Give an account of the variational method for establishing an upper bound on the ground-state energy of a Hamiltonian $H$ with a discrete spectrum $H|n\rangle=E_{n}|n\rangle$, where $E_{n} \leqslant E_{n+1}, n=0,1,2 \ldots$

A particle of mass $m$ moves in the three-dimensional potential

$$V(r)=-\frac{A e^{-\mu r}}{r}$$

where $A, \mu>0$ are constants and $r$ is the distance to the origin. Using the normalised variational wavefunction

$$\psi(r)=\sqrt{\frac{\alpha^{3}}{\pi}} e^{-\alpha r}$$

show that the expected energy is given by

$$E(\alpha)=\frac{\hbar^{2} \alpha^{2}}{2 m}-\frac{4 A \alpha^{3}}{(\mu+2 \alpha)^{2}}$$

Explain why there is necessarily a bound state when $\mu<A m / \hbar^{2}$. What can you say about the existence of a bound state when $\mu \geqslant A m / \hbar^{2}$ ?

[Hint: The Laplacian in spherical polar coordinates is

$$\left.\nabla^{2}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial}{\partial r}\right)+\frac{1}{r^{2} \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{r^{2} \sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}\right]$$