Describe briefly the variational approach to determining approximate energy eigenvalues for a Hamiltonian $H$.

Consider a Hamiltonian $H$ and two states $\left|\psi_{1}\right\rangle,\left|\psi_{2}\right\rangle$ such that

$$\begin{array}{cl} \left\langle\psi_{1}|H| \psi_{1}\right\rangle=\left\langle\psi_{2}|H| \psi_{2}\right\rangle=\mathcal{E}, & \left\langle\psi_{2}|H| \psi_{1}\right\rangle=\left\langle\psi_{1}|H| \psi_{2}\right\rangle=\varepsilon \\ \left\langle\psi_{1} \mid \psi_{1}\right\rangle=\left\langle\psi_{2} \mid \psi_{2}\right\rangle=1, & \left\langle\psi_{2} \mid \psi_{1}\right\rangle=\left\langle\psi_{1} \mid \psi_{2}\right\rangle=s \end{array}$$

Show that, by considering a linear combination $\alpha\left|\psi_{1}\right\rangle+\beta\left|\psi_{2}\right\rangle$, the variational method gives

$$\frac{\mathcal{E}-\varepsilon}{1-s}, \quad \frac{\mathcal{E}+\varepsilon}{1+s}$$

as approximate energy eigenvalues.

Consider the Hamiltonian for an electron in the presence of two protons at $\mathbf{0}$ and $\mathbf{R}$,

$$H=\frac{\mathbf{p}^{2}}{2 m}+\frac{e^{2}}{4 \pi \epsilon_{0}}\left(\frac{1}{R}-\frac{1}{|\mathbf{r}|}-\frac{1}{|\mathbf{r}-\mathbf{R}|}\right), \quad R=|\mathbf{R}|$$

Let $\psi_{0}(\mathbf{r})=e^{-r / a} /\left(\pi a^{3}\right)^{\frac{1}{2}}$ be the ground state hydrogen atom wave function which satisfies

$$\left(\frac{\mathbf{p}^{2}}{2 m}-\frac{e^{2}}{4 \pi \epsilon_{0}|\mathbf{r}|}\right) \psi_{0}(\mathbf{r})=E_{0} \psi_{0}(\mathbf{r}) .$$

It is given that

$$\begin{aligned} &S=\int \mathrm{d}^{3} r \psi_{0}(\mathbf{r}) \psi_{0}(\mathbf{r}-\mathbf{R})=\left(1+\frac{R}{a}+\frac{R^{2}}{3 a^{2}}\right) e^{-R / a} \\ &U=\int \mathrm{d}^{3} r \frac{1}{|\mathbf{r}|} \psi_{0}(\mathbf{r}) \psi_{0}(\mathbf{r}-\mathbf{R})=\frac{1}{a}\left(1+\frac{R}{a}\right) e^{-R / a} \end{aligned}$$

and, for large $R$, that

$$\int \mathrm{d}^{3} r \frac{1}{|\mathbf{r}-\mathbf{R}|} \psi_{0}(\mathbf{r})^{2}-\frac{1}{R}=\mathrm{O}\left(e^{-2 R / a}\right)$$

Consider the trial wave function $\alpha \psi_{0}(\mathbf{r})+\beta \psi_{0}(\mathbf{r}-\mathbf{R})$. Show that the variational estimate for the ground state energy for large $R$ is

$$E(R)=E_{0}+\frac{e^{2}}{4 \pi \epsilon_{0} R}(S-R U)+\mathrm{O}\left(e^{-2 R / a}\right) .$$

Explain why there is an attractive force between the two protons for large $R$.