave packet of electromagnetic fields is not the position
> wavefunction of photons.

Fine, but what *is* the position wave function?
And how do we show that it cannot be localized
for a nonzero interval of time?

> For photons it is not even possible to define a two component
> position wave function which transforms locally under rotations
> because one cannot define a real basis of transverse polarizations
> for all directions of the momentum k.

OK, perhaps we should forget about photons. The
thread was about the Dirac equation anyway!

...

>> Is there a proof that one cannot avoid the
>> Schroedinger equation? Is probability
>> conservation not possible in another way?
>
> In my lectures on quantum mechanics I derive the Schroedinger equation
> from the assumed linearity of the time evolution
>
> Psi(t) = U(t) Psi(0)

Dirac made the same requirement.

> the requirement that the evolution is differentiable and the
> requirement that it preserves the overall probability
>
> < Psi(t) | Psi(t)> =< Psi(0) | Psi(0)> for all Psi(0)

This requirement was especially dear to Dirac!

> Consequently U(t) is unitary and d_t U antihermitian. This yields
>
> i d_t Psi = H Psi
>
> with some hermitian H. The Hamitonian H, which generates the time
> evolution is by definition the energy in terms of momenta
> (which generate spatial translations) and position. So
>
> i d_t Psi = sqrt(m^2 + p^2) Psi
>
> is inevitable in relativistic quantum mechanics.

But Dirac found another equation! Didn't he use
the exact requirements that you mentioned?!

Are the negative energy solutions the problem?
But the interpretation as positrons solves
this.. Of course then the conserved probability
changes into electron probability minus
positron probability, but with particle pair
creation as a possibility we would not even
want the separate probability densities to
be conserved, I would say.. So OK, there isn't
a *positive-definite* conserved probability
density then, but there shouldn't be one!

--
Jos

[Hendrik van Hees]

On 29/04/12 02:34, Jos Bergervoet wrote:
> On Apr 27, 12:39 am, Norbert Dragon<dra...@itp.uni-hannover.de>
> wrote:
> ..
>> The problem is, that the position wave function Psi and its time
>> derivative
>>
>> i d_t Psi = sqrt(m^2 +p^2) Psi
>>
>> cannot both be initially restricted to a bounded domain. The
>> Schroedinger equation is an uncertainty relation for both.
>
> Only if you use the Schroedinger equation (which
> has this time derivative). But this thread was
> about the Dirac equation, which has another time
> derivative.

Right, neither the solutions of the equation above (which is NOT the
Schr?dinger equation, which reads in the most simple case (no magnetic
fields, no spin)

i \partial_t psi=(-\Delta/(2m)+V) psi

and is non-relativistic) nor the solutions of the Klein-Gordon or Dirac
equation have a proper interpretation as single-particle wave functions
as the position representation of single-particle states has in
nonrelativistic quantum mechanics. Only in low-energy scattering
processes and under the condition of binding energies small compared to
the mass of the particle in question for the bound-state problem, such
an interpretation makes an approximate sense, but let's stick to the
more simple case of scattering processes here.

The reason is simply that at high enough energies there's always a
non-zero probability to create new particles and you may even have a
final state, where the original particle doesn't appear anymore (like
pair annihilation e^+ e^- -> 2 gamma in QED).

Thus, the single-particle equations should be interpreted as equations
for quantized field operators in the Heisenberg picture of the time
evolution. A certain subclass of such field equations admit unitary
representations of the Poincare group, which act in a local way as in
classical field theory. Such quantum-field theoretical realizations of
relativistic quantum theory leads to the very successful local
quantum-field theoretical models as used, e.g., in the Standard Model of
elementary particles and various effective hadronic theories dealing
with high-energy particle or nuclear (heavy-ion) physics although even
this most simple realization of a relativistic quantum theory is not
fully understood mathematically yet.

The above pseudo-Schroedinger equation (which I wouldn't call so) falls
not into this category of local quantum-field theories, and it is thus
not clear to me, whether you can find a proper physical interpretation
for them, except (to a certain extent) as the description of a free
particle. In the interacting case, I think nobody has found a working
physical interpretation for such non-local relativistic field equations,
yet.

> This is not clear.. You apparently do not want
> to call the solution to these equations position
> wave functions? In that case, how can they lead
> to any conclusion about the position wave function
> at all?

For photons (and any massless quantum field with spin >=1) there's not
even an unanimous definition for a position operator at all, let alone a
position wave function. Even a plane-wave Fock basis for photons has its
difficulties (infrared problems) since they are not the appropriate
asymptotic states for soft photons, but you have to either resum
soft-photon ladder diagrams (see, e.g., Weinberg, QT of Fields, Vol. I)
or use the appropriate coherent states as the proper asymptotic states.
Both (perturbative) solutions of the infrared problem are more or less
equivalent. This tells you that even a Fock-space representation of
photons has its difficulties, let alone particularly single-photon
states. It's by far not trivial to prepare states with a fixed photon
number at all. Single-photon states are available nowadays by making
entangled photon pairs by parametric downconversion in a appropriate
birefridgerent crystal and using one as an idler photon to prepare
single-photon states on demand.

> Fine, but what *is* the position wave function?

There is non.

> And how do we show that it cannot be localized
> for a nonzero interval of time?

Of course, there exist wave-packet solutions in classical
electrodynamics, but they cannot be interpreted as wave-functions in a
quantum-theoretical sense.

> But Dirac found another equation! Didn't he use
> the exact requirements that you mentioned?!

Of course, I don't understand Norbert's arguments. Perhaps one should
read his manuscripts carefull, but I've not the time for this now.

>
> Are the negative energy solutions the problem?

In a sense yes, since if you take out the antiparticles from the Dirac
quantum field you necessarily end up with a non-local realization of the
Poincare group with all the problems for interpreting these models in
the case of interacting particles. It's not clear to me, how far you can
get with such non-local models for free particles, but that's anyway an
academic problem, since non-interacting particles are non-observable.

Note, that of course, there is the possibility to build a local-QFT
model for strictly neutral (massive or massless) spin-1/2 particles,
known as Majorana particles. It is not yet decided whether the neutrinos
are Dirac or Majorana particles.

> But the interpretation as positrons solves
> this.. Of course then the conserved probability
> changes into electron probability minus
> positron probability, but with particle pair
> creation as a possibility we would not even
> want the separate probability densities to
> be conserved, I would say.. So OK, there isn't
> a *positive-definite* conserved probability
> density then, but there shouldn't be one!

Precisely that's the point! Thus, the modern interpretation of the Dirac
equation is that of an equation for the quantum-field operator in the
Heisenberg field of a local QFT (like the Standard Model).

[Norbert Dragon]

================= Moderator's note =============================

I disagree with the claims below, but I think it's interesting enough to
discuss these issues to approve this posting.

HvH.
================================================================

* Hendrik van Hees writes:

>* Jos Bergervoet wrote:

>>* Norbert Dragon had written

>>> The problem is, that the position wave function Psi and its time
>>> derivative

>>> i d_t Psi =3D sqrt(m^2 +p^2) Psi

>>> cannot both be initially restricted to a bounded domain. The
>>> Schroedinger equation is an uncertainty relation for both.

>> Only if you use the Schroedinger equation (which
>> has this time derivative). But this thread was
>> about the Dirac equation, which has another time
>> derivative.

> Right, neither the solutions of the equation above (which is NOT the=20
> Schr=F6dinger equation, which reads in the most simple case (no magneti=
c=20
> fields, no spin)

> i \partial_t psi=3D(-\Delta/(2m)+V) psi

Horrible.=20

By your definition there is no quantum mechanical relativistic=20
particle and no Rabi oscillation of Kaons or neutrinos.

You should consider the relativistic energy momentum relation

E(p) =3D sqrt(m^2 + p^2)

and the Schroedinger equation

i d_t Psi =3D H Psi

which holds in all quantum mechanical systems as consequence of the
linearity and unitarity of the time evolution.


> nor the solutions of the Klein-Gordon or Dirac=20
> equation have a proper interpretation as single-particle wave functions=
=20
> as the position representation of single-particle states has in=20
> nonrelativistic quantum mechanics.=20

Your inability to relate solutions of the Klein-Gordon-equation to the=20
space of square integrable position wave functions is not a property
of the Klein-Gordon-equation. Position wave functions of relativistic=20
particles are positive energy solutions of the Klein-Gordon-equation.

> Only in low-energy scattering=20
> processes and under the condition of binding energies small compared to=
=20
> the mass of the particle in question for the bound-state problem, such=20
> an interpretation makes an approximate sense, but let's stick to the=20

> more simple case of scattering processes here.

Proof by lack of knowledge and imagination?

> The reason is simply that at high enough energies there's always a=20
> non-zero probability to create new particles and you may even have a=20
> final state, where the original particle doesn't appear anymore (like=20

> pair annihilation e^+ e^- -> 2 gamma in QED).

Do you claim that positions of high energy electrons cannot be=20
measured?=20

> Thus, the single-particle equations should be interpreted as equations=20
> for quantized field operators in the Heisenberg picture of the time=20
> evolution.=20

Thus? And you even have an argument for the correct "picture"?

Theorem: There cannot be any argument about the picture because
its choise amounts to the choise of a basis. But quantum mechanics
is independent of the chosen basis of the Hilbert space of states.

> A certain subclass of such field equations admit unitary=20
> representations of the Poincare group, which act in a local way as in=20
> classical field theory.=20

There are even equations which admit both local and non-local=20
unitary representations of the Poincare group, e.g. the=20
Klein-Gordon-equation.

To start a reasonably precise discussion of the position wave=20
functions, you should specify its relation to the momentum wave
function.

I take it for granted, that spatial momentum generates spatial
translations

[ X^i, P_j] =3D i \delta^i_j

Moreover, the transformation of position wave functions is determined
by the Poicare group to be

(U Psi~)(Lambda k) =3D sqrt(k^0/Lambda^0_n k^n) Psi~(k)

Do you agree? Do you agree that _therefore_ the position wave function=20
trasnforms nonlocally?

> The above pseudo-Schroedinger equation (which I wouldn't call so) falls=
=20
> not into this category of local quantum-field theories,=20

You write about your private categories.

Quantum field theory deals with the Fock space of relativistic=20
particles. The one particle states constitute the vector space of=20
square integrable momentum wave functions. Their Fourier transform
gives the position wave function of a particle -- if anything holds=20
but your arbitrary rules to allow of disallow this or that concept.

> In the interacting case, I think nobody has found a working=20
> physical interpretation for such non-local relativistic field equations=
,=20

Asymptotic states are non-interacting and have their usual meaning:
momentum wave functions are probability amplitudes for momenta, their
Fourier transform gives their position wave function.

If you deny this relation how do you derive the relation between=20
cross sections and S-matrix amplitudes? Do your cross sections
exist only for nonrelativistic particles, but not for neutrinos?

> For photons (and any massless quantum field with spin >=3D1) there's no=
t=20
> even an unanimous definition for a position operator at all, let alone =
a=20
> position wave function.=20

The problem with the position operator for a photon is _not_, that=20
there is none, but that there are as many as there are maps from=20
S^2 --> S^1 and that within this set, there is no point, which is=20
mapped by SO(3) as the points of S^2, i.e. for no choise do (X,Y,Z)=20
transform as a three vector under rotations.

The problem is related to the transversality condition a the helicity=20
states but is way beyond the level of the discussion here.

> It's by far not trivial to prepare states with a fixed photon=20
> number at all.=20

Psi =3D Integral d^3k psi~(k) a^dagger(k) |vacuum>

Moreover: the violation of Bell's inequalities is demonstrated by the
investigation of photon pairs. At least some experimentalists claim
to be able to prepare photon states with definite photon number and
they claim to be able to measure photon position.

>> Fine, but what *is* the position wave function?

> There is non.

What do you think is measured by detectors and photographic films?

If there is no position wave function how do you derive the relation=20
between cross sections and S-matrix amplitudes?=20

>> But Dirac found another equation! Didn't he use
>> the exact requirements that you mentioned?!

The Dirac equation for mass M is solved by spinors=20

psi =3D (Psi, sigma^n d_n Psi / M)

where Psi is the two component position wave function which satifies

i d_t Psi =3D sqrt(m^2+p^2) Psi

You only have to require this equation as restriction of the initial
values, then the Dirac equation guarantees that Psi is a solution to=20
the Schroedinger equation at all times.

--=20

[Hendrik van Hees]

On 29/04/12 21:24, Norbert Dragon wrote:
> ================= Moderator's note =============================
>
> I disagree with the claims below, but I think it's interesting enough to
> discuss these issues to approve this posting.
>
> HvH.
> ================================================================
>
> * Hendrik van Hees writes:
>
>> * Jos Bergervoet wrote:
>
>>> * Norbert Dragon had written
>
>>>> The problem is, that the position wave function Psi and its time
>>>> derivative
>
>>>> i d_t Psi =3D sqrt(m^2 +p^2) Psi
>
>>>> cannot both be initially restricted to a bounded domain. The
>>>> Schroedinger equation is an uncertainty relation for both.
>
>>> Only if you use the Schroedinger equation (which
>>> has this time derivative). But this thread was
>>> about the Dirac equation, which has another time
>>> derivative.
>
>> Right, neither the solutions of the equation above (which is NOT the=20
>> Schr=F6dinger equation, which reads in the most simple case (no magneti=
> c=20
>> fields, no spin)
>
>> i \partial_t psi=3D(-\Delta/(2m)+V) psi
>
> Horrible.

Why should the non-relativistic Schr?dinger equation for a particle in a
potential be possible? It leads to well-interpretable wave functions,
which is not the case for the relativistic equation.

>
> By your definition there is no quantum mechanical relativistic=20
> particle and no Rabi oscillation of Kaons or neutrinos.
>
> You should consider the relativistic energy momentum relation
>
> E(p) =3D sqrt(m^2 + p^2)

This is the dispersion relation for single-particle states of
(asymptotically) free relativistic particles. Nobody denies this.

>
> and the Schroedinger equation
>
> i d_t Psi =3D H Psi
>
> which holds in all quantum mechanical systems as consequence of the
> linearity and unitarity of the time evolution.

It is very well known that this equation has many difficulties, which
are solved by using quantum-field theory to define a many particle
theory. This you can read, e.g., in the first pages of Peskin/Schroeder.
The argument goes as follows:

If this equation defines the time evolution of a wave function, then the
time-evolution operator in position space is, in natural units with
hbar=c=1, given by

U(t,x,x')=<x|exp(-i t sqrt(p^2+m^2)|x'>,

where x are the position eigenvectors. It turns out to be a Bessel
function. The argument against the single-particle interpretation of
this "wave function" is already seen by evaluating the Fourier transform
for |x|>>t (i.e., the deep space-like region), leading to

U(t,x,x') ~ exp[-m sqrt(x^2-t^2)]

with some non-zero rational function as a factor. A single-particle
interpretation would thus imply a violation of causality since there
should be no signal propagation outside the light cone.

In QFT this problem is solved by the addition of negative-frequency
modes with annihilation operators in the momentum-mode expansion of the
quantum field (Feynman-Stueckelberg trick) and the possibility to invoke
the micro-causality condition (here x,y denote four vectors)

[phi(x),phi(y)]=0 for (x-y) space-like.

> Proof by lack of knowledge and imagination?
>
>> The reason is simply that at high enough energies there's always a=20
>> non-zero probability to create new particles and you may even have a=20
>> final state, where the original particle doesn't appear anymore (like=20
>> pair annihilation e^+ e^- -> 2 gamma in QED).
>
> Do you claim that positions of high energy electrons cannot be=20
> measured?

No, I only claim, what you've quoted above. Of course, the positions of
high-energy electrons can be measured to a certain degree of accuracy.
The difference to non-relativistic quantum theory is that the
localizability of a relativistic particle is more limited since you have
to probe the electron with other high-energy particles (e.g., photons)
to reach a high position resolution. This, on the other hand, doesn't
necessarily lead to a more precise measurement of the electron's
position, but to the creation of more particles.

>
>> Thus, the single-particle equations should be interpreted as equations=20
>> for quantized field operators in the Heisenberg picture of the time=20
>> evolution.=20
>
> Thus? And you even have an argument for the correct "picture"?
>
> Theorem: There cannot be any argument about the picture because
> its choise amounts to the choise of a basis. But quantum mechanics
> is independent of the chosen basis of the Hilbert space of states.

I haven't claimed to have a proof for the "right picture", but that the
"wave equations" for single-particle "wave functions" can be
successfully reinterpreted as equations of motion for Heisenberg-field
operators. There's no problem to formulate QFT in other pictures (modulo
the formal problems with the interaction picture, reflected by Haag's
theorem, which are usually ignored in the perturbative formulation of
the theory).

>
>> A certain subclass of such field equations admit unitary=20
>> representations of the Poincare group, which act in a local way as in=20
>> classical field theory.=20
>
> There are even equations which admit both local and non-local=20
> unitary representations of the Poincare group, e.g. the=20
> Klein-Gordon-equation.

Nobody denied this.

>
> To start a reasonably precise discussion of the position wave=20
> functions, you should specify its relation to the momentum wave
> function.
>
> I take it for granted, that spatial momentum generates spatial
> translations
>
> [ X^i, P_j] =3D i \delta^i_j

This is possible within local quantum field theories for massive
particles of any spin and for massless particles with spin 0 and 1/2.

>
> Moreover, the transformation of position wave functions is determined
> by the Poicare group to be
>
> (U Psi~)(Lambda k) =3D sqrt(k^0/Lambda^0_n k^n) Psi~(k)
>
> Do you agree? Do you agree that _therefore_ the position wave function=20
> trasnforms nonlocally?

One-particle momentum states of (asymptotically) free scalar particles
transform under LTs as

U(L) |p>=|Lp>,

where I have chosen the Wigner basis in the standard construction of the
unitary irreducible representations of the proper orthochronous Poincare
group, and p runs over the four-momenta fulfilling the "on-shell
condition". Unitarity of this transformation leads to the demand

<p|p'> ~ E(\vec{p}) \delta^{(3)}(\vec{p}-\vec{p}')

Usually the proportionality factor is chosen to be (2 pi)^3 2
E(\vec{p}), leading to the invariant scalar product in momentum space
representations of single-particle states,

<phi|psi>=int d^3 p/[(2 pi)^3 2 E(\vec{p})] phi^*(\vec{p}) \psi(\vec{p}).

This implies the behavior of the single-particle state's momentum-space
representation, quoted by you above.

>
>> The above pseudo-Schroedinger equation (which I wouldn't call so) falls=
> =20
>> not into this category of local quantum-field theories,=20
>
> You write about your private categories.

No, these are the categories used for some decades of successful
applications of local quantum field theories.

>
> Quantum field theory deals with the Fock space of relativistic=20
> particles. The one particle states constitute the vector space of=20
> square integrable momentum wave functions. Their Fourier transform
> gives the position wave function of a particle -- if anything holds=20
> but your arbitrary rules to allow of disallow this or that concept.

I don't understand, where there should be a contradiction. Of course,
all this is standard-quantum field theory, and I haven't denied this. It
has nothing to do with the fact that a single-particle wave function
interpretation for solutions of the c-number Klein-Gordon or Dirac
equation contradicts causality and thus has been given up early on.

As far as I know, the historically first successful interpretation of a
relativistic quantum theory has been Dirac's hole theory which is,
despite its somewhat complicated practical handling, equivalent to the
more modern formulation as quantum electrodynamics on Fock space.

