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The energy eigenstates of two particles in a 2d isotropic harmonic oscillator can be described in terms of a product basis of one particle states (I'm using the angular momentum basis with angular momentum quantum number $m \in \mathbb{Z}$) $|k_1, m_1\rangle \otimes |k_2, m_2\rangle$ (with energy $2(k_1+k_2)+|m_1|+|m_2|)$ or likewise by separating the problem into a center of mass and a relative motion part, i.e. $|K, M\rangle \otimes |k, m\rangle$ (with energy $2(K+k)+|M|+|m|)$).

The corresponding wavefunctions are $\langle x_1|k_1,m_1\rangle = \Psi(r_1; k_1, m_1) \sim L_{k_1}^{|m_1|}(r_1^2)r^{|m_1|}e^{-r_1^2/2}$ and $\langle x|k,m\rangle = \Psi(r/\sqrt{2}; k, m), \langle X|K,M\rangle = \Psi(\sqrt{2}R; K, M)$. Here $L_k^m$ denotes the generalized Laguerre polynomial, furthermore $X = (x_1 + x_2)/2$ and $x = x_1 -x_2$, s.t. $R^2 = |X|^2 = (r_1^2 + r_2^2 + 2r_1r_2\cos(\phi_1-\phi_2))/4$ and $r^2 = |x|^2 = r_1^2 + r_2^2 -r_1r_2\cos(\phi_1-\phi_2)$.

I'm interested in calculating the overlap between these bases, i.e $(\langle K, M|\otimes \langle k, m|)\left(|k_1, m_1\rangle \otimes |k_2, m_2\rangle\right)$. I have the feeling that people already came across this problem (and solved it), however, I can't find anything online. Maybe anyone has seen this done somewhere before? Otherwise, I'd also be glad about some identities about Laguerre polynomials (they're the main cause of trouble) that could come in handy here.

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  • $\begingroup$ It's not obvious to me that angular momentum is a good quantum number for the relative motion part. The forces on the objects are not directed towards each other, but rather towards the lowest point of the potential they both feel. I.e the interparticle potential is not central. $\endgroup$
    – mike stone
    Feb 27, 2021 at 17:44
  • $\begingroup$ In relative coordinates the problem seperates into two isotropic harmonic oscillators, one for the center of mass motion and one for the relative motion part. They just differ by a their oscillator length. $\endgroup$ Feb 27, 2021 at 18:00
  • $\begingroup$ Ah I see. Neat: $x_1^2+y_1^2+x_2^2+y_2^2= \frac 12 (x_1+x^2)^2+\frac 12 (y_1+y_2)^2 +\frac 12 (x_1-x_2)^2+\frac 12 (y_1-y_2)^2$.. $\endgroup$
    – mike stone
    Feb 27, 2021 at 18:08

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There is Bateman's formula: $$ \exp{\left(x y e^{i \phi}\right)} L_n\left(x^2 + y^2 - 2 x y \cos{\phi}\right) = \sum_{k = 0}^\infty \left(x y e^{i\phi}\right)^{(k - n)} \frac{n!}{k!} L_n^{(k - n)}\left(x^2\right) L_n^{(k - n)}\left(y^2\right). $$ I found it by Googling. I have no idea how to prove it.

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