Chapter 6 Problem Solution (Townsend)

Problem 6.1

Solution

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:

$E_n = \frac{(n_x^2 + n_y^2 + n_z^2)\hbar^2\pi^2}{2mL^2}, \quad n_x, n_y, n_z = 1, 2, 3, \dots$

Given lengths, this is equal to:

$E_n = \frac{\hbar^2\pi^2}{2mL^2}\left(\frac{n_x^2}{a^2} + \frac{n_y^2}{b^2} + \frac{n_z^2}{c^2}\right), \quad n_x, n_y, n_z = 1, 2, 3, \dots$

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

$E_n = \frac{\hbar^2\pi^2}{2mL^2}\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right), \quad n_x, n_y, n_z = 1, 2, 3, \dots$

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.

$E_{2,1,1} = \frac{\hbar^2\pi^2}{2mL^2}\left(\frac{4}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right), \quad n_x, n_y, n_z = 1, 2, 3, \dots$

For the two other cases, we get:

$E_{1,2,1} = \frac{\hbar^2\pi^2}{2mL^2}\left(\frac{1}{a^2} + \frac{4}{b^2} + \frac{1}{c^2}\right), \quad n_x, n_y, n_z = 1, 2, 3, \dots$

and

$E_{1,1,2} = \frac{\hbar^2\pi^2}{2mL^2}\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{4}{c^2}\right), \quad n_x, n_y, n_z = 1, 2, 3, \dots$

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

Problem 6.2

Solution

By the Schrödinger equation, we know that:

$-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi = E\psi$

Using separable integration, we suppose:

$\psi(x, y, z) = X(x)Y(y)Z(z)$

This gives us:

$-\frac{\hbar^2}{2m}\left(YZ\frac{d^2X}{dx^2} + XZ\frac{d^2Y}{dy^2} + XY\frac{d^2Z}{dz^2}\right) + \frac{1}{2}m\omega^2(x^2 + y^2 + z^2)XYZ = EXYZ$

Divide both sides by $XYZ$ , we get:

$-\frac{\hbar^2}{2m}\left(\frac{1}{X}\frac{d^2X}{dx^2} + \frac{1}{Y}\frac{d^2Y}{dy^2} + \frac{1}{Z}\frac{d^2Z}{dz^2}\right) + \frac{1}{2}m\omega^2(x^2 + y^2 + z^2) = E$

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:

$-\frac{\hbar^2}{2m}\frac{d^2X}{dx^2} + \frac{1}{2}m\omega^2x^2 = E_xX$

$-\frac{\hbar^2}{2m}\frac{d^2Y}{dy^2} + \frac{1}{2}m\omega^2y^2 = E_yY$

$-\frac{\hbar^2}{2m}\frac{d^2Z}{dz^2} + \frac{1}{2}m\omega^2z^2 = E_zZ$

Problem 6.3

Solution

Using separation of variable, we know that energy $E$ is given by:

$E = E_x + E_y + E_z$

with

$E_x = (n_x + 1/2)\hbar\omega, \, E_y = (n_y + 1/2)\hbar\omega$

and

$E_z = (n_z + 1/2)\hbar\omega.$

Thus, the ground state energy $E$ is equal to:

$E = (n_x + n_y + n_z + \frac{3}{2})\hbar\omega$

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

$\psi_{0x}(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^2/2\hbar}$

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

(1) $\psi_0 = \psi_{0x}(x)\psi_{0y}(y)\psi_{0z}(z)$

(2) $= \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^2/2\hbar}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega y^2/2\hbar}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega z^2/2\hbar}$

(3) $= \left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^2/2\hbar}$

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 $L=0$ .

Problem 6.4

Solution

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

Problem 6.5

Solution

By the Schrödinger equation, we know that:

Problem 6.6

Solution

(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:

$P_0=\frac{2}{3} \quad P_1=\frac{1}{3}$

Problem 6.7

Solution

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,

$<span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno"> </span><img src="http://bootcampforscience.com/wp-content/ql-cache/quicklatex.com-48b8b5e8d0d4380bcd5fbd556ab22057_l3.png" height="219" width="307" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*}& -\sqrt\frac{15}{32\pi}\hbar^2\Big[2e^{2i\phi}(-1+3\cos^2\theta)-4e^{2i\phi}\Big] \\ & =-2e^{2i\phi}\sqrt\frac{15}{32\pi}\hbar^2\Big[-3+3(1-\sin^2\theta)\Big]\\ & =-2e^{2i\phi}\sqrt\frac{15}{32\pi}\hbar^2\Big[-3+3-3\sin^2\theta)\Big]\\ & =6\hbar^2\sqrt\frac{15}{32\pi}e^{2i\phi}\sin^2\theta\\ & =6\hbar^2Y_{22}(\theta,\phi)\end{align*}" title="Rendered by QuickLaTeX.com"/>$

Thus, the eigenvalue is $6\hbar^2$ .

Problem 6.13

Solution

The complete form of the wave function is given by:

$\psi(\theta,\phi, t) = \sum^\infty_{n=0}\psi(\theta,\phi)e^{-iE_nt/\hbar}$

Given the Hamiltonian and spherical harmonics, we can find corresponding energy eigenvalues. The general expression of energy eigenvalues is given by:

$E = \frac{L^2}{2I} + \omega_0 L_z = \frac{l(l+1)\hbar^2}{2I} + \omega_0 m\hbar$

So for specific states:

$E_{1,1} = \frac{\hbar^2}{I} + \omega_0 \hbar$

$E_{1,0} = \frac{\hbar^2}{I}$

$E_{1,-1} = \frac{\hbar^2}{I} - \omega_0\hbar$

Thus, the complete wave function is:

$\begin{aligned} \psi(\theta,\phi, t) &= \frac{1}{2}Y_{1,1}e^{-it(\frac{\hbar}{I}+\omega_0)} + \frac{i}{\sqrt{2}}Y_{1,0}e^{-i\frac{\hbar}{I}t} - \frac{1}{2}Y_{1,-1}e^{-i(\frac{\hbar}{I}-\omega_0)t} \\ &= e^{i\frac{\hbar}{I}t}\left(\frac{1}{2}Y_{1,1}e^{-i\omega_0 t} + \frac{i}{\sqrt{2}}Y_{1,0} - \frac{1}{2}Y_{1,-1}e^{i\omega_0 t}\right) \end{aligned}$

Problem 6.15

Solution

(a) Energy Eigenvalue Equation

The energy eigenvalue equation is given by:

$E\psi = \hat{H}\psi$

where $E$ is the eigenvalue, $L$ is the total angular momentum, and $L_z$ is the $z$ -component of angular momentum. From quantum mechanics in three dimensions, we know that: