# (1/2,1/2) representation and photon

1. Aug 26, 2008

### ismaili

I'm considering that, the $$(1/2,1/2)$$ representation of Lorentz group is the four-vector representation, and this four-vector representation has spin 0,1.
As we know, this four-vector representation can serve for the gauge field $$A_\mu$$, i.e. photon. However, the photon has spin 1. So we face a problem---how to get rid of the spin 0 part of (1/2,1/2) representation.

I found in a QFT book by Maggiore, after quantizing the EM field, she computed the angular momentum operator from Noether theorem, then extracting out the intrinsic spin part (throwing away the orbital angular momentum part); finally, computing the eigenvalues of intrinsic spin operator on one particle state, she showed that the circular polarized one particle states are eigenstates of the intrinsic spin operator and the eigenvalues are 1 and -1. This shows that the spin of photon is 1.

My question is, is there a concrete way to realize that the spin-0 part of (1/2,1/2) field representation is ruled out by some mechanism? (gauge symmetry or...) Because by explicit quantizing the theory and compute the spin eigenvalues of one particle state is a tough work. Moreover, in other theory, we also take the four-vector field $$A_\mu$$, i.e. (1/2,1/2) representation as a spin-1 particle, so probably there is a way to understand that spin-0 part is ruled out?

Any ideas or instructions will be appreciated.
Sincerely

2. Aug 26, 2008

### Avodyne

Take a look at volume I of Weinberg. This has a good discussion of this issue. I don't have it in front of me, but if IIRC the spin-zero part is removed by writing the kinetic term as $F^{\mu\nu}F_{\mu\nu}$ rather than as a more general linear combination of $\partial^\mu A^\nu\partial_\mu A_\nu$ and $\partial^\mu A^\nu\partial_\nu A_\mu$; note that $F^{\mu\nu}$ transforms as $(3,1)\oplus(1,3)$, so it only includes spin one. Gauge symmetry then removes the $m{=}0$ component of the spin-one part, so that only helicities $m{=}{\pm}1$ remain.

3. Aug 27, 2008

### ismaili

I glanced over volumeI of Weinberg's, the subject seems to be at the end of chap 5. I will study it soon. Thank you!

However, I have another related question.
Before I carefully study the field representation of Lorentz group, I always obtain the spin of a certain field by counting the number of Lorentz indices of that field, for example, I thought $$\partial_\mu A_{\nu\rho}$$ is a spin-3 field.
But things seem to be more subtle than that.
Is the counting of spin by the number of Lorentz indices correct?

4. Aug 27, 2008

### Avodyne

No. You need to know the rep in the (m,n) format, where m and n are integers or half-integers. Then the allowed spins are |m-n|, ..., m+n. So the vector, which is (1/2,1/2), has spins 0 and 1. The symmetric two-index tensor, which is in (0,0)+(1,1), has spins 0,1,2.

Oh, and when I wrote (3,1)+(1,3) before, I was using the notation (2n+1,2m+1), so I should have written (1,0)+(0,1) in (m,n) notation.

5. Aug 27, 2008

### samalkhaiat

6. Aug 30, 2008

### ismaili

Apologize that I don't have enough time now, so I can only reply your nice discussion shortly.
I thought (1/2,1/2) is a reducible representation of SO(1,3), since it is the direct product of two SU(2)'s. Therefore, (1/2,1/2) representation can be divided into total spin = 1 and 0, right?
Where did I got lost?
Actually I finished the study of Weinberg's section 5.9 just now, and he said that for massless particles, the representation (K,J) can only describe massless particles with helicity (eigenvalue of $$S_z$$) J - K. That's the reason why (1/2,1/2) representation, i.e. four vector representation can not describe photon directly, sine photon is a massless particle with helicity $$\pm 1$$. The way he solved this is, constructing the antisymmetric fields, $$\partial_\mu A_\nu - \partial_\nu A_\mu$$.

But he mentioned that, higher spin representation would produce short distance interaction. So, in order to satisfy the inverse squared law, we still choose the four-vector representation to describe the photon.

Actually, I didn't get it much.
By the way, many thanks for your nice article.

7. Sep 9, 2008