>
>> In the interacting case, I think nobody has found a working=20
>> physical interpretation for such non-local relativistic field equations=
> ,=20
>
> Asymptotic states are non-interacting and have their usual meaning:
> momentum wave functions are probability amplitudes for momenta, their
> Fourier transform gives their position wave function.

Asymptotic states are asymptotically non-interacting. That's a crucial
detail! Non-interacting particles, of course never scatter and are thus
unobservable.

>
> If you deny this relation how do you derive the relation between=20
> cross sections and S-matrix amplitudes? Do your cross sections
> exist only for nonrelativistic particles, but not for neutrinos?

As is usually done in modern textbooks of quantum-field theory (of
course one has to use wave packets in the initial state rather than
momentum eigenstates, as detailed again quite nicely in
Peskin-Schroeder's textbook).

>
>> For photons (and any massless quantum field with spin>=3D1) there's no=
> t=20
>> even an unanimous definition for a position operator at all, let alone =
> a=20
>> position wave function.=20
>
> The problem with the position operator for a photon is _not_, that=20
> there is none, but that there are as many as there are maps from=20
> S^2 --> S^1 and that within this set, there is no point, which is=20
> mapped by SO(3) as the points of S^2, i.e. for no choise do (X,Y,Z)=20
> transform as a three vector under rotations.

Voila!

>
> The problem is related to the transversality condition a the helicity=20
> states but is way beyond the level of the discussion here.
>
>> It's by far not trivial to prepare states with a fixed photon=20
>> number at all.=20
>
> Psi =3D Integral d^3k psi~(k) a^dagger(k) |vacuum>

Sure, and all I said is that it is not easy to prepare such a state!

>
> Moreover: the violation of Bell's inequalities is demonstrated by the
> investigation of photon pairs. At least some experimentalists claim
> to be able to prepare photon states with definite photon number and
> they claim to be able to measure photon position.

Yes, sure. Where is the contradiction?

>
>>> Fine, but what *is* the position wave function?
>
>> There is non.
>
> What do you think is measured by detectors and photographic films?

What's measured is an interaction of a particle with the detector
material with a finite ("macroscopic") resolution. There's no
contradiction between such "localizations" of particles and the formal
developments of QFT we are discussing here (to the contrary, QFT
predicts such outcomes of measurements very successfully).

>
> If there is no position wave function how do you derive the relation=20
> between cross sections and S-matrix amplitudes?

I don't need a position-wave function to measure the "position" of
particles with detectors and photographic films, even of photons. Not
even in theory I need position-wave functions to derive S-matrix
elements (via the LSZ-reduction formalism). See my QFT manuscript or my
"H?here Quantenmechanik" lecture notes (in German) on my home page:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf
http://fias.uni-frankfurt.de/~hees/publ/hqm.pdf

In the first manuscript, I however use the finite quantization-volume
regularization to derive transition probabilities, which now I'd not use
anymore since the wave-packet description of scattering processes is
much more natural and even better suited to discuss our dispute here,
namely what's measured as "scattering cross sections".

>
>>> But Dirac found another equation! Didn't he use
>>> the exact requirements that you mentioned?!
>
> The Dirac equation for mass M is solved by spinors=20
>
> psi =3D (Psi, sigma^n d_n Psi / M)
>
> where Psi is the two component position wave function which satifies
>
> i d_t Psi =3D sqrt(m^2+p^2) Psi
>
> You only have to require this equation as restriction of the initial
> values, then the Dirac equation guarantees that Psi is a solution to=20
> the Schroedinger equation at all times.

Yes, nobody denies these mathematical facts, but this doesn't mean that
you can interprete these solutions as single-particle wave functions in
the sense of non-relativistic quantum mechanics since this violates
causality even for non-interacting particles (as quoted from
Peskin-Schroeder at the beginning of this posting).

[Jos Bergervoet]

On Apr 29, 9:24 pm, Norbert Dragon <dra...@itp.uni-hannover.de> wrote:

[Off-topic discussion of moderation policy snipped.]

=> Norbert: Can we first discuss the mathematical items
that gave confusion, before discussing physical meaning.
Notably these two:
1) Are there localized solutions that evolve locally
in time?
2) Is your (Norbert's) relativistic Schroedinger equation
indeed "inevitable", or are others possible?
See remarks below..

...

> Moreover, the transformation of position wave functions is determined
> by the Poicare group to be
>

> (U Psi~)(Lambda k) = sqrt(k^0/Lambda^0_n k^n) Psi~(k)

>
> Do you agree? Do you agree that _therefore_ the position

> wave function trasnforms nonlocally?

This question initially brought us to item 1), the
time evolution! You claimed then, that a localized
Schroedinger solution cannot also have a localized
time derivative. That is wrong! If it concerns your
own relativistic version! It is true for the
non-relativistic version (since infinite speed of
propagation occurs) but *not* for your one:

i d_t Psi = sqrt(m^2 + p^2) Psi

See proof below..

>>> But Dirac found another equation! Didn't he use
>>> the exact requirements that you mentioned?!
>
> The Dirac equation for mass M is solved by spinors
>

> psi = (Psi, sigma^n d_n Psi / M)

>
> where Psi is the two component position wave function which satifies
>

> i d_t Psi = sqrt(m^2+p^2) Psi

Exactly! Solutions of the Dirac equation *also*
satisfy your relativistic Schroedinger equation.
In fact each individual scalar component of the
spinor does!
So this resolves the paradox. You are right if
you say that it is "inevitable" that they satisfy
your equation. But you wouldn't be right in
claiming that it is inevitable to use only your
equation (Dirac & K-G also give the solutions!)

> You only have to require this equation as restriction of the initial
> values, then the Dirac equation guarantees that Psi is a solution to

> the Schroedinger equation at all times.

Glad to agree! And since the time derivative
of the Dirac solution is computable simply in
coordinate space, we can now immediately see
that this time derivative is *also localized*
if the solution itself is! And the time
evolution will show you a perfect finite-speed
behavior so it will be localized at all times.

This also solves the problem of the Lorentz
transformation: a moving observer will see
the solution localized as well! (Contrary to
earlier claims here, which mistakenly were
based on the non-relativistic equation, I
guess..)

Remaining question: are there solutions to
the relativistic Schroedinger equation that
are *not* obtainable as Dirac solutions by
just augmenting the spinor? And are these
perhaps non-local in their time evolution?
(As far as I see: No..)

--
Jos

[Norbert Dragon]

* Hendrik van Hees writes:

>* Norbert Dragon wrote:

>> [Excessive self quotation of a moderator's note snipped]

>>> Right, neither the solutions of the equation above (which is NOT the

>>> Schroedinger equation, which reads in the most simple case (no magnet=
ic
>>> fields, no spin)

>>> i \partial_t psi =3D (-\Delta/(2m)+V) psi

>> Horrible.

> Why should the non-relativistic Schroedinger equation for a particle in=
a=20
> potential be possible?=20

Can you explain your logic? I commented as "horrible" your narrow=20
minded restriction, to call=20

i d_t Psi =3D H Psi

the Schroedinger equation only in case that H is the Hamiltonian of
nonrelativistic motion in a potential. Much more generally it is the=20
time evolution equation of all quantum systems.

> It leads to well-interpretable wave functions,=20

> which is not the case for the relativistic equation.

There is no problem with the _interpretation_ of the position wave
function of relativistic particles: its modulus squared at x is the
probability density to find the particle at x.

It is only that the position wave function has an unwanted=20
property which you would like to disregard.

>> By your definition there is no quantum mechanical relativistic=3D20

>> particle and no Rabi oscillation of Kaons or neutrinos.

>> You should consider the relativistic energy momentum relation

>> E(p) =3D sqrt(m^2 + p^2)

> This is the dispersion relation for single-particle states of=20

> (asymptotically) free relativistic particles. Nobody denies this.

Then we agree? The position wave function of a relativistic particle
evolves in time with

i d_t Psi =3D H Psi with H =3D sqrt(m^2 + p^2) ?

You cannot have yes and no. If you choose no, you should tell us,
what you think the time evolution of the _position_wave_function_ is.
Please do not change the subject by speaking of the time evolution=20
of something else.

>> and the Schroedinger equation

>> i d_t Psi =3D H Psi

>> which holds in all quantum mechanical systems as consequence of the
>> linearity and unitarity of the time evolution.

> It is very well known that this equation has many difficulties,=20

I am not aware of any _difficulties_ of the _equation_. I only know
conclusions which are unwelcome to you -- but they are unavoidable.

> A single-particle interpretation would thus imply a violation of=20
> causality since there should be no signal propagation outside the=20
> light cone.

The Schroedinger equation propagates probability amplitudes, not=20
signals. A particle is not found at a place, be_cause_ it was=20
at some other place before. Your claimed violation of causality is
actually a nonlocal propagation of a probability amplitude.
This applies to the nonrelativistic Schroedinger equation and it=20
applies also to the relativistic Schroedinger equation,
only that in one case you find the property acceptable and in the
second case shocking.

No matter, what you feel, you have to accept the result: the
position wave function in relativistic physics as in=20
nonrelativistic physics cannot be restricted together with its
time derivative to a localized region.

Nowhere have you shown that the failure to strictly localize a=20
quantum particle leads to the violation of causality. I would=20
like to know at least your gedanken experiment.

> In QFT this problem is solved by the addition of negative-frequency=20
> modes=20

What you call "problem solved" is shutting your eyes and considering
questions with answers which please you more. But the problem is not
gone in quantum field theory. Quantum field theory is about operators
which act in a Hilbert space of states. The (massive) one particle=20
states satisfy

i d_t Psi =3D H Psi with H =3D sqrt(m^2+p^2) (1)

and therefore cannot be strictly localized together with their time=20
derivative.

>>> Thus, the single-particle equations should be interpreted as equation=
s

>>> for quantized field operators in the Heisenberg picture of the time

>>> evolution.

Relativistic Quantum mechanics does not leave freedom for=20
interpretation. The Poincar=E9 algebra contains the generators of=20
translations, which on one particle states can be chosen to act=20
multiplicatively on momentum wave functions. At p these are the=20
probability amplitudes to find the particle with momentum p,=20
if their scalar product is

< chi | phi > =3D Integral d^3k chi~^*(k) phi~(k) .

Their Fourier transform is the position wave function and satisfies (1)
whether this pleases you or not.

>>> A certain subclass of such field equations admit unitary

>>> representations of the Poincare group, which act in a local way as in

>>> classical field theory.

>>
>> There are even equations which admit both local and non-local

>> unitary representations of the Poincare group, e.g. the

>> Klein-Gordon-equation.

> Nobody denied this.

So at last, you agree that from the Klein-Gordon or Dirac equation=20
you cannot conclude whether its solutions transform this way or that
way. The solutions can transform locally or nonlocally.=20

Position wave functions transform nonlocally under Lorentz boosts.

>> To start a reasonably precise discussion of the position wave

>> functions, you should specify its relation to the momentum wave
>> function.

>> I take it for granted, that spatial momentum generates spatial
>> translations

>> [ X^i, P_j] =3D i \delta^i_j

> This is possible within local quantum field theories for massive=20

> particles of any spin and for massless particles with spin 0 and 1/2.

Nice that you agree. Now, please, choose the basis, in which the=20
position operators act multiplicatively on position wave functions
with scalar product

< chi | phi > =3D Integral d^3x chi^*(x) phi(x)

Do you deny that their time evolution is given by (1)? Do you
deny that therefore one cannot localize phi(x) and its time=20
derivative?

>>> The above pseudo-Schroedinger equation (which I wouldn't call so)=20

Have you any colleagues who hesitate to call

i d_t Psi =3D H Psi

the Schroedinger equation in case that H =3D sqrt(m^2+p^2)? Which name=20
do you use then?

> I don't understand, where there should be a contradiction. Of course,=20
> all this is standard-quantum field theory, and I haven't denied this. I=
t=20
> has nothing to do with the fact that a single-particle wave function=20
> interpretation for solutions of the c-number Klein-Gordon or Dirac=20

> equation contradicts causality and thus has been given up early on.

No one can "give up" (1), because it holds throughout relativistic
quantum mechanics. You can read (1) also as a restriction to the=20
initial data of the Klein-Gordon-equation or the Dirac equation.
Then these equations are satisfied and you cannot them "give up".=20
They are simply satisfied whether it pleases you or not.

You can chose to disregard the logical consequences because they do not
fit your prejudices. But this is bad science.

>> If you deny this relation how do you derive the relation between

>> cross sections and S-matrix amplitudes? Do your cross sections
>> exist only for nonrelativistic particles, but not for neutrinos?

> As is usually done in modern textbooks of quantum-field theory (of=20
> course one has to use wave packets in the initial state rather than=20
> momentum eigenstates, as detailed again quite nicely in=20
> Peskin-Schroeder's textbook).

Itzykson Zuber (page 200) need the position wave function psi(x) for
their derivation of the cross section. Do you seriously suggest that
one can avoid this concept?

>>> It's by far not trivial to prepare states with a fixed photon

>>> number at all.

>> Psi =3D Integral d^3k psi~(k) a^dagger(k) |vacuum>

> Sure, and all I said is that it is not easy to prepare such a state!

You say much irrelevant. Could you please stick to the point.
The point is, that position wave functions can not be localized
together with their time derivative. This property is not changed
by disregarding position wave functions.

>> If there is no position wave function how do you derive the relation

>> between cross sections and S-matrix amplitudes?

> I don't need a position-wave function to measure the "position" of=20
> particles with detectors and photographic films, even of photons. Not=20
> even in theory I need position-wave functions to derive S-matrix=20
> elements (via the LSZ-reduction formalism).=20

Your answer is not to the point.
I asked for the relation between _cross_section_ and S-matrix but you
answer about the relation between _Green_function_ and S-matrix.=20

> Yes, nobody denies these mathematical facts, but this doesn't mean that=
=20
> you can interpret these solutions as single-particle wave functions in=20
> the sense of non-relativistic quantum mechanics since this violates=20
> causality even for non-interacting particles=20

What you call violation of causality holds equally well in=20
nonrelativistic quantum mechanics. There you do not mind.
Your understanding of physics is rather selective.

--=20
Note to moderators: I deny you the right to add your notes to my text
or to snip parts of my posts.
I am old enough, sufficiently knowledgeable and consider the changing
of my contributions by a moderator a disrespectfulness.

[Jos Bergervoet]

On May 3, 1:13 pm, Norbert Dragon <dra...@itp.uni-hannover.de> wrote:
> * Hendrik van Hees writes:

...

> The Schroedinger equation propagates probability amplitudes, not

> signals.

Can you explain why it only should have positive
frequencies?

> No matter, what you feel, you have to accept the result: the
> position wave function in relativistic physics as in

> nonrelativistic physics cannot be restricted together with its
> time derivative to a localized region.

Why do you add "together with its time derivative"? Wouldn't
the claim hold for the function alone in any interval of time?
(And if you mean the situation at one point in time then it
actually does *not* hold for the non-relativistic case, even
with the addition..)

..

>> In QFT this problem is solved by the addition of negative-frequency

>> modes

>
> What you call "problem solved" is shutting your eyes and considering
> questions with answers which please you more.

But why isn't it allowed to have negative frequency
components in the probability amplitude?

> Relativistic Quantum mechanics does not leave freedom for

> interpretation. The Poincar=E9 algebra contains the generators of

> translations, which on one particle states can be chosen to act

> multiplicatively on momentum wave functions. At p these are the

> probability amplitudes to find the particle with momentum p,

> if their scalar product is
>

> < chi | phi> = Integral d^3k chi~^*(k) phi~(k) .

>
> Their Fourier transform is the position wave function and satisfies (1)
> whether this pleases you or not.

It would please me very much if you could say why
probability amplitude components should all be of
positive frequency!

--
Jos

[Hendrik van Hees]

Moderation comment: (a) First of all let me apologize that this posting
again got distorted due to encoding problems. This time it seems to be
the line breaks. I don't know, how to cure this with my system to
moderate the postings. Even with recode from utf8 to ascii I cannot get
rid of this disturbing effect.

(b) I make this note in my news-group reply to this posting since
Norbert Dragon doesn't like moderators' comments in his postings...

=========================================================================

On 03/05/12 13:13, Norbert Dragon wrote:
> * Hendrik van Hees writes:
>
>> * Norbert Dragon wrote:
>
>>> [Excessive self quotation of a moderator's note snipped]
>
>>>> Right, neither the solutions of the equation above (which is NOT the
>>>> Schroedinger equation, which reads in the most simple case (no magnet=
> ic
>>>> fields, no spin)
>
>>>> i \partial_t psi =3D (-\Delta/(2m)+V) psi
>
>>> Horrible.
>
>> Why should the non-relativistic Schroedinger equation for a particle in=
> a=20
>> potential be possible?=20
>
> Can you explain your logic? I commented as "horrible" your narrow=20
> minded restriction, to call=20
>
> i d_t Psi =3D H Psi
>
> the Schroedinger equation only in case that H is the Hamiltonian of
> nonrelativistic motion in a potential. Much more generally it is the=20
> time evolution equation of all quantum systems.

It is precisely, because I do not want to foster the idea, such an
equation can be interpreted as a relativistic one-particle wave function
with the same interpretation as a probability amplitude as in
non-relativistic physics since this equations shows non-local acausal
behavior, i.e., when you have a localized initial state, the
corresponding time evolved wave function has support outside the forward
light cone. This has been known already in the very early history of
quantum theory and has lead to the reinterpretation of local
relativistic field equations in terms of equations for quantized fields.
Historically, the first formulation of this reinterpretation was in
terms of Dirac's hole theory of his equation. The point is (a) that you
reinterpret the local relativistic wave equations (like the Klein-Gordon
or Dirac equation) in terms of a many-body theory for particles and
antiparticles, for the free-particle case represented by the
momentum-eigenmodes with positive AND negative frequencies,
respectively. This lead somewhat later (in the late 1940ies) to the
development of propagator theory, where the negative-frequency mode has
been reinterpreted as an anti-particle energy-momentum eigenstate with
positive energy. That's how you get within propagator theory to the
notion of the time-ordered propagator, where the positive-frequency
modes propagate in forward and the negative-frequency modes in backward
time direction. For a thorough representation of these more intuitive
ideas towards a (perturbative) relativistic quantum (scattering) theory,
see Bjorken/Drell volume 1. Finally, thanks to the work of Dyson's,
these heuristic arguments have been brought into a systematic
quantum-field theoretical framework, leading after all to the very
successfull Standard Model of elementary particles (and perhaps soon to
extensions of the standard model when really "new physics" is found at
the LHC).

>
>> It leads to well-interpretable wave functions,=20
>> which is not the case for the relativistic equation.
>
> There is no problem with the _interpretation_ of the position wave
> function of relativistic particles: its modulus squared at x is the
> probability density to find the particle at x.

However, there IS the problem of time-like evolution, which lead to the
reinterpretation of the "wave functions" in terms of a many-body theory
or local relativistic qft (in the following, I'll stick to the modern
way of speaking about local quantum field theories only).

>
> It is only that the position wave function has an unwanted=20
> property which you would like to disregard.

To the contrary! I don't want to disregard this unwanted property, but
use the standard interpretation.

