I Exchange symmetry of two particles on a sphere

[Siupa]

Consider a system of two identical spin zero particles on a sphere. Let $\vec{L} = \vec{L}_1 + \vec{L}_2$ be the total orbital angular momentum of the two particles, and $l_1, l_2$ be the orbital angular momentum quantum numbers corresponding to particle 1 and particle 2.

Consider the simultaneous eigenstates of $L^2, L_z$ with $l_1 + l_2 = 1$.

Are these states symmetric or anti-symmetric under particle exchange?

Now, the combined angular momentum quantum number for these states is $L = 1$.

On one hand, $(-1)^L$ would tell me that they're antisimmetric for $L = 1$ (although this actually refers to parity, not particle exchange. But often these are related)

On the other hand, the symmetry properties of Clebsch-Gordan coefficients would tell me that any state in the highest weight tuple is always symmetric under particles exchange, by $(-1)^{j_1 + j_2 - J}$ with $j_1 + j_2 = l_1 + l_2 = 1$ and $J = L = 1$.

Which one is the correct approach?

There's a similar question here, but it only deals with the case $l_1 = l_2$, where the argument from parity and the argument from Clebsch-Gordan agree and give the same symmetry properties. However, in this case we must have $l_1 \neq l_2$, and this seems to make the two approaches disagree.

 

[Paul Colby]

The states are defined in terms of the tensor product of the single particle states, $$\mathbf{l}_1\otimes\mathbf{l}_2$$ This space has $(2l_1+1)(2l_2+1)$ vectors. Selecting combinations that are eigenvectors of $L^2$ and $L_z$ is done with the Clebsch-Gordan coefficients. The symmetry under interchange of the two particles is therefore determined by the symmetry of these coefficients. With more than two particles, this symmetry may be mixed.

That said, the symmetry under parity is determined by the interchange of $x \rightarrow -x$ for all particles at once. This should be $(-)^{l_1}(-)^{l_2}$.