Two spin d.o.f. for massless gauge bosons

[tom.stoer]

It is well know that massless spin-1 gauge bosons have two spin states s³=+1 and s³=-1. There are two independent approaches how this can be shown:
1) via the representations of the Lorentz group for p²=0
2) via fixing / eliminating unphysical gauge d.o.f., e.g. via elimination of the longitudinal polarization by implemetation of the Gauss law constraintin the A°(x)=0 gauge

It is interesting that these two approaches seem to be unrelated - even so we know that A(x) lives in a vector bundle and therefore gauge and space-time symmetry are not independent.

My question is whether there is a deeper relation between (1) and (2).

 

[strangerep]

It is well know that massless spin-1 gauge bosons have two spin states s³=+1 and s³=-1. There are two independent approaches how this can be shown:
1) via the representations of the Lorentz group for p²=0
2) via fixing / eliminating unphysical gauge d.o.f., e.g. via elimination of the longitudinal polarization by implemetation of the Gauss law constraintin the A°(x)=0 gauge

It is interesting that these two approaches seem to be unrelated - even so we know that A(x) lives in a vector bundle and therefore gauge and space-time symmetry are not independent.

My question is whether there is a deeper relation between (1) and (2).

I'd say the answer is "yes", because...

In (1), we must assume arbitrarily that the particle transforms trivially under the
two translation-like generators of the little group E(2). (Without this assumption,
the analysis predicts unphysical phenomena.)

Some authors (YS Kim and collaborators) maintain that these translation-like
generators in E(2) are the origin of the gauge freedom which method (2) eliminates.

 

[tom.stoer]

interesting - any references?

 

[Bill_K]

Hard to see how they could be related. The translation-like generators lead to states with arbitrarily large helicity, while the unphysical polarization states have helicity zero.

 

[dextercioby]

Needless to say, I'm also interested in an answer to this question.

 

[strangerep]

any references?

The argument is explained in section 7.3 of this book:

Y.S.Kim, M.E.Noz,
"Phase Space Picture of QM -- Group Theoretical Approach",
World Scientific, 1991

I think it's also in various papers by these authors, and maybe
Han as well, but I don't have time right now to hunt them down.

Basically, when transformations corresponding to the "N1, N2"
translation-like generators of E(2) are applied to the EM potential,
the result is the addition of a quantity proportional to the
4-momentum, i.e., a gauge transformation. But -- now that I read
it again -- they seem to restrict to a plane wave and impose the
Lorenz condition before doing this. They do it to simplify the form
of the N1, N2 matrices, so maybe this doesn't affect the main point.

CAVEAT: I've always experienced uneasiness when reading Kim's work.
Nothing I can put my finger on, just a vague distrust.
I'd be interested to hear what others think.

 

[Bill_K]

http://www.arxiv.org/abs/hep-th/9712009

From glancing at this, I think he's used a representation of the translation-like generators which is not gauge invariant and erroneously concluded that they are identical to gauge transformations.

 

[strangerep]

http://www.arxiv.org/abs/hep-th/9712009

From glancing at this, I think he's used a representation of the translation-like generators which is not gauge invariant and erroneously concluded that they are identical to gauge transformations.

Can you be more specific? I'm guessing you're referring to section VI (4-vectors
and Gauge Transformations), but afaict he's only assuming a particular coordinate
system (i.e., light cone coordinates). His transformations in eq(6.10) appear to use
the matrix $$D(\xi,\eta,0)$$ derived in an earlier section, and I don't see where
any gauge-noninvariance of $$D(\xi,\eta,0)$$ enters.

Cheers.

 

[tom.stoer]

My conclusion that it is possible to construct gauge transformations as certain "contracted" rotations, which means that there is an equivalence in the infinite momentum frame. But that does not uncover the deep relation in general.

Nevertheless: nice paper

 

[strangerep]

My conclusion that it is possible to construct gauge transformations as certain "contracted" rotations, which means that there is an equivalence in the infinite momentum frame. But that does not uncover the deep relation in general.

How do you mean?

The "contracted rotations" are only a manifestation of the fact that SO(3) can
be contracted to E(2). So this is just an aspect of the procedures for finding
the massless Poincare irreps.

What bothers me more is that Kim's technique seems totally silent about gauge
transformations of the electron field. It doesn't make the connection between
electron gauge transformations and photon gauge transformations.