You write this even in your own manuscript on quantum mechanics (only in
the German version),

http://www.itp.uni-hannover.de/~dragon/stonehenge/qm.pdf

you even explain and mathematically prove these unwanted properties
(non-locality, acausality) in great detail right after Eq. (9.41) on p.
109 (see also your Appendix D, where the statement in Peskin-Schroeder,
which I mentioned in my previous posting, is proven in terms of Bessel
and modified Bessel functions and their asymptotic expansion. The very
fact of the there discussed non-locality of the wave function's time
evolution leads to the violation of Einstein causality, if interpreted
as a single-particle wave function as in non-relativistic QM, and thus a
many-body interpretation of inevitable.

In addition, for interacting fields you cannot restrict the time
evolution to the positive-frequency modes, as has been already found in
the very early days of the development of relativistic quantum theory by
Schroedinger, who thus restricted himself to the non-relativistic case,
and also Dirac realized this with his first-order equation for spin-1/2
particles (now known as the Dirac equation). He thus switched to a
many-body interpretation in terms of his Dirac sea, nowadays known as
the hole-theoretical representation of QED. This you can read in compact
and concise form in Weinberg's Quantum Theory of Fields, Vol. I
(introductory chapter on history of QFT).

>
>>> By your definition there is no quantum mechanical relativistic=3D20
>>> particle and no Rabi oscillation of Kaons or neutrinos.
>
>>> You should consider the relativistic energy momentum relation
>
>>> E(p) =3D sqrt(m^2 + p^2)
>
>> This is the dispersion relation for single-particle states of=20
>> (asymptotically) free relativistic particles. Nobody denies this.
>
> Then we agree? The position wave function of a relativistic particle
> evolves in time with
>
> i d_t Psi =3D H Psi with H =3D sqrt(m^2 + p^2) ?

This is not the equation for wave function but for a positive-frequency
mode of the Klein-Gordon equation for a quantized free scalar field. As
explained above this wording makes an important difference in terms of
the physical interpretation!

>
> You cannot have yes and no. If you choose no, you should tell us,
> what you think the time evolution of the _position_wave_function_ is.

There is no single-particle position-wave function. That's my whole point!

> Please do not change the subject by speaking of the time evolution=20
> of something else.

The important point is that you must speak about the time evolution of
something else, namely many-body quantum theory in terms of a
relativistic local QFT.

>
>>> and the Schroedinger equation
>
>>> i d_t Psi =3D H Psi
>
>>> which holds in all quantum mechanical systems as consequence of the
>>> linearity and unitarity of the time evolution.
>
>> It is very well known that this equation has many difficulties,=20
>
> I am not aware of any _difficulties_ of the _equation_. I only know
> conclusions which are unwelcome to you -- but they are unavoidable.

They are not unavoidable, but solved already by Dirac with his hole
theory (btw. after the important correction by Oppenheimer, that the
holes in electron theory cannot represent protons but must have
precisely the same mass as electrons; that's why ironically Dirac has
missed the prediction of the existence of antiparticles although he
already had them at his finger tips with the idea of hole theory :-))
and finally put in a systematic framework in terms of a QFT by Dyson.

>
>> A single-particle interpretation would thus imply a violation of=20
>> causality since there should be no signal propagation outside the=20
>> light cone.
>
> The Schroedinger equation propagates probability amplitudes, not=20
> signals. A particle is not found at a place, be_cause_ it was=20
> at some other place before. Your claimed violation of causality is
> actually a nonlocal propagation of a probability amplitude.
> This applies to the nonrelativistic Schroedinger equation and it=20
> applies also to the relativistic Schroedinger equation,
> only that in one case you find the property acceptable and in the
> second case shocking.

Sure, in the nonrelativistic case, there is no problem, since there
faster-than light propagation is none. If you interpret the relativistic
positive-frequence-mode solutions (or better the corresponding wave
packets) in terms of a probability amplitude (and thus its square as a
position-probability distribution), you can in principle have
faster-than light propagation (of course only in a probabilistic sense,
and the corresponding probability distributions are exponentially damped).

This is avoided in a QFT interpretation by the well-known fact that
there the physically observable propagation of causal signals by
construction is by the retarded response functions. This is built into
QFT by construction by the choice of commutation (bosons) or
anti-commutation (fermions) relations for the field operators that
vanish at space-like distances between the space-time arguments of these
field operators.

>
> No matter, what you feel, you have to accept the result: the
> position wave function in relativistic physics as in=20
> nonrelativistic physics cannot be restricted together with its
> time derivative to a localized region.

Precisely that's my point!

>
> Nowhere have you shown that the failure to strictly localize a=20
> quantum particle leads to the violation of causality. I would=20
> like to know at least your gedanken experiment.

It was not me but Schroedinger, Dirac, Heisenberg, Born, Jordan, and
many other early "quantum mechanists" in the early twentieth century who
have found these problems and solved in terms of a reinterpretation as
many-body theories and the prediction of the existence of antiparticles.

>
>> In QFT this problem is solved by the addition of negative-frequency=20
>> modes=20
>
> What you call "problem solved" is shutting your eyes and considering
> questions with answers which please you more. But the problem is not
> gone in quantum field theory. Quantum field theory is about operators
> which act in a Hilbert space of states. The (massive) one particle=20
> states satisfy
>
> i d_t Psi =3D H Psi with H =3D sqrt(m^2+p^2) (1)
>
> and therefore cannot be strictly localized together with their time=20
> derivative.

Now you switch to the Schroedinger picture. That's fine with me, but I
don't see, how you want to save the interpretation of the states in
terms single-particle-wave functions. The Hilbert-space of states is
usually taken to be a Fock space of quantum field theory (which has its
not completely solved problems either, but we are not yet at a point in
the discussion that we could talk about these now).

>
>>>> Thus, the single-particle equations should be interpreted as equation=
> s
>>>> for quantized field operators in the Heisenberg picture of the time
>>>> evolution.
>
> Relativistic Quantum mechanics does not leave freedom for=20
> interpretation. The Poincar=E9 algebra contains the generators of=20
> translations, which on one particle states can be chosen to act=20
> multiplicatively on momentum wave functions. At p these are the=20
> probability amplitudes to find the particle with momentum p,=20
> if their scalar product is
>
> < chi | phi> =3D Integral d^3k chi~^*(k) phi~(k) .
>
> Their Fourier transform is the position wave function and satisfies (1)
> whether this pleases you or not.

No, the interpretation as a position-wave function for this quantity
leads to contradictions with causality as explained at length above.

>
>>>> A certain subclass of such field equations admit unitary
>>>> representations of the Poincare group, which act in a local way as in
>>>> classical field theory.
>>>
>>> There are even equations which admit both local and non-local
>>> unitary representations of the Poincare group, e.g. the
>>> Klein-Gordon-equation.
>
>> Nobody denied this.
>
> So at last, you agree that from the Klein-Gordon or Dirac equation=20
> you cannot conclude whether its solutions transform this way or that
> way. The solutions can transform locally or nonlocally.=20

Of course, quantum-field theory can be formally built as representation
theory of the proper orthochronous Poincare group (Wigner 1939, in great
detail given by Weinberg in QT of Fields, Vol. I). Among those are also
the local representations with their characteristic feature that the
field operators contain both annihiliation operators (in front of
positive-frequency modes) and creation operators (in front of
negative-frequency) modes in order to give local representations of
proper orthochronous Poincare transformations. These local QFTs are the
so far successful ones in high-energy particle and nuclear physics.

>
> Position wave functions transform nonlocally under Lorentz boosts.

I'm not aware that any non-local representations have lead to successful
physical models. If you know an example in the literature, please let us
know!

>
>>> To start a reasonably precise discussion of the position wave
>>> functions, you should specify its relation to the momentum wave
>>> function.
>
>>> I take it for granted, that spatial momentum generates spatial
>>> translations
>
>>> [ X^i, P_j] =3D i \delta^i_j
>
>> This is possible within local quantum field theories for massive=20
>> particles of any spin and for massless particles with spin 0 and 1/2.
>
> Nice that you agree. Now, please, choose the basis, in which the=20
> position operators act multiplicatively on position wave functions
> with scalar product
>
> < chi | phi> =3D Integral d^3x chi^*(x) phi(x)
>
> Do you deny that their time evolution is given by (1)? Do you
> deny that therefore one cannot localize phi(x) and its time=20
> derivative?

Why should I deny this? This precisely is an argument FOR and not
against of what I'm saying.

It's clear also that the non-localizability of relativistic particles
has a very physical reason: If you want to localize a particle you have
to measure its position. The better you wish to localize it the higher
momentum-transfers in scattering have to be applied. If you come with
your resolution in the order of magnitude of the Compton wavelength 2 pi
hbar/m you come close or above the threshold for pair production of the
particle. Thus rather than getting a better localization of your
particle to be measured you make new particles. This has been shown in
an early famous paper by Bohr and Rosenfeld by studying the uncertainty
relation in the relativistic realm (see Landau/Lifshits Vol. IV for a
discussion).

>
>>>> The above pseudo-Schroedinger equation (which I wouldn't call so)=20
>
> Have you any colleagues who hesitate to call
>
> i d_t Psi =3D H Psi
>
> the Schroedinger equation in case that H =3D sqrt(m^2+p^2)? Which name=20
> do you use then?

It's the evolution equation for positive-frequency modes in the
decomposition of a free-field operator in terms of annihilation and
creation operators wrt. the single-particle momentum-eigenbasis of the
corresponding Fock space. I never had problems in discussions with any
colleague, using this language. This is found at the very beginning of
nearly any introductory lecture on relativistic QFT when the most simple
case of a free scalar (neutral or charged) field is discussed.

>
>> I don't understand, where there should be a contradiction. Of course,=20
>> all this is standard-quantum field theory, and I haven't denied this. I=
> t=20
>> has nothing to do with the fact that a single-particle wave function=20
>> interpretation for solutions of the c-number Klein-Gordon or Dirac=20
>> equation contradicts causality and thus has been given up early on.
>
> No one can "give up" (1), because it holds throughout relativistic
> quantum mechanics. You can read (1) also as a restriction to the=20
> initial data of the Klein-Gordon-equation or the Dirac equation.
> Then these equations are satisfied and you cannot them "give up".=20
> They are simply satisfied whether it pleases you or not.

No, they are not satisfied for the usual local QFTs, where the field
operator is composed of both positive- and negative-frequency modes.
E.g., the neutral scalar field, just to quote the most simple example, reads

Phi(t,x)=int d^3 p [u_p(t,x) a(p) + u_p^*(t,x) a^{dagger}(p)],

with

u_p(t,x)=1/sqrt[(2 pi)^3 E(p)] exp[-i E(p) t + i p.x],

where I've chosen the normalization convention such that

[a(p),a^dagger(p')]=delta^{(3)}(p-p')

>
> You can chose to disregard the logical consequences because they do not
> fit your prejudices. But this is bad science.

It's not a prejudice but the experience of several decades of brillant
physicists who have given us local relativistic QFT as a very successful
method to describe experimental facts of high-energy particle and
nuclear physics.

>
>>> If you deny this relation how do you derive the relation between
>>> cross sections and S-matrix amplitudes? Do your cross sections
>>> exist only for nonrelativistic particles, but not for neutrinos?
>
>> As is usually done in modern textbooks of quantum-field theory (of=20
>> course one has to use wave packets in the initial state rather than=20
>> momentum eigenstates, as detailed again quite nicely in=20
>> Peskin-Schroeder's textbook).
>
> Itzykson Zuber (page 200) need the position wave function psi(x) for
> their derivation of the cross section. Do you seriously suggest that
> one can avoid this concept?

There I don't see anything else than what I see in Peskin Schroeder,
namely a wave-packet description of S-matrix theory as a thorough
discussion of cross sections within a local relativistic QFT demands.
There is no contradiction between Itzykson/Zuber and my point of view.
How could it, since this is the link between the QFT formalism with
phenomena in particle physics, making all the success of QFT in terms of
comparison between theory and experiment!

>
>>>> It's by far not trivial to prepare states with a fixed photon
>>>> number at all.
>
>>> Psi =3D Integral d^3k psi~(k) a^dagger(k) |vacuum>
>
>> Sure, and all I said is that it is not easy to prepare such a state!
>
> You say much irrelevant. Could you please stick to the point.
> The point is, that position wave functions can not be localized
> together with their time derivative. This property is not changed
> by disregarding position wave functions.

Ok, then please don't mention massless particles at this point of the
discussion anymore. This is even more complicated than what we discuss
right now, since there the plane-wave modes and the corresponding
Fock-space concept for the corresponding asymptotic states becomes
problematic.

>
>>> If there is no position wave function how do you derive the relation
>>> between cross sections and S-matrix amplitudes?
>
>> I don't need a position-wave function to measure the "position" of=20
>> particles with detectors and photographic films, even of photons. Not=20
>> even in theory I need position-wave functions to derive S-matrix=20
>> elements (via the LSZ-reduction formalism).=20
>
> Your answer is not to the point.

It is! There is no necessity to use a wave-function interpretation for
the plane-wave modes in asymptotic states of local relativistic QFT and
the calculation of S-matrix elements, which lead to observable
quantities like scattering cross sections, and this is precisely to the
point of what we discuss here!

> I asked for the relation between _cross_section_ and S-matrix but you
> answer about the relation between _Green_function_ and S-matrix.=20

Just read one page further in Itzykson Zuber. On page 201 you find the
relation between the S-matrix elements cross sections. Of course, I
considered that you and anybody else following the discussion here are
aware of this relation.

>
>> Yes, nobody denies these mathematical facts, but this doesn't mean that=
> =20
>> you can interpret these solutions as single-particle wave functions in=20
>> the sense of non-relativistic quantum mechanics since this violates=20
>> causality even for non-interacting particles=20
>
> What you call violation of causality holds equally well in=20
> nonrelativistic quantum mechanics. There you do not mind.
> Your understanding of physics is rather selective.

There is no causility violation in non-relativistic physics, and
action-at-a-distance laws as well as faster-than-light signal or
particle propagation are no problems at all. Why should it be? In
relativistic physics, of course, these are problems and have been solved
by the reinterpretation of the single-particle solutions of relativistic
field equations for local fields as mode functions in the plane-wave
decomposition of local field operators, i.e., within a many-body
interpretation as explained repeatedly above.

>
> --=20
> Note to moderators: I deny you the right to add your notes to my text
> or to snip parts of my posts.

Point taken.

> I am old enough, sufficiently knowledgeable and consider the changing
> of my contributions by a moderator a disrespectfulness.

I never implied any disrespectfulness against any poster to this
newsgroups when moderating their postings. I also do not change any
posters' texts.

[Norbert Dragon]

* Jos Bergervoet writes:

>* Norbert Dragon wrote:

> [Off-topic discussion of moderation policy snipped.]

The text of my contributions is mine. If the moderators
of this group claim the right, to distort my contributions
by inserting their comments into my postings, I will stop
posting to this group. I will also stop to post if this
statement is snipped.

>=> Norbert: Can we first discuss the mathematical items
> that gave confusion, before discussing physical meaning.

> Notably these two:
> 1) Are there localized solutions that evolve locally
> in time?
> 2) Is your (Norbert's) relativistic Schroedinger equation
> indeed "inevitable", or are others possible?

> You claimed [...], that a localized

> Schroedinger solution cannot also have a localized
> time derivative. That is wrong! If it concerns your
> own relativistic version!

What do you mean with my "own relativistic version"?

Do you suggest not to call

i d_t Psi = H Psi (1)

the Schroedinger equation or not to call

H = sqrt(m^2 + p^2) (2)

the Hamitonian of a massive relativistic particle?

How would you react, if I claimed these equations to be "my own"?
I am flattered. Should we call it the Dragon-Witten-equation?

H = P^0 is the generator of time evolution throughout
quantum mechanics. Not to call (1) the Schroedinger equation
proves ignorance of quantum mechanics and not to call (2) the
Hamiltonian of a free, massive relativistic particle proves ignorance
of relativistic physics and of the one-particle Fock space of
quantum field theory.

> It is true for the non-relativistic version (since infinite speed
> of propagation occurs) but *not* for your one:

> i d_t Psi = sqrt(m^2 + p^2) Psi

> See proof below..

I suspect that this is your proof:

> And since the time derivative
> of the Dirac solution is computable simply in
> coordinate space, we can now immediately see
> that this time derivative is *also localized*
> if the solution itself is! And the time
> evolution will show you a perfect finite-speed
> behavior so it will be localized at all times.

No. To construct a Dirac spinor

psi = (Psi, sigma^n d_n psi / M)

out of the two component position wave function Psi you need a
two component function Psi, which at least at time t=0 solves

i d_t Psi = sqrt(m^2 + p^2) Psi

If Psi is localized, then d_t Psi and the Dirac spinor psi is _not_!
Proof by Fourier transformation and the convolution theorem. In
particular, sqrt(m^2+p^2) is not the Fourier transformation of a
localized function plus some delta-functions and their derivatives.

Therefore the position wave function of a spin-1/2 particle psi(t=0, x)
is never localized.

>> The Dirac equation for mass M is solved by spinors

>> psi = (Psi, sigma^n d_n Psi / M)

>> where Psi is the two component position wave function which satifies

>> i d_t Psi = sqrt(m^2+p^2) Psi

> So this resolves the paradox.

Not at all: the Dirac equation, the wave equation and Maxwell's
equation have the property, that their solutions at some time t
and some point x depend only on the initial values in the backward
lightcone of (t,x) but this does not mean, that one can choose the
initial values of the position wave functions to be localized.
One cannot. Position wave functions are contained in the positive
energy solutions of the Dirac equation and they cannot be localized
strictly.

So in nonrelativistic quantum mechanics one cannot localized a position
wave function because of the instanteneous propagation and in
relativistic physics the Schroedinger equation involves a Hamiltonian
which acts nonlocally on one particle position wave functions.

> You are right if you say that it is "inevitable" that they satisfy
> your equation. But you wouldn't be right in
> claiming that it is inevitable to use only your
> equation (Dirac & K-G also give the solutions!)

Not all solutions of the Dirac or Klein-Gordon equation can be
position wave functions, because the Hamitonian is

H = sqrt(m^2 + p^2)

In particular, H is bounded from below.
This applies also for antiparticles.

>> You only have to require this equation as restriction of the initial
>> values, then the Dirac equation guarantees that Psi is a solution to
>> the Schroedinger equation at all times.

> This also solves the problem of the Lorentz transformation:

No.

> Remaining question: are there solutions to
> the relativistic Schroedinger equation that
> are *not* obtainable as Dirac solutions by
> just augmenting the spinor?

^^^^^^^^^^
projecting the spinor to its first two components

Clearly no. I pointed out repeatedly, that each two component
solution Psi of the Schroedinger equation yields a Dirac spinor

psi = (Psi, sigma^n d_n Psi / M)

which solves the Dirac equation.

--

[Hendrik van Hees]

On 05/05/12 18:36, Jos Bergervoet wrote:
> On May 3, 1:13 pm, Norbert Dragon<dra...@itp.uni-hannover.de> wrote:
>> * Hendrik van Hees writes:
> ...
>> The Schroedinger equation propagates probability amplitudes, not
>> signals.
>
> Can you explain why it only should have positive
> frequencies?

If you want to interpret the Klein-Gordon equation as describing the
time evolution of a relativistic wave function for a (free) particle,
then you must assume that the Hamiltonian has only positive eigenvalues,
because you want an energy spectrum which is bounded from below, i.e.,
that there exists a stable ground state (state of minimal) energy. Since
at the same time you want a Lorentz-invariant model (i.e. a theory that
is (at least) invariant under proper orthochronous Poincare
transformations), you have to demand positivity of the energy since
energy and momentum together build a time-like (or light-like) four
vector, and under proper orthochronous Poincare transformations the sign
of the 0-component of such a four-vector is invariant.

