# Photon Isospin

1. Feb 26, 2004

### bkfizz02

I'm curious about the isospin of a photon. I have been told it is a linear combination of I=1 and I=0. If I=0, then I3 must be 0, but if I=1, what values of I3 are allowed?

I don't really have a good idea of how to think about the isospin of this boson because I can't break it down into quarks like I can for pions.

Any clarifications are appreciated.

2. Feb 26, 2004

### xeguy

For I = 1, the allowed values of the isospin projection I3 are -1,0,1.

3. Feb 26, 2004

### arivero

bkfizz02 probably your informant was thinking from the point of view of electroweak theory. There you have four particles, W+ W- Z and photon. It is obvious that W+ and W- change isospin, and one could think that Z is the 0 projection of the I=1 triplet. It is not; because such projection combines with the I=0 singlet to give both the Z and the photon. The mixing angle is called Weinberg angle.

In any case it is very rare to think of electroweak bosons in this way.

Last edited: Feb 26, 2004
4. Feb 26, 2004

### bkfizz02

xeguy: Thanks, I guess I should have been more specific. I was curious what the specfic I3 value(s) is for the photon if it has I=1 because there are other exchange bosons that probably help span the spin 1 isospin space.

arivero:
Ok, this makes sense. I think I have heard it postulated that there exist to different particles (W0 and B0) which are the isospin states |1,0> and |0,0> respectively, but they are for book keeping and not observables. The observables are the Z and photon which are the usual linear combinations for the I3=0 states.

So, I think that I can think of the photon as a linear combination of |1,0> and |0,0> in isopsin space. My main motivation is for figuring out allowable isospin channels for a wide range of scattering problems, so this has given me a good starting point.

Thanks both for getting my thinking back on track.

Cheers.