Also, the whole technique of finding such EM gauge transformations within Poincare
is impotent (afaict) when faced with the question of the origin of the other gauge
transformations in the Standard Model.

 

[tom.stoer]

The gauge transformations of the fermion fields are not relevant for reducing the number of d.o.f.. Usually one would start with a global fermion gauge trf., make it local and derive the existence and properties of the gauge potential. One could do it the other way round, namely start with the gauge potential and construct the Dirac operator which immediately gives us the gauge trf. of the fermion fields. But this step is missing.

I agree that the gauge symmetries SU(2) and US(3) are both missing.

For me this paper is just another hint towards a deeper connection between Poincare and gauge invariance, but w/o uncovering these hidden aspects.

 

[Lapidus]

It is well know that massless spin-1 gauge bosons have two spin states s³=+1 and s³=-1. There are two independent approaches how this can be shown:
1) via the representations of the Lorentz group for p²=0
2) via fixing / eliminating unphysical gauge d.o.f., e.g. via elimination of the longitudinal polarization by implemetation of the Gauss law constraintin the A°(x)=0 gauge

It is interesting that these two approaches seem to be unrelated - even so we know that A(x) lives in a vector bundle and therefore gauge and space-time symmetry are not independent.

My question is whether there is a deeper relation between (1) and (2).

I don't understand the question.

We have a massless (parity conserving) spin-1 particle, which has two degrees of freedoms, its two helicity states.

The smallest Lorentz tensor that describes spin-1 particles is the four-vector. So we have a redundant description, a Lorentz four vector describing two physical degrees of freedom.

What is then here unrelated??

 

[samalkhaiat]

It is well know that massless spin-1 gauge bosons have two spin states s³=+1 and s³=-1. There are two independent approaches how this can be shown:
1) via the representations of the Lorentz group for p²=0
2) via fixing / eliminating unphysical gauge d.o.f., e.g. via elimination of the longitudinal polarization by implemetation of the Gauss law constraintin the A°(x)=0 gauge

It is interesting that these two approaches seem to be unrelated - even so we know that A(x) lives in a vector bundle and therefore gauge and space-time symmetry are not independent.

My question is whether there is a deeper relation between (1) and (2).

2) applies to a classical field A satisfying the Maxwell equation which is (as you might know) incompatible with manifest covariance if A is a nontrivial quantized field (i.e., Maxwell equation has to be modified in QFT).
1) says that E(2) is the little group of massless representations. Contrary to the naïve expectation, the field A ( connection defined on the trivial principal bundle $P(M,U(1))$) does not belong to the massless vector representation $(1/2,1/2)$. Indeed, such representation is carried by the trivial vector field $A_{a}(x) = \partial_{a}\phi (x)$ which leads to $F=dA=0$.
The massless spin-1 representation appropriate for electromagnetic field is given by the tensor representation

$$F = dA \sim (0,1) \oplus (1,0)$$

When we construct the QFT in terms of the field A, we also explicitly demonstrate (by considering the action of E(2) generators on the field A) the impossibility of quantizing A in the positive-metric Hilbert space. Therefore, in order to have a sensible theory, we introduce a subsidiary condition, which is a condition selecting the physical subspace of the indefinite-metric Hilbert space. This is how, in QFT, we get rid off the unwanted DoF in the field A.
So, asking about the relation between (1) and (2) is just like asking about the relation between the classical gauge field A and the quantized gauge field A.

Regards

sam

 

[tom.stoer]

The smallest Lorentz tensor that describes spin-1 particles is the four-vector. So we have a redundant description, a Lorentz four vector describing two physical degrees of freedom.

What is then here unrelated??

You have two completely different ways to reduce the unphysical to the physical degrees of freedom. You know from gauge symmetry that two degrees of freedom are unphysical w/o ever referring to Poincare symmetry and vice versa. The two approaches must be related somehow, but mathematically they aren't.

 

[dextercioby]

It is well know that massless spin-1 gauge bosons have two spin states s³=+1 and s³=-1. There are two independent approaches how this can be shown:
1) via the representations of the Lorentz group for p²=0
2) via fixing / eliminating unphysical gauge d.o.f., e.g. via elimination of the longitudinal polarization by implemetation of the Gauss law constraintin the A°(x)=0 gauge

It is interesting that these two approaches seem to be unrelated - even so we know that A(x) lives in a vector bundle and therefore gauge and space-time symmetry are not independent.

My question is whether there is a deeper relation between (1) and (2).

Hi Tom,

Sam's reply above kind of confirmed my ideas regarding your issue. Indeed, 2) is entirely a classical theory issue, as this procedure is needed to build a consistent Hamiltonian formulation. Once this is achived, thus the classical theory is <cured> from the troubling gauge-invariance, a canonical quantization is made in terms of field commutators, just like in the Dirac field's case, which strongly resembles the case of the normal scalar field.

Incidentally, the Hamiltonian formulation of the em-field establishes that the nr. of d.o.f. of the e-m field is 2 for each space-time point. One can show that these 2 dof are related to the intrinsic angular momentum of the field.

While your 1) 'sees' the problem from a quantum perspective in which the representations of (restricted) Poincare's group must be taken into account. But this can be done, only if the manifest covariance is kept, which is not possible by going with point 2) in the quantum theory as described in my 1st paragraph. As Sam said, one can try a Gupta-Bleuler quantization or a covariant BRST quantization with the selection of the physical state space by imposing

$$\partial^{\mu} \hat{A}_{\mu} = 0$$.

 

[tom.stoer]

2) applies to a classical field A satisfying the Maxwell equation which is (as you might know) incompatible with manifest covariance if A is a nontrivial quantized field (i.e., Maxwell equation has to be modified in QFT).

No, neither do you have to modify Maxwell's equations in the canonical quantization with A°=0 gauge, nor does missing manifest covariance bother you as long as the Poincare algebra still closes - which can be proven explicitly.

When we construct the QFT in terms of the field A, we also explicitly demonstrate ... the impossibility of quantizing A in the positive-metric Hilbert space. Therefore, in order to have a sensible theory, we introduce a subsidiary condition, which is a condition selecting the physical subspace of the indefinite-metric Hilbert space.

We do not introduce a subsidiary condition; we set A°=0 by chosing a gauge and we let the Gauss law act as a constraint on the physical Hilbert space; this Gauss law is not introduced by hand, but appears automatically. After A°=0 even in the enlarged Hilbert space the norm is always non-negative; negative norm states do appear in covariant gauges, not in physical gauges.

So, asking about the relation between (1) and (2) is just like asking about the relation between the classical gauge field A and the quantized gauge field A.

No, the entire mechanism works in a full QFT setup; you can first quantize - then reduce the unphysical d.o.f.; there's no reason to reduce these states classically.

http://www.adsabs.harvard.edu/abs/1994AnPhy.233...17L
http://www.adsabs.harvard.edu/abs/1994AnPhy.233..317L

My question is the following:
1) you can show based on Poincare invariance and w/o ever referring to gauge symmetry that one is able to construct a representation with two physical helicity states.
2) you can show the same thing based on gauge invariance and w/o ever referring to Poincare invariance.

So the two approaches which are formally completely unrelated give you exactly the same result. Of course this is required by consistency, but beyond this coincidence there seems to be a deeper relation between gauge and Poincare invariance which I still don't see.

 

[Lapidus]

You have two completely different ways to reduce the unphysical to the physical degrees of freedom. You know from gauge symmetry that two degrees of freedom are unphysical w/o ever referring to Poincare symmetry and vice versa. The two approaches must be related somehow, but mathematically they aren't.

I found Lisa Randall's explanation of gauge symmetry in her Warped passages book quite lovely. Hope no one is offended since it is from a popular book and it can add something to this discussion!

Each individual photon has different possible polarizations, but not all imaginable polarizations are allowed. It turns out that when a photon travels in any particular direction, the wave can oscillate only in directions that are perpendicular to its direction of motion. This wave acts like water waves on the ocean, which also oscillate perpendicularly. That is why you see a buoy or a boat bob up and down as a water wave passes by.

The wave associated with a photon can oscillate in any direction perpendicular to its direction of motion. Really, there is an infinite number of such directions: imagine a circle perpendicular to the line of motion, and you can see that the wave is able to oscillate in any radial direction (from the center to the outside of the circle), and there are an infinite number of such directions.

