A simple model of a helium-like atom with electron-electron interaction replaced by the Hooke's law is described by the Hamiltonian
$$H = \left[ \frac{- \hbar^2}{2m} ( \nabla_1^2 + \nabla_2^2 )+ \frac{1}{2}m\omega^2 (\ {r_1}^2+\ {r_2}^2 )-\frac{\lambda}{4}m{\omega^2}|\vec{r}_1-\vec{r}_2 |^2\right] $$
What is the exact ground state energy level?
I know that for a three-dimensional harmonic oscillator with the Hamiltonian
$$H = \left[\frac{- \hbar^2}{2m} \nabla^2 + \frac{1}{2}m {\omega^2}{\ r^2}\right] $$
the energy eigenvalues are given by
$$E_n = \left( \ {n_x}+\ {n_y}+\ {n_z}+\frac{3}{2} \right) $$
So how to work with the interaction term?
Best Answer
In this specific case, the hamiltonian separates into the three dimensions separately. Writing $\mathbf r_j=\sum_kx_k^{(j)}\hat{\mathbf e}_k$, your hamiltonian reads $$ H=\sum_k\left[\frac{(p_k^{(1)})^2+(p_k^{(2)})^2}{2m} + \frac12m\omega^2\left((x_k^{(1)})^2+(x_k^{(2)})^2\right)-\frac14\lambda m\omega^2(x_k^{(1)}-x_k^{(2)})^2\right]. $$ Thus you just treat this like a collection of three pairs of coupled 1D harmonic oscillators, each with hamiltonian $$ H_k=\frac{p_{(1)}^2+p_{(2)}^2}{2m} + \frac12m\omega^2\left(x_{(1)}^2+x_{(2)}^2\right)-\frac14\lambda m\omega^2(x_{(1)}-x_{(2)})^2, $$ and you just combine the two quadratic forms to give $$ H_k=\frac{p_{(1)}^2+p_{(2)}^2}{2m} + \frac12m\omega^2\left(\left(1-\frac\lambda2\right)x_{(1)}^2+\lambda x_{(1)}x_{(2)}+\left(1-\frac\lambda2\right)x_{(2)}^2\right). $$ You then need to rotate over into equal and even linear combinations of $x_{(1)}$ and $x_{(2)}$ to separate out the linear coupling term $x_{(1)}x_{(2)}$, so using $$ y_{(1)}=\frac{x_{(1)}+x_{(2)}}{\sqrt{2}} \quad\text{and}\quad y_{(2)}=\frac{-x_{(1)}+x_{(2)}}{\sqrt{2}} $$ (with a similar rotation on the momenta, which leaves them unchanged) the above reduces to $$ H_k=\frac{p_{(1)}^2+p_{(2)}^2}{2m} + \frac12m\omega^2\left(\left(1-\lambda\right)y_{(1)}^2+y_{(2)}^2\right) $$ with $[p_{(i)},y_{(j)}]=i\hbar\delta_{ij}$ and $[y_{(i)},y_{(j)}]=0=[p_{(i)},p_{(j)}]$. You can then just read off the spectrum from there.