A one-dimensional harmonic oscillator has Hamiltonian

$$H=\frac{1}{2 m} \hat{p}^{2}+\frac{1}{2} m \omega^{2} \hat{x}^{2}=\hbar \omega\left(a^{\dagger} a+\frac{1}{2}\right),$$

where

$$a=\left(\frac{m \omega}{2 \hbar}\right)^{1 / 2}\left(\hat{x}+\frac{i}{m \omega} \hat{p}\right), \quad a^{\dagger}=\left(\frac{m \omega}{2 \hbar}\right)^{1 / 2}\left(\hat{x}-\frac{i}{m \omega} \hat{p}\right) \quad \text { obey } \quad\left[a, a^{\dagger}\right]=1$$

Assuming the existence of a normalised state $|0\rangle$ with $a|0\rangle=0$, verify that

$$|n\rangle=\frac{1}{\sqrt{n !}} a^{\dagger n}|0\rangle, \quad n=0,1,2, \ldots$$

are eigenstates of $H$ with energies $E_{n}$, to be determined, and that these states all have unit norm.

The Hamiltonian is now modified by a term

$$\lambda V=\lambda \hbar \omega\left(a^{r}+a^{\dagger r}\right)$$

where $r$ is a positive integer. Use perturbation theory to find the change in the lowest energy level to order $\lambda^{2}$ for any $r$. [You may quote any standard formula you need.]

Compute by perturbation theory, again to order $\lambda^{2}$, the change in the first excited energy level when $r=1$. Show that in this special case, $r=1$, the exact change in all energy levels as a result of the perturbation is $-\lambda^{2} \hbar \omega$.