Consider a system of two identical particles. The combined system of these two particles has a total orbital angular momentum quantum number $\ell = 1$.
I am aware that the spatial wavefunction of a system with $\ell = 1$ has odd parity, as this can be shown from the spherical harmonics. However, I am not sure if anything can be said about the exchange symmetry of the spatial wavefunction in this case.
I.e. is the spatial wavefunction of the combined system symmetric or antisymmetric under exchange of the particle labels? And could you explain why.
Best Answer
One cannot know in general on the basis of $L=1$ alone.
Combining two states with the same $j$ will produce $L=0,1\ldots,2j$. The resulting states will be of the form \begin{align} \vert LM\rangle =\sum_{m_1m_2} C_{j m_1;j m_2}^{LM} \vert j m_1\rangle \vert j m_2\rangle \end{align} where $C_{j m_1;j_2 m_2}^{LM}$ is a Clebsch-Gordan coefficient.
The CGs satisfy the general relation $C_{j_1 m_1;j_2 m_2}^{LM}=(-1)^{j_1+j_2-L}C_{j_2 m_2;j_1 m_1}^{LM}$, and applying to your situation where $j_1=j_2=j$ yields $$ C_{j m_1;j m_2}^{LM}=(-1)^{2j-L} C_{j m_2;j m_1}^{LM}\, . \tag{1} $$ In other words, the CGs of Eq.(1) are even under permutation if $2j-L$ is even, and odd if $2j-L$ is odd.
Thus, if your states have $j=1/2$, the resulting $L=1$ states are even while the $L=0$ state will be odd. They are respectively the triplet and singlets states.
If, on the other hand, $j=1$, then the $L=2$ and $L=0$ states will be even, while the $L=1$ states will be odd.
In fact it is clear that if $j$ is half-integer, then the $L=1$ states are even under permutation, while if $j$ is integer, the $L=1$ states are odd.