A one-dimensional harmonic oscillator has Hamiltonian

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

where $\left[A, A^{\dagger}\right]=1$. Show that $A|n\rangle=\sqrt{n}|n-1\rangle$, where $H|n\rangle=\left(n+\frac{1}{2}\right) \hbar \omega|n\rangle$ and $\langle n \mid n\rangle=1$.

This oscillator is perturbed by adding a new term $\lambda X^{4}$ to the Hamiltonian. Given that

$$A=\frac{m \omega X-i P}{\sqrt{2 m \hbar \omega}}$$

show that the ground state of the perturbed system is

$$\left|0_{\lambda}\right\rangle=|0\rangle-\frac{\hbar \lambda}{4 m^{2} \omega^{3}}\left(3 \sqrt{2}|2\rangle+\sqrt{\frac{3}{2}}|4\rangle\right)$$

to first order in $\lambda$. [You may use the fact that, in non-degenerate perturbation theory, a perturbation $\Delta$ causes the first-order shift

$$\left|m^{(1)}\right\rangle=\sum_{n \neq m} \frac{\langle n|\Delta| m\rangle}{E_{m}-E_{n}}|n\rangle$$

in the $m^{\text {th }}$energy level.]