Now, Norbert Dragon takes as the wave equation not the local
Klein-Gordon equation but the non-local equation,

i \partial_t phi(x)=sqrt(-Delta+m^2) phi(x).

The solutions of this equation obviously are uniquely determined by
giving an initial condition of the wave function alone since it's of
first order in the time derivative. The general solution can be given in
terms of a Fourier transformation as

phi(x)=\int d^3 \vec{p}/(2 pi)^3 A(\vec{p}) \exp[-i
x.p]|_{p0=sqrt{\vec{p}^2+m^2}}.

Obviously one has

A(\vec{p})=\int d^3 \vec{x} phi(t=0,\vec{x}) exp(-i \vec{x} . \vec{p}),

and thus the solution of the equation is uniquely determined by giving
the solution at t=0.

Now, it is easy to show that the invariant (retarded) propagator of this
equation has support for space-like distances (see, e.g., Peskin
Schroeder). Thus, any solution which has finite support in space at t=0,
has at later times support also outside the forward lightcone.

If you now interpret such a solution as a probability distribution for
position as a function of time, you have a finite probability to find a
particle to propagate with faster-than-light speed, which is not
acceptable as a causal model for free-particle propagation within the
Special Theory of Relativity.

That's why this interpretation of the positive-frequency modes of the
Klein-Gordon equation (or, equivalently, the above non-local first-order
equation) has been abandoned as a model for free-particle motion in
relativistic quantum theory, and the usual "microcausal" local QFTs have
been developed by the early quantum theorists like Pauli, Jordan, Born,
Heisenberg, and of course Dirac.

Dirac looked your a linear and local Poincare covariant framework and
has found his famous equation. But also there, you have the same trouble
with a single-particle wave-function interpretation of the free equation
when admitting only positive energies.

For interacting particles, and after all only theories describing
interacting particles give observable consequences since we cannot
detect particles that are not interacting with detectors in the lab, the
trouble is even worse: At least with local interactions, usually these
interactions lead out of the subspace of positive-frequency solutions.

Dirac then had the genious idea to interpret the negative-energy states
as occupied by a sea of particles, and then call this state the vacuum.
This lead finally to the prediction of anti-particles and the
formulation of QED in it's representation as the "hole theory", since in
Dirac's picture positrons (antielectrons) are holes in the filled Dirac
sea of negative-energy electrons, leading to an antiparticle with
positive energy and the opposite charge of an electron. Finally, all
this has been put together by Tomonaga, Schwinger, Feynman, and finally
Dyson in terms of local, microcausal quantum field theory, and that's
the way we learn about relativistic quantum theory nowadays, and thus
nobody discusses the above non-local first-order equation.

For a complete discussion of these issues, see

S. Weinberg, Quantum Theory of Fields, Vol. 1

>
>> No matter, what you feel, you have to accept the result: the
>> position wave function in relativistic physics as in
>> nonrelativistic physics cannot be restricted together with its
>> time derivative to a localized region.
>
> Why do you add "together with its time derivative"? Wouldn't
> the claim hold for the function alone in any interval of time?
> (And if you mean the situation at one point in time then it
> actually does *not* hold for the non-relativistic case, even
> with the addition..)

This is only the local description of the above statement that there are
no Einstein-causal solutions of this equation.

[Norbert Dragon]

* Jos Bergervoet writes:

>* Norbert Dragon wrote:

> Can you explain why it only should have positive
> frequencies?

Energies of free, relativistic particles are positive and bounded from
below.

In particular, the Hamitonian in relativistic quantum field theory
acts on one-particle states as

H = sqrt(M^2 + p^2) (1)

because H = P^0 acts in Fock space as

H = Integral d^3k sqrt(m^2 + k^2) a^dagger(k) a(k)

Even if you do not deal with quantum field theory, the Hamiltonian
has to be given by H = sqrt(m^2 + p^2), because H = P^0 is the
time component of a four vector under Lorentz transformations.

[M_mn , P_k ] = i (eta_mk P_n - eta_nk P_n)

enforces (1) in case that the spectrum for p=0 contains the discrete
value M.

>> No matter, what you feel, you have to accept the result: the
>> position wave function in relativistic physics as in
>> nonrelativistic physics cannot be restricted together with its
>> time derivative to a localized region.

> Why do you add "together with its time derivative"? Wouldn't
> the claim hold for the function alone in any interval of time?
> (And if you mean the situation at one point in time then it
> actually does *not* hold for the non-relativistic case, even
> with the addition..)

You can choose the initial value of the position wave function to be
localized to a bounded domain, but, in the relativistic case, not
its time derivative.

In the nonrelativistic case, I have to correct my statement:
d_t Psi(t=0) = i Delta / 2 M Psi(t=0) is also localized, but not
Psi(t=epsilon) for any epsilon > 0 .

>> What you call "problem solved" is shutting your eyes and considering
>> questions with answers which please you more.

> But why isn't it allowed to have negative frequency
> components in the probability amplitude?

All particles have positive energy. Their Hamiltonian is given by (1).

If you allowed also particles with negative energies, this would
not help, but lead to completely unacceptable models. Negative
energy particles are distinguishable from positive energy
particles. A one-particle state would therefore be given by momentum
wave functions for two species of particles and their Fourier transform
would yield position wave functions with doubled number of components.
Their modulus square is the the probability density to find the positive
or negative energy species of the particle at some point in space.
Non of these these position wave functions would be localizable.

Worse: in each model with positive and negative energy particles the
vacuum is unstable against the decay into pairs of positive and
negative energy particles -- if there is any interaction among these
two species of particles.

--

[Norbert Dragon]

* Hendrik van Hees schreibt:

>* Jos Bergervoet wrote:

>* Norbert Dragon had written:

>> Can you explain why it only should have positive
>> frequencies?

> If you want to interpret the Klein-Gordon equation as describing the
> time evolution of a relativistic wave function for a (free) particle,

> [...]

> Now, Norbert Dragon takes as the wave equation not the local
> Klein-Gordon equation but the non-local equation,

Do not tell others, what I take. I do not "interpret the Klein-Gordon
equation" and choose this or that equation but I deduce from the

linearity of the time evolution

Psi(t) = U(t) Psi(0)

and the fact, that the overall probability is conserved for any initial
Psi(0) that U(t) is unitary. Therefore, if the time evolution is
differentiable, the Schroedinger equation

i d_t Psi = H Psi , H = i (d_t U) U^dagger = H^dagger

is inevitable. Therefore:

Theorem: The time evolution of each quantum system is generated by a
hermitean operator H, called the Hamiltonian.

Do not tell others that you can choose something else.

Definition: A quantum system is relativistic, if its Hamitonian H = P^0
is the the time component of the momentum 4 vector and element of a
Hermitian representation of the Poicare Lie algebra acting on the
Hilbert space.

Therfore, on one particle states H = sqrt(m^2+p^2) . Their time
evolution is therefore

i d_t Psi = (sqrt(m^2 + p^2) Psi

Do not tell others that you can choose something else.
You only choose to disregard the equation and its consequences.

This equation is not taken by me but inavoidable in relativistic
quatum mechanics -- it holds for one particle states in each
relativistic quantum field theory.

> Now, it is easy to show that the invariant (retarded) propagator of this
> equation has support for space-like distances (see, e.g., Peskin
> Schroeder).

Which, by the way, contains in this connection a serious mathematical
error, though they cite Gradsteyn Ryshik (they seem to be unaware of
what a factor i or -1 does in the exponent): The propagator is _not_,
contrary to their claim and their citation, a Bessel function, but the
second derivative of a product of a Bessel function with step functions
-- their derivatives yield also delta-functions.

A careful derivation of the propagator can be found at Bogoliubov
Shirkov which I used in my notes on quantum mechanics.

> If you now interpret such a solution as a probability distribution for
> position as a function of time, you have a finite probability to find a
> particle to propagate with faster-than-light speed, which is not
> acceptable as a causal model for free-particle propagation within the
> Special Theory of Relativity.

There is no freedom of interpretation: given any measuring device A:
the probability to obtain its (discrete) result a_i if you measure
the state Psi, is

p(a_i, A, Psi) = |<Lambda_i|Psi>|^2

This is the heart of quantum mechanics which you cannot interpret away.

The position operator X^i is defined to generate translations of the
momentum and vice versa,

[X^i, P_j] = i delta^i_j .

Therefore the Fourier transformation of the momentum wave functions is
the position wave function of the particle.

> That's why this interpretation of the positive-frequency modes of the
> Klein-Gordon equation (or, equivalently, the above non-local first-order
> equation) has been abandoned as a model for free-particle motion in
> relativistic quantum theory,

The properties of the position wave function follow from first
principles, which I have enumerated. Rather than to mumble about
"interpretation" you should either accept the inevitable or state
which of the fundamental structures of quamtum mechanics you want
to sacrify.

You already argued, that there is no position wave function and you
also conceded that for massive relativistic particles there exist
position operators. They commute, satisfy the Heisenberg algebra,
so they can be represented as multiplicative operators acting on
square integrable position wave functions.

It seems that you do not care to contradict yourself.

> and the usual "microcausal" local QFTs have
> been developed by the early quantum theorists like Pauli, Jordan, Born,
> Heisenberg, and of course Dirac.

The microcausality condition does not apply to the position wave
function but to operators which have different properties. Among others,
they transform locally while position wave functions cannot.
Microcausality does not tell anything about the possibility to
localize a _state_.

You should distinguish between a state and an operator.

> Dirac looked your a linear and local Poincare covariant framework and
> has found his famous equation. But also there, you have the same trouble
> with a single-particle wave-function interpretation of the free equation
> when admitting only positive energies.

Which trouble? Is is a conclusion, not a trouble, an undeniable result,
that position wave functions cannot be localized over a time intervall.

--

[Jos Bergervoet]

On May 5, 6:38=A0pm, Norbert Dragon <dra...@itp.uni-hannover.de> wrote:
...

> Do you suggest not to call
>
> i d_t Psi = H Psi (1)
>
> the Schroedinger equation

If it is explicitly written as a wave equation for
the position wave function:

i d_t Psi(r'') = \int H(r'',r') Psi(r') d^3r'

with H the convolutor of the type K_2(|r|)/r^2
where r = |r''-r'| and K_2 a modified Bessel,
then I would hesitate! Probably, the equation

i d_t Psi(r'') = -1/2m \nabla^2 Psi

deserves the title more.. However, if someone
else extends the usage to the relativistic case
it would go along with it (as I immediately did
in this thread).

..

>> And since the time derivative
>> of the Dirac solution is computable simply in
>> coordinate space, we can now immediately see
>> that this time derivative is *also localized*
>> if the solution itself is! And the time
>> evolution will show you a perfect finite-speed
>> behavior so it will be localized at all times.
>
> No. To construct a Dirac spinor
>
> psi = (Psi, sigma^n d_n psi / M)
>
> out of the two component position wave function Psi you need a
> two component function Psi, which at least at time t=0 solves
>
> i d_t Psi = sqrt(m^2 + p^2) Psi
>
> If Psi is localized, then d_t Psi and the Dirac spinor psi is _not_!

Yes. I overlooked the difficulty in the step of
constructing the "small" components to go with
"large" components localized at t=0. (Of course at
t>0 I saw it.. but couldn't cancel the post :^) )

..

> So in nonrelativistic quantum mechanics one cannot localized a position
> wave function because of the instanteneous propagation and in
> relativistic physics the Schroedinger equation involves a Hamiltonian
> which acts nonlocally on one particle position wave functions.

I wouldn't blame the Hamiltonian. It's the solution
space that is to blame (the restriction to positive
frequencies). We could blame the Fourier transform!
It is not possible to construct a function which is
localized together with its time derivative by summing
only positive-frequency plane waves. That is just a
plane-wave summation property. Regardless of any
time evolution equation you are going to apply!

So, strict selection of "positive frequencies only" makes
non-local wavefunctions a mathematically inevitable fact.

--
Jos

[Norbert Dragon]

* Hendrik van Hees writes

>* Norbert Dragon wrote:

>> Can you explain your logic? I commented as "horrible" your narrow

>> minded restriction, to call
>>
>> i d_t Psi = H Psi

>>
>> the Schroedinger equation only in case that H is the Hamiltonian of
>> nonrelativistic motion in a potential. Much more generally it is the

>> time evolution equation of all quantum systems.

> It is precisely, because I do not want to foster the idea, such an
> equation can be interpreted as a relativistic one-particle wave function
> with the same interpretation as a probability amplitude as in
> non-relativistic physics since this equations shows non-local acausal
> behavior,

Laughing out loudly.

What is your time evolution of a quantum state? Anything different from

Psi(t) = U(t) Psi(0) ?

Do you sacrify unitary time evolution?

> i.e., when you have a localized initial state, the corresponding time
> evolved wave function has support outside the forward light cone.

Such is life. The position wave function cannot be localized strictly.

No observation conflicts with this theoretical result, that the
localization is not strict but that only exponential falloff in units
of the Compton wave length is possible.

Do you seriously suggest that states in quantum mechanics do not change
in time via a unitary time evolution? Do you seriously suggest that the
generator of time evolution of one particle states is not
H = sqrt(m^2 + p^2) ?

> [Historical detour snipped]

Your historical detour diverts from simple arguments.

Do you suggest to deviate from the time evolution

Psi(t) = exp(- i H t) Psi(0) with H = sqrt(m^2 + p^2)

for _one_particle_states_?

>> There is no problem with the _interpretation_ of the position wave
>> function of relativistic particles: its modulus squared at x is the
>> probability density to find the particle at x.

> However, there IS the problem of time-like evolution, which lead to the
> reinterpretation of the "wave functions" in terms of a many-body theory
> or local relativistic qft (in the following, I'll stick to the modern
> way of speaking about local quantum field theories only).

You cannot "reinterpret" the wave function unless you leave quantum
mechanics and quantum field theory.

Do you agree, that one particle states are characterized by momentum
wave functions and that their Fourier transform gives the position
wave function?

If you disagree, spare us a historical detour, but tell us where you
disagree.

>> It is only that the position wave function has an unwanted=20
>> property which you would like to disregard.

> To the contrary! I don't want to disregard this unwanted property, but
> use the standard interpretation.

What do you accept?

Is the modulus squared |Psi~(k)|^2 of the momentum wave function the
probability density to find the particle with momentum k?

Does momentum generate spatial translation and vice versa,

[X^i, P_j] = i delta^i_j ?

Then the Fourier transformation of Psi~ is the position wave function.

> You write this even in your own manuscript on quantum mechanics (only in
> the German version),

> http://www.itp.uni-hannover.de/~dragon/stonehenge/qm.pdf

> you even explain and mathematically prove these unwanted properties

So what? You cannot cite me for a counter-argument. I prove the
unwanted properties and I argue that they cannot be avoided.

The objects with nice properties are other objects though they are
given similar names, e.g. quantum fields are called psi as if they
were wave functions. They are not.

The similarity of names does not "reinterpret" wave functions. The
authors rather consider other objects and disregard the property
that postition wave functions cannot be localized.

> (non-locality, acausality) in great detail right after Eq. (9.41) on p.

> 109 (see also your Appendix E, where the statement in Peskin-Schroeder,

> which I mentioned in my previous posting, is proven

Peskin Schroeder claim something wrong and wrongly cite Gradsteyn
Ryshik. But we should not care too much about their missing delta
function and its derivative. At least we agree qualitatively. The
book is logically shallow whenever it comes to subtle questions --
e.g. you will not find (as you do in Boboliubov Shirkov) that the
divergencies of Feynman diagrams are related to the fact that Wicks
theorem holds only for noncoinciding space time points (because there
the time order is undefined) or that the zero point energy of H is
uniquely fixed to vanish by Lorentz invariance.

> In addition, for interacting fields you cannot restrict the time
> evolution to the positive-frequency modes,

Do you seriously argue that interaction improves the localizability?
At the same time, you work on the hypothethis, that quantum field
theory is formally analytic in the coupling.

For vanishing coupling you cannot localize the position wave function.
How can a perturbation change this?

> [Historical detour snipped]

Why do you evade my arguments? They are simple enough.

>> i d_t Psi = H Psi with H = sqrt(m^2 + p^2) ?

> This is not the equation for wave function but for a positive-frequency
> mode of the Klein-Gordon equation for a quantized free scalar field.

Stick to some logic! Your "but" is logically wrong.

If the Schroedinger equation applies to the positive frequency
part of a free scalar field, this does not exculde it to apply
to one-particle states. It applies to one-particle states.

Question: Do in your mind one particle states in the Schroedinger
picture change in the course of time according to something
different from

Psi(t) = exp(-i H t) Psi(0) with H = sqrt(m^2 + p^2)

>> You cannot have yes and no. If you choose no, you should tell us,
>> what you think the time evolution of the _position_wave_function_ is.

> There is no single-particle position-wave function. That's my whole point!

You agreed already that there are position operators X^i, which are part of
a Heisenberg algebra. So in a suitable basis they act multiplicatively on
position wave functions. Do you withdraw your statement about position
operators or about "the whole point"? You cannot have both.

By the way, granted there were no one particle position wave function then
one could not localize the particle. So where do you disagree?

>> Please do not change the subject by speaking of the time evolution

>> of something else.

> The important point is that you must speak about the time evolution of
> something else, namely many-body quantum theory in terms of a
> relativistic local QFT.

Why do I have to speak about something else? Are there dementors around
or even moderators? Who can force me not to analyse and conclude that
one cannot strictly localize a relativistic particle?

And what does it help to consider many particles if already one
particle cannot be localized? Can two be localized or three?

But the longer you argue the more I get the impression that you
thoroughly agree: one cannot localize a particle because acoording
to your revolutionarily fuzzy ideas there are no particles at all.

I allow to argue that this is not what standard quantum field theory
and high energy physics is about and that readers should be warned that
your statements may be beyond the borderline of this scientific group.

In relativistic physics there is a one particle momentum wave function
which yields the probability density for momentum. Or do you deny even
that?

I stop at this point. We should clarify the simple questions first
before we clarify the rest.

--
I do not like the fool's cap put on the head of my contribution by a
well-meaning moderator who tries to shield the participants of this
news group from arguments which he considers potentially harmful to
their the mental health.

[Hendrik van Hees]

On 06/05/12 11:58, Norbert Dragon wrote:

> Do not tell others, what I take. I do not "interpret the Klein-Gordon
> equation" and choose this or that equation but I deduce from the
> linearity of the time evolution
>
> Psi(t) = U(t) Psi(0)
>
> and the fact, that the overall probability is conserved for any initial
> Psi(0) that U(t) is unitary. Therefore, if the time evolution is
> differentiable, the Schroedinger equation
>
> i d_t Psi = H Psi , H = i (d_t U) U^dagger = H^dagger
>
> is inevitable. Therefore:
>
> Theorem: The time evolution of each quantum system is generated by a
> hermitean operator H, called the Hamiltonian.
>
> Do not tell others that you can choose something else.
>
> Definition: A quantum system is relativistic, if its Hamitonian H = P^0
> is the the time component of the momentum 4 vector and element of a
> Hermitian representation of the Poicare Lie algebra acting on the
> Hilbert space.
>
> Therfore, on one particle states H = sqrt(m^2+p^2) . Their time
> evolution is therefore
>
> i d_t Psi = (sqrt(m^2 + p^2) Psi
>
> Do not tell others that you can choose something else.
> You only choose to disregard the equation and its consequences.
>

I never said that I've chosen something else, I only do not interpret
the postive-frequency modes of the KG equation as single-particle wave
functions. That's the common way relativistic quantum field theory is
done nowadays, namely in terms of a relativistic quantum field theory
with particles and antiparticles (or the special case of strictly
neutral particles), where one uses CAUSAL local quantum-field theory to
describe particles in the relativistic realm. See any modern textbook on
QFT like Peskin Schroeder, or the in my opinion best one by S. Weinberg,
where it is explained in a very concise way, why one uses causal local
QFT and not something more general, which from a mathematical point of
view may be possible, but hasn't lead to any success in the description
of scattering processes in high-energy particle physics, while the local
qft picture (even with the further restriction to derive the dynamics
from the action principle and with canonical quantization) has this
phenomenological success on its side!

