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The photon doesn't have a well-defined quantity such as spin. Instead, it is characterized by helicity $h$.

Let's assume the state of two photons in CM frame (with $\mathbf k$ being the momentum of one of the photons), each of which has definite helicity $h_{i}$: $$ |\Psi \rangle = |h_{1},\mathbf k;h_{2},-\mathbf k\rangle $$ How to calculate the total spin (NOT total helicity) of such a system? Or, if it is possible, how to make a conclusion about the spin of an arbitrary two-photon state by having its helicity?

For example, suppose the following basis of two-photon states (here I omit momenta labels): $$ \tag 1 |L,R\rangle , \quad |R,L\rangle, \quad |L,L\rangle \pm |R,R\rangle $$ The first two states have total helicity $\pm 2$, so it seems that they correspond to a spin 2 state (naively, there can't be a higher spin for two photons from spin group representation point of view). The last two states have zero total helicity, so it seems that they correspond to spin zero. But I'm not sure about this statement. I can then suppose the last two states have zero helicity, but their spin is non-zero in general (in the sense that they belong to a non-zero spin representation of the Lorentz group).

Also, there is the problem with treating a two-photon system as a system with a well-defined spin. The total spin of a two-photon system must be calculated just as the sum of spins of each photon. The spin of a single photon as angular momentum at rest, however, isn't defined, so, from this point of view, there is no quantity of total spin for a two-photon system. Also, in some sense, the spin can be introduced as the quantity, which determines the non-coordinate transformation properties of the quantity under rotation transformation. So, for such a tensor, which is irrep of the spin group (say, the Lorentz group), the number $N$ of its independent components defines the spin $S$ by the relation $N = 2S+1$. Unfortunately, this interpretation quickly fails when we take into account gauge symmetry (or just use the true massless representation, which is $F_{\mu\nu}$, not $A_{\mu}$). Instead of spin, such a quantity is rather helicity.

P.S. This topic is relevant if we want to determine selection rules for orbital angular momentum of a two-photon state for a given total angular momentum $J$ and (suppose that it is defined) total spin $S$. Allowed values of $L$ are $$ \tag 2 |J-S|\leqslant L\leqslant J+S $$ But if we know only the helicity $h$ of the state, then we can't use $(2)$ for getting selection rules for $L$. In any case, independently of the fact of the definition of spin, there are selection rules for basis $(1)$, and I'm interested in them.

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  • $\begingroup$ The spin 0 neutral pion decays into two photons, $\pi^0\rightarrow \gamma \gamma$. Go to the pion's rest frame, so the center of momentum you wrote, and look at the total $J_z$ on that decay axis. The two photons, spin 1, are symmetric w.r.t. each other. So they cannot add up to spin 1. What can they add up to? $\endgroup$ – Cosmas Zachos Jun 28 '16 at 22:31
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    $\begingroup$ @CosmasZachos : in this case the only possible final two photon state is negative parity eigenstate with zero total helicity $$ \tag 1 |\Psi\rangle = |RR\rangle - |L,L\rangle $$ Here $L/R$ are left/right circular polarizations. The total angular momentum $J = 0$ is the sum of orbital and spin angular momenta, $$ J = L+S = 0 $$ How can I be sure, that for state $(1)$ $L = S = 0$, but not $L+S = 0$, while $L = 2, S = 2$? $\endgroup$ – Name YYY Jun 28 '16 at 22:48
  • $\begingroup$ The tensor meson f'(1525) with spin 2 has a 2γ decay mode, so the two photons also combine symmetrically to J=2. By Yang's theorem/symmetry, they cannot combine to J=1, so no vector meson decays to 2 photons. $\endgroup$ – Cosmas Zachos Jun 29 '16 at 0:14
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    $\begingroup$ @CosmasZachos : But which is two photon state in this case? $|L,R\rangle$ or $|LL\rangle + |RR\rangle$? And what is the spin of the second state? Only zero, or it is two? $\endgroup$ – Name YYY Jun 29 '16 at 7:12
  • $\begingroup$ If the spin of the f '(1525) is along the decay product axis, pointing to the left, on that axis $J_z=2$ , so, of course, the decay photons are $\mid R,L\rangle$ . I am not sure, but I suspect the interaction term in the effective lagrangian, for $h_{\mu \nu}$ representing the f ', is something like $e^2 h_{\mu\nu}F^{\mu\lambda}F^\nu_\lambda$ . $\endgroup$ – Cosmas Zachos Jun 29 '16 at 14:35
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It seems that I've derived an answer.

