# How to couple the angular momenta of a massive particle and a massless one

1. Oct 5, 2011

### kof9595995

Say, I want to couple the angular momenta of a electron spin angular momentum, and a photon's momentum. I guess in terms of representations it's:
$SU(2) \otimes [U(1) \oplus U(1)] = [SU(2) \otimes U(1)] \oplus [SU(2) \otimes U(1)]$
But I'm not at all certain if this is correct, one thing for example, U(1) represents liittle group SO(2), but SU(2) represents SO(3), I probably shouldn't just put them together by a direct product, etc.

2. Oct 8, 2011

### Bill_K

Don't worry about the little group - a photon is spin one. It's true that rotations will include all three polarizations, but the longitudinal mode can always be removed by a gauge transformation. Coupling a photon spin to an electron spin is just a case of multiplying spin one times spin one-half.

3. Oct 8, 2011

### kof9595995

Thanks for reply, but I don't get it, isn't photon helicity 1 instead of spin 1? The representation is very different and it transforms very differently with a spin.
BTW let me rephrase my question since I found my original post a bit vague: Say for a spin 1/2 and a spin 1 massive, the angular momenta add as
$Spin(\frac{1}{2}) \otimes Spin(1) = Spin(\frac{1}{2}) \oplus Spin(\frac{3}{2})$
And that's why we can go from |j1j2m1m2> to a new set of quantum number system |JMj1j2>, my question is for the coupling of a electron and a photon do we also have a new set of quantum numbers related to rotations, and what does it look like?

4. Oct 9, 2011

### Bill_K

Angular momentum is what describes the mixing of the states of a system under a three-dimensional rotation R(x). Everything has angular momentum in this same sense, even photons. Photons are described by a 3-vector A, and they have three states, and when you rotate them they behave like spin one.

What's different about them is that they also have gauge transformations. You can follow the rotation with a gauge transformation that alters the longitudinal component. And therefore you can keep the longitudinal component zero, and in this way define a combined transformation that acts on only helicity 1 states. This combined transformation represents the little group P(2), the two-dimensional Poincare group.

Nevertheless, the valid approach is to keep the two transformations (rotation and gauge transformation) separate. If you want to couple a photon spin to an electron spin it works the usual way: you get spin 1/2 and spin 3/2. AND, you still have the gauge freedom which can be applied to the coupled states.

Last edited: Oct 9, 2011
5. Oct 9, 2011

### kof9595995

Emm, I understand helicity is also angular momentum, but I don't quite understand the rest of your words. My understandings and questions have grown out of Weinberg's QFT Vol1, chap 2 so far, so I'm not clear about how is the issue related to gauge transformation. Could you explain it within the language of Weinberg Chap2? If it's not quite possible I guess I'll have to put it off for the moment and come back to the issue later.

6. Oct 9, 2011

### Bill_K

Briefly, Weinberg's remark at the bottom of p.72: "We shall see in Section 5.9 that electromagnetic gauge invariance arises from the part of the little group parametrized by α and β."

7. Oct 9, 2011

### kof9595995

Ok thanks, so that means I shall come back to the issue later :)