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Matrix rep of Lorentz transformation in Dirac spinor space
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More options Apr 1 2012, 6:50 am
Newsgroups: sci.physics.relativity, sci.physics.research
From: iuval <clejan.iu...@gmail.com>
Date: 01 Apr 2012 11:50:41 +0100 (BST)
Local: Sun, Apr 1 2012 6:50 am
Subject: Matrix rep of Lorentz transformation in Dirac spinor space
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?

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More options Apr 3 2012, 12:58 pm
Newsgroups: sci.physics.research
From: Norbert Dragon <dra...@itp.uni-hannover.de>
Date: Tue, 03 Apr 2012 12:58:00 EDT
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space

*  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

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.

--

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More options Apr 7 2012, 10:15 am
Newsgroups: sci.physics.research
From: iuval <clejan.iu...@gmail.com>
Date: Sat, 07 Apr 2012 10:15:41 EDT
Local: Sat, Apr 7 2012 10:15 am
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space

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.

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.

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More options Apr 7 2012, 2:06 pm
Newsgroups: sci.physics.relativity, sci.physics.research
From: Hendrik van Hees <h...@fias.uni-frankfurt.de>
Date: 07 Apr 2012 19:06:32 +0100 (BST)
Local: Sat, Apr 7 2012 2:06 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
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.

to add the explicit treatment with Dirac spinors. Unfortunately my QFT
manuscript is in German, but the formulae are all there. Look at

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
D-60438 Frankfurt am Main
http://fias.uni-frankfurt.de/~hees/

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More options Apr 8 2012, 12:21 pm
Newsgroups: sci.physics.research
From: Norbert Dragon <dra...@itp.uni-hannover.de>
Date: Sun, 08 Apr 2012 12:21:52 EDT
Local: Sun, Apr 8 2012 12:21 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space

*  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).

>> 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)

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

> 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.

--

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More options Apr 12 2012, 6:03 pm
Newsgroups: sci.physics.research
From: iuval <clejan.iu...@gmail.com>
Date: 12 Apr 2012 23:03:15 +0100 (BST)
Local: Thurs, Apr 12 2012 6:03 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space

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' (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.

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More options Apr 13 2012, 12:45 pm
Newsgroups: sci.physics.research
From: Norbert Dragon <dra...@itp.uni-hannover.de>
Date: Fri, 13 Apr 2012 12:45:01 EDT
Local: Fri, Apr 13 2012 12:45 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space

*  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.

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.

--

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More options Apr 14 2012, 8:17 am
Newsgroups: sci.physics.research
From: Hendrik van Hees <h...@fias.uni-frankfurt.de>
Date: Sat, 14 Apr 2012 14:17:01 +0200 (CEST)
Local: Sat, Apr 14 2012 8:17 am
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
On 13/04/12 00:03, iuval wrote:

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.

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

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More options Apr 15 2012, 3:50 am
Newsgroups: sci.physics.research
From: iuval <clejan.iu...@gmail.com>
Date: Sun, 15 Apr 2012 09:50:58 +0200 (CEST)
Local: Sun, Apr 15 2012 3:50 am
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
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.

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?

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More options Apr 15 2012, 5:48 pm
Newsgroups: sci.physics.research
From: Hendrik van Hees <h...@fias.uni-frankfurt.de>
Date: 15 Apr 2012 22:48:41 +0100 (BST)
Local: Sun, Apr 15 2012 5:48 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
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:

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

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:

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

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More options Apr 16 2012, 11:35 am
Newsgroups: sci.physics.research
From: Norbert Dragon <dra...@itp.uni-hannover.de>
Date: Mon, 16 Apr 2012 11:35:22 EDT
Local: Mon, Apr 16 2012 11:35 am
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
*  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.

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.

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More options Apr 16 2012, 4:06 pm
Newsgroups: sci.physics.research
From: Norbert Dragon <dra...@itp.uni-hannover.de>
Date: 16 Apr 2012 21:06:17 +0100 (BST)
Local: Mon, Apr 16 2012 4:06 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space

*  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.

