Problem Sheet: Week 16 - Quantum Mechanics

Exercise 9.1

In this exercise we will find the left-moving energy eigenfunction of energy $E$ . Our starting point is the general form identified in the main text,

u_E^\rL(x) = \begin{cases} B'e^{-ikx} &\text{ if } x < 0, \\ C' e^{ik'x} + D'e^{-ik'x} &\text{ if } x \geq 0. \end{cases}

(9.12)

Use continuity of the eigenfunction at $x = 0$ to show that $B' = C' + D'$ . Use this to eliminate $B'$ .
Calculate $\frac{\partial}{\partial x} u_E^\rL(x)$ .
Use continuity of the first derivative of the eigenfunction at $x = 0$ to show that $kB' = k'(D'-C')$ .
Using part 1 and part 3, show that $C' = \left(\frac{k'-k}{k'+k}\right)D'$ and $B' = \left(\frac{2k'}{k'+k}\right)D'$ .
$D'$ is an overall normalisation constant which we choose to be $D' = 1$ . Find a final expression for the left-moving energy eigenfunction of energy $E$ .

Answers

n/a (show that).
$\frac{\partial u_E^\rL(x)}{\partial x} = \begin{cases} -ikB'e^{-ikx} &\text{ if } x < 0, \\ ik'(C' e^{ik'x} - D'e^{-ik'x}) &\text{ if } x \geq 0. \end{cases}$
(9.25)
n/a (show that)
n/a (show that)
$u_E^\rL(x) = \begin{cases} \frac{2k'}{k'+k}e^{-ikx} &\text{ if } x < 0, \\ \frac{k'-k}{k'+k} e^{ik'x} + e^{-ik'x} &\text{ if } x \geq 0. \end{cases}$
(9.26)

Exercise 9.2

In this exercise we will find the energy eigenfunction when $0 < E < V_0$ . In Region I the situation is identical to before, given by (9.4). We will therefore begin by focusing on region II, $x \geq 0$ .

Starting from the TISE in Region II (i.e. (9.6)), substitute in for $\op{P}^2$ , and re-arrange to show that it can be written as
$\vph\frac{\partial^2 u_E^\rII}{\partial x^2} = \frac{2M}{\hbar^2}(V_0 - E)u_E^\rII(x).$
(9.27)
Since $(V-E_0)$ is now positive, define $\lambda^2 = 2M(V_0 - E)/\hbar^2$ . Show that $e^{\lambda x}$ and $e^{-\lambda x}$ are both solutions to the TISE from part 1. We will discount the solution $e^{\lambda x}$ on physical grounds: this would correspond to an exponentially increasing probability density to find the particle beyond the step, which is unreasonable. We will therefore take the solution in Region II to be $u_E^\rII(x) = Ce^{-\lambda x}$ .
Use continuity of the energy eigenfunction at $x = 0$ to show that $A + B = C$ .
Calculate $\frac{\partial}{\partial x} u_E(x)$ .
Use continuity of the first derivative of the eigenfunction at $x = 0$ to show that $ik(A-B) = -\lambda C$ .
Using part 3 and part 5, show that $B = -\left(\frac{\lambda + ik}{\lambda - ik}\right)A$ and $C = -\left(\frac{2ik}{\lambda - ik}\right)A.$
$A$ is an overall normalisation constant which we choose to be $A = 1$ . Find a final expression for the energy eigenfunction of energy $E < V_0$ .
Calculate the reflection coefficient $R = |B|^2$ . Does your answer make physical sense?

Answers

n/a (show that).
n/a (show that).
n/a (show that).
$\frac{\partial u_E(x)}{\partial x} = \begin{cases} ik(A e^{ikx} - Be^{-ikx}) &\text{ if } x < 0, \\ -\lambda C e^{-\lambda x} &\text{ if } x \geq 0. \end{cases}$
(9.28)
n/a (show that).
n.a (show that).
$u_E(x) = \begin{cases} e^{ikx} - \frac{\lambda + ik}{\lambda - ik}e^{-ikx} &\text{ if } x < 0, \\ -\frac{2ik}{\lambda - ik}e^{-\lambda x} &\text{ if } x \geq 0. \end{cases}$
(9.29)
$R = 1$ . This makes sense: the particle is definitely reflected (although it tunnels into the wall a little first!).

Chapter 10: Quantum mechanics in three dimensions