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?


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.

  • $\begingroup$ I think the eigenvalue is $E_0 = \frac{3}{2}{\hbar}{\omega} \sqrt{(1+(1-{\lambda})} $ I'm really having troubles working with the notations. I have as references Schiff, Sakurai and Gasiorowicz. Can you suggest any other book(s) that have discussions on two-body systems? $\endgroup$ – Sayontön Vöttacharjo Feb 13 '17 at 20:29
  • $\begingroup$ Nope, that's still not it. I'm not sure which references would improve the situation for you, but it's a good opportunity to learn to take a resource in a different notation to yours and either (i) take it in full and do it from the ground up in its notation, or (ii) appropriating its ideas into what you had before. That's a crucial skill for later. $\endgroup$ – Emilio Pisanty Feb 13 '17 at 20:33

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


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