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

  • Thread starter tom.stoer
  • Start date
  • #26
tom.stoer
Science Advisor
5,766
161
It's then a further consequence that the photon field does not transform covariantly.
Again: does this imply local gauge symmetry?
 
  • #27
strangerep
Science Advisor
3,113
938
strangerep said:
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. :-)
 
  • #28
tom.stoer
Science Advisor
5,766
161
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.
 
  • #29
samalkhaiat
Science Advisor
Insights Author
1,693
935
negative norm states do appear in covariant gauges, not in physical gauges.
It is impossible to formulate the quantized theory of [itex]A_{a}[/itex] 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 [itex]D(\Lambda)[/itex] of the Lorentz group;

[tex]
U^{\dagger}(\Lambda)\phi_{r}(\bar{x})U(\Lambda)= D_{r}{}^{s}(\Lambda) \phi_{s}(x)
[/tex]

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 [itex]D(\Lambda)[/itex] for [itex]A_{a}[/itex]. Indeed, under Lorentz transformation, the gauge potential transforms as

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

Clearly, this is not how a vector field transform under the Lorentz transformation [itex]\Lambda[/itex]. Also, apart from the trivial transformation [itex]\Lambda^{a}_{c}= \delta^{a}_{c}[/itex], 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 [itex]W(0,u,v)[/itex] on the polarization “vector” [itex]\epsilon_{a}(p)[/itex].
[tex]
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.
[/tex]
Consider the Maxwell equation,
[tex]\partial^{2} A^{a}= \partial^{a}\partial_{c}A^{c},[/tex]
with the gauge transformation,
[tex]A^{a}\rightarrow A^{a} + \partial^{a}f(x).[/tex]
Inserting
[tex]A^{a}= \epsilon^{a}(p)e^{ipx},[/tex]
we find
[tex]p^{2}\epsilon^{a}= (p.\epsilon) p^{a}, \ \ (2)[/tex]
and
[tex]\epsilon^{a}\rightarrow \epsilon^{a} + if(p)p^{a}, \ \ (3)[/tex]
where
[tex]f(p)e^{ip.x}= f(x).[/tex]
Now, for [itex]p^{2}\neq 0[/itex], the polarization “vector” is proportional to [itex]p^{a}[/itex];
[tex]\epsilon^{a}= \frac{p.\epsilon}{p^{2}}p^{a}.[/tex]
However, this “massive” mode is not physical because, it can be gauged away by the choice;
[tex]f(p) = i \frac{(p.\epsilon)}{p^{2}}.[/tex]
For massless mode, eq(2) implies the Lorentz condition [itex]p^{a}\epsilon_{a}=0[/itex]. So, in the frame [itex]p=(\omega , 0,0,\omega)[/itex], the polarization vector is
[tex]
\epsilon^{a}= (\epsilon^{0},\epsilon^{1},\epsilon^{2},\epsilon^{0}).
[/tex]
Again, the [itex]\epsilon^{0}[/itex] component is not physical, we gauge it away by choosing
[tex]f(p) = i\frac{\epsilon^{0}}{\omega}.[/tex]
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:
[tex]\epsilon^{a}= (0,\epsilon^{1},\epsilon^{2},0). \ \ (4)[/tex]
The totality of Lorentz transformations which leave [itex]p^{a}[/itex] 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
[tex]W(0,u_{1},v_{2})W(0,u_{2},v_{2})= W(0,u_{1}+u_{2},v_{1}+v_{2}),[/tex]
and
[tex][X_{1},X_{2}]=0,[/tex]
where
[tex]X_{1}= \frac{\partial}{\partial u}W(0,u,0), \ X_{2}= \frac{\partial}{\partial v}W(0,0,v)[/tex]
So, the action of T(2) on the physical polarization vector, eq(4), is
[tex]
\epsilon^{a}\rightarrow W^{a}_{c}(0,u,v)\epsilon^{c}= \epsilon^{a} + \frac{u\epsilon^{1}+v\epsilon^{2}}{\omega}p^{a}.
[/tex]

This looks very much like the gauge transformation of eq(3). Ok, so we have managed to show that the group [itex]T(2) \subset E(2)[/itex], 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
 
Last edited:
  • #30
tom.stoer
Science Advisor
5,766
161
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.
 

Related Threads on Two spin d.o.f. for massless gauge bosons

  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
2
Views
752
Replies
2
Views
1K
Replies
6
Views
5K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
2K
Top