[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

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More options Apr 17 2012, 10:02 pm
Newsgroups: sci.physics.research
From: Hendrik van Hees <h...@fias.uni-frankfurt.de>
Date: Tue, 17 Apr 2012 22:02:49 EDT
Local: Tues, Apr 17 2012 10:02 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
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.

On 16/04/12 17:35, Norbert Dragon wrote:

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

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More options Apr 17 2012, 10:03 pm
Newsgroups: sci.physics.research
From: Jos Bergervoet <jos.r.bergerv...@gmail.com>
Date: Tue, 17 Apr 2012 22:03:22 EDT
Local: Tues, Apr 17 2012 10:03 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
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

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More options Apr 18 2012, 12:38 pm
Newsgroups: sci.physics.research
From: Norbert Dragon <dra...@itp.uni-hannover.de>
Date: 18 Apr 2012 17:38:50 +0100 (BST)
Local: Wed, Apr 18 2012 12:38 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
*  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.

--

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More options Apr 19 2012, 3:26 pm
Newsgroups: sci.physics.research
From: Jos Bergervoet <jos.r.bergerv...@gmail.com>
Date: 19 Apr 2012 20:26:28 +0100 (BST)
Local: Thurs, Apr 19 2012 3:26 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
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

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More options Apr 22 2012, 4:17 am
Newsgroups: sci.physics.research
From: Norbert Dragon <dra...@itp.uni-hannover.de>
Date: 22 Apr 2012 09:17:47 +0100 (BST)
Local: Sun, Apr 22 2012 4:17 am
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
*  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
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.

--

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More options Apr 25 2012, 5:05 pm
Newsgroups: sci.physics.research
From: Jos Bergervoet <jos.r.bergerv...@gmail.com>
Date: 25 Apr 2012 22:05:07 +0100 (BST)
Local: Wed, Apr 25 2012 5:05 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
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

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More options Apr 26 2012, 6:39 pm
Newsgroups: sci.physics.research
From: Norbert Dragon <dra...@itp.uni-hannover.de>
Date: 26 Apr 2012 23:39:31 +0100 (BST)
Local: Thurs, Apr 26 2012 6:39 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
*  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

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)

--

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More options Apr 28 2012, 8:34 pm
Newsgroups: sci.physics.research
From: Jos Bergervoet <jos.r.bergerv...@gmail.com>
Date: Sat, 28 Apr 2012 20:34:17 EDT
Local: Sat, Apr 28 2012 8:34 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
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

...

>> 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)

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

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More options Apr 29 2012, 12:50 pm
Newsgroups: sci.physics.research
From: Hendrik van Hees <h...@fias.uni-frankfurt.de>
Date: Sun, 29 Apr 2012 12:50:04 EDT
Local: Sun, Apr 29 2012 12:50 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
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).

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

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More options Apr 29 2012, 3:24 pm
Newsgroups: sci.physics.research
From: Norbert Dragon <dra...@itp.uni-hannover.de>
Date: Sun, 29 Apr 2012 21:24:54 +0200 (CEST)
Local: Sun, Apr 29 2012 3:24 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
================= 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:

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

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

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More options Apr 29 2012, 4:21 pm
Newsgroups: sci.physics.research
From: Hendrik van Hees <h...@fias.uni-frankfurt.de>
Date: Sun, 29 Apr 2012 16:21:47 EDT
Local: Sun, Apr 29 2012 4:21 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
On 29/04/12 21:24, Norbert Dragon wrote:

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

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!

...

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More options May 1 2012, 5:08 pm
Newsgroups: sci.physics.research
From: Jos Bergervoet <jos.r.bergerv...@gmail.com>
Date: 01 May 2012 22:08:08 +0100 (BST)
Local: Tues, May 1 2012 5:08 pm
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
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*
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

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More options May 3 2012, 7:13 am
Newsgroups: sci.physics.research
From: Norbert Dragon <dra...@itp.uni-hannover.de>
Date: Thu, 3 May 2012 13:13:57 +0200 (CEST)
Local: Thurs, May 3 2012 7:13 am
Subject: Re: Matrix rep of Lorentz transformation in Dirac spinor space
*  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

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

>> 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

I asked for the relation between _cross_section_ and S-matrix but you

> 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.