# Symmetry of a spatial wavefunction independent of $M_L$?


The symmetry does not depend on $M_L$. The simplest way to see this is as follows. Start with $\vert L,M_L\rangle$ as a product state.
If it is a product of only two states it would be written as $$\vert L M_L\rangle = \sum_{m_1m_2}C_{\ell m_1;\ell m_2}^{LM_L} \vert \ell m_1\rangle \vert \ell m_2\rangle \tag{1}$$ where $C_{\ell m_1,\ell m_2}^{LM_L}$ is a Clebsch-Gordan coefficient. Permute particles $1$ and $2$, which is same as permuting $m_1$ and $m_2$ and you get $$C_{\ell m_2;\ell m_1}^{L M_L}=(-1)^{2\ell-L} C_{\ell m_1;\ell m_2}^{L M_L}$$ showing that the phase $(-1)^{2\ell -L}$ and thus the symmetry character does not depend on $M_L$.
If you have more than two particles the job is a little more complicated. Start with $\vert L,L\rangle$ as a product state, generalizing (1) to more than one constituent. The coefficients in the linear combinations are no longer CGs but this doesn't matter for now.
From the state $\vert L,M_L\rangle$ you reach the state $\vert L,M_L-1\rangle$ by application of the lowering operator $$\hat L_-= \sum_i \hat L_{i,-}= \hat L_{1,-}+\hat L_{2,-}+\hat L_{3,-}\ldots$$ Note that this sum is symmetric under permutation of particle numbers, so that, for instance: \begin{align} P_{12}\vert L,M_L-1\rangle &= P_{12}{\cal N}_{M_L}\left(\hat L_{1,-}+\hat L_{2,-}+\ldots \right)\vert L,M_L\rangle\, ,\\ &={\cal N}_{M_L}\left(\hat L_{1-}+\hat L_{2-}+\ldots\right)P_{12}\vert L,M_L\rangle \end{align} where ${\cal N}_{M_L}$ is a normalization constant. This shows that the symmetry under permutation of $\vert L,M_L-1\rangle$ is that of $\vert L,M_L\rangle$, and independent of this $M_L$.