Orbital angular momentum selection rules for three identical particles

Question

I'm trying to figure out if there are selection rules for the total orbital angular momentum for a system of three identical particles, say bosons.

For two identical bosons one can argue that the exchange symmetry implies that the state must be parity-even. From the parity properties of the spherical harmonics, this forces the state to have even orbital angular momentum.

How does this work with a system of three identical particles? We may assign two orbital angular momenta: $\mathbf{L}_{1}$ between two of the particles, and $\mathbf{L}_2$ corresponding to the relative angular momentum between those two particles (as a reduced one-particle system) and the third. The total orbital angular momentum is $\mathbf L=\mathbf L_1+\mathbf L_2$.

I can use symmetry arguments to say that the $\mathbf L_1$ quantum number $\ell_1$ is even. But I don't know how to use exchange symmetry to constrain $\mathbf L_2$ since we're now comparing a single particle and a two-particle state.

I'm hoping to be able to use this to argue that a scattering process in some definite orbital angular momentum, total spin, and $CP$ state going to three identical bosons should be $s$-wave or $p$-wave.

Thanks!

Answer 1

For 3-particle systems the relevant permutation group is $S_3$, which is significantly different from $S_2$ in that not all representations of $S_3$ are 1-dimensional.

Whereas 2-particle states can always be written so they transform back to themselves under permutation of labels, i.e. always possible to write $$ \psi_a(x_1)\psi_b(x_2)=\frac{1}{2}\psi^+_{ab}(x_1,x_2) + \frac{1}{2}\psi^-_{ab}(x_1,x_2) $$ with $$ P_{12}\psi^{\pm}_{ab}(x_1,x_2)=\pm \psi^{\pm}_{ab}(x_1,x_2) $$ the symmetric and antisymmetric combinations, there are 3-particle states that cannot be made to transform back to a multiple of themselves. Already this is apparent for the triple coupling of spin-1/2 particles: there are two copies of total spin $1/2$ in the decomposition and you cannot arrange the states in $S=1/2$ to be symmetric or antisymmetric under all permutations of particles.

The situation does not improve for 3 spin-1 particles. The fully symmetric states will have total $L=3,1$, the fully antisymmetric states will have $L=0$, but there are states of mixed symmetry with $L=1,2$ (there are two copies of these mixed symmetry states).

Thus, you have no obvious way of knowing if a state with $L=1$ is symmetric or of mixed symmetry.

Of course the spin-statistics theorem does not forbid spin-states of mixed symmetry: the must be combined with spatial stages (also of mixed symmetry) to produce a properly symmetrized total state.