Define the Rayleigh-Ritz quotient $R[\psi]$ for a normalisable state $|\psi\rangle$ of a quantum system with Hamiltonian $H$. Given that the spectrum of $H$ is discrete and that there is a unique ground state of energy $E_{0}$, show that $R[\psi] \geqslant E_{0}$ and that equality holds if and only if $|\psi\rangle$ is the ground state.

A simple harmonic oscillator (SHO) is a particle of mass $m$ moving in one dimension subject to the potential

$$V(x)=\frac{1}{2} m \omega^{2} x^{2}$$

Estimate the ground state energy $E_{0}$ of the SHO by using the ground state wavefunction for a particle in an infinite potential well of width $a$, centred on the origin (the potential is $U(x)=0$ for $|x|<a / 2$ and $U(x)=\infty$ for $|x|>a / 2)$. Take $a$ as the variational parameter.

Perform a similar estimate for the energy $E_{1}$ of the first excited state of the SHO by using the first excited state of the infinite potential well as a trial wavefunction.

Is the estimate for $E_{1}$ necessarily an upper bound? Justify your answer.

$\left[\right.$ You may use : $\int_{-\pi / 2}^{\pi / 2} y^{2} \cos ^{2} y d y=\frac{\pi}{4}\left(\frac{\pi^{2}}{6}-1\right) \quad$ and $\left.\quad \int_{-\pi}^{\pi} y^{2} \sin ^{2} y d y=\pi\left(\frac{\pi^{2}}{3}-\frac{1}{2}\right) \cdot\right]$