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

In summary, the angular momentum of a photon is spin 1. When coupled with an electron, the angular momentum of the photon is also spin 1/2 and spin 3/2. These angular Momenta are related through a combined transformation that is parametrized by α and β.
  • #1
kof9595995
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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:
[itex]SU(2) \otimes [U(1) \oplus U(1)] = [SU(2) \otimes U(1)] \oplus [SU(2) \otimes U(1)][/itex]
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.
 
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  • #2
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
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
[itex]Spin(\frac{1}{2}) \otimes Spin(1) = Spin(\frac{1}{2}) \oplus Spin(\frac{3}{2})[/itex]
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
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.
 
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  • #5
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
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
Ok thanks, so that means I shall come back to the issue later :)
 

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

1. How do you calculate the total angular momentum of a system involving a massive and massless particle?

The total angular momentum of a system involving a massive and massless particle can be calculated by adding the individual angular momenta of the particles. The angular momentum of a massive particle is given by L = Iω, where I is the moment of inertia and ω is the angular velocity. For a massless particle, the angular momentum is simply L = pr, where p is the linear momentum and r is the distance from the axis of rotation.

2. What is the difference between the angular momenta of a massive and massless particle?

The main difference between the angular momenta of a massive and massless particle is that the angular momentum of a massive particle is dependent on its moment of inertia, while the angular momentum of a massless particle is solely dependent on its linear momentum and distance from the axis of rotation. This means that a massless particle can have angular momentum even if it has zero mass.

3. Can the angular momenta of a massive and massless particle cancel each other out?

Yes, the angular momenta of a massive and massless particle can cancel each other out if they have equal and opposite values. This is known as angular momentum cancellation and can occur when the two particles are rotating in opposite directions with equal angular velocities.

4. How does the coupling of angular momenta affect the overall motion of a system?

The coupling of angular momenta can affect the overall motion of a system by changing the direction and speed of the system's rotation. This is because the total angular momentum of a system is conserved, meaning that any changes in the angular momenta of individual particles will affect the overall angular momentum of the system.

5. Can the coupling of angular momenta result in a system with no overall angular momentum?

No, the coupling of angular momenta will always result in a system with non-zero overall angular momentum. This is because the total angular momentum of a system is always conserved, so even if individual angular momenta cancel each other out, the total angular momentum will still be present in the system.

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