A particle of mass $m$ moving in a one-dimensional harmonic oscillator potential satisfies the Schrödinger equation

$$H \Psi(x, t)=i \hbar \frac{\partial}{\partial t} \Psi(x, t),$$

where the Hamiltonian is given by

$$H=-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}}+\frac{1}{2} m \omega^{2} x^{2}$$

The operators $a$ and $a^{\dagger}$ are defined by

$$a=\frac{1}{\sqrt{2}}\left(\beta x+\frac{i}{\beta \hbar} p\right), \quad a^{\dagger}=\frac{1}{\sqrt{2}}\left(\beta x-\frac{i}{\beta \hbar} p\right)$$

where $\beta=\sqrt{m \omega / \hbar}$ and $p=-i \hbar \partial / \partial x$ is the usual momentum operator. Show that $\left[a, a^{\dagger}\right]=1$.

Express $x$ and $p$ in terms of $a$ and $a^{\dagger}$ and, hence or otherwise, show that $H$ can be written in the form

$$H=\left(a^{\dagger} a+\frac{1}{2}\right) \hbar \omega$$

Show, for an arbitrary wave function $\Psi$, that $\int d x \Psi^{*} H \Psi \geq \frac{1}{2} \hbar \omega$ and hence that the energy of any state satisfies the bound

$$E \geq \frac{1}{2} \hbar \omega$$

Hence, or otherwise, show that the ground state wave function satisfies $a \Psi_{0}=0$ and that its energy is given by

$$E_{0}=\frac{1}{2} \hbar \omega$$

By considering $H$ acting on $a^{\dagger} \Psi_{0},\left(a^{\dagger}\right)^{2} \Psi_{0}$, and so on, show that states of the form

$$\left(a^{\dagger}\right)^{n} \Psi_{0}$$

$(n>0)$ are also eigenstates and that their energies are given by $E_{n}=\left(n+\frac{1}{2}\right) \hbar \omega .$