Describe briefly the variational approach to the determination of an approximate ground state energy $E_{0}$ of a Hamiltonian $H$.

Let $\left|\psi_{1}\right\rangle$ and $\left|\psi_{2}\right\rangle$ be two states, and consider the trial state

$$|\psi\rangle=a_{1}\left|\psi_{1}\right\rangle+a_{2}\left|\psi_{2}\right\rangle$$

for real constants $a_{1}$ and $a_{2}$. Given that

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

and that $\epsilon<s \mathcal{E}$, obtain an upper bound on $E_{0}$ in terms of $\mathcal{E}, \epsilon$ and $s$.

The normalized ground-state wavefunction of the Hamiltonian

$$H_{1}=\frac{p^{2}}{2 m}-K \delta(x), \quad K>0,$$

$$\psi_{1}(x)=\sqrt{\lambda} e^{-\lambda|x|}, \quad \lambda=\frac{m K}{\hbar^{2}} .$$

Verify that the ground state energy of $H_{1}$ is

$$E_{B} \equiv\left\langle\psi_{1}|H| \psi_{1}\right\rangle=-\frac{1}{2} K \lambda .$$

Now consider the Hamiltonian

$$H=\frac{p^{2}}{2 m}-K \delta(x)-K \delta(x-R)$$

and let $E_{0}(R)$ be its ground-state energy as a function of $R$. Assuming that

$$\psi_{2}(x)=\sqrt{\lambda} e^{-\lambda|x-R|},$$

use $(*)$ to compute $s, \mathcal{E}$ and $\epsilon$ for $\psi_{1}$ and $\psi_{2}$ as given. Hence show that

$$E_{0}(R) \leqslant E_{B}\left[1+2 \frac{e^{-\lambda R}\left(1+e^{-\lambda R}\right)}{1+(1+\lambda R) e^{-\lambda R}}\right]$$

Why should you expect this inequality to become an approximate equality for sufficiently large $R$ ? Describe briefly how this is relevant to molecular binding.