> This equation is not taken by me but inavoidable in relativistic
> quatum mechanics -- it holds for one particle states in each
> relativistic quantum field theory.

As you well know, that's not true. In relativistic QFT, for the most
simple case of a strictly neutral scalar field, you have

phi(x)=\int d^3 \vec{p} N(\vec{p}) [a(\vec{p}) exp(-i p.x)
+a^{dagger}(\vec{p}) exp(+i p.x)]_{p^0=+\sqrt{m^2+\vec{p}^2}.

Herel, of course, phi is a field OPERATOR, and a(\vec{p}) is an
annihilation operator for a single-particle momenum-eigenstate. Using
the renormalization factor N(\vec{p})=1/[(2 \pi)^3 2
sqrt(m^2+\vec{p}^2)], these fulfill the commutation relations

[a(\vec{p}),a^{\dagger}(\vec{p}')]=\delta^{(3)}(\vec{p}-\vec{p}').

The Hilbert space of states is the Fock space, describing multi-particle
situations, and the Hamiltonian is given by the integral over the
time-time component of the normal-ordered energy-momentum tensor of the
KG field.

The QFT is constructed(!!!) in such a way that local observables,
particularly the energy-density operator, T^{00}(x), commute at
space-like distances, and this is one possibility to construct a causal
S-matrix theory (see Weinberg QT of Fields, Vol. 1).

>
>> Now, it is easy to show that the invariant (retarded) propagator of this
>> equation has support for space-like distances (see, e.g., Peskin
>> Schroeder).
>
> Which, by the way, contains in this connection a serious mathematical
> error, though they cite Gradsteyn Ryshik (they seem to be unaware of
> what a factor i or -1 does in the exponent): The propagator is _not_,
> contrary to their claim and their citation, a Bessel function, but the
> second derivative of a product of a Bessel function with step functions
> -- their derivatives yield also delta-functions.
>
> A careful derivation of the propagator can be found at Bogoliubov
> Shirkov which I used in my notes on quantum mechanics.

Ok, this I have to check. Indeed, Peskin-Schroeder is not always very
careful with the mathematics. In Bjorken Drell, Vol. II Appendix C (I've
only the German edition of this book, but I hope the chapters are as in
the American original), they give the commutator function as

Delta(x)=1/4 pi r \partial_r {\pm J_0[m sqrt(t^2-r^2)] for t>r, t<-r,
0 for t<r}

This shows that the usual retarded propagator of QFT,

Delta_ret(x)=-Theta(t) Delta(x),

which is the response function to perturbations with external fields in
free-particle approximation of linear-response theory, has the right
causality feature, namely to vanish outside the light-cone.

However, for the non-local equation, you cut the negative-frequency
contributions, and then also the even function,

Delta_1(x)=1/(4 pi r) \partial_r [Y_0[m sqrt(t^2-r^2)] for |t|>r,
-2/pi K_0[m sqrt(r^2-t^2) for |t|<r],

which does not vanish at spacelike x.

In the perturbative evaluations of the S-matrix of course you have the
time ordered (Feynman) propgators, which are retarded (advanced) for the
postive- (negative-)frequency eigenmodes of the KG equation,
reinterpreting the negative-frequency solutions as positive-energy
excitations moving forward in time (for the charged KG field, these are
the antiparticle modes).

>
>> If you now interpret such a solution as a probability distribution for
>> position as a function of time, you have a finite probability to find a
>> particle to propagate with faster-than-light speed, which is not
>> acceptable as a causal model for free-particle propagation within the
>> Special Theory of Relativity.
>
> There is no freedom of interpretation: given any measuring device A:
> the probability to obtain its (discrete) result a_i if you measure
> the state Psi, is
>
> p(a_i, A, Psi) = |<Lambda_i|Psi>|^2
>
> This is the heart of quantum mechanics which you cannot interpret away.

But that's what's done in everyday-life in terms of local causal QFT!
I'm sure, you don't want to imply that this very successful formulation
of relativistic quantumt theory is wrong.

>
> The position operator X^i is defined to generate translations of the
> momentum and vice versa,
>
> [X^i, P_j] = i delta^i_j .
>
> Therefore the Fourier transformation of the momentum wave functions is
> the position wave function of the particle.

Fine with me, as long as you can give such operators. This is possible
for massive particles of any spin and for massless particles with spin 0
and spin 1/2.

>
>> That's why this interpretation of the positive-frequency modes of the
>> Klein-Gordon equation (or, equivalently, the above non-local first-order
>> equation) has been abandoned as a model for free-particle motion in
>> relativistic quantum theory,
>
> The properties of the position wave function follow from first
> principles, which I have enumerated. Rather than to mumble about
> "interpretation" you should either accept the inevitable or state
> which of the fundamental structures of quamtum mechanics you want
> to sacrify.
>
> You already argued, that there is no position wave function and you
> also conceded that for massive relativistic particles there exist
> position operators. They commute, satisfy the Heisenberg algebra,
> so they can be represented as multiplicative operators acting on
> square integrable position wave functions.
>
> It seems that you do not care to contradict yourself.

I only say that these functions are not causal and thus are not used in
everyday PHYSICS applications of relativistic QFT.

>
>> and the usual "microcausal" local QFTs have
>> been developed by the early quantum theorists like Pauli, Jordan, Born,
>> Heisenberg, and of course Dirac.
>
> The microcausality condition does not apply to the position wave
> function but to operators which have different properties. Among others,
> they transform locally while position wave functions cannot.

So, you finally agree to that, but that's what you've also written in
your own manuscript.

> Microcausality does not tell anything about the possibility to
> localize a _state_.

No, it doesn't, but it's more important to the consistency of
relativistic quantum theory and to observations than the localizability
of relativistic particles! That's why the single-particle interpretation
of the dicussed wave packets, constructed with only positive frequency
modes, have been given up since the early times of relativistic quantum
theory.

>
> You should distinguish between a state and an operator.

That's what I do all the time!

>
>> Dirac looked your a linear and local Poincare covariant framework and
>> has found his famous equation. But also there, you have the same trouble
>> with a single-particle wave-function interpretation of the free equation
>> when admitting only positive energies.
>
> Which trouble? Is is a conclusion, not a trouble, an undeniable result,
> that position wave functions cannot be localized over a time intervall.

It's an experimental fact that antiparticles exist in accordance with
the local causal QFT formulation, in some cases with an overwhelming
accuracy with the prediction of this formulation!

[Jos Bergervoet]

On May 6, 12:31 am, Norbert Dragon <dra...@itp.uni-hannover.de> wrote:

> * Jos Bergervoet writes:
>
>> Can you explain why it only should have positive
>> frequencies?
>
> Energies of free, relativistic particles are positive and bounded from
> below.

OK, but now you directly make the step from frequencies
to energies. And even then the requirement seems to be
more desirable than inevitable.. Any other theory would
seem to have instability, or some other kind of strange
behavior.. But can we be sure of it?

...

> Even if you do not deal with quantum field theory, the Hamiltonian
> has to be given by H = sqrt(m^2 + p^2), because H = P^0 is the
> time component of a four vector under Lorentz transformations.
>
> [M_mn , P_k ] = i (eta_mk P_n - eta_nk P_n)
>
> enforces (1) in case that the spectrum for p=0 contains the discrete
> value M.

But couldn't we imagine a second branch of states on
which the Hamitonian acts as:

H = -sqrt(M^2 + p^2) (1b)

In particular, how can we ever construct a Majorana
spinor wave function, which should essentially be
real? (You probably say they do not have a position
wave function, just like the photon?)

...

>> Why do you add "together with its time derivative"? Wouldn't
>> the claim hold for the function alone in any interval of time?
>> (And if you mean the situation at one point in time then it
>> actually does *not* hold for the non-relativistic case, even
>> with the addition..)
>
> You can choose the initial value of the position wave function to be
> localized to a bounded domain, but, in the relativistic case, not
> its time derivative.

Only if "relativistic case" means the Hamiltonian of (1)
and not the Dirac Hamiltonian. In the latter case, if
the spinor is bounded (all components) then the time
derivative is automatically bounded as well and the time
evolution exp(itH) nicely constrained to the light cone.
But then negative frequencies will be present!

Of course you can construct purely positive-frequency
solutions for the Dirac equation but then not all
components can be simultaneously localized. Not even
at one time t.

..

>>> What you call "problem solved" is shutting your eyes and considering
>>> questions with answers which please you more.
>
>> But why isn't it allowed to have negative frequency
>> components in the probability amplitude?
>
> All particles have positive energy. Their Hamiltonian is given by (1).

It certainly would please me more if they have positive
energy. But the only proof I see is that mixing the two
signs would, well.. not "please" us!

> If you allowed also particles with negative energies, this would
> not help, but lead to completely unacceptable models. Negative
> energy particles are distinguishable from positive energy
> particles. A one-particle state would therefore be given by momentum
> wave functions for two species of particles and their Fourier transform
> would yield position wave functions with doubled number of components.

Yes, Dirac for instance. But why is that a completely
unacceptable model? The increased number of components
also is acceptable if we introduce color, hypercharge,
or ordinary spin.

> Their modulus square is the the probability density to find the positive
> or negative energy species of the particle at some point in space.
> Non of these these position wave functions would be localizable.

This is incorrect. Dirac spinors are localizable and light-
cone preserving, precisely by allowing negative frequency
solutions! (Whithout that they are not.) And in particular
Majorana spinor wave functions are localizable.

> Worse: in each model with positive and negative energy particles the
> vacuum is unstable against the decay into pairs of positive and
> negative energy particles -- if there is any interaction among these
> two species of particles.

Wouldn't that depend on the precise details of the interaction?
Also, there is no decay into photons (or Majorana Fermions)
even though we do allow negative frequencies. What is the
Majorana wave function if it is not the probability
amplitude? (I mean the "its-own-antiparticle" Majorana..)

And what is the meaning of the Dirac probability (all components
squared), so nicely constructed to be a conserved quantity?

--
Jos

[Hendrik van Hees]

On 06/05/12 00:31, Norbert Dragon wrote:

> H = sqrt(M^2 + p^2) (1)
>
> because H = P^0 acts in Fock space as
>
> H = Integral d^3k sqrt(m^2 + k^2) a^dagger(k) a(k)

> All particles have positive energy. Their Hamiltonian is given by (1).

Within local causal QFT the free Hamiltonian is indeed given by the
formula below Eq. (1), which indeed is defined in Fock space. It's the
spatial integral over the normal ordered 00 component of the (for the
free case) conserved energy-momentum tensor.

A free (or asymptotically free) single-particle momentum eigenstate is
at the same time an eigenstate of the non-interacting Hamiltonian with
the eigenvalue E(\vec{p})=sqrt(\vec{p}^2+m^2). That's the usual meaning
of Eq. (1) within QFT.

>
> If you allowed also particles with negative energies, this would
> not help, but lead to completely unacceptable models. Negative
> energy particles are distinguishable from positive energy
> particles. A one-particle state would therefore be given by momentum
> wave functions for two species of particles and their Fourier transform
> would yield position wave functions with doubled number of components.
> Their modulus square is the the probability density to find the positive
> or negative energy species of the particle at some point in space.
> Non of these these position wave functions would be localizable.

Of course, all particles (and antiparticles) have positive energy.
That's why you reinterpret the negative-frequency modes as antiparticles
(or more generally the hole excitations of the asymptotic vacuum, if you
have a strictly neutral particle, as implied by the free-particle
Hamiltonian above).

>
> Worse: in each model with positive and negative energy particles the
> vacuum is unstable against the decay into pairs of positive and
> negative energy particles -- if there is any interaction among these
> two species of particles.

That's a feature, not a bug of quantum field theory!

The vacuum for itself is of course not unstable but to the contrary the
stable state of minimal energy, representing that state of the QFT, that
no particles or fields are present. That antiparticles exist and that in
relativistic collisions particle-antiparticle pairs are produced and
annihilated all the time as long as such processes are "allowed" by the
observed conservation laws for energy, momentum, angular momentum, and
various charges, is a well-established fact.

What to my knowledge has not been observed so far is the instability of
the vacuum due to strong electric and/or magnetic fields, predicted in
famous papers by Sauter (within the hole-theoretical formulation) and
Schwinger within the modern QFT formulation (Schwinger-pair production).
This is an example for interesting active research; of course not with
static fields but with strong laser sources.

[Jos Bergervoet]

On 5/6/2012 3:01 PM, Jos Bergervoet wrote:
...

> It is not possible to construct a function which is
> localized together with its time derivative by summing
> only positive-frequency plane waves.

If they are on mass-shell, I meant of course! (If
not, you could just give each of them frequency +1)

[Hendrik van Hees]

On 06/05/12 15:49, Norbert Dragon wrote:

> Do you agree, that one particle states are characterized by momentum
> wave functions and that their Fourier transform gives the position
> wave function?

I stick to the well-established and successful formulation of
relativistic quantum theory in form of local causal quantum field
theory. I do not interpret single-particle modes as wave functions since
this contradicts causality. I don't need to repeat what I've written in
detail in several previous postings and what you can read in any solid
textbook of QFT. I recommend Weinberg's QT of Fields, where this is
discussed in a modern way and going from the most general possibilities
to construct unitary irreps. of the Poincare group to the so far only
physically successful realization of a relativistic quantum theory,
namely that through local causal quantum fields.

>
> If you disagree, spare us a historical detour, but tell us where you
> disagree.

The history of science can help a great deal when it comes to discuss,
why certain ansatzes, that have been made in the past, have been
rejected, and your ansatz to interpret solutions of the non-local
Schroedinger-like equation, i.e., those solutions of the Klein-Gordon
equation that are restricted to the positive-frequency Fourier modes
modes, has been rejected from the first moment by the founders of
qunantum theory (Schroedinger, Heisenberg, Dirac,...).

>
>>> It is only that the position wave function has an unwanted=20
>>> property which you would like to disregard.
>
>> To the contrary! I don't want to disregard this unwanted property, but
>> use the standard interpretation.
>
> What do you accept?
>
> Is the modulus squared |Psi~(k)|^2 of the momentum wave function the
> probability density to find the particle with momentum k?

Why shouldn't I accept this?

>
> Does momentum generate spatial translation and vice versa,
>
> [X^i, P_j] = i delta^i_j ?

If there exists such a position operator, of course yes.

>
> Then the Fourier transformation of Psi~ is the position wave function.

In which sense? For me a position wave function should not give acausal
results. Thus I wouldn't accept a non-causal solution as a position wave
function.

Also we know, what happens when we try to localize a particle with an
accuracy in the order of magnitude of this particle's Compton
wavelength: You create new particles rather than get this precision in
localization. This is the resolution of these problems.

>
>> You write this even in your own manuscript on quantum mechanics (only in
>> the German version),
>
>> http://www.itp.uni-hannover.de/~dragon/stonehenge/qm.pdf
>
>> you even explain and mathematically prove these unwanted properties
>
> So what? You cannot cite me for a counter-argument. I prove the
> unwanted properties and I argue that they cannot be avoided.

They ARE avoided by not interpreting them as wave functions and using
local causal QFT instead!

>
> The objects with nice properties are other objects though they are
> given similar names, e.g. quantum fields are called psi as if they
> were wave functions. They are not.

Yes! That's the very point!

>
> The similarity of names does not "reinterpret" wave functions. The
> authors rather consider other objects and disregard the property
> that postition wave functions cannot be localized.

They do not disregard this fact but use it to argue for the use of local
causal quantum field theories, and thus a many-body reinterpretation of
the equations in question.

>
>> (non-locality, acausality) in great detail right after Eq. (9.41) on p.
>> 109 (see also your Appendix E, where the statement in Peskin-Schroeder,
>> which I mentioned in my previous posting, is proven
>
> Peskin Schroeder claim something wrong and wrongly cite Gradsteyn
> Ryshik. But we should not care too much about their missing delta
> function and its derivative. At least we agree qualitatively. The
> book is logically shallow whenever it comes to subtle questions --
> e.g. you will not find (as you do in Boboliubov Shirkov) that the
> divergencies of Feynman diagrams are related to the fact that Wicks
> theorem holds only for noncoinciding space time points (because there
> the time order is undefined) or that the zero point energy of H is
> uniquely fixed to vanish by Lorentz invariance.

I fully agree. Peskin/Schroeder is in fact a very sloppy book, which I
don't like too much. It was only the first one, I had at hand to point
to these very common arguments against a single-particle interpretation
of relativistic wave equations in the same sense as done in the
"first-quantization formulation" of non-relativistic QT. Much better is
Weinberg. I'm not famliar with Bogoliubov/Shirkov, but I guess, I can
have a look at this at the library, but I'm pretty sure that
particularly these authors refrain from such an interpretation.

>
>> In addition, for interacting fields you cannot restrict the time
>> evolution to the positive-frequency modes,
>
> Do you seriously argue that interaction improves the localizability?
> At the same time, you work on the hypothethis, that quantum field
> theory is formally analytic in the coupling.

I never said this. It's most probably not the case. Nobody has found
exact solutions to realistic interacting QFTs so far. Even the proof of
existence is not yet achieved.

>
> For vanishing coupling you cannot localize the position wave function.
> How can a perturbation change this?

I never that that one can localize the "position wave function", and
that's precisely why I don't want to call them position wave functions.

>
>> [Historical detour snipped]
>
> Why do you evade my arguments? They are simple enough.
>
>>> i d_t Psi = H Psi with H = sqrt(m^2 + p^2) ?
>
>> This is not the equation for wave function but for a positive-frequency
>> mode of the Klein-Gordon equation for a quantized free scalar field.
>
> Stick to some logic! Your "but" is logically wrong.

I don't understand what is wrong with this standard statement about the
meaning of these solutions? You use them together with the
negative-frequency modes to construct local causal quantum fields, but
you do not interpret them as one-particle wave functions in the sense of
"first quantization".

>
> If the Schroedinger equation applies to the positive frequency
> part of a free scalar field, this does not exculde it to apply
> to one-particle states. It applies to one-particle states.
>
> Question: Do in your mind one particle states in the Schroedinger
> picture change in the course of time according to something
> different from
>
> Psi(t) = exp(-i H t) Psi(0) with H = sqrt(m^2 + p^2)

In the Schroedinger Picture by definition the states transform in this
way. Why should I deny this? Of course, you can formulate local causal
QFT also in terms of the Schroedinger Picture, but a simple change of
the picture of time evolution doesn't change the physical content of the
theory. The Hamiltonian for a neutral scalar field reads

H=int d^3 p a^+(p) a(p) E(p)

with E(p)=sqrt(p^2+m^2) with annihilation operators a(p) and creation
operators a^+(p) fulfilling the canonical Bose commutation relations

[a(p),a^+(p')]=delta^{(3)}(p-p').