The spin of one photon in some sense is determined as the quantity, which characterizes the non-coordinate transformation of 4-potential $A_{\mu}$ under spatial rotation (i.e., mixing components without changing the coordinates, and hence without change the momentum of photon). Let's fix $A_{0} = 0$ (therefore we eliminate one unphysical degree of freedom). In this sense, photon has spin 1 because 4-potential is the vector with three non-trivial components. Difference in compare with massive representations is obvious: there is also one remaining unphysical component of $\mathbf A$, which has to be eliminated (for example, by Coulomb gauge condition $\nabla \cdot \mathbf A = 0$). This, of course, is in great agreement with the fact that massless states haven't defined spin (let's forget here, that $A_{\mu}$ isn't even Lorentz 4-vector).

Suppose we want to determine the spin of two photon states $|\Psi\rangle$ , which are given by $(1)$. Precisely, in terms of creation operators $a_{h}^{\dagger}(\pm\mathbf k)$, where $h = L,R$, they have the form $$ |\Psi\rangle = \text{Symmetrized}\left(\hat{a}^{\dagger}_{h_{1}}(\mathbf k)\hat{a}^{\dagger}_{h_{2}}(-\mathbf k)|0\rangle \right) $$ Here "symmetrized" means parity (anti)symmetrization.

Now, the recipe of determining the spin of two photon state at rest frame (let's label one of their momenta as $\mathbf k$) is very simple. We need to take the tensor product of 3-potentials $\hat{\mathbf A}(\pm\mathbf k)$ (with Coulomb gauge condition $\mathbf k \cdot \hat{\mathbf A}(\pm \mathbf k) = 0$), $$ \hat{A}_{ij} = \hat{A}_{i}(\mathbf k)\otimes \hat{A}_{j}(-\mathbf k), $$ then to rewrite components of $\hat{A}_{ij}|0\rangle$ in terms of states $(1)$, and, finally, to expand this $A_{ij}$ in terms of irreducible representations of $SO(3)$ group (this is since we are dealing with vector representations of the Lorentz group; let's remind that we ignore the fact that $A_{\mu}$ isn't Lorentz 4-vector!) for determining the spin $S$ of states $(1)$. Note, that because of transversality of $\hat{\mathbf A}$, tensor $\hat{A}_{ij}$ is automatically transverse, i.e., $$ k_{i}\hat{A}_{ij}(\mathbf k) = k_{j}\hat{A}_{ij}(\mathbf k) = 0 $$ So, the only possible irreps with definite spin are those, which satisfy transversality condition, are $$ A^{1}_{ij} = \delta_{ij}-n_{i}n_{j}, \quad A^{2}_{ij} =\epsilon_{ijk}n_{k},\quad A^{3}_{ij} = s_{ij}, $$ where $\mathbf n \equiv \frac{\mathbf k}{|\mathbf k|}$, and $s_{ij}$ is traceless symmetric transverse tensor. $A^{1/2}_{ij}$ denotes, respectively, parity even and parity odd spin 0 irreps, while $A^{3}_{ij}$ denotes parity even (since it is symmetric) spin 2 irrep. Note that such approach explains naturally why spin 1 (not total angula momentum) two photon state doesn't appear: it is associated with irrep $A^{4}_{ij} = b_{i}n_{j}+b_{j}n_{i}$ (with $\mathbf b$ being some 3-vector), which is forbidden because of transversality condition.

Now it can be easily obtained, that $|\Psi_{\pm}\rangle$ are parity even/odd spin 0 states, while $|\Psi_{LR/RL}\rangle$ are parity even spin 2 states.

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