Problem 1.

(a) Calculate $\langle V\rangle$ for the ground state of hydrogen. Show that $E=\langle V\rangle / 2$. What is $\langle K\rangle$, the expectation value of the kinetic energy, for the ground state? Show that these expectation values obey the virial theorem from classical mechanics.
(b) Calculate $\langle V\rangle$ for the ground state of the isotropic three-dimensional harmonic oscillator. How are $\langle K\rangle$ and $\langle V\rangle$ related for the oscillator? What do you expect based on the virial theorem? Explain.

Proof .

$$\langle V\rangle=4\left(\frac{1}{a_{0}}\right)^{3} \int_{0}^{\infty} d r r^{2} e^{-2 r / a_{0}}\left(-\frac{e^{2}}{r}\right)=-\frac{e^{2}}{a_{0}}=-\mu c^{2} \alpha^{2}$$
Since
$$E_{1}=-\frac{1}{2} \mu c^{2} \alpha^{2}$$
we see that $E_{1}=\langle V\rangle / 2$. Using $E_{1}=\langle K\rangle+\langle V\rangle$, we see that
$$\langle K\rangle=E_{1}-\langle V\rangle=\frac{1}{2} \mu c^{2} \alpha^{2}$$

From Problem $10.13$, we expect
$$\langle K\rangle=-\frac{1}{2}\left\langle\mathrm{r} \cdot \nabla \frac{e^{2}}{r}\right\rangle=\frac{1}{2}\left\langle\frac{e^{2}}{r}\right\rangle=-\frac{1}{2}\langle V\rangle=\frac{1}{2} \mu c^{2} \alpha^{2}$$
(b) From Problem $10.13$, for the isotropic harmonic oscillator we expect
$$\langle K\rangle=\frac{1}{2}\left\langle\mathbf{r} \cdot \nabla \frac{1}{2} \mu \omega^{2} r^{2}\right\rangle=\frac{1}{2}\left\langle\mu \omega^{2} r^{2}\right\rangle=\langle V\rangle$$
Evaluating $\langle V\rangle$ explicitly
$$\langle V\rangle=\frac{1}{2}\left\langle\mu \omega^{2} r^{2}\right\rangle=\frac{1}{2} \mu \omega^{2}\left\langle r^{2}\right\rangle=\frac{1}{2} \mu \omega^{2}\left(\left\langle x^{2}\right\rangle+\left\langle y^{2}\right\rangle+\left\langle z^{2}\right\rangle\right)=\frac{3}{4} \hbar \omega$$
since for each one dimensional oscillator
$$\frac{1}{2} \mu \omega^{2}\left\langle x^{2}\right\rangle=\frac{1}{4} \hbar \omega$$
See Chapter 7 . For the ground state $E_{0}=3 \hbar \omega / 2$, and therefore $\langle K\rangle=\langle V\rangle$ for the oscillator, in accord with the virial theorem.

Problem 2.

Suppose that nucleons within the nucleus are presumed to move independently in a potential energy well in the form of an isotropic harmonic oscillator. What are the first five nuclear “magic numbers” within such a model?

Proof .

$$E_{n}=\left(n+\frac{3}{2}\right) \hbar \omega, \quad n=2 n_{r}+l \quad n=0,1,2 \ldots$$
Thus the ground state has $n=0$, which means $n_{r}=0$ and $l=0 .$ Since there is a single state with $l=0$ and we can put two identical spin-1/2 particles (neutrons or protons) in such a state, the ground-state energy level is filled with two neutrons or two protons.

The first-excited state has $n=1$, and hence $n_{r}=0$ and $l=1 .$ Since there are three $m$ values for $l=1$, we can put $3 \times 2$ identical nucleons in this level, meaning that ground-state and first-excited state levels are filled with $2+6=8$ nucleons.

The second-excited state has $n=2$, and hence $n_{r}=0$ and $l=2$ or $n_{r}=1$ and $l=0$ Since there are five $m$ values for $l=2$, we can put $5 \times 2+1 \times 2=12$ identical nucleons in this level, meaning that ground-state, first-excited state, and second-excited levels are filled with $2+6+12=20$ nucleons.

The third-excited state has $n=3$, and hence $n_{r}=0$ and $l=3$ or $n_{r}=1$ and $l=1$. Since there are seven $m$ values for $l=3$, we can put $7 \times 2+3 \times 2=20$ identical nucleons in this level, meaning that first four levels are filled with $2+6+12+20=40$ nucleons. Finally, the fourth-excited state has $n=4$, and hence $n_{r}=0$ and $l=4$ or $n_{r}=1$ and $l=2$ or $n_{r}=2$ and $l=0 .$ Since there are nine $m$ values for $l=4$, we can put $9 \times 2+5 \times 2+1 \times 2=30$ identical nucleons in this level, meaning that first five levels are filled with $2+6+12+20+30=70$ nucleons. Thus the first five magic numbers for the pure isotropic harmonic oscillator are $2,8,20,40$, and $70 .$ Compare with the results of

Problem 3.

A negatively charged pion (a spin-0 particle) is bound to a proton forming a pionic hydrogen atom. At time $t=0$ the system is in the state
$$|\psi\rangle=\frac{1}{2}|1,0,0\rangle+\frac{1}{\sqrt{2}}|2,1,1\rangle+\frac{1}{2}|2,1,0\rangle$$
In an external magnetic field in the $z$ direction, the Hamiltonian is given by
$$\hat{H}=\frac{\hat{\mathbf{p}}^{2}}{2 \mu}-\frac{e^{2}}{|\hat{\mathbf{r}}|}+\omega_{0} \hat{L}{z}$$ where $\mu$ is the reduced mass of the pion-proton system. (a) What is $|\psi(t)\rangle$, the state of the system at time $t$ ? What is $\langle E\rangle$ at time $t$ for this state? (b) What are $\left\langle L{x}\right\rangle$ and $\left\langle L_{z}\right\rangle$ at time $t$ for this state?

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