For free (and only for free!) particles the occupation numbers for each
momentum mode is conserved, and thus a single-particle state remains a
single-particle state.

>
>>> You cannot have yes and no. If you choose no, you should tell us,
>>> what you think the time evolution of the _position_wave_function_ is.
>
>> There is no single-particle position-wave function. That's my whole point!
>
> You agreed already that there are position operators X^i, which are part of
> a Heisenberg algebra. So in a suitable basis they act multiplicatively on
> position wave functions. Do you withdraw your statement about position
> operators or about "the whole point"? You cannot have both.

My whole point is that you cannot interpret<x|Psi(t)> (in the
Schroedinger picture, which you introduced into the discussion, I'd
prefer the Heisenberg picture, because that's more convenient, but
anyway let's use the Schroedinger picture since this doesn't matter
here) as a single-particle wave function. To make it absolutely clear,
in my notation,

|Psi(t)>=int d^3 p A(p) a^+(p) exp[-i E(p) t]|0>

with A(p) a square-integrable function.

>
> By the way, granted there were no one particle position wave function then
> one could not localize the particle. So where do you disagree?

So now, you agree that there are no one-particle wave functions. Then we
don't disagree, but you insist to interpret the above quantity as such a
position-wave function.

>
>>> Please do not change the subject by speaking of the time evolution
>>> of something else.
>
>> The important point is that you must speak about the time evolution of
>> something else, namely many-body quantum theory in terms of a
>> relativistic local QFT.
>
> Why do I have to speak about something else? Are there dementors around
> or even moderators? Who can force me not to analyse and conclude that
> one cannot strictly localize a relativistic particle?

That's perfectly fine. By the way, moderators are not like dementors but
try their best to keep newgroups readable. You know yourself the sad
example of the German physics newsgroup...

>
> And what does it help to consider many particles if already one
> particle cannot be localized? Can two be localized or three?

No, but a many-particle formulation in its specific form of a local
causal QFT leads not only to a causal description of relativistic
quantum physics but also to the prediction of the nowadays well
established fact of the existence of antiparticles, the connection
between spin and statistics (also well establishde), the CPT theorem
(quite well established too).

>
> But the longer you argue the more I get the impression that you
> thoroughly agree: one cannot localize a particle because acoording
> to your revolutionarily fuzzy ideas there are no particles at all.

How do you come to this conclusion? Local causal QFTs describe (among
many more entities) particles very well.

>
> I allow to argue that this is not what standard quantum field theory
> and high energy physics is about and that readers should be warned that
> your statements may be beyond the borderline of this scientific group.

Then, any modern textbook like Weinberg's is beyond the borderline of
this scientific group! That's a ridiculous statement.

To the contrary, I think your point of view to revive very old ideas on
the interpretation of single-particle wave functions that have been
rejected in favor of the standard formulation of relativistic QT as a
local causal QFT, is close to the borderline of what should be rejected
from being posted in this moderated newgroup.

>
> In relativistic physics there is a one particle momentum wave function
> which yields the probability density for momentum. Or do you deny even
> that?

Why should I?

>
> I stop at this point. We should clarify the simple questions first
> before we clarify the rest.

I don't understand, why you insist on a single-particle interpretation
of solutions of the Klein-Gordon equations in the first-quantized sense.
That's all. I think, however, it is hopeless to come to an agreement. So
I guess that's my last contribution to this fruitless discussion.

[Norbert Dragon]

* Hendrik van Hees schreibt:

>* Norbert Dragon wrote:

>> i d_t Psi = (sqrt(m^2 + p^2) Psi

>> Do not tell others that you can choose something else.
>> You only choose to disregard the equation and its consequences.

> I never said that I've chosen something else, I only do not interpret
> the postive-frequency modes of the KG equation as single-particle wave
> functions.

Do not tell others that you can choose something else for the position
wave function.

That the equation also holds for other objects is irrelevant. It holds

for the position wave function.

> See S. Weinberg, where it is explained in a very concise way, why

> one uses causal local QFT and not something more general,

You do not understand. The equation

i d_t Psi = (sqrt(m^2 + p^2) Psi (1)

holds throughout relativistic quantum mechanics, in particular it holds
for the position wave function of one particle states in quantum field
theory.

> which from a mathematical point of
> view may be possible, but hasn't lead to any success in the description
> of scattering processes in high-energy particle physics, while the local
> qft picture (even with the further restriction to derive the dynamics
> from the action principle and with canonical quantization) has this
> phenomenological success on its side!

Your knowledge about relativistic quantum field theory is very
limited. Qft does not contradict (1) for one particle states, on the
contrary, (1) is satisfied for all one particle states in all quantum
field theories.

>> This equation is not taken by me but unavoidable in relativistic
>> quantum mechanics -- it holds for one particle states in each

>> relativistic quantum field theory.

> As you well know, that's not true.

Your objection does not become true by repetition.

H = integral d^3 k sqrt(m^2 + k^2) a^dagger(k) a(k)

acts on one particle states

Psi = integral d^3 k Psi~(k) a^dagger(k) | Vacuum >

as the operator

H = sqrt(m^2 + P^2)

where

P = integral d^3 k k a^dagger(k) a(k)

acts multiplicatively on Psi~.

> In relativistic QFT, for the most
> simple case of a strictly neutral scalar field, you have

> phi(x)=\int d^3 \vec{p} N(\vec{p}) [a(\vec{p}) exp(-i p.x)

> +a^{dagger}(\vec{p}) exp(+i p.x)]_{p^0=+\sqrt{m^2+\vec{p=
}^2}.

The quantized scalar field is _not_ the position wave function.
The books which you praise cannot have mislead you to that extent.

> Here, of course, phi is a field OPERATOR, and a(\vec{p}) is an
> annihilation operator for a single-particle momentum-eigenstate.

As you stress, you speak of something else. What you speak of is
irrelevant for the localization of the position wave function.

Operators such as the quantized scalar fields tell you nothing about
probabilities of the results of measurements and have nothing to do
with the unitarity of the time evolution or of the Poincar=E9
transformations. E.g. field operators transform locally but _their_
transformation is not unitary because the operators do not span a
Hilbert space but _act_ on a Hilbert space.

> The QFT is constructed(!!!) in such a way that local observables,
> particularly the energy-density operator, T^{00}(x), commute at

> space-like distances, and this is one possibility to construct a causal=

> S-matrix theory

That T^00 commutes at spacelike separations does not tell you that one
can localize a state. To decide this, you have to analyze the state.
The result is: one cannot localize states strictly over a finite time.

> Indeed, Peskin-Schroeder is not always very

> careful with the mathematics. In Bjorken Drell, Vol. II Appendix C (I'v=
e
> only the German edition of this book, but I hope the chapters are as in=

> the American original), they give the commutator function [...]

You could simply agree that your quoted reference is in serious
error.

But you prefer to fluff up your argument by the irrelevant comment that
Bjroken Drell correctly give some other function as if I had stated
that all authors are mathematically unreliable. Peskin and Schroeder
are, at least in connection to (1).

> This shows that the usual retarded propagator of QFT,

> Delta_ret(x)=-Theta(t) Delta(x),

> which is the response function to perturbations with external fields in=

> free-particle approximation of linear-response theory, has the right
> causality feature, namely to vanish outside the light-cone.

So what? I do not deny that there are two-point functions which vanish
for spacelike separation. I only argue that position wave function
cannot be localized.

> However, for the non-local equation, you cut the negative-frequency
> contributions, and then also the even function,

> Delta_1(x)=1/(4 pi r) \partial_r [Y_0[m sqrt(t^2-r^2)] for |t|>r,

> -2/pi K_0[m sqrt(r^2-t^2) for |t|<r]=

,

> which does not vanish at spacelike x.

So what? I do not deny that there are two-point functions which do not
vanish for spacelike separation. I only argue that position wave
function cannot be localized.

>>> If you now interpret such a solution as a probability distribution fo=
r
>>> position as a function of time, you have a finite probability to find=

a
>>> particle to propagate with faster-than-light speed, which is not
>>> acceptable as a causal model for free-particle propagation within the
>>> Special Theory of Relativity.

Proof by repetition of the claim.

I do not interpret a wave function as a probability. I take it for
what it is, a probability amplitude and I ask you to show me how an
exponential decrease for spacelike separation leads to a measurable
violation of causality.

>> There is no freedom of interpretation: given any measuring device A:
>> the probability to obtain its (discrete) result a_i if you measure
>> the state Psi, is

>> p(a_i, A, Psi) = |<Lambda_i|Psi>|^2

>> This is the heart of quantum mechanics which you cannot interpret away=

.

> But that's what's done in everyday-life in terms of local causal QFT!

Take A to be the position operator X. Then you agree that

w(Delta, X, Psi) = Integral_Delta d^3 x |<Lambda_x|Psi>|^2

is the probability to find the particle within Delta -- which you
denied in a parallel posting.

>> The position operator X^i is defined to generate translations of the
>> momentum and vice versa,

>> [X^i, P_j] = i delta^i_j .

>> Therefore the Fourier transformation of the momentum wave functions is
>> the position wave function of the particle.

> Fine with me, as long as you can give such operators. This is possible

> for massive particles of any spin and for massless particles with spin =
0
> and spin 1/2.

To find position operators is possible for all massive particles.

The difficulties with massless particles are beyond our elementary
discussion. They are more subtle than Weinberg's discussion of helicity
states. In particular the transversality condition of massless states
can be discussed only in the momentum space R^3 - {0} = S^2 x R, which
has a nontrivial topology. To deal with that space, one cannot use
spherical coordinates for S^2, as Weinberg does. Rather one has to
employ several coordinate patches and seriously consider their
transition functions (which is comparatively easy with stereographic
coordinates.)

>> It seems that you do not care to contradict yourself.

> I only say that these functions are not causal and thus are not used in=

> everyday PHYSICS applications of relativistic QFT.

If you do not say more, we do not disagree.

Position wave functions are not used in everyday physics application.

This does not contradict what I stated before up to the minor exception
that whenever you use a formula for the cross section and its relation
to S-matrix amplitudes then you use the fact that the Fourier
transform of the momentum wave function is the position wave function.

The only used properties of these position wave functions are, that the
incoming current has a homogeneous spatial distribution on the
extension of the target -- which is why scattering experiments never
see anything of the fact, that localization of states decreases
exponentially rather than to vanish strictly.

>> The microcausality condition does not apply to the position wave

>> function but to operators which have different properties. Among other=

s,
>> they transform locally while position wave functions cannot.

> So, you finally agree to that,

I never disagreed.

I consistently stressed that on cannot strictly localize position
wave functions and that strictly localized objects and locally
transforming object cannot be position wave functions.

>> Microcausality does not tell anything about the possibility to
>> localize a _state_.

> No, it doesn't, but it's more important to the consistency of

> relativistic quantum theory and to observations than the localizability=

> of relativistic particles! That's why the single-particle interpretatio=
n
> of the discussed wave packets, constructed with only positive frequency=

> modes, have been given up since the early times of relativistic quantum=

> theory.

Leave your objections. I never discussed whether position wave
functions are important. I stressed that one cannot localize them and
that they transform nonlocally.

You disagreed. You preferred to deny the undeniable and even denied
one particle states their Schroedinger equation and their existence.

At the very end your objection dwindles down to your classification
that position wave functions are unimportant.

>> You should distinguish between a state and an operator.

> That's what I do all the time!

Not at all. You tried throughout the discussion to "reinterpret"
the position wave function as a quantum field, a four parameter set of
operators.

Since about 1930 this is obsolete. It is not, that today one does not
know what the objects of quantum field theory are and what equations
they satisfy though you had your difficulties accepting that the
position wave function of one particle states are subject to (1).

>> Which trouble? Is is a conclusion, not a trouble, an undeniable result=
,
>> that position wave functions cannot be localized over a time interval.

> It's an experimental fact that antiparticles exist in accordance with
> the local causal QFT formulation, in some cases with an overwhelming
> accuracy with the prediction of this formulation!

So what? It is also a fact that the sun rises each day. This fact is
as unrelated to our discussion as are antiparticles.

Antiparticles have positive energy and satisfy (1) just as particles.

If you include antiparticles, then they do not make position wave
functions any more localizable. Antiparticles are nothing but a second
kind of particles with the same mass and with opposite charge. If you
allow for antiparticles you double the components of the position wave
function which still satisfy (1) and still are not strictly localizable.

[Norbert Dragon]

* Jos Bergervoet writes:

>* Norbert Dragon wrote:

>> Do you suggest not to call

>> i d_t Psi = H Psi (1)

>> the Schroedinger equation

> If it is explicitly written as a wave equation for
> the position wave function:

> i d_t Psi(r'') = \int H(r'',r') Psi(r') d^3r'

> with H the convolutor of the type K_2(|r|)/r^2
> where r = |r''-r'| and K_2 a modified Bessel,
> then I would hesitate!

(1) and H = sqrt(m^2 + p^2) is independent of the basis of the
Hilbert space in which you use to represent Psi. If you represent
Psi by its momentum wave function Psi~(k), then (1) reads

i d_t Psi~(t,k) = sqrt(m^2 + k^2) Psi~(t,k)

and has the solution

Psi~(t, k) = exp( -i t sqrt(m^2 + k^2)) Psi~(0, k)

In terms of the position wave function (1) is

i d_f Psi(t,x) = sqrt(m^2 - Laplace) Psi(t,x)

with a solution of the form

Psi(t,x) = Integral d^3 x' U(t,x-x') Psi(0,x^')

> Of course at t>0 I saw it.. but couldn't cancel the post :^)

Not even (1) allows for non-local error correction.

>> So in nonrelativistic quantum mechanics one cannot localized a position
>> wave function because of the instanteneous propagation and in
>> relativistic physics the Schroedinger equation involves a Hamiltonian
>> which acts nonlocally on one particle position wave functions.

> I wouldn't blame the Hamiltonian. It's the solution
> space that is to blame (the restriction to positive
> frequencies). We could blame the Fourier transform!
> It is not possible to construct a function which is
> localized together with its time derivative by summing
> only positive-frequency plane waves. That is just a
> plane-wave summation property. Regardless of any
> time evolution equation you are going to apply!

> So, strict selection of "positive frequencies only" makes
> non-local wavefunctions a mathematically inevitable fact.

The number of my diciples seems to have increased by one.

[Jos Bergervoet]

On 5/7/2012 7:17 PM, Norbert Dragon wrote:
> * Jos Bergervoet writes:
..

>> It is not possible to construct a function which is
>> localized together with its time derivative by summing
>> only positive-frequency plane waves. That is just a
>> plane-wave summation property. Regardless of any
>> time evolution equation you are going to apply!
>
>> So, strict selection of "positive frequencies only" makes
>> non-local wavefunctions a mathematically inevitable fact.
>

> The number of my diciples seems to have increased by one.

We agree about what happens if we select positive
frequencies only. But should we make this requirement?
We know that a localized function would never be an
eigenstate of momentum. Also it would not be an
energy eigenstate. There is clearly uncertainty
about its energy. It would have a spectrum ranging
from m to +infinity if we follow the above rule.
If not, we would extend the energy spectrum with
negative values.

Without using the requirement, we'd simply construct
a state with more uncertainty about the energy but
better localization in space. Where is the problem?

In addition, for particles that are their own
antiparticle the wave function is real (or at least
essentially so, e.g. sqrt(i) times a real function,
using the Majorana equation). That is equivalent
of requiring equal amounts of positive and negative
frequencies! So it is in that case not even allowed
to use the earlier requirement made above..

So what should we do in that case?

--
Jos

[Jos Bergervoet]

On 5/7/2012 3:23 PM, Norbert Dragon wrote:
> * Hendrik van Hees schreibt:
>> * Norbert Dragon wrote:
>
>>> i d_t Psi = (sqrt(m^2 + p^2) Psi

...

> If you include antiparticles, then they do not make position wave
> functions any more localizable. Antiparticles are nothing but a second
> kind of particles with the same mass and with opposite charge.

But, suppose that the particle is its own antiparticle.
Would the position wave function be localizable then?
Necessarily so? or only in some cases?

--
Jos

[Hendrik van Hees]

On 08/05/12 14:14, Jos Bergervoet wrote:
> On 5/7/2012 7:17 PM, Norbert Dragon wrote:
>> * Jos Bergervoet writes:
> ..
>>> It is not possible to construct a function which is
>>> localized together with its time derivative by summing
>>> only positive-frequency plane waves. That is just a
>>> plane-wave summation property. Regardless of any
>>> time evolution equation you are going to apply!
>>
>>> So, strict selection of "positive frequencies only" makes
>>> non-local wavefunctions a mathematically inevitable fact.
>>
>> The number of my diciples seems to have increased by one.
>
> We agree about what happens if we select positive
> frequencies only. But should we make this requirement?

Again, the standard way of thinking about this is (against Norbert
Dragon's interpretation) that a single-particle interpretation of the
solutions of the non-local equation,

i \partial_t psi(t,x)=sqrt(-Delta+m^2) psi(t,x)

as a "wave function" in the sense of non-relativistic QT is only
approximately tenable, i.e., if you do not try to localize the particle
at precision scales of its Compton wavelength. That's why, in my
opinion, one should not talk about "wave functions" in the
single-particle picture and "relativistic quantum mechanics", but start
from the well-formulated local causal QFT picture. BTW even this very
successful has still its formal mathematical problems, which are not
completely solved even after several decades of hard work of axiomatic
quantum-field theorists, but this goes way beyond the level of
discussion we are here at the moment.

That's, why one usually calculates S-matrix elements (asymptotic states)
and constructs the S-matrix from local causal QFTs, i.e., with
interaction Hamiltonian densities that commute at spacelike distances.

If Norbert Dragon "believes" in Bogoliubov-Shirkov, which I've just
looked at (BTW it's a brillant book, written much in the spirit of my
favorite Weinberg's QT of Fields, but highlighting the same physics from
a different point of view and with many mathematical elaborations which
are complementary to Weinberg) now in connection with this discussion,
he must agree that this is the way to construct a physically sensible
formulation (and interpretation!) of relativistic quantum theory.

> We know that a localized function would never be an
> eigenstate of momentum. Also it would not be an
> energy eigenstate. There is clearly uncertainty
> about its energy. It would have a spectrum ranging
> from m to +infinity if we follow the above rule.
> If not, we would extend the energy spectrum with
> negative values.

Of course not. No "generalized eigenstate" with the "eigenvalue" within
the continuous spectrum of a self-adjoint operator can be a
Hilbert-space vector and thus cannot be a state. It's a distribution to
be used in the spectral representations of true Hilbert-space vectors,
which represent (pure) states.

>
> Without using the requirement, we'd simply construct
> a state with more uncertainty about the energy but
> better localization in space. Where is the problem?

Without using the requirement of only positive energy values for a free
particle within a single-particle wave function interpretation, you run
into the problem that the Hamiltonian is not bounded from below anymore,
since you'd have not only the continuum E>m but also the one E<m. So
there wouldn't be a stable ground state anymore, and matter couldn't be
stable at all.

>
> In addition, for particles that are their own
> antiparticle the wave function is real (or at least
> essentially so, e.g. sqrt(i) times a real function,
> using the Majorana equation). That is equivalent
> of requiring equal amounts of positive and negative
> frequencies! So it is in that case not even allowed
> to use the earlier requirement made above..
>
> So what should we do in that case?

