Problem Sheet: Week 17 - Quantum Mechanics

Exercise 11.1

Confirm explicitly that the energy eigenfunctions of the 3D box potential

\begin{align*} \vph u_\mathbf{n}(\rvec) &= u_{n_x}(x)u_{n_y}(y)u_{n_z}(z),\\ &= \begin{cases} \sqrt{\frac{8}{L^3}} \sin \frac{n_x \pi x}{L}\sin \frac{n_y \pi y}{L}\sin \frac{n_z \pi z}{L} &\text{ if } 0 \leq x,y,z \leq L, \\ 0 &\text{ otherwise,} \end{cases} \end{align*}

satisfy the TISE (inside the box)

\vph-\frac{\hbar^2}{2M}\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right)u_E^\rI(\rvec) = Eu_E^\rI(\rvec),

and that the energy eigenvalues are given by

\vph E_\mathbf{n} = \frac{\hbar^2 \pi^2}{2ML^2} \left(n_x^2 + n_y^2 + n_z^2\right).

Answer

n/a (show that)

Exercise 11.2

In this exercise we will confirm that two degenerate energy eigenstates are nevertheless orthogonal.

Write down the relationship between an energy eigenstate $\ket{E_\mathbf{n}}$ and the corresponding energy eigenfunction $u_\mathbf{n}(\rvec)$ .
Use part 1 to show that scalar product in terms of wavefunctions is given by

\inner{E_{\mathbf{n}}}{E_{\mathbf{n}'}} = \int_V u_{\mathbf{n}}^*(\rvec) u_{\mathbf{n}'}(\rvec) d^3\rvec.

Use the fact that $u_\mathbf{n}(\rvec) = u_{n_x}(x)u_{n_y}(y)u_{n_z}(z)$ (where $u_{n_x}(x)$ are the energy eigenfunctions of the 1D infinite square well, etc), and part 2 (along with its 1D analogue), to show that
$\inner{E_{\mathbf{n}}}{E_{\mathbf{n}'}} = \inner{E_{n_x}}{E_{n_x'}}\inner{E_{n_y}}{E_{n_y'}}\inner{E_{n_z}}{E_{n_z'}},$
(11.23)
where $\ket{E_{n_x}}$ is the energy eigenstate of the 1D infinite square well associated to $u_{n_x}(x)$ , etc.
Thus explain, using the orthogonality of the 1D eigenstates of the infinite square well, why energy eigenstates of the 3D box potential are orthogonal unless $\mathbf{n} = \mathbf{n}'$ (even if $\mathbf{n}$ and $\mathbf{n}'$ have component(s) in common).

Answers

$\ket{E_\mathbf{n}} = \int_V u_\mathbf{n}(\rvec) \ket{\rvec} d^3\rvec$ .
n/a (show that)
n/a (show that)
(11.23) shows that $\inner{E_{\mathbf{n}}}{E_{\mathbf{n}'}} \neq 0$ only if $n_x = n_x'$ , $n_y = n_y'$ and $n_z = n_z'$ simultaneously, i.e. $\mathbf{n} = \mathbf{n}'$ . Even if one or two components are same, remaining component in product will vanish.

Exercise 11.3

In this exercise we will consider an asymmetric box with walls at $x = 0$ and $x = L_x$ , $y = 0$ and $y = L_y$ and $z = 0$ and $z = L_z$ .

Write down the energy eigenfunctions and eigenvalues of a 1D infinite square well, with walls at $x = 0$ and $x = L_x$ .
Use part 1 to explain why a reasonable educated guess for the energy eigenfunctions of the asymmetric 3D box is
$\begin{align*} \vph u_\mathbf{n}(\rvec) &= &= \begin{cases} \sqrt{\frac{8}{L_x L_y L_z}} \sin \frac{n_x \pi x}{L_x}\sin \frac{n_y \pi y}{L_y}\sin \frac{n_z \pi z}{L_z} &\text{ if } 0 \leq x\leq L_x, 0\leq y \leq L_y, 0 \leq z \leq L_z, \\ 0 &\text{ otherwise,} \end{cases} \end{align*}$
(11.24)
Confirm that (11.24) satisfies the TISE inside the asymmetric box, and determine the corresponding energy eigenvalues of the 3D box.
If $L_x = L$ , $L_y = \sqrt{2}L$ and $L_z = \sqrt{3}L$ , write down the first 5 lowest energy levels. Are there any degeneracies?

