(a) Since
$$
\begin{aligned}
\hat{x} &=\sqrt{\frac{\hbar}{2 m \omega}}\left(\hat{a}+\hat{a}^{\dagger}\right) \
\hat{x}^{2} &=\frac{\hbar}{2 m \omega}\left(\hat{a}^{2}+\hat{a}^{\dagger} \hat{a}+\hat{a} \hat{a}^{\dagger}+\hat{a}^{+^{2}}\right)
\end{aligned}
$$
The first-order shift in the energy is given by
$$
\begin{aligned}
E_{n}^{(1)} &=\left\langle n\left|H_{1}\right| n\right\rangle=b\left(\frac{\hbar}{2 m \omega}\right)^{2}\left\langle n\left|\left(\hat{a}^{2}+\hat{a}^{\dagger} \hat{a}+\hat{a} \hat{a}^{\dagger}+\hat{a}^{\dagger^{2}}\right)^{2}\right| n\right\rangle \
&=b\left(\frac{\hbar}{2 m \omega}\right)^{2}\left\langle n\left|\left(\hat{a}^{2}+2 \hat{a}^{\dagger} \hat{a}+1+\hat{a}^{\dagger^{2}}\right)^{2}\right| n\right\rangle
\end{aligned}
$$
Since $\hat{a}|n\rangle=\sqrt{n}|n-1\rangle$ and $\hat{a}^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle$, the only terms that can contribute are
$$
E_{n}^{(1)}=b\left(\frac{\hbar}{2 m \omega}\right)^{2}\left\langle n\left|\left[\hat{a}^{2} \hat{a}^{\dagger^{2}}+\hat{a}^{\dagger^{2}} \hat{a}^{2}+\left(2 \hat{a}^{\dagger} \hat{a}+1\right)^{2}\right]\right| n\right\rangle
$$
Now
$$
\begin{aligned}
\hat{a}^{2}|n\rangle=\sqrt{n} \sqrt{n-1}|n-2\rangle & \hat{a}^{2}|n+2\rangle=\sqrt{n+2} \sqrt{n+1}|n\rangle \
\hat{a}^{\dagger^{2}}|n\rangle=\sqrt{n+2} \sqrt{n+1}|n+2\rangle & \hat{a}^{\dagger^{2}}|n-2\rangle=\sqrt{n} \sqrt{n-1}|n\rangle
\end{aligned}
$$
Therefore
$$
\begin{aligned}
E_{n}^{(1)} &=b\left(\frac{\hbar}{2 m \omega}\right)^{2}\left[(n+1)(n+2)+n(n-1)+(2 n+1)^{2}\right] \
&=b\left(\frac{\hbar}{2 m \omega}\right)^{2}\left[6 n^{2}+6 n+3\right]=\frac{3 \hbar^{2} b}{4 m^{2} \omega^{2}}\left(1+2 n+2 n^{2}\right)
\end{aligned}
$$
(b) Since $E_{n}^{(0)}=\left(n+\frac{1}{2}\right) \hbar \omega$,
$$
E_{n}^{(0)} \underset{n \rightarrow \infty}{\longrightarrow} n \hbar \omega, \quad E_{n}^{(1)} \underset{n \rightarrow \infty}{\longrightarrow} \frac{3}{2} \frac{b \hbar^{2} n^{2}}{m^{2} \omega^{2}} \text { and } \frac{E_{n}^{(1)}}{E_{n}^{(0)}} \stackrel{\longrightarrow}{n \rightarrow \infty} \frac{3}{2} \frac{b \hbar n}{m^{2} \omega^{3}}
$$
Thus $E_{n}^{(1)} / E_{n}^{(0)}$ will be greater than one for sufficiently large $n$. Of course, for sufficiently large $n$ we shouldn’t expect perturbation theory to work, since states of high $n$ have wave functions that extend into the large $x$ region, and for sufficiently large $x$, the perturbation, which varies like $x^{4}$, should dominate the unperturbed potential energy, which varies like $x^{2}$

$$
\hat{H}{0}=\frac{\hat{p}{x}^{2}}{2 m}+\frac{1}{2} m \omega^{2} \hat{x}^{2} \text { and } \hat{H}{1}=\frac{1}{2} m \omega{1}^{2} \hat{x}^{2}
$$
the first-order correction to the energy is given by
$$
\begin{aligned}
E_{n}^{(1)} &=\left\langle n\left|\hat{H}{1}\right| n\right\rangle=\frac{1}{2} m \omega{1}^{2}\left\langle n\left|\hat{x}^{2}\right| n\right\rangle \
&=\frac{1}{2} m \omega_{1}^{2} \frac{\hbar}{2 m \omega}\left\langle n\left|\left(\hat{a}+\hat{a}^{\dagger}\right)^{2}\right| n\right\rangle=\frac{\hbar \omega_{1}^{2}}{4 \omega}\left\langle n\left|\left(\hat{a} \hat{a}^{\dagger}+\hat{a}^{\dagger} \hat{a}\right)\right| n\right\rangle \
