The position and momentum for a harmonic oscillator can be written

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

where $m$ is the mass, $\omega$ is the frequency, and the Hamiltonian is

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

Starting from the commutation relations for $a$ and $a^{\dagger}$, determine the energy levels of the oscillator. Assuming a unique ground state, explain how all other energy eigenstates can be constructed from it.

Consider a modified Hamiltonian

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

where $\lambda$ is a dimensionless parameter. Calculate the modified energy levels to second order in $\lambda$, quoting any standard formulas which you require. Show that the modified Hamiltonian can be written as

$$H^{\prime}=\frac{1}{2 m} \alpha \hat{p}^{2}+\frac{1}{2} m \omega^{2} \beta \hat{x}^{2}$$

where $\alpha$ and $\beta$ depend on $\lambda$. Hence find the modified energies exactly, assuming $|\lambda|<\frac{1}{2}$, and show that the results are compatible with those obtained from perturbation theory.