But in the physical description of these oscillations, we need only two independent perpendicular oscillations to account for them all. In physics terminology they are called transverse polarizations.

The important thing is that, in principle, there could have been a third polarization direction, one that oscillates along the direction in which the wave travels (had it existed, it would have been called the longitudinal polarization). That is how sound waves travel, for example. But no such polarization of the photon exists. Only two of the three conceivable independent polarization directions exist in nature. A photon never oscillates along its direction of motion or in the time direction: it oscillates only along the directions perpendicular to its motion.

Even if we didn't already know from independent theoretical considerations that the longitudinal polarization was spurious, quantum field theory would have told us not to include it. If a physicist were to make calculations using a theory of forces that mistakenly included all three polarization directions, the theory's predictions of their properties wouldn't make sense. For example, she would predict ridiculously high gauge boson interactions rates. In fact, she would predict gauge bosons that interacted more often than always—that is, more than 100% of the time. Any theory that makes such nonsensical predictions is clearly wrong, and both nature and quantum field theory make it clear that this nonperpendicular polarization does not exist.

Unfortunately, the simplest theory of forces that physicists could formulate includes this spurious polarization direction. That is not so surprising because a theory that would work for any photon can't possibly contain information about one particular photon traveling in one particular direction. And without such information, special relativity would not distinguish any direction. In a theory that preserves the symmetries of special relativity (including rotational symmetry), you would need three directions—not two—to describe all the directions in which a photon could oscillate; in such a description, the photon could oscillate in any direction of space.

But we know that isn't true. For any particular photon, its direction of motion is singled out and oscillation in that direction is forbidden. But you wouldn't want to have to make a different theory for each and every photon, all with their own directions of travel. You would want a theory that works no matter which way a photon is travelling.

Although you could try to make a theory that didn't include the spurious polarization direction at all, it is far simplerand cleaner to respect rotational symmetry and eliminate the bad polarization in some other way. Physicists, aiming for simplicity, have recognized that quantum field theory works best when they include the spurious longitudinal polarization in their theory but add an extra ingredient to filter out the good, physically relevant predictions from the bad.

This is where internal symmetries enter the picture. The role of internal symmetries in the theory of forces is to eliminate the contradictions that the unwanted polarization would create without making us forfeit the symmetries of special relativity. Internal symmetries are the simplest way to filter out the polarization along the direction of travel that independent theoretical considerations and experimental observations tell us does not exist. They classify polarizations into good and bad categories, those that are consistent with the symmetries and those that are not.

 

[Lapidus]

My question is the following:
1) you can show based on Poincare invariance and w/o ever referring to gauge symmetry that one is able to construct a representation with two physical helicity states.
2) you can show the same thing based on gauge invariance and w/o ever referring to Poincare invariance.

So the two approaches which are formally completely unrelated give you exactly the same result. Of course this is required by consistency, but beyond this coincidence there seems to be a deeper relation between gauge and Poincare invariance which I still don't see.

I guess the key to your problem must be that gauge symmetry is not a symmetry in the usual sense, but a redundancy in the description of a massless spin-1 particle. Gauge transformations of a state does not change the state, just give a different description of the same state of the physical system. That is of course not the case for global symmetries like Poincare transformations, which produce a different state that is physical equivalent but not identical to the original state.

 

[tom.stoer]

... that gauge symmetry is not a symmetry in the usual sense, but a redundancy in the description of a massless spin-1 particle. Gauge transformations of a state does not change the state, just give a different description of the same state of the physical system. That is of course not the case for global symmetries like Poincare transformations, which produce a different state that is physical equivalent but not identical to the original state.

I fully agree, but I can't see how to construct a "key to my problem ..." :-(

 

[vanhees71]

The deeper reason is the analysis of the unitary (ray) irreducible representations (irreps) of the Poincare group. As it turns out, due to the structure of the Poincare group, there are no non-trivial central charges, and one can restrict oneself to true unitary irreps of the covering group, i.e., the proper orthochronous Lorentz transformations one has to consider its universal covering group, SL(2,C) (which is very important since otherwise one would miss the possibility of half-integer spin).

The next step is that one characteristic of the irrep must be the Casimir $$p^2=m^2=\text{const}$$, where $$p$$ are the four-momentum operators, generating the space-time translations. As it turns further out, one finds causal local QFTs of interacting particles only for $$m^2 \geq 0$$.

