2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.