(a) Discuss the variational principle that allows one to derive an upper bound on the energy $E_{0}$ of the ground state for a particle in one dimension subject to a potential $V(x)$.

If $V(x)=V(-x)$, how could you adapt the variational principle to derive an upper bound on the energy $E_{1}$ of the first excited state?

(b) Consider a particle of mass $2 m=\hbar^{2}$ (in certain units) subject to a potential

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

(i) Using the trial wavefunction

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

with $a>0$, derive the upper bound $E_{0} \leqslant E(a)$, where

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

(ii) Find the zero of $E(a)$ in $a>0$ and show that any extremum must obey

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

(iii) By sketching $E(a)$ or otherwise, deduce that there must always be a minimum in $a>0$. Hence deduce the existence of a bound state.

(iv) Working perturbatively in $0<V_{0} \ll 1$, show that

$$-V_{0}<E_{0} \leqslant-\frac{1}{2} V_{0}^{2}+\mathcal{O}\left(V_{0}^{3}\right)$$

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