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;
<br />
U^{\dagger}(\Lambda)\phi_{r}(\bar{x})U(\Lambda)= D_{r}{}^{s}(\Lambda) \phi_{s}(x)<br />
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
<br />
U^{\dagger}A^{a}(\bar{x})U= \Lambda^{a}{}_{b}A^{b}(x) + \partial^{a}f(x;\Lambda) \ \ (1)<br />
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).
<br />
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.<br />
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
<br />
\epsilon^{a}= (\epsilon^{0},\epsilon^{1},\epsilon^{2},\epsilon^{0}).<br />
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
<br />
\epsilon^{a}\rightarrow W^{a}_{c}(0,u,v)\epsilon^{c}= \epsilon^{a} + \frac{u\epsilon^{1}+v\epsilon^{2}}{\omega}p^{a}.<br />
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
