Ground State of a Helium-like Atom with Electronic Interaction Replaced by Hooke's Law

Question

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?

Answer 1

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.

Answer 2

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 $$E'_0 = 3\hbar\omega$$