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?

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.

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

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

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.

Nothing wrong with this.

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) .

Each Dirac representation looks like this in a suitable basis.

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

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

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

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.

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

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

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

--
Superstition brings bad luck.

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

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

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).

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

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.

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?

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).

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.

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

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

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

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.

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?

Sorry, I don't know anything about that.

On 01/04/12 12:50, iuval wrote:

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

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

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

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

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

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

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).

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

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 )

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

You should read

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

really.

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.

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

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.

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.

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

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.

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.

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.

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

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.

Not at all: if Psi were to commute, then

Psi^bar Psi = - Psi^bar Psi

(because the matrix epsilon is antisymmetric)

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.

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.

--
Superstition brings bad luck.

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

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

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

Your confusion results from confused ideas about transformations.

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

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

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

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

The gamma-matrices satisfy

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

and the bar-operation is constructed such that

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

which is why

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

transforms like a vector field.

Also devivatives do not transform

T_Lambda d_m  = d_m T_Lambda ,

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

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

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

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

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

--
Superstition brings bad luck

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

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.

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

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

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:

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?

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).

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

The by far best textbook on this subject is

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

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

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

On 15/04/12 09:50, iuval wrote:

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

*  Hendrik van Hees writes:

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))

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

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).

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.

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

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.

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.

--
Superstition brings bad luck

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

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)

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.

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.

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.

Not at all. Position wave funtions yield probability desities

|psi_i(x)|^2 ,   i in {up, down}

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

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

I shall answer your remaining questions separately.

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

--
Aberglaube bringt Unglück

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

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

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

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

*  Jos Bergervoet writes:

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.

--
Superstition brings bad luck

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

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

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* 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

*  Jos Bergervoet writes:

Side comment:

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

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

Moreover, in relativistic theories, there is no local current

j^m(Psi, d Psi)

which completes the position probability density

j^0 = |Psi|^2

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

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.

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.

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.

--
Superstition brings bad luck

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

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

 ...

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?

 ..

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

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

 ..

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

--
Jos

*  Jos Bergervoet writes:

page 85 and page 91

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

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

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

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

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

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.

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.

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

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.

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

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

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

Then the transformation

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

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

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

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

--
Supersition brings bad luck

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

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

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.

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?

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

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

 ...

Dirac made the same requirement.

This requirement was especially dear to Dirac!

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

On 29/04/12 02:34, Jos Bergervoet wrote:

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.

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.

There is non.

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

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.

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.

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

================= 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.

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.

Proof by lack of knowledge and imagination?

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

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?

You write about your private categories.

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

,=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?

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.

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.

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

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
Superstition brings bad luck

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

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.

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

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.

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.

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).

Nobody denied this.

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

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.

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

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.

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

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).

Voila!

...

read more »

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..

...

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..

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

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

*  Hendrik van Hees writes:

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

i d_t Psi =3D H Psi

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

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.

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.

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

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.

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.

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.

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.

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?

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?

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

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

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?

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.

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

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
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