Answers

$u_n(x) = \sqrt{\frac{2}{L_x}}\sin \frac{n \pi x}{L_x}$ when $0 \leq x \leq L_x$ , and 0 otherwise; $E_n = \frac{\hbar^2 \pi^2 n^2}{2ML_x^2}$ .
This is the direct generalisation of the symmetric case from notes, with same product structure.
n/a (show that) for first part; $E_{\mathbf{n}} = \frac{\hbar^2 \pi^2}{2M}\left(\frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} + \frac{n_z^2}{L_z^2}\right)$ .
In order: $E_{(1,1,1)}$ , $E_{(1,1,2)}$ , $E_{(1,2,1)}$ , $E_{(1,2,2)}$ , $E_{(1,1,3)}$ . No degeneracies.

Exercise 11.5

Starting from the wavefunction
$\psi(\rvec,t) = \frac{1}{\sqrt{2}}\left(e^{-3i\Ebox t/\hbar}u_{(1,1,1)}(\rvec) + e^{-9i\Ebox t/\hbar} u_{(2,2,1)}(\rvec)\right).$
(11.14)
show that the marginal probability density over $x$ and $y$ ,
$\pd(x,y,t) = \vph\infint \pd(\rvec,t) dz,$
(11.25)
is given by
$\pd(x,y,t) = \frac{1}{2}\left(|u_1(x)|^2 |u_1(y)|^2 + |u_2(x)|^2 |u_2(y)|^2 + 2\cos\frac{6\Ebox t}{\hbar} u_1(x)u_2(x)u_1(y)u_2(y)\right).$
(11.26)
Starting again from (11.14), show that the marginal probability density over $z$ ,
$\pd(z,t) = \vph \infint \infint \pd(\rvec,t) dx dy,$
(11.27)
is given by
$\pd(z,t) = |u_1(z)|^2.$
(11.28)
Write down an expression relating the marginal densities $\pd(y,t)$ and $\pd(x,y,t)$ .
Use part 3 to show that, starting from (11.26),
$\pd(y,t) = \frac{1}{2}\left(|u_1(y)|^2 + |u_2(y)|^2\right)$
(11.29)

Answers

Exercise 11.6

In this exercise we will explore a measure of correlation known as the covariance. It is defined as

\mathrm{cov}(x,y) = \av{xy}-\av{x}\av{y},

where

\av{xy} = \vph \infint \infint xy \pd(x,y,t) dx dy.

The covariance is the difference between the expectation value of the product, and the product of the expectation values.

For the probability density
$\pd(x,t) = \frac{1}{2}\left(|u_1(x)|^2 + |u_2(x)|^2\right),$
(11.19)
making reference to Figure 7.5, explain why $\av{x} = \frac{L}{2}$ .

From (11.20), it is evident that $\pd(x,t)$ and $\pd(y,t)$ are identical in form, hence $\av{y} = \frac{L}{2}$ also.

Show that for
$\pd(x,y,t) = \frac{1}{2}\left(|u_1(x)|^2 |u_1(y)|^2 + |u_2(x)|^2 |u_2(y)|^2 + 2\cos\frac{6\Ebox t}{\hbar} u_1(x)u_2(x)u_1(y)u_2(y)\right).$
(11.26)
we have
$\av{xy} = \frac{L^2}{4} + \cos \frac{6\Ebox t}{\hbar} \infint x u_1(x) u_2(x) dx \infint y u_1(y) u_2(y) dy.$
(11.32)
Show that that
$\frac{2}{L}\int_0^Lx \sin \frac{\pi x}{L} \sin \frac{2 \pi x}{L} = - \frac{16L}{9\pi^2}.$
(11.33)
Hence calculate $\mathrm{cov}(x,y)$ (as a function of time).

Answers

$\pd(x)$ is the average of the prob. density of $u_1(x)$ and $u_2(x)$ . The average position of $\pd(x)$ is thus average position in ground state and first excited state. From figure, can see that prob. density is symmetric about centre of well for all eigenfunctions, so $\av{x} = \frac{L}{2}$ .
n/a (show that).
n/a (show that).
$\mathrm{cov}(x,y) = \left(\frac{16L}{9\pi^2}\right)^2 \cos \frac{6\Ebox t}{\hbar}$