Ground State of a Helium-like Atom with Electronic Interaction Replaced by Hooke's Law 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?
 A: 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.
A: Hamiltonian of the k-th HO can be separated into two parts
$H_k = H_k^{(1)}+H_k^{(2)}$
Where,
$$H_k^{(1)} = \frac{(p_k^{(1)})^2}{2m}+\frac12m\omega^2(y_k^{(1)})^2$$
and
$$H_k^{(2)} = \frac{(p_k^{(2)})^2}{2m}+\frac12m\omega^2(1-\lambda)(y_k^{(2)})^2$$
With the substitutions
$$Y_k^{(1)}= \sqrt\frac{m\omega}{2\hbar}y_k^{(1)}$$
$$Y_k^{(2)}= \sqrt\frac{m\omega(1-\lambda)}{2\hbar}y_k^{(2)}$$
$$P_k^{(1)}= \frac{1}{\sqrt{2m\hbar\omega}}p_k^{(1)}$$
and
$$P_k^{(2)}= \frac{1}{\sqrt{2m\hbar\omega}}p_k^{(2)}$$
$$H_k^{(1)}= \hbar\omega\left[(P_k^{(1)})^2+(Y_k^{(1)})^2\right]$$
$$H_k^{(2)}= \hbar\omega\left[(P_k^{(2)})^2+(Y_k^{(2)})^2\right]$$
The eigenvalue equation is
$$H_k|u_k\rangle = E_k|u_k\rangle$$
With the operators
$$a_{(1)}=Y_k^{(1)}+iP_k^{(1)}$$
and
$$a_{(1)}^\dagger=Y_k^{(1)}-iP_k^{(1)}$$
$$H_k^{(1)}= \hbar\omega\left(a_{(1)}a_{(1)}^\dagger+\frac12\right)$$
So that
$$\hbar\omega\left(a_{(1)}a_{(1)}^\dagger+\frac12\right)|u_k^{(1)}\rangle = E_k^{(1)}|u_k^{(1)}\rangle$$
By repeated application of $a_{(1)}^\dagger$, $k$ times in this case,
$$|u_k^{(1)}\rangle = a_{(1)}^\dagger|u_{(k-1)}^{(1)}\rangle$$
The energy eigenvalue is thus
$$E_k^{(1)}= \hbar\omega\left(k+\frac12\right)$$
Similarly 
$$E_k^{(2)}= \hbar\omega\left(k+\frac12\right)\sqrt{1-\lambda}$$
So
$$ E_k = E_k^{(1)}+ E_k^{(2)}= \hbar\omega\left(k+\frac12\right)(1+\sqrt{1-\lambda})$$
Then, for the whole atom
$$E = 3E_k = 3\hbar\omega\left(k+\frac12\right)(1+\sqrt{1-\lambda})$$
And the ground-state energy, $k=0$ is given by
$$E_0 = \frac32\hbar\omega(1+\sqrt{1-\lambda})$$
The interaction term when absent, the ground-state would then be
$$E'_0 = 3\hbar\omega$$
