Two-body problem with identical particles

Question

In the two body problem we have two particles interacting via a potential $V(| r_1 - r_2 |)$: $$ H = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + V(|r_1 - r_2|) \, .$$ It is well known that, with a canonical transformation, the latter can be rewritten as $$ H = \frac{P^2}{2M} + \frac{p^2}{2\mu} + V(|r|) \, ,$$ where $P = p_1 + p_2$, $M = m_1 + m_2$, $ \frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}$. This is the Hamiltonian of a free particle and a particle subject to a central potential.

Question: Is it possible to apply this scheme if the initial particles (1 and 2) are identical (bosons or fermions)? How can we symmetrize / anti-symmetrize the resulting wave function?

Answer 1

Let $\Psi(\mathbf{r}_1, \mathbf{r}_2)$ be the wave function of the original Hamiltonian. If the particles are identical bosons/fermions, it means that $$\Psi(\mathbf{r}_1, \mathbf{r}_2) = \pm \Psi(\mathbf{r}_2, \mathbf{r}_1).$$ Introducing the center-of-mass coordinate and the relative coordinate as (now $m_1 = m_2$) $$\mathbf{R} = \frac{\mathbf{r}_1 + \mathbf{r}_2}{2}, \mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2$$ we obtain $$\Phi(\mathbf{R},\mathbf{r})=\Psi\left(\mathbf{R} +\frac{\mathbf{r}}{2}, \mathbf{R} -\frac{\mathbf{r}}{2}\right)= \pm \Psi\left(\mathbf{R} -\frac{\mathbf{r}}{2}, \mathbf{R} +\frac{\mathbf{r}}{2}\right) =\pm\Phi(\mathbf{R},-\mathbf{r}).$$ In other words, the only thing that changes is how the wave function transforms under a permutation of the two particles.