Why does a system being in an $s$-wave mean that the spatial wavefunction is symmetric? Does it? Probably being silly here, but me and my fellow undergrads can't seem to come up with an exact answer to why, if a system is in an $s$-wave, edit: or any orbital with an even value of $\ell$, the spacial wf must be symmetric under exchange of like particles.
Also, what is the relationship between the symmetry of a particles wavefunction and its parity? Another one that doesn't seem to have a clear-cut answer among us!
 A: The relationship between exchange symmetry and parity is something that has also greatly confused me. Definitions:

*

*Parity refers to whether a quantum state picks up a minus sign when
all the spatial coordinates are inverted, rather like an even or
odd function.

*Exchange symmetry refers to whether the combined state for a system
of two particles picks up a minus sign when the particle labels are
exchanged.

At first, there seems to be no obvious link between these two symmetries.
Let's begin with parity. In spherical polar coordinates, a parity transformation is equivalent to $\theta \rightarrow \pi-\theta$, $\phi \rightarrow \pi+\phi$ with $r$ remaining the same. In any system with spherical symmetry, the part of the spatial wavefunction which depends on $\theta$ and $\phi$ is given by the spherical harmonics $Y^M_L(\theta, \phi)$. It can be shown that the spherical harmonics have parity
$$Y^M_L(-{\bf r})=(-1)^LY^M_L({\bf r})$$
(see https://en.wikipedia.org/wiki/Spherical_harmonics#Symmetry_properties). Thus, the spatial wavefunction will have parity $(-1)^L$.
Exchange symmetry is more complicated. Since you are interested in the exchange symmetry of the spatial wavefunction, we can ignore spin and only think about orbital angular momentum. Let the two individual particles have orbital angular momentum eigenstates $\vert j_1 m_1 \rangle$ and $\vert j_2 m_2\rangle$. The eigenstates of the combined system of two particles are then
$$\vert LM\rangle = \sum_{m_1}\sum_{m_2}C^{LM}_{j_1m_1;j_2m_2} \vert j_1 m_1 \rangle \vert j_2 m_2 \rangle \tag{1}$$
where $C^{LM}_{j_1 m_1;j_2 m_2}$ are the Clebsch-Gordan coefficients. The Clebsch-Gordan coefficients have well-defined exchange symmetry given by
$$C^{LM}_{j_2 m_2; j_1 m_1}=(-1)^{j_1+j_2-L}C^{LM}_{j_1 m_1;j_2 m_2}$$
(see https://en.wikipedia.org/wiki/Clebsch%E2%80%93Gordan_coefficients#Symmetry_properties). Let's assume that the two particles have the same $j_1 = j_2 = j$. The exchange symmetry factor then becomes
$$(-1)^{2j-L}.$$
This factor can be taken outside the sum in equation (1), so also gives the exchange symmetry of $\vert LM\rangle$. Since $2j$ is even, the exchange symmetry of the spatial wavefunction is
$$(-1)^L$$
AHA! We have found that both the parity and exchange symmetry of the spatial wavefunction are given by $(-1)^L$. However, this has arisen from the assumption that $j_1=j_2$, so there doesn't seem to be a deep connection between these two symmetries.