In terms of a relativistic local QFT there's no problem since you simply
use local quantum fields, composed with annihilation operators and
positive-frequency modes and creation operators and negative-frequency
modes (e.g., the here discussed momentum-eigenmodes), i.e.,

phi(x)=int d^3 p/sqrt[(2pi)^3 2 p_0] [a(\vec{p})
exp(-ip.x)+b^{dagger}(\vec{p}) exp(+i p.x)]_{p0=+sqrt(m^2+\vec{p}^2)},

where the a, b and their adjoints obey the usual bosonic (commutation)
relations like

[a(\vec{p}),a^{\dagger}(\vec{p}')]=[b(\vec{p},b^{\dagger}(\vec{p}')] =

\delta^{(3)}(\vec{p}-\vec{p}').

That's known as the Feynman-Stueckelberg trick. For the strictly neutral
KG quantum field you simply set b(\vec{p})=a(\vec{p}), which makes the
particles to be the same as their antiparticles (like the neutral pion).

One can show that necessarily you need both positive and
negative-frequency modes to construct fields that

(a) transform locally under proper orthochronous Poincare
transformations, i.e., analogous to a classical field in special relativity.

(b) have a Hamiltonian bounded from below (existence of a stable ground
state)

(c) the Hamiltonian densities commute at space-like distances leading to
a causal Poincare covariant S-matrix (that's a sufficient but perhaps
not a necessary condition; I know no working example of a realistic
non-local QFT to describe (interacting) particles with relativistic QT).

This leads to at least three successfully predicted general properties
of local causal QFTs

(a) the existence of antiparticles

(b) the relation between spin and statistics (integer spin: bosons;
half-integer spin: fermions; in 3D and higher spatial dimensions there
are only these two statistics; in 2D you can also have anyons, but
that's another story)

(c) The PCT theorem: Any local causal QFT that admits a representation
of the proper orthochronous Lorentz group must be also covariant under
the PCT reflection symmetry (parity, charge conjugation, time reversal).
Neither P nor CP alone are conserved in the standard model (broken by
the weak interactions), but PCT is since the standard model is THE
example for a successful local causal QFT.

[Norbert Dragon]

* Jos Bergervoet schreibt:

> We agree about what happens if we select positive
> frequencies only. But should we make this requirement?

We should require that all normalized momentum wave functions
are possible. Their Fourier transformation are the position
wave functions. No restriction apart from

H = sqrt(m^2+p^2)

is known to be compatible with the unitary action of the Lorentz group.

> We know that a localized function would never be an
> eigenstate of momentum. Also it would not be an
> energy eigenstate. There is clearly uncertainty
> about its energy.

There is no way around

H = sqrt(m^2 + p^2)

in a relativistic theory if the energy is bounded from below
and if there are particles with a discrete mass.

> It would have a spectrum ranging from m to +infinity if we follow
> the above rule. If not, we would extend the energy spectrum with
> negative values.

> Without using the requirement, we'd simply construct
> a state with more uncertainty about the energy but
> better localization in space. Where is the problem?

If you allow for states with both signs of the energy and if
they interact, then the vacuum would be unstable against decay into
pairs of positive-negative energy particles.

Moreoever, negative energy particles would not help at all with the
localizability of the states. Negative energy particles would be an
additional, distinguishable species of particles with their own
position wave function which would fail to be localizable because of

H' = - sqrt(m^2+p^2)

Quantum field theory and each acceptable model of relativistic
quantum mechanics includes the assumption that the vacuum is
unique (i.e. invariant under Poincaré transformations and therefore
has vanishing energy and momentum) and the lowest energy state.

> In addition, for particles that are their own
> antiparticle the wave function is real (or at least
> essentially so, e.g. sqrt(i) times a real function,
> using the Majorana equation).

No. You are not free to choose this or that equation for the
position wave function.

If the position wave function can be defined, it is the Fourier
transform of the momentum wave function where the momentum wave
function is arbitrary up to the fact that it is normalized.

Some of the position wave functions may turn out to be real but
generally they are not and generally they do not solve the
Majorana equation whether they are position wave functions of
Majorana particles or not.

> That is equivalent of requiring equal amounts of positive
> and negative frequencies! So it is in that case not even allowed
> to use the earlier requirement made above..

> So what should we do in that case?

You should just accept the fact that one cannot strictly localize
a particle. In the analysis of scattering processes one only needs
that the target is small as compared to the the size of the incomming
beam. The scattering is the coherent superposition of scattering events
everywhere in the interaction region which for simplicity is taken to
be infinite ( ~ the size of an atom).

One never needs strictly localized states. What one needs are
local interactions, which is a concept completely different from
localized states, e.g. from local interactions you cannot tell where
the interaction had taken place no more than you can tell from the
position at the screen which slit of a double slit the particle had
passed.

--
for, he reasons pointedly,
that which must not, can not be.

Morgenstern on how to deny the undeniable

[Norbert Dragon]

* Jos Bergervoet writes:

>* Norbert Dragon wrote:

>> i d_t Psi = (sqrt(m^2 + p^2) Psi

>> If you include antiparticles, then they do not make position wave
>> functions any more localizable. Antiparticles are nothing but a second
>> kind of particles with the same mass and with opposite charge.

> But, suppose that the particle is its own antiparticle.

That is the case which I considered up to now.

> Would the position wave function be localizable then?

No.

There seems to linger the misconception that antiparticles have
negative energies and that this would help any for the localization.

Antiparticles are just ordinary particles which are degenerate in mass
with their particles and have opposite charge such that they can be
pair produced even if their charge is conserved.

The picturesque description of loop corrections as being due to virtual
particles, which, to make it more impressive, run back in time blurs
the simple structures of quantum field theory. For example: there are
no negative energy states though loop integrals extend over the momenta
in all R^4, including arbitrarily negative energies. This does not mean
that there are corresponding states with negative energies.

--
Schubert on how to win a discussion

Doch endlich ward dem Diebe Die Zeit zu lang. Er macht
Das Bächlein tückisch trübe, Und eh ich es gedacht,So zuckte seine
Rute

[Roland Franzius]

Local quantum operator theory eg.

Because the 3-space foliation of space-time is changing by application
of a Lorentz boost, working in operator representations over a fixed
Hilbert space of states based on a fixed R^3 with space-wave functions
as pure states is not a truely convincing concept. Alternative
formulations were given by Wightman and Haag.

The standard book list of quantum mathematical physics is given here:

http://www.lqp.uni-goettingen.de/bibliography/books.html

--

Roland Franzius

[Hendrik van Hees]

On 09/05/12 01:28, Norbert Dragon wrote:
> * Jos Bergervoet schreibt:
>
>> We agree about what happens if we select positive
>> frequencies only. But should we make this requirement?
>
> We should require that all normalized momentum wave functions
> are possible. Their Fourier transformation are the position
> wave functions. No restriction apart from
>
> H = sqrt(m^2+p^2)
>
> is known to be compatible with the unitary action of the Lorentz group.

Yes, it's compatible with the unitary action of the Lorentz group but
has trouble with locality since this is not a local operator.

That's why one formulates relativistic quantum theory as local causal
QFT, and there the Hamiltonian is given by the space integral of a
Hamilton density operator field that commutes at space like distances of
the arguments,

[H(x),H(y)]=0 for (x-y)^2<0 (west-coast metric).

Thats a sufficient (I don't know whether also necessary) condition for
covariance of the S-matrix wrt. proper orthochronous Poincare
transformations. This is also written in the by you recommended textbook
by Bogoliubov and Shirkov and explained there in great mathematical
detail. The same line of thought can also be found in better
comprehensible form in Weinberg, QT of fields vol. 1.

[Hendrik van Hees]

On 09/05/12 01:28, Norbert Dragon wrote:

> No.
>
> There seems to linger the misconception that antiparticles have
> negative energies and that this would help any for the localization.
>
> Antiparticles are just ordinary particles which are degenerate in mass
> with their particles and have opposite charge such that they can be
> pair produced even if their charge is conserved.
>
> The picturesque description of loop corrections as being due to virtual
> particles, which, to make it more impressive, run back in time blurs
> the simple structures of quantum field theory. For example: there are
> no negative energy states though loop integrals extend over the momenta
> in all R^4, including arbitrarily negative energies. This does not mean
> that there are corresponding states with negative energies.

Finally, I can agree with you :-). The "Feynman-Stueckelberg trick" to
solve the here discussed problem indeed uses simply the
negative-frequency modes to put a creation operator to the local quantum
field in addition to a annihilation operator in the mode expansion of
this field. It is easy to show that only with such a combination, be it
for strictly neutral particles (that are their own antiparticles so to
say) or with charged particles, where the antiparticles are
distinguisable by at least one conserved charge, which arises from
invariance under multiplication of the local field operators with a
phase factor. The corresponding charges for particles and antiparticles
have different sign.

[Hendrik van Hees]

[[Mod. note -- I apologise to the author and to readers for the long
delay in processing this article, which was originally submitted at
Wed, 02 May 2012 10:00:46 +0200.
-- jt]]

On 01/05/12 23:08, Jos Bergervoet wrote:
> On Apr 29, 9:24 pm, Norbert Dragon<dra...@itp.uni-hannover.de> wrote:
>
> [Off-topic discussion of moderation policy snipped.]
>
> => Norbert: Can we first discuss the mathematical items
> that gave confusion, before discussing physical meaning.
> Notably these two:
> 1) Are there localized solutions that evolve locally
> in time?

The very reason, why such wave packets with positive frequency modes can
NOT be interpreted as single-particle wave functions, is the problem of
non-causal signal propagation. As I quoted from Peskin and Schroeder's
textbook in another post (that appeared on April/29 in the newsgroup)
the corresponding single-particle propagator would have support in
space-like regions, leading to trouble with causality even for free
particles, when interpreted as a single-particle wave function.

In the interacting case, as far as I know, there are no examples for a
time evolution sticking to the positive-frequency solutions at all
times, i.e., when you include interactions, you always have a finite
probability to also excite the negative-frequency modes in the outgoing
asymptotically free state.

As is well known, this problem is cured by interpreting the theory as a
many-body theory, which is most transparently formulated as a quantum
field theory. From this formalism it turns out that any observable
causally connected events must be time-like (or light-like when massless
particles are involved, but that's an even more delicate issue due to
infrared problems and the appropriate choice of asymptotic states) with
respect to each other.

> 2) Is your (Norbert's) relativistic Schroedinger equation
> indeed "inevitable", or are others possible?

It is "inevitable" in the sense that any single-particle mode must obey
the on-shell condition

(\Box+m^2) \psi(x,\sigma)=0,

where \sigma are the spin- or helicity quantum numbers. The reason is
that any irreducible unitary representation of the Poincare group must
have a well-defined value for all Casimir operators of the Poincare
group, among which is the four-momentum operator squared, giving the
mass squared of the corresponding particle.

If you know restrict yourself to solutions with positive frequency,
which you have to do if you try to interpret the solutions as
single-particle wave functions in the same sense as in non-relativistic
quantum mechanics in order to have a stable ground state. As stated
above, then you inevitably have trouble with causality and completeness
for the case of interacting particles.

[Hendrik van Hees]

On 09/05/12 01:29, Roland Franzius wrote:

> Local quantum operator theory eg.
>
> Because the 3-space foliation of space-time is changing by application
> of a Lorentz boost, working in operator representations over a fixed
> Hilbert space of states based on a fixed R^3 with space-wave functions
> as pure states is not a truely convincing concept. Alternative
> formulations were given by Wightman and Haag.
>
> The standard book list of quantum mathematical physics is given here:
>
> http://www.lqp.uni-goettingen.de/bibliography/books.html

In this list, I'm missing the Epstein-Glaser approach, or is it
somewhere hidden in one of the books listed? A book, nicely readable for
the mathematics-oriented physicist, about this is

Scharf, Finite QED, Springer Verlag.

What do you think about this approach as a mathematician? Is this more
rigorous than the usual treatment of perturbation theory? As far as I
know, the full problem of the existence of QED in a strict mathematical
sense of axiomatic QFT is still not solved (be it in the positive or the
negative). As far as I understand of this very formal and complicated
things is that this is due to (a) IR problems and (b) the Landau-pole
problem.

Is it clear yet that at least the latter is solved in non-abelian gauge
theories that are asymptotically free? This is one of the famous Clay
millenium-prize problems, and as far as I can see on their homepage it's
not solved:

http://www.claymath.org/millennium/Yang-Mills_Theory/

[Norbert Dragon]

* Hendrik van Hees writes:

>* Norbert Dragon wrote:

>> Do you agree, that one particle states are characterized by momentum
>> wave functions and that their Fourier transform gives the position
>> wave function?

> I stick to the well-established and successful formulation

> [...]

You have not answered the question.

Do you agree, that one particle states are characterized by momentum
wave functions and that their Fourier transform gives the position
wave function?

> I recommend Weinberg's QT of Fields, where this is
> discussed in a modern way

You love to cite books rather than to cite to the point.

Where does Weinberg discuss position wave functions, an object
of which you at times argue that is does not exist and at other
times argue that it violates causality.

Your only precise quotation Peskin Schroeder, page 14 lead us to a
serious mathematical blunder of the authors in an equation which they
had already corrected once.


Please, answer the simple question:

Do you agree, that one particle states are characterized by momentum
wave functions and that their Fourier transform gives the position
wave function?

[Hendrik van Hees]

On 11/05/12 17:44, Norbert Dragon wrote:
> * Hendrik van Hees writes:
>
>> * Norbert Dragon wrote:
>
>>> Do you agree, that one particle states are characterized by momentum
>>> wave functions and that their Fourier transform gives the position
>>> wave function?
>
>> I stick to the well-established and successful formulation
>> [...]
>
> You have not answered the question.

I gave my point of view several times. After one of your last postings,
I think we agree in principle, and that the whole "fight" is about
wording. I think the exposition of the issue in Bogoliubov-Shirkov is
the standard one, I also follow, and as I understand you agree with this
book. But let me try again.

>
> Do you agree, that one particle states are characterized by momentum
> wave functions and that their Fourier transform gives the position
> wave function?

I agree that the only successful relativistic relativistic QFT, so far
developed, is local causal quantum field theory.

For the free case, there is a single-particle basis of generalized
simultaneous eigenstates of mass squared m^2>=0, three-momentum, spin
and spin component (or helicity for massless particles). A Fock space
basis is given by the N-particle boson (integer spin) or fermion
(half-integer spin) states with N \in {0,1,2,3,...}.

>
>> I recommend Weinberg's QT of Fields, where this is
>> discussed in a modern way
>
> You love to cite books rather than to cite to the point.

I don't understand, why I should quote things at length in this
newsgroup which are well explained in the literature.

>
> Where does Weinberg discuss position wave functions, an object
> of which you at times argue that is does not exist and at other
> times argue that it violates causality.

Of course, he doesn't discuss them. Why should he? I don't see that you
can interpret them in a physical meaningful way.

>
> Your only precise quotation Peskin Schroeder, page 14 lead us to a
> serious mathematical blunder of the authors in an equation which they
> had already corrected once.

I don't understand your point. They don't give the full function, but
the asymptotic expansion. The "position wave function" indeed goes
propto exp(-m sqrt(r^2-t^2)) for r>>t^2 (deep space-like region), and
more is not needed for the argument made. I guess, the
Bogoliubov-Shirkov expression is the exact and correct solution.

>
>
> Please, answer the simple question:
>
> Do you agree, that one particle states are characterized by momentum
> wave functions and that their Fourier transform gives the position
> wave function?

Thei Fourier transform cannot be interpreted as position-wave function
as I already wrote several times.

In Bogoliubov-Shirkov it is shown in great detail, why one needs a local
causal QFT description to obtain an S matrix that is Lorentz covariant
and causal: A sufficient (I don't know whether even a necessary)
condition is to express the Hamiltonian as the integral of a Hamilton
density operator H(x) that obeys the microcausality condition,

[H(x),H(y)]=0 for all x,y with (x-y)^2<0 (space-like separation).

[Norbert Dragon]

* Hendrik van Hees writes:

>* Norbert Dragon wrote:

>> We should require that all normalized momentum wave functions
>> are possible. Their Fourier transformation are the position
>> wave functions. No restriction apart from

>> H = sqrt(m^2+p^2)

>> is known to be compatible with the unitary action of the Lorentz group.

> Yes, it's compatible with the unitary action of the Lorentz group but
> has trouble with locality since this is not a local operator.

> That's why one formulates relativistic quantum theory as local causal
> QFT, and there the Hamiltonian is given by the space integral of a
> Hamilton density operator field that commutes at space like distances of
> the arguments,

> [H(x),H(y)]=0 for x spacelike to y

The operator, which you call nonlocal, is the operator which you call
local in the next line. Or do you deny that in quantum field theory
the Hamiltonian

Integral d^3x H(x)

acting on one particle states is

H = sqrt(m^2+p^2)?

Please: Yes or No.

If No, what Hamiltonian do you take as the hermitean operator, which
in the Schroedinger picture generates the time evolutino of massive,
relativistic one-particle states.

[Norbert Dragon]

* Hendrik van Hees writes:

>* Norbert Dragon wrote:

>>>> Do you agree, that one particle states are characterized by momentum
>>>> wave functions and that their Fourier transform gives the position
>>>> wave function?

>>> I stick to the well-established and successful formulation
>>> [...]

>> You have not answered the question.

> I gave my point of view several times.

You repeatedly stated something, but you never anwered the question.

> After one of your last postings, I think we agree in principle, and
> that the whole "fight" is about wording.

You still have not answered the question.

Up to now I never came about someone who like you claims to know
the subject but shows himself incapable to answer a very basic
question.

>> Do you agree, that one particle states are characterized by momentum
>> wave functions and that their Fourier transform gives the position
>> wave function?

> I agree that the only successful relativistic relativistic QFT, so far
> developed, is local causal quantum field theory.

I have not asked your opinion about something else. I asked:

Do you agree that one particle states are characterized by momentum

wave functions and that their Fourier transform gives the position
wave function?

> For the free case, there is a single-particle basis of generalized
> simultaneous eigenstates of mass squared m^2>=0, three-momentum, spin
> and spin component (or helicity for massless particles). A Fock space
> basis is given by the N-particle boson (integer spin) or fermion
> (half-integer spin) states with N \in {0,1,2,3,...}.

Do you agree that one particle states are characterized by momentum

wave functions and that their Fourier transform gives the position
wave function?

>>> I recommend Weinberg's QT of Fields, where this is
>>> discussed in a modern way

>> You love to cite books rather than to cite to the point.

> I don't understand, why I should quote things at length in this
> newsgroup which are well explained in the literature.

If you participate in this group you should care for the questions
which are raised and discussed.

Therefore I repeat:

Do you agree that one particle states are characterized by momentum

wave functions and that their Fourier transform gives the position
wave function?

>> Where does Weinberg discuss position wave functions, an object
>> of which you at times argue that is does not exist and at other
>> times argue that it violates causality.

> Of course, he doesn't discuss them. Why should he?

You cited him as your witness in our discussion and now you concede
that he does not even discuss the question.

Moderators note: Also I have emotions. I state my attitude to such a
style of discussion in my signature.

>> Please, answer the simple question:

>> Do you agree, that one particle states are characterized by momentum
>> wave functions and that their Fourier transform gives the position
>> wave function?