The next step is to realize that the irrep must act transitively on the mass-shell in momentum space, i.e., for each momentum eigenstate state with momentum $$p$$ on the mass shell, there must exist a Poincare transformation such that you reach any other momentum state with the momentum on the same mass shell. For $$m>0$$, you can choose as the "standard momentum" $$p_0=(m,0,0,0)$$ and then with a Lorentz boost get to any other momentum on the mass shell.

The further characterization of the irrep. is then the irrep. of the subgroup which leaves $$p_0$$ constant (the "little group" of the irrep). For $$m>0$$ this is the rotation group with the well-known representations, leading to spins $$s=0,1/2,1,3/2,\ldots$$.

For massless particles, you have to choose as standard momentum $$p_0=(\Lambda,0,0,\Lambda)$$, and then you can reach any other momentum by boosts. The little group of this irrep is generated by null rotations and normal rotations around the three-axis. The latter is the helicity. In order to induce the correct irreps for the full rotation group, the helicities themselves can also only be integer or half-integer numbers $$h=0,\pm 1/2,\pm 1,\ldots$$.

The null-rotations, however lead to a continuous sub group of the little group, and since there's no indication of massless particles with a continuous spin-like structure, one must assume that the null-rotations are represented by trivial representations, and this leads to gauge invariance, if you try to represent these representations by massless relativistic fields.

You find this in my QFT manuscript on my homepage:

http://theorie.physik.uni-giessen.de/~hees/publ/lect.pdf

 

[tom.stoer]

The deeper reason is the analysis of the unitary (ray) irreducible representations (irreps) of the Poincare group. As it turns out, due to the structure of the Poincare group, there are no non-trivial central charges, and one can restrict oneself to true unitary irreps of the covering group, i.e., the proper orthochronous Lorentz transformations one has to consider its universal covering group, SL(2,C) (which is very important since otherwise one would miss the possibility of half-integer spin).

The next step is that one characteristic of the irrep must be the Casimir $$p^2=m^2=\text{const}$$, where $$p$$ are the four-momentum operators, generating the space-time translations. As it turns further out, one finds causal local QFTs of interacting particles only for $$m^2 \geq 0$$.

The next step is to realize that the irrep must act transitively on the mass-shell in momentum space, i.e., for each momentum eigenstate state with momentum $$p$$ on the mass shell, there must exist a Poincare transformation such that you reach any other momentum state with the momentum on the same mass shell. For $$m>0$$, you can choose as the "standard momentum" $$p_0=(m,0,0,0)$$ and then with a Lorentz boost get to any other momentum on the mass shell.

The further characterization of the irrep. is then the irrep. of the subgroup which leaves $$p_0$$ constant (the "little group" of the irrep). For $$m>0$$ this is the rotation group with the well-known representations, leading to spins $$s=0,1/2,1,3/2,\ldots$$.

For massless particles, you have to choose as standard momentum $$p_0=(\Lambda,0,0,\Lambda)$$, and then you can reach any other momentum by boosts. The little group of this irrep is generated by null rotations and normal rotations around the three-axis. The latter is the helicity. In order to induce the correct irreps for the full rotation group, the helicities themselves can also only be integer or half-integer numbers $$h=0,\pm 1/2,\pm 1,\ldots$$.

...

What you describe here is standard textbook QFT, except for

...

The null-rotations, however lead to a continuous sub group of the little group, and since there's no indication of massless particles with a continuous spin-like structure, one must assume that the null-rotations are represented by trivial representations, and this leads to gauge invariance, if you try to represent these representations by massless relativistic fields.

Are you saying that Poincare invariance for massless vector fields enforces gauge symmetry?

 

[dextercioby]

[...]
The null-rotations, however lead to a continuous sub group of the little group, and since there's no indication of massless particles with a continuous spin-like structure, one must assume that the null-rotations are represented by trivial representations, and this leads to gauge invariance, if you try to represent these representations by massless relativistic fields.[...]

Professor van Hees, do you have a reference for your statement, especially on how null rotations lead to gauge invariance ?

Thank you!

 

[vanhees71]

First of all, that's not my finding (I'd be glad, if I had discovered something so fundamental and beautiful myself :-)). It's all due to Wigner's famous analysis of the unitary representations of the Poincare group.

