Infinite sum of products of four Bessel functions

Question

The discrete Schrödinger equation for two interacting electrons in 1D under an electric field reads $$ E\psi_{mn}=[(m+n)F+U\delta_{mn}]\psi_{mn}-\psi_{m+1,n}-\psi_{m-1,n} -\psi_{m,n+1}-\psi_{m,n-1}\ . $$ Since Bessel functions solve the one-electron problem [see,e.g., Eur. J. Phys. 31 (2010) 639], I looked for a solution to the two-electron problem of the form $$ \psi_{mn}=\sum_{pq}C_{pq}J_{m-p}(x)J_{n-q}(x)\ , \qquad x=2/F\ . $$ Inserting the ansatz in the Schrödinger equation yields an equation for the coefficients $C_{pq}$. I then found the following infinite sum of products of Bessel functions $$ \sum_{n=-\infty}^{\infty}J_{n}(x)J_{n-k_1}(x)J_{n-k_2}(x)J_{n-k_3}(x) $$ where $k_i$ ($i=1,2,3$) are arbitrary integers. Any chance to get a close expression for the summation? Thanks!

Answer 1

From Neumann's formula: $$ J_\nu(x)\,J_\mu(x)=\frac{2}{\pi}\int_{0}^{\pi/2}J_{\nu+\mu}(2x\cos\psi)\cos((\mu-\nu)\psi)\,d\psi$$ we have that: $$ J_{n}(x)\,J_{n-k_1}(x) = \frac{2}{\pi}\int_{0}^{\pi/2}J_{2n-k_1}(2x\cos\psi)\cos(k_1 \psi)\,d\psi$$ $$ J_{n-k_2}(x)\,J_{n-k_3}(x) = \frac{2}{\pi}\int_{0}^{\pi/2}J_{2n-k_2-k_3}(2x\cos\psi)\cos((k_3-k_2) \psi)\,d\psi$$ so: $$J_n(x)\,J_{n-k_1}(x)\,J_{n-k_2}(x)\,J_{n-k_3}(x)\\=\frac{4}{\pi^2}\iint_{(0,\pi/2)^2}J_{2n-k_1}(2x\cos\psi)J_{2n-k_2-k_3}(2x\cos\phi)\cos(k_1\psi)\cos((k_2-k_3)\phi)\,d\psi\,d\phi$$ and we just need to evaluate: $$ \sum_{n\in\mathbb{Z}}J_{2n-k_1}(u)\,J_{2n-(k_2+k_3)}(v) $$ through: $$ e^{i z\cos\theta} = \sum_{n\in\mathbb{Z}} i^n J_n(z)\, e^{ni\theta}.$$ Then integrate over $\left(0,\frac{\pi}{2}\right)^2$. Continues.