Chapter 6 Problem Solution (Townsend)

In section 6.1, we show that the allowed energy values (energy eigenvalues) for a particle confined in a three-dimensional box is given by:

Given lengths, this is equal to:

For ground level, the quantum numbers are all equal to one. The ground level energy is equal to:

For the first excited state, the quantum numbers change such that one of them is equal to two and the rest remain equal to one. The smallest possible energy is achieved when the quantum number corresponding to the smallest dimension is increased to two.

For the two other cases, we get:

and

All of which are distinct first excited energy eigenvalues. Thus, the degeneracy is zero.

By the Schrödinger equation, we know that:

Using separable integration, we suppose:

This gives us:

Divide both sides by \(XYZ\), we get:

The left-hand side of the equation is a function of \(x\), \(y\), and \(z\). The right-hand side of the equation is a constant. In order for the relation to hold true for all \(x\), \(y\), and \(z\), it must be true that each expression for \(x\), \(y\), and \(z\) is a constant. This gives us:

Using separation of variable, we know that energy \( E \) is given by:

with

and

Thus, the ground state energy \( E \) is equal to:

For a simple harmonic oscillator, the wave function \(\psi_x(x)\) is given by:

Then, the ground state wave function/energy eigenfunction is given by:

Orbital angular momentum is given by \(L=l(l+1)\hbar\), where \(l=0,1,2,\dots\). For ground state, \(l=1\). Thus, orbital momentum \(L=0\).

By separation of variable, we know that energy $E=E_x+E_y+E_z$, such that:
$$
E=(n_x+n_y+n_z+\frac{3}{2})\hbar\omega
$$
For the first excited state, one of the quantum numbers is one. The energy is equal to:
$$
E_{100}=E_{010}=E_{001}=\frac{5}{2}\hbar\omega
$$
with a degeneracy of 3

By the Schrödinger equation, we know that:

(a) There are different ways to show that the wave function is properly normalized. One could start from scratch by integrating the wave function form 0 to \(2\pi\), or we can take advantage of the “discrete” form of the wave packet, where \(\Psi(\phi)\) is a superposition of eigenstates. Recall that the angular component of the wave function \(\Psi(\phi)=\Phi_m(\phi)\) is given by (see chapter note for derivation):
$$
\Psi(\phi)=\frac{1}{\sqrt{2\pi}}e^{im\phi}
$$
Using the formula, we rewrite $\Psi(\phi)$, giving us:
$$
\Psi(\phi)=\sqrt\frac{2}{3}e^{0\phi}+\sqrt\frac{1}{3}\sqrt\frac{1}{2\pi}e^{i\phi}=\sqrt\frac{2}{3}\Phi_0(\phi)+\sqrt\frac{1}{3}\Phi_1(\phi)
$$
If \(\Psi(\phi)\) is properly normalized, the sum of probability must sum to one. Since the probability of finding a particle with a quantum number $m$ is proportional to the coefficient squared \(c_m^2\), we get:
$$
\frac{2}{3}+\frac{1}{3}=1
$$
Thus, $\Psi(\phi)$ is properly normalized.
The $z$ component of the angular momentum is given by:
$$
l_z=m\hbar, m=0,\pm 1,\pm 2,\dots
$$
From the wave function, we can see that there are two possible values of $m$: $m=0$ or $m=1$. Thus, possible measurements of $L_z$ are:
$$
l_z=0 \quad l_z=\hbar
$$
The probabilities of obtaining those results are:
$$ P_0=\frac{2}{3} \quad P_1=\frac{1}{3} $$

Similar problem: Example 6.2

We want to verify that $Y_{2,2}$ is an eigenfunction of $L_{op}^2$. To do this, we use the eigenvalue equation:
$$
L_{op}^2Y_{l,m}=l(l+1)\hbar^2Y_{l,m}, l=0,1,2 \dots
$$
and
$$
L_{op}^2=-\hbar^2\Big[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\frac{\partial}{\partial\theta})+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\Big]
$$
Computing the left hand side of the equation, we get
$$
-\hbar^2\Big[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\frac{\partial}{\partial\theta})+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\Big]Y_{2,2}
$$

$$
-\sqrt\frac{15}{32\pi}\hbar^2\Big[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\frac{\partial}{\partial\theta}(\sin^2\theta e^{2i\phi}))+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}(\sin^2\theta e^{2i\phi})\Big]
$$

$$
-\sqrt\frac{15}{32\pi}\hbar^2\Big[\frac{e^{2i\phi}}{\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\frac{\partial}{\partial\theta}(\sin^2\theta ))+\frac{\partial^2}{\partial\phi^2}(e^{2i\phi})\Big]
$$

Using chain rule, we know that
$$
\frac{\partial^2}{\partial\phi^2}(e^{2i\phi})=\frac{\partial}{\partial\phi}(2ie^{2i\phi})=4i^2e^{2i\phi}=-4e^{2i\phi}
$$

$$
\frac{\partial}{\partial\theta}(\sin\theta\frac{\partial}{\partial\theta}(\sin^2\theta)=\frac{\partial}{\partial\theta}(2\sin^2\theta\cos\theta)=2\frac{\partial}{\partial\theta}((1-\cos^2\theta)\cos\theta)=2(-\sin\theta+3\cos^2\theta\sin\theta)
$$

Then,
$$
\begin{align}& -\sqrt\frac{15}{32\pi}\hbar^2\Big[2e^{2i\phi}(-1+3\cos^2\theta)-4e^{2i\phi}\Big] \\ & =-2e^{2i\phi}\sqrt\frac{15}{32\pi}\hbar^2\Big[-3+3(1-\sin^2\theta)\Big]\\ & =-2e^{2i\phi}\sqrt\frac{15}{32\pi}\hbar^2\Big[-3+3-3\sin^2\theta)\Big]\\ & =6\hbar^2\sqrt\frac{15}{32\pi}e^{2i\phi}\sin^2\theta\\ & =6\hbar^2Y_{22}(\theta,\phi)\end{align}
$$

Thus, the eigenvalue is \(6\hbar^2\).