I tried to write this up in my manuscript about QFT, which you find here:

http://theorie.physik.uni-giessen.de/~hees/publ/lect.pdf

The formal theory about the unitary irreps of the Poincare group is given in Appendix B (p. 267ff).

 

[tom.stoer]

It's all due to Wigner's famous analysis of the unitary representations of the Poincare group.

Yes, I know, but I don't know how to relate this to gauge invariance.

 

[strangerep]

What [Hendrik] describes here is standard textbook QFT, except for

The null-rotations, however lead to a continuous sub group of the little group, and since there's no indication of massless particles with a continuous spin-like structure, one must assume that the null-rotations are represented by trivial representations, and this leads to gauge invariance, if you try to represent these representations by massless relativistic fields.

Actually, the last part is also standard textbook QFT. See (eg) Weinberg vol 1, pp69-74,
especially his remarks at the top of p72.

The key phrase in what Hendrik said is "one must assume that the null-rotations are represented by trivial representations". Weinberg says a similar thing in the form:

Massles particles are not observed to have any continuous degree of freedom
[...corresponding to the translation-like generators of ISO(2) little group...]; to avoid such a continuum of states, we must require that physical states are eigenvectors with [... trivial transformation behaviour...]

It's then a further consequence that the photon field does not transform covariantly. See sect 5.9 of Weinberg, especially the discussion following eq(5.9.16) leading to eq(5.9.22).

 

[tom.stoer]

It's then a further consequence that the photon field does not transform covariantly.

Again: does this imply local gauge symmetry?

 

[strangerep]

It's then a further consequence that the photon field does not transform covariantly.

Again: does this imply local gauge symmetry?

Sounds like you don't have a copy of Weinberg at hand?

Here's the gist of his argument: if we try to construct a quantum field from the massless Poincare irreps of spin 1, we find that the equations are inconsistent. In Weinberg's words (top of p250): "No 4-vector field can be constructed from the annihilation and creation operators for a particle of mass zero and helicity +\- 1."

Then he shows what happens if we ignore this difficulty and try to proceed anyway. It turns out that a Lorentz transformation on such a field has an extra piece. See eqns (5.9.30,31). This extra piece has exactly the form of the usual gauge transformation of the EM potential. Later he shows how this embarrassment can be avoided by coupling to a conserved current in the interaction term of the Lagrangian.

BTW, it doesn't hurt to keep in mind that Poincare transformations are physical, but gauge transformations are not. So getting one from the other is a delicate business, since we must note where unphysical assumptions enter -- which is not always obvious. :-)

 

[tom.stoer]

Unfortunately I really do not have Weinberg's book here at hand.

I have to check asap; sounds as if this deep relation I am looking for does indeed exist.

 

[samalkhaiat]

negative norm states do appear in covariant gauges, not in physical gauges.

It is impossible to formulate the quantized theory of $A_{a}$ in the axial, light-cone, and temporal gauges in a satisfactory way. Indefinite-metric is indispensable in those gauges, unless one accepts the violation of translational invariance. This has been proven on general ground in;
Nakanishi, N., Phys. Lett. 131B, 381(1983).

My question is the following:
1) you can show based on Poincare invariance and w/o ever referring to gauge symmetry that one is able to construct a representation with two physical helicity states.

Yes, as I said before, this applies to fields which themselves form a representation, i.e., transform according to some matrix representation $D(\Lambda)$ of the Lorentz group;

$$U^{\dagger}(\Lambda)\phi_{r}(\bar{x})U(\Lambda)= D_{r}{}^{s}(\Lambda) \phi_{s}(x)$$

I also said that the gauge potential (which describes the gauge bosons) does not form a representation of the Lorentz group, i.e., there exists no nontrivial representation matrix $D(\Lambda)$ for $A_{a}$. Indeed, under Lorentz transformation, the gauge potential transforms as

$$U^{\dagger}A^{a}(\bar{x})U= \Lambda^{a}{}_{b}A^{b}(x) + \partial^{a}f(x;\Lambda) \ \ (1)$$

