Consider a composite particle state $|\psi\rangle$ (like a hadron or a meson) that is an eigenstate of some Hamiltonian (e.g. the QCD Hamiltonian). Since the Hamiltonian is invariant under rotations and parity this particle state is also an eigenstate of the angular momentum and parity operator: $$L^2 |\psi\rangle = l(l+1)|\psi\rangle$$ $$P |\psi\rangle = (-1)^{a}|\psi\rangle$$
where $a$ is an integer number. Why is $a = l$?
For two particles one can use the 'trick' to transform into relative coordinates and then find that in relative coordinates the eigenfunction is $\sim Y_{lm}$. The parity of the spherical harmonics then leads to $(-1)^l$.
I don't see how to extend this to 3 or more particles.
EDIT: I had the following idea how to extend to 3 particles:
For 3 particles the Hamiltonian looks like: $$H = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \frac{p_3^2}{2m_3} + V_1(|x_2-x_1|) + V_2(|x_3-x_1|) + V_3(|x_3-x_2|)$$.
Now choose new coordinates by $$R = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}$$ $$y = x_2 - x_1$$ $$z = x_3 - x_1$$
The Hamiltonian becomes: $$H = \frac{p_R^2}{2M} + \frac{p_y^2}{2\mu_{12}} + \frac{p_z^2}{2\mu_{13}} + \frac{p_y\cdot p_z}{m_1} + V_1(|y|) + V_2(|z|) + V_3(|z-y|)$$ where $\frac{1}{\mu_{ij}} = \frac{1}{m_i} + \frac{1}{m_j}$ are reduced masses
The total angular momentum is given by $$L = x_1 \times p_1 + x_2 \times p_2 + x_3 \times p_3 = R \times p_R + y \times p_y + z \times p_z$$.
The $l$ in parity $ = (-1)^l$ is given by the internal angular momentum $$L_i = y \times p_y + z \times p_z$$ which commutes with the Hamiltonian.
Therefore an eigenfunction is given by $$|\psi(y,z)\rangle = |f(|y|,|z|)\rangle |L M\rangle_{\hat{y}\hat{z}}$$. Using Clebsch-Gordan coefficients this angular can be written as: $$|L M\rangle_{\hat{y}\hat{z}} = \sum_{m m'} \langle lm,l'm'|LM\rangle Y_{lm}(\hat{y}) Y_{l'm'}(\hat{z})$$ for some $l$ and $l'$
The overall parity is given by $(-1)^{l + l'}$ which does not necessary equals $(-1)^L$. For example $l = l' = 1$ would lead to a (from my point of view valid) solution: $$|10\rangle_{\hat{y}\hat{z}} = \frac{1}{\sqrt{2}} \left(Y_{11}(\hat{y})Y_{1-1}(\hat{z}) - Y_{1-1}(\hat{y})Y_{11}(\hat{z})\right)$$ with parity $(-1)^{l+l'} = (-1)^{1+1} = 1 \neq (-1)^1 = (-1)^L$.
There must be something which excludes such combinations. Why is this solution not valid?