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In page 71 Weinberg's QFT,

$$A\Psi^{\theta }_{a,b} ~=~(a\cos{(\theta )}-b\sin{(\theta )})\Psi^{\theta }_{a,b}.$$

He says that massless particles represented by $\Psi ^{\theta }_{a,b}$ are not observed to have the continuous degree of freedom $\theta$. I do not understand why.

These $\Psi ^{\theta }_{a,b}$ are eigenstates of $A$ with eigenvalue $a$ where

$$ U\left [ W\left ( \alpha ,\beta ,\theta \right ) \right ]~=~ 1+i\alpha A+i\beta B+i\theta J{3}. $$

$W$ is an element of the little group with a faithful representation $D(W)$.

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You should replicate the necessary formulae; not everyone is going to have access to the text. –  genneth Mar 25 '12 at 22:09
    
I may have found the reason , It may be because there does not exist a reference frame in which the photon is at rest so it is impossible to have SO(2) to be an invariance group for it . Is this correct ? –  PhysicsGuy Mar 26 '12 at 7:54
    
This is the reason for the little group--- the little group would be SO(3) if the particle could stop (rotations in the rest frame). This question made me look for my copy of Weinberg--- can you please provide the symbol context? What is he doing here? This is probably an argument that says that if you rotate a massless representation, the different angles are not a continuum of indistinguishable particles, but the details of the argument are what you are asking about, and everyone makes up an ideosyncratic argument for this stuff. –  Ron Maimon Mar 26 '12 at 18:49
    
very nice, carefully observed! –  Arnold Neumaier Mar 26 '12 at 19:09
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2 Answers 2

Weinberg uses an empirical fact (''are not observed to have'') to eliminate this case in his analysis. He says that there are such representation, but that they are irrelvant as they don't match observation.

Indeed, such unitary representations occur in Wigner's classification; they are the so-called continuous spin representations. But Weinberg doesn't want to do more representation theory than necessary and hence refers to experience to be able to take a shortcut.

In fact, one can eliminate the continuous spin representation also by causality arguments, and a student of Weinberg did this; see Abbot, Phys. Rev. D 13 (1976), 2291-2294. But these arguments are lengthy. Since they don't lead to causal quantum fields, there is little reason to give such arguments in a basic QFT textbook.

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The answer is simply because observed massless particles always obey $E = pc$

It helps to first see what is going on here: A and B are boost/rotation transformations which leave invariant the momentum vector which represents a massless particle propagating in the z direction. Now, why don't they change the momentum?

Well, the A and B, acting on a momentum in the z-direction, both represent a centrifugal acceleration K (orthogonal to the z-momentum) compensated by a counter rotation J. The total effect of A (as well as B) is therefor simply zero.

One can actually always compensate an orthogonal boost operation in this way with a rotation in the other direction, also for particles with mass, so this group here is actually not uniquely for mass less particles only.

What links it to massless particles is the ratio between the K and the J and the fact that this ratio works for any momentum in the z direction. This brings us to the answer: The J and the K nullify each other because of the fixed ratio between E and p, a property of massless particles.

The general case is. $A\cos \theta ~+~B\sin\theta$

Where $\theta$ is the angle which determines the direction in the x-y plane of the centrifugal acceleration. Small nitpick about his signs: In a right-handed coordinate system they should be:

$J_3 ~,~~~~ A ~:=~ J_2 - K_1 ~,~~~~ B ~:= -J_1 - K_2$

Note that if you reverse the direction of the momentum, (k,0,0-k) instead of (k,0,0,k), that the signs of the generators also change. In this case you get indeed.

$J_3 ~,~~~~ A ~:=~ J_2 + K_1 ~,~~~~ B ~:= -J_1 + K_2$

One should expect this because under spatial inversion K behaves like a vector and J like a pseudo vector.

Hans.

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