Clearly, this is not how a vector field transform under the Lorentz transformation $\Lambda$. Also, apart from the trivial transformation $\Lambda^{a}_{c}= \delta^{a}_{c}$, eq(1) is not a gauge transformation.
So, what you say in (1) does not apply naturally to the gauge potentials. We pretend it does because we want to do QFT; there is no other possibility available to us.
More on the massive/massless representations of the Poicare’ group can be found in:
www.physicsforums.com/showthread.php?t=315387

Ok, let me say something regarding the business of generating “gauge transformation” from the action of the Wigner’s (null-rotation) matrix $W(0,u,v)$ on the polarization “vector” $\epsilon_{a}(p)$.
$$W^{a}{}_{c}(0,u,v)= \left( \begin{array}{cccc} 1+a^{2} & u & v & a^{2} \\ u & 1 & 0 & u \\ v & 0 & 1 & v \\ -a^{2} & -u & -v & 1-a^{2} \end{array} \right), \ \ \ a^{2}= (u^{2}+v^{2})/2.$$
Consider the Maxwell equation,
$$\partial^{2} A^{a}= \partial^{a}\partial_{c}A^{c},$$
with the gauge transformation,
$$A^{a}\rightarrow A^{a} + \partial^{a}f(x).$$
Inserting
$$A^{a}= \epsilon^{a}(p)e^{ipx},$$
we find
$$p^{2}\epsilon^{a}= (p.\epsilon) p^{a}, \ \ (2)$$
and
$$\epsilon^{a}\rightarrow \epsilon^{a} + if(p)p^{a}, \ \ (3)$$
where
$$f(p)e^{ip.x}= f(x).$$
Now, for $p^{2}\neq 0$, the polarization “vector” is proportional to $p^{a}$;
$$\epsilon^{a}= \frac{p.\epsilon}{p^{2}}p^{a}.$$
However, this “massive” mode is not physical because, it can be gauged away by the choice;
$$f(p) = i \frac{(p.\epsilon)}{p^{2}}.$$
For massless mode, eq(2) implies the Lorentz condition $p^{a}\epsilon_{a}=0$. So, in the frame $p=(\omega , 0,0,\omega)$, the polarization vector is
$$\epsilon^{a}= (\epsilon^{0},\epsilon^{1},\epsilon^{2},\epsilon^{0}).$$
Again, the $\epsilon^{0}$ component is not physical, we gauge it away by choosing
$$f(p) = i\frac{\epsilon^{0}}{\omega}.$$
So, for electromagnetic wave moving in the z-direction, the physical polarization vector is confined in the xy-plane and has only two degrees of freedom:
$$\epsilon^{a}= (0,\epsilon^{1},\epsilon^{2},0). \ \ (4)$$
The totality of Lorentz transformations which leave $p^{a}$ invariant is called a little group on p. For massless field, it is isomorphic to the two dimensional Euclidean group E(2); the group of translations, T(2), and rotations, SO(2), in 2-dimensional plane perpendicular to p. It is easy to see that the group of all Wigner matrices is isomorphic to T(2). Indeed
$$W(0,u_{1},v_{2})W(0,u_{2},v_{2})= W(0,u_{1}+u_{2},v_{1}+v_{2}),$$
and
$$[X_{1},X_{2}]=0,$$
where
$$X_{1}= \frac{\partial}{\partial u}W(0,u,0), \ X_{2}= \frac{\partial}{\partial v}W(0,0,v)$$
So, the action of T(2) on the physical polarization vector, eq(4), is
$$\epsilon^{a}\rightarrow W^{a}_{c}(0,u,v)\epsilon^{c}= \epsilon^{a} + \frac{u\epsilon^{1}+v\epsilon^{2}}{\omega}p^{a}.$$

This looks very much like the gauge transformation of eq(3). Ok, so we have managed to show that the group $T(2) \subset E(2)$, generates the U(1) gauge transformation. Does this mean that we have found the origin of gauge invariance? No, I don’t think so.

Regards

sam

 

[tom.stoer]

Unfortunately your post is nearly unreadable.

I don't believe in the statement "It is impossible to formulate the quantized theory ... in the axial, light-cone, and temporal gauges in a satisfactory way. Indefinite-metric is indispensable in those gauges, unless one accepts the violation of translational invariance. This has been proven on general ground in;
Nakanishi, N., Phys. Lett. 131B, 381(1983)."
There were many papers (it started approx. 20 years ago) were many papers on non-perturbative quantization of QCD using physical gauges have been published.