Problem 1.

Consider a perturbation $\hat{H}{1}=b \hat{x}^{4}$ to the simple harmonic oscillator Hamiltonian $$\hat{H}{0}=\frac{\hat{p}_{x}^{2}}{2 m}+\frac{1}{2} m \omega^{2} \hat{x}^{2}$$
This is an example of an anharmonic oscillator, one with a nonlinear restoring force.

(a) Show that the first-order shift in the energy is given by
$$E_{n}^{(1)}=\frac{3 \hbar^{2} b}{4 m^{2} \omega^{2}}\left(1+2 n+2 n^{2}\right)$$
(b) Argue that no matter how small $b$ is, the perturbation expansion will break down for some sufficiently large $n$. What is the physical reason?

Proof .

(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 $x^{2}$

Problem 2.

For the simple harmonic oscillator, for which

$$\hat{H}_{0}=\frac{\hat{p}{x}^{2}}{2 m}+\frac{1}{2} m \omega^{2} \hat{x}^{2}$$

take the perturbing Hamiltonian to be

$$\hat{H}_{1}=\frac{1}{2} m \omega{1}^{2} \hat{x}^{2}$$

where $\omega_{1} \ll \omega .$ Calculate the energy shifts through second order and compare with the exact eigenvalues.

Proof .

$$\hat{H}{0}=\frac{\hat{p}{x}^{2}}{2 m}+\frac{1}{2} m \omega^{2} \hat{x}^{2} \text { and } \hat{H}{1}=\frac{1}{2} m \omega{1}^{2} \hat{x}^{2}$$
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.

Problem 3.

(a) Calculate the exact energy eigenstates of the Hamiltonian (11.7) of the ammonia molecule in an external electric field.
(b) Assuming that $\mu_{e}|\mathbf{E}| \ll A$, use perturbation theory to determine the first-order correction to the unperturbed eigenstates $|I\rangle$ and $|I I\rangle$ and compare with the results of (a).

Proof .

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 $\hat{L}{z}|2,1,1\rangle=$


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