&=\frac{\hbar \omega_{1}^{2}}{4 \omega}(2 n+1)=\frac{1}{2}\left(\frac{\omega_{1}}{\omega}\right)^{2}\left(n+\frac{1}{2}\right) \hbar \omega
\end{aligned}
$$
The second-order correction to the energy is given by
$$
\begin{aligned}
E_{n}^{(2)} &=\sum_{k \neq n} \frac{\left|\left\langle k\left|\hat{H}{1}\right| n\right\rangle\right|^{2}}{E{n}^{(0)}-E_{k}^{(0)}}=\left(\frac{1}{2} m \omega_{1}^{2}\right)^{2} \sum_{k \neq n} \frac{\left|\left\langle k\left|\hat{x}^{2}\right| n\right\rangle\right|^{2}}{\left(n+\frac{1}{2}\right) \hbar \omega-\left(k+\frac{1}{2}\right) \hbar \omega} \
&=\left(\frac{1}{2} m \omega_{1}^{2}\right)^{2}\left(\frac{\hbar}{2 m \omega}\right)^{2} \sum_{k \neq n} \frac{\left|\left\langle k\left|\left(\hat{a}+\hat{a}^{\dagger}\right)^{2}\right| n\right\rangle\right|^{2}}{(n-k) \hbar \omega} \
&=\left(\frac{\hbar \omega_{1}^{2}}{4 \omega}\right)^{2} \sum_{k \neq n} \frac{(n+1)(n+2)|\langle k \mid n+2\rangle|^{2}+(n)(n-1)|\langle k \mid n-2\rangle|^{2}}{(n-k) \hbar \omega} \
&=\left(\frac{\hbar \omega_{1}^{2}}{4 \omega}\right)^{2}\left(\frac{(n+1)(n+2)}{-2 \hbar \omega}+\frac{n(n-1)}{2 \hbar \omega}\right)=-\frac{1}{8}\left(\frac{\omega_{1}}{\omega}\right)^{4}\left(n+\frac{1}{2}\right) \hbar \omega
\end{aligned}
$$
The eigenvalues of the exact Hamiltonian
$$
\hat{H}=\frac{\hat{p}{x}^{2}}{2 m}+\frac{1}{2} m\left(\omega^{2}+\omega{1}^{2}\right) \hat{x}^{2}
$$
are given by
$$
\begin{aligned}
E_{n} &=\left(n+\frac{1}{2}\right) \hbar \sqrt{\omega^{2}+\omega_{1}^{2}} \
&=\left(n+\frac{1}{2}\right) \hbar \omega \sqrt{1+\left(\frac{\omega_{1}}{\omega}\right)^{2}} \
&=\left(n+\frac{1}{2}\right) \hbar \omega\left(1+\frac{1}{2}\left(\frac{\omega_{1}}{\omega}\right)^{2}-\frac{1}{8}\left(\frac{\omega_{1}}{\omega}\right)^{4}+\cdots\right)
\end{aligned}
$$
Thus the exact eigenvalues and the perturbative results agree to the order calculated.

For the pionic hydrogen atom
$$
E_{n}=-\frac{\mu c^{2} \alpha^{2}}{2 n^{2}}
$$
where the reduced mass $\mu$ is given by
$$
\mu=\frac{m_{\pi} m_{p}}{m_{\pi}+m_{p}}
$$
(a)
$$
\begin{aligned}
|\psi(t)\rangle &=e^{-i \hat{H} t / \hbar}|\psi(0)\rangle \
&=\frac{e^{-i E_{1} t / \hbar}}{2}|1,0,0\rangle+e^{-i E_{2} t / \hbar}\left(\frac{e^{-i \omega_{0} t / \hbar}}{\sqrt{2}}|2,1,1\rangle+\frac{1}{2}|2,1,0\rangle\right)
\end{aligned}
$$
Therefore
$$
\begin{aligned}
\langle E\rangle &=\left|\frac{e^{-i E_{1} t / \hbar}}{2}\right|^{2} E_{1}+\left|\frac{e^{-i E_{2} t / \hbar} e^{-i \omega_{0} t}}{\sqrt{2}}\right|^{2} E_{2}+\left|\frac{e^{-i E_{2} t / \hbar}}{2}\right|^{2} E_{2} \
&=\frac{1}{4} E_{1}+\frac{3}{4} E_{2}=\left(\frac{1}{4}+\frac{3}{16}\right) E_{1}=\frac{7}{16} E_{1}
\end{aligned}
$$
(b)
$$
\left\langle L_{z}\right\rangle=\left|\frac{e^{-i E_{2} t / \hbar} c^{-i \omega_{0} t}}{\sqrt{2}}\right|^{2} \hbar=\frac{\hbar}{2}
$$
since $|1,0,0\rangle$ and $|2,1,0\rangle$ are eigenstates of $\hat{L}{z}$ with eigenvalue 0 and $\hat{L}{z}|2,1,1\rangle=$
$$