> Their Fourier transform cannot be interpreted as position-wave function

> as I already wrote several times.

You agreed that there is a position operator X^i acting on one particle
states.

Repeated question; How are the functions, on which the position
operator acts multiplicatively related to the momentum wave functions?

Next question: How do you derive the relation between the
_cross_section_ and the S-matrix amplitudes without the fact that the
Fourier transform of the momentum wave function is the position wave
function.

Can you give any book (please specify the page number) where the
cross section is derived without using the fact that the Fourier
transform of the momentum wave function is the position wave
function? Weinberg needs it and and also Peskin Schroeder and I need
it but you deny the relation. So how do you derive a cross section?

--
According to my standards it is dishonest to cite a reference which
does not contain an argument pertaining to the discussion.
I regard such a style of discussion a disgrace.

[Jos Bergervoet]

On 5/11/2012 9:01 PM, Hendrik van Hees wrote:
> On 11/05/12 17:44, Norbert Dragon wrote:

...

>> Do you agree, that one particle states are characterized by momentum
>> wave functions and that their Fourier transform gives the position
>> wave function?
>
> I agree that the only successful relativistic relativistic QFT, so far
> developed, is local causal quantum field theory.
>
> For the free case, there is a single-particle basis of generalized
> simultaneous eigenstates of mass squared m^2>=0, three-momentum, spin
> and spin component (or helicity for massless particles).

Yes! So we could sum over these states. This would give an
integral over the creation operators a^\dag(p) using some
coefficient function (of p) which would deserve the name
"momentum wave function". Or not?

If so, then presumably the Fourier transform would deserve
the name "position wave function". If *that* would be the
defining description, then positive frequencies, non-local
evolution, and everything we discussed earlier would indeed
be properties of this "position wave function". For me this
would solve those (previous) problems!

>>> I recommend Weinberg's QT of Fields, where this is
>>> discussed in a modern way
>>
>> You love to cite books rather than to cite to the point.
>
> I don't understand, why I should quote things at length in this
> newsgroup which are well explained in the literature.

Good idea. But where does Weinberg answer our questions?

...

>> Do you agree, that one particle states are characterized by momentum
>> wave functions and that their Fourier transform gives the position
>> wave function?

If not, then we have no definition of what "momentum wave
functions" are, isn't it?

> Thei Fourier transform cannot be interpreted as position-wave function
> as I already wrote several times.

I do not immediately see a discussion of this concept in
Weinberg. I had expected it in 5.1 ("Free fields") but
there he never constructs an arbitrary superposition of
one-particle states. He does use "coefficient functions"
in momentum space, but they later turn out to be just
exp(ipx) and are only used for the construction of the
field operator in point x in coordinate space.

There still seems to be no reason, however, to exclude
using a coefficient function which is more general, but
also a quick scan through "Relativistische Quantenfeld-
theorie" (B&D) does not shed light on it.. Do I overlook
it? Where is it discussed?

--
Jos

[Hendrik van Hees]

On 12/05/12 00:42, Norbert Dragon wrote:

> Do you agree that one particle states are characterized by momentum
> wave functions and that their Fourier transform gives the position
> wave function?

I don't agree with the word "wave function" in the context of
relativistic quantum theory. I said very clearly what I agree with,
namely that there is the standard generalized
mass-momentum-spin-spin/helicity basis for the one-particle part of the
Fock space, which then is "spanned" by the fully symmetrized or
antisymmetrized N-particle product states:

>
>> For the free case, there is a single-particle basis of generalized
>> simultaneous eigenstates of mass squared m^2>=0, three-momentum, spin
>> and spin component (or helicity for massless particles). A Fock space
>> basis is given by the N-particle boson (integer spin) or fermion
>> (half-integer spin) states with N \in {0,1,2,3,...}.

> You agreed that there is a position operator X^i acting on one particle
> states.
>
> Repeated question; How are the functions, on which the position
> operator acts multiplicatively related to the momentum wave functions?

Even in non-relativistic quantum mechanics a position or momentum
"eigenfunction" is not a wave function but a distribution.

>
> Next question: How do you derive the relation between the
> _cross_section_ and the S-matrix amplitudes without the fact that the
> Fourier transform of the momentum wave function is the position wave
> function.

You can see, how I derive it in my notes:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

page 222.

There I use a finite quantization volume to overcome the problem with
the energy-momentum conserving delta-distribution when taking the square
of the S-matrix element. This manuscript is some years old, and nowadays
I prefer the derivation with wave packets in the initial state, i.e.,
with square-integrable asymptotically free superpositions of
momentum-eigenstates which is narrowly peaked at the incoming momenta.
Nowhere do I need the interpretation of the Fourier transform of
asymptotically free one-particle states as a wave function.

>
> Can you give any book (please specify the page number) where the
> cross section is derived without using the fact that the Fourier
> transform of the momentum wave function is the position wave
> function? Weinberg needs it and and also Peskin Schroeder and I need
> it but you deny the relation. So how do you derive a cross section?

As Peskin Schroeder, p.99 onwards. Nowhere there he uses the
wave-function interpretation of the in state given in Eq. (4.68). In
scattering theory the particles are not localized in position but in
momentum space. In addition, you interpret only the S-matrix as
transition probability between asymptotic states. Nowhere are transient
states interpreted physically!

Let's look at Weinberg. Here the derivation is found on pages 134ff. He
even uses the quantization-volume regularization as I do in my
manuscript. Again, here the usual interpretation is used, namely that
the cross section is given by the S-matrix element squared, which is a
transition-probability amplitude that an asymptotically free initial
state (usually a one-body or two-body state for the case of decays or
scattering measurements) to an asymptotically free final state (which
can have any particle number). Again, nowhere a wave-packet
interpretation of single-particle modes is used.

[Hendrik van Hees]

On 12/05/12 00:42, Norbert Dragon wrote:

> The operator, which you call nonlocal, is the operator which you call
> local in the next line. Or do you deny that in quantum field theory
> the Hamiltonian
>
> Integral d^3x H(x)
>
> acting on one particle states is
>
> H = sqrt(m^2+p^2)?
>
> Please: Yes or No.

There is no yes or no. One has to be precise enough to discuss these issues.

The Hamiltonian for free neutral Klein-Gordon particles is given by

H=int d^3 p sqrt(p^2+m^2) a^+(p) a(p).

Acting on a generalized momentum eigenvector gives

H |p>=E_p |p> with E_p=sqrt(p^2+m^2).

This shows that the generalized momentum eigenvectors are also
generalized energy-eigenvectors with the on-shell value of the energy.

which is given by the spacial integral over the normal-ordered
Hamilton-density operator, which is formulated in terms of local causal
field operators (that are in this special case of neutral particles
Hermitian):

H(x)=1/2 :[\dot{Phi}^2(x)+[\nabla Phi(x)]^2+m^2 Phi(x)^2]:

From the canonical bosonic commutation relations of the field
operators, H(x) inherits the microcausality condition

[H(x),H(y)]=0 for (x-y)^2<0.

Nowhere is the interpretation of position wave functions used here!

>
> If No, what Hamiltonian do you take as the hermitean operator, which
> in the Schroedinger picture generates the time evolutino of massive,
> relativistic one-particle states.

This I've specified clearly above. You can transform this into the
Schroedinger picture if you wish. I prefer to use the Heisenberg picture
which is more convenient to use in relativistic QFT. Heisenberg and
Schroedinger picture are equivalent.

[Hendrik van Hees]

On 13/05/12 00:41, Jos Bergervoet wrote:

> I do not immediately see a discussion of this concept in
> Weinberg. I had expected it in 5.1 ("Free fields") but
> there he never constructs an arbitrary superposition of
> one-particle states. He does use "coefficient functions"
> in momentum space, but they later turn out to be just
> exp(ipx) and are only used for the construction of the
> field operator in point x in coordinate space.

Of course, Weinberg doesn't discuss these "position wave functions"
since they are not needed and cannot be interpreted in a proper physical
way as in nonrelativistic quantum theory. That's why one doesn't usually
call them "wave functions" at all.

This is the very point, I'm trying to make the whole time, and it's very
easy to see that Peskin-Schroeder is correct on the principal issue
although he is missing some subtleties in the proper definition of the
invariant distribution, i.e., the propagator of the non-local equation

i\partial_t psi=sqrt(-\Delta+m^2) \psi.

The point however is that this propagator has support in the space like
region, and this makes a proper interpretation of the corresponding
wave-packet-like solutions impossible.

Of course, momentum eigenstates are distributions too and not
representing states, because they are not normalizable to 1 but only to
a delta distribution. This is so even in nonrelativistic quantum
physics. You have to smear these states with a square integrable
function as you said in your mail.

As Norbert Dragon has said, the textbook by Bogoliubov-Shirkov gives
very carefully derived definitions of the various invariant
distributions of free-particle motion, but this very same textbook also
sticks to the standard formulation of relativistic QT as a local causal
QFT, leading to a causal Lorentz covariant S matrix from the
microcausality condition of the Hamilton density, which in turn is built
with local causal quantum fields and not from positive-frequency modes
alone.

>
> There still seems to be no reason, however, to exclude
> using a coefficient function which is more general, but
> also a quick scan through "Relativistische Quantenfeld-
> theorie" (B&D) does not shed light on it.. Do I overlook
> it? Where is it discussed?

The whole first volume of Bjorken/Drell (at least in the German
translation) gives all the oldfashioned ideas based on Feynman's
propagator approach, using handwaving arguments of reinterpretations of
the single-particle wave functions in the spirit of Dirac's hole theory.
I've had always the impression that one can spare oneself the hurdle to
understand this stuff. It all becomes way more clear with the modern
local causal QFT approach to formulate all this. That's how finally
progress has been made in the aftermath of the Shelter Island conference
in 1948 and other conferences at this time, when Schwinger and Feynman
discussed their approaches towards relativstic quantum theory and the
renormalization problem. Dyson brought all this into a quite coherent
picture in terms of relativistic QFT, and that's the way we should learn
it nowadays, and that's how it's taught in the standard theory courses
at our universities!

Another older standard text which discusses the space-time distributions
is Schweber, Introduction to QFT.

An application to modern (perturbative) renormalization theory in the
space-time domain is given in Collins, Renormalization.

[Jos Bergervoet]

On 5/13/2012 12:43 AM, Hendrik van Hees wrote:
> On 12/05/12 00:42, Norbert Dragon wrote:
>

>> ... Or do you deny that in quantum field theory

>> the Hamiltonian
>>
>> Integral d^3x H(x)
>>
>> acting on one particle states is
>>
>> H = sqrt(m^2+p^2)?
>>
>> Please: Yes or No.
>
> There is no yes or no. One has to be precise enough to discuss these issues.
>
> The Hamiltonian for free neutral Klein-Gordon particles is given by
>
> H=int d^3 p sqrt(p^2+m^2) a^+(p) a(p).
>
> Acting on a generalized momentum eigenvector gives
>
> H |p>=E_p |p> with E_p=sqrt(p^2+m^2).
>
> This shows that the generalized momentum eigenvectors are also
> generalized energy-eigenvectors with the on-shell value of the energy.
>
> which is given by the spacial integral over the normal-ordered
> Hamilton-density operator, which is formulated in terms of local causal
> field operators (that are in this special case of neutral particles
> Hermitian):
>
> H(x)=1/2 :[\dot{Phi}^2(x)+[\nabla Phi(x)]^2+m^2 Phi(x)^2]:
>
> From the canonical bosonic commutation relations of the field
> operators, H(x) inherits the microcausality condition
>
> [H(x),H(y)]=0 for (x-y)^2<0.
>
> Nowhere is the interpretation of position wave functions used here!

The question asked about "one particle states"! (Not
requiring interpretation as "position wave functions".)

And your answer says: yes for "momentum eigenvectors".
It might be possible to solve the game of words now.

We agree, I hope, that the momentum eigenvectors meant
here, are a subset of "one particle states". We agree,
probably, also that the same Hamiltonian still is valid
for any combination of momentum eigenvectors with
coefficient functions in momentum space.

But do we also agree that well behaved coefficient
functions in momentum space deserve the name "momentum
wave function"?

And if so, after Fourier transforming these coefficient
functions to coordinate space, would it be allowed then
to call them "position wave function"?

And if we *don't* agree on calling them wave functions,
do we agree then that:
"Coefficient functions of momentum eigenvectors after
Fourier transformation to coordinate space evolve in
time with the Feynman propagator which has a nonzero
imaginary exponential tail outside the lightcone."
Then we would at least agree on what happens, even if
not on the precise words..

The term "wave function" could then be entirely dropped
(as in a complete collapse! Some would say "I knew it..")

--
Jos

[Hendrik van Hees]

On 14/05/12 22:47, Jos Bergervoet wrote:

> The question asked about "one particle states"! (Not
> requiring interpretation as "position wave functions".)
>
> And your answer says: yes for "momentum eigenvectors".
> It might be possible to solve the game of words now.

Yes, with the qualification that "momentum eigenvectors" are not
representing states but generalized eigenstates in the sense of
distributions as for any spectral value of an essentially self-adjoint
operator in the continuous part of the spectrum.

>
> We agree, I hope, that the momentum eigenvectors meant
> here, are a subset of "one particle states". We agree,
> probably, also that the same Hamiltonian still is valid
> for any combination of momentum eigenvectors with
> coefficient functions in momentum space.

Sure, the Hamiltonian doesn't depend on the states, it's valid as a
linear operator on a dense subspace of the Hilbert space. There's no
general difference in this respect between relativistic and
non-relativistic quantum theory.

>
> But do we also agree that well behaved coefficient
> functions in momentum space deserve the name "momentum
> wave function"?

You can build wave packets out of the momentum eigenstates, which form
true single-particle Hilbert-space vectors,

|f>=\int d^3 p f(\vec{p}) |\vec{p}>,

with square integrable complex functions f(\vec{p}).

>
> And if so, after Fourier transforming these coefficient
> functions to coordinate space, would it be allowed then
> to call them "position wave function"?

I wouldn't like to call them position wave functions for the reason of
noncausality.

>
> And if we *don't* agree on calling them wave functions,
> do we agree then that:
> "Coefficient functions of momentum eigenvectors after
> Fourier transformation to coordinate space evolve in
> time with the Feynman propagator which has a nonzero
> imaginary exponential tail outside the lightcone."
> Then we would at least agree on what happens, even if
> not on the precise words..
>
> The term "wave function" could then be entirely dropped
> (as in a complete collapse! Some would say "I knew it..")

Yes, that's what I've been advocating all the time. You don't need to
confuse yourself with single-particle interpretations, but can stick to
the well-established formalism of (perturbative) local causal QFT and
accepting the many-body nature of relativistic quantum theory!

[iuval]

So would you all agree that the standard interpretation of the first two
components of the Dirac Spinor (for momentum/energy eigenstates)
representing the amplitude for the electron and the bottom two
representing the amplitude for the positron is wrong? Would it be
correct to say that for positive energies the top two represent the
amplitude for an electron and the bottom two represent nothing, whereas
for negative energies the top two represent nothing and the bottom two
represent the amplitude for the positron? And we can form combinations
of electrons and positrons from the (properly antisymmetrized, which
happens naturally in a fock space construction due to anti-commutation
of creation/annihilation operators, but has to be done by hand
otherwise)) one particle wave functions?

Iuval

[Hendrik van Hees]

Of course, in the standard Dirac representation of the Clifford algebra,
the upper (lower) two components are representing particles
(antiparticles), associated with the positive- (negative-) frequency
mode part in the expansion of the Dirac-field *operator* in terms of
annihilation and creation operators wrt. the momentum basis.

As with the here lengthyly discussed scalar fields, an interpretation of
a c-number valued Dirac field as a wave equation in the sense of a
single-particle interpretation is valid only approximately in
situations, where the particle in question is close to a
non-relativistic situation. Particularly, you cannot localize the
particle at scales smaller than its Compton wave length since then you
inevitably create particle-antiparticle pairs, and the single-particle
interpretation is inconsistent with observations.

[iuval]

Dear Norbert,
Could you please spell out the mathematics for your claim below explicitly?=
All that I see now is that the scalar product in position space is not the=
standard integration over 3 space one because of the factor of 1/sqrt(m^2 =
+ k^2), and because we still have a 4 space integration. Normally (if it we=
ren't for the extra k-dependent factor), after substituting the definition =
of chi(k) and phi(k) in terms of their position space fourier transforms (c=
hi(x1) and phi(x2)), one would be able to do the integration over d^4k resu=
lting in a delta function of d^4(x1-x2), which would eliminate one of the x=
integrals and make x1=x2 in f(x1)g(x2) (but one would still be left with=
a 4 space integral). Where does the convolution theorem come in?

BTW, George Jones, in the thread "Lorentz invariance of inner products" in =
this group, seems to make the error of neglecting the extra mass-shell cons=
traint factor in the integration measure on going from momentum to position=
space. Here is the quote, with the slight complication of spin 1/2 diracol=
ogy (but see the original thread):
"<f|g> = |f(p)^+ gamma^0 g(p) d mu ,
/

where, as above, the integration measure is given by

d mu delta( p^mu p_mu - m^2) theta(p^0) dp^0 dp^1 dp^2 dp^3 .

Now define the action of an arbitrary Lorentz transformation B on an
arbitrary spinor wave funtion by
_ _
f'(p) = | L 0 | f(B^(-1) p) ,
|_0 (L^-1)^+_|

where L is an SL(2,C) matrix and B is the Lorentz derived from L.
_ _
f'(p)^+ gamma^0 = f(B^(-1) p)^+| L^+ 0 | gamma^0
|_ 0 L^(-1)_|
_ _
= f(B^(-1) p)^+ gamma^0 | L^(-1) 0 |
|_ 0 L^+_|

Therefore,

/
<f'|g'> = | f(B^(-1) p)^+ gamma^0 g(B^(-1) p) d mu
/
/
= |f(p)^+ gamma^0 g(p) d mu
/

= <f|g> ,

since d mu is Lorentz-invariant. Taking Fourier transforms gives the
configuration space inner product

/
<f|g> = |f(x)^+ g(x) dx^1 dx^2 dx^3 ,
/

where f(x) and g(x) are both evaluated at any fixed x^0."

This last equation seems incorrect to me if the scalar product is defined a=
s above in momentum space.

On Sunday, April 8, 2012 12:21:52 PM UTC-4, Norbert Dragon wrote:

> If the scalar product of momentum wavefunctions is (stick for
> simplicity with scalar particles)
>
> < chi | phi > = int d^3 k 1/sqrt(m^2 + k^2) chi^*(k) phi(k)
>
> then their Lorentz transformed function is simply
>
> chi'(Lambda k) = chi(k)
>
> But given the above scalar product, |phi(k)|^2 cannot be the
> probability density to find the particle, which is in the state phi,
> in a momentum range d^3 k around k.

Perhaps, but why can't |phi(x)|^2 be the probability density of finding the=
particle in range d^4 x around x?
>
> With the above scalar product, phi/sqrt(sqrt(m^2 + k^2)) is the
> momentum wave function. It transforms nonlocally because of the
> convolution theorem, because its fouriertransform is the product
> of a locally transforming factor with the function
> sqrt(sqrt(m^2 + k^2)).
Are you confusing Fourier transforms with Lorentz transformations? Please s=
pell this out. How does the convolution theorem apply at all?
>
>
>
> --
> Superstition brings bad luck.
>
> www.itp.uni-hannover.de/~dragon