Give an account of the variational principle for establishing an upper bound on the ground-state energy, $E_{0}$, of a particle moving in a potential $V(x)$ in one dimension.

Explain how an upper bound on the energy of the first excited state can be found in the case that $V(x)$ is a symmetric function.

A particle of mass $2 m=\hbar^{2}$ moves in the potential

$$V(x)=-V_{0} e^{-x^{2}}, \quad V_{0}>0$$

Use the trial wavefunction

$$\psi(x)=e^{-\frac{1}{2} a x^{2}}$$

where $a$ is a positive real parameter, to establish the upper bound $E_{0} \leqslant E(a)$ for the energy of the ground state, where

$$E(a)=\frac{1}{2} a-V_{0} \frac{\sqrt{a}}{\sqrt{1+a}} .$$

Show that, for $a>0, E(a)$ has one zero and find its position.

Show that the turning points of $E(a)$ are given by

$$(1+a)^{3}=\frac{V_{0}^{2}}{a}$$

and deduce that there is one turning point in $a>0$ for all $V_{0}>0$.

Sketch $E(a)$ for $a>0$ and hence deduce that $V(x)$ has at least one bound state for all $V_{0}>0$.

For $0<V_{0} \ll 1$ show that

$$-V_{0}<E_{0} \leqslant \epsilon\left(V_{0}\right),$$

where $\epsilon\left(V_{0}\right)=-\frac{1}{2} V_{0}^{2}+\mathrm{O}\left(V_{0}^{4}\right)$.

[You may use the result that $\int_{-\infty}^{\infty} e^{-b x^{2}} d x=\sqrt{\frac{\pi}{b}}$ for $b>0 .$ ]