Addition of spin angular momentum for massless particles

Question

How do I add the spin angular momentum of massless particles, like photons, where only the transverse polarizations are allowed?

If all three polarizations were allowed, this would be an easy exercise: you'd get $S=0,1,2$ states. But some of these states clearly aren't allowed for purely transverse photons: For example, with two photons moving in the same direction as quantization axis, the state

$$\big|S=2,M=1\big\rangle=\frac{1}{\sqrt{2}}\big(|m_1=1,m_2=0\rangle+|m_1=0,m_2=1\rangle)$$

contains the $m=0$ state which is not allowed.

Edit: In the spectroscopic notation, which states are allowed?

Answer 1

The situation is simpler than you think. Basically you don't need to do Clebsch-Gordan at all!

The massless irreducible representations of the (proper orthochronous) Lorentz group are $1$-dimensional, labelled by their helicity $h$. Since $h = \pm 1$ are related by parity we group them as the $\pm$ helicities of the photon.

To compute the helicity of multiparticle states you must take a tensor product of the possible helicity configurations. Then you should decompose these into irreducible representations, according to the Clebsch-Gordan procedure.

But in this case, the $+1$ and $-1$ irreps are $1$-dimensional. Therefore their tensor product is automatically an irrep! So there's no Clebsch-Gordan decomposition to do.

Thus we are left with four possibilities, namely

$$1 \otimes 1, \ 1\otimes -1,\ -1 \otimes 1,\ -1 \otimes -1$$

Considering the action of the helicity operator in each representation, you can verify that these have helicity $2,0,0,-2$ respectively.

Combining the $1\otimes -1$ and $-1\otimes 1$ representations symmetrically and antisymmetrically, you reproduce the two $0$ helicity states of Rob's answer.

Let me know if you want some more explicit detail of the calculation!

Answer 2

Here's the algorithm for finding Clebsch-Gordan coefficients:

1. Start with a maximally aligned state like $\big|S,M = 2,2\big> = \big| m_1,m_2 = 1,1\big>$

2. Operate repeatedly with the lowering operator to generate the remaining states with the same $S$, such as $$\big|S,M=2,1\big> = \frac{\left|1,0\right\rangle + \left|0,1\right>}{\sqrt2}$$

3. Start the train for the next $S$ down by finding the remaining orthogonal combination of individual states. Here that'd be $$\big|S,M=1,1\big> = \frac{\left|1,0\right\rangle - \left|0,1\right>}{\sqrt2}$$

4. Continue until exhausted.

The question is how to apply this algorithm to a system where one of the lowered states, $\left|0\right\rangle$, is not physical.

One possibility, given by KoObO, is that you use the standard Clebsch-Gordan coefficients but set the coefficients of all the $\left|0\right\rangle$ terms to zero, and renormalize. However, without $\left|0\right\rangle$ there are not enough degrees of freedom to have distinct states for the $m=0$ projections of $S=0,1,2$. In fact, without $\left|0\right\rangle$ there are only four two-photon states, three symmetric and one antisymmetric.

This makes me suspect that the four states with $|M|=1$ are unphysical, since they all require some $\left|0\right\rangle$ in every term.

For the states with $M=0$, we're now in a pickle. For the usual spin-2 algebra, there are two symmetric states with $S=0$ and $S=2$, and one antisymmetric state with $S=1$. But without the $\left|0\right\rangle$ projection for each photon there is nothing to distinguish between $S=0$ and $S=2$. Is the symmetric, $M=0$ state a mixture of $S=0,2$? That wouldn't make any sense: you can distinguish between states of different total $S$ by rotating your coordinate system. We have already constructed states with $S=2,M=2$. Maybe $S=0$ is forbidden? That would be surprising.

The other way out of this pickle is to assert that the algebra of angular momentum addition is based on the multiplicity of of the states involved, rather than the total spins. In that case the two-state photon should have the same Clebsch-Gordan coefficients as the two-state electron, with a factor of two difference in the total spin everywhere, and there will be no two-photon state with $S=1$. This possibility seems more concordant with a scenario where there are no allowed states with $M=1$.

I don't know of a theoretical way to distinguish between these, so I turn to experimental data. I know that two-photon wavefunctions must be even under exchange. For now, assume that two-photon processes prefer to carry zero orbital angular momentum. Electromagnetic decays conserve parity, so negative-parity initial states should decay to photon pairs with antisymmetric spins, while positive parity initial states should decay to photon pairs with symmetric spins.

First let's look in the PDG meson table:

1. The $\pi^0$, $\eta$, and $\eta'$, with $J^P=0^-$, decay predominately to two photons. This suggests that the antisymmetric spin state carries no angular momentum.

2. For the $f_0(600)$, with $J^P=0^+$, the two-photon decay is listed as "seen". That would be consistent with a symmetric spin state with zero angular momentum. However I note that the decay width $\Gamma$ for the $f_0$ is enormous, and there are people who believe that the $f_0$ is two-pion state and not a "real" particle. I'll leave other very wide mesons and other mesons where the two-photon decay is described as "seen" without a branching ratio.

3. For the $\rho$ and $\omega$ mesons, with $J^P=1^-$, there are no two-photon decays listed. This is inconsistent with the existence of an antisymmetric two-photon state with $S=1$.

4. The $b(1235)$ and $f_1(1285)$, with $J^P=1^+$, have no branching ratio for two photons. This is inconsistent with the existence of a symmetric state with $S=1$.

5. The $f_2(1270)$, with $J^P=2^+$, has a small but well-measured branching ratio to two photons, consistent with a symmetric two-photon wavefunction for $S=2$.

6. In the strange sector, the $K^0$ ($J^P=0^-$) decays to two photons at least 1000 times more often than to three photons.

This list of decay modes is most consistent with a world where the two-photon spin wavefunction is symmetric for $S=2$, antisymmetric for $S=0$, and does not exist for $S=1$.

Let's leave the strongly-interacting world and look at positronium.

The positronium spin singlet may decay to two photons, suggesting that two photons may carry zero angular momentum. The positronium spin triplet may decay only into three or five photons, suggesting that there is no way for two photons to carry unit spin.

Unfortunately for my argument here, both positronium states are even under parity, and the singlet is even under charge exchange as well (in order to be totally antisymmetric under exchange symmetry). The symmetries involved then tell us that the photon wavefunction must be even under $C$, $P$, and therefore under exchange of spins.

We can weasel out of this conundrum if we relax our assumptions about orbital angular momentum. If the positronium spin singlet decays to two photons with $L=2$, $S=2$, $\vec L+\vec S=\vec J=0$, we can still conserve $C$, $P$, angular momentum, and exchange symmetry, while still keeping good explanations for the symmetric and antisymmetric two-photon spin states. The extra two units of $L$ might also serve as a handwavy reason why positronium decay is five orders of magnitude slower than the equivalent electromagnetic quarkonium decay in the $\pi^0$.

Based on this logic, I conclude that the photon is a two-state spin system with spin 1. Pure photon states may combine antisymmetrically to make $S=0$, or symmetrically to make an $S=2$ triplet; I think that there is no pure two-photon spin state with $M=1$. But I'd love to learn of a more authoritative source.

Answer 3

I think (but I'm not sure) that if you do it like if the photon has three physical polarization states then you get, as you said $S=0,1,2$. You can compute easily all of these states. But remember, you can express the states of the coupled basis in the uncoupled basis (via a Clebsch-Gordan decomposition). Then, you can delete the terms containing a non physical state (typically $\left|1,0\right>\times\left|1,m_2\right>$) and then you get all your physical states. Note that you will probably have to (re)normalize the states.