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.

A particle of unit mass moves in the potential

$$V(x)= \begin{cases}\infty & x \leqslant 0 \\ \lambda x & x>0\end{cases}$$

with $\lambda$ a positive constant. Explain why it is important that any trial wavefunction used to derive an upper bound on $E_{0}$ should be chosen to vanish for $x \leqslant 0$.

Use the trial wavefunction

$$\psi(x)= \begin{cases}0 & x \leqslant 0 \\ x e^{-a x} & x>0\end{cases}$$

where $a$ is a positive real parameter, to establish an upper bound $E_{0} \leqslant E(a, \lambda)$ for the energy of the ground state, and hence derive the lowest upper bound on $E_{0}$ as a function of $\lambda$.

Explain why the variational method cannot be used in this case to derive an upper bound for the energy of the first excited state.