Problem Sheet: Week 14 - Quantum Mechanics

Exercise 4.1

In this exercise we will show that in quantum mechanics, the kinetic energy and the potential energy are incompatible observables. In this question, we will consider an arbitrary wavefunction $\psi(x)$ .

Calculate $\op{V}\op{K}\psi(x).$
Calculate $\op{K}\op{V}\psi(x).$
Using parts 1. and 2. show that

\vph[\op{K},\op{V}]\psi = -\frac{\hbar^2}{2M}\left(\frac{\partial^2 V}{\partial x^2}\psi(x) + 2 \frac{\partial V}{\partial X} \frac{\partial}{\partial x}\psi(x) \right)

(4.15)

We will denote by $\hat{V}'$ and $\hat{V}''$ the operators satisfying

\begin{align*} \hat{V}'\ket{\psi} &= \infint \frac{\partial V}{\partial X} \psi(x) \ket{x} dx,\\ \hat{V}''\ket{\psi} &= \infint \frac{\partial^2 V}{\partial X^2} \psi(x) \ket{x} dx. \end{align*}

(4.16)

Hence, show that

\vph[\hat{K},\hat{V}] = -\frac{\hbar^2}{2M}\hat{V}'' - \frac{i\hbar}{M}\hat{V}'\hat{P}.

(4.17)

The operators $\hat{V}'$ and $\hat{V''}$ can vanish (can satisfy $\hat{V}'\ket{\psi} = 0$ ). When does this happen? This is the only case in which $[\hat{K},\hat{V}] = 0$ . Does this make sense physically?

Answers

$-\frac{\hbar^2}{2M}V(x)\frac{\partial^2 \psi}{\partial x^2}$
$-\frac{\hbar^2}{2M}\left(\frac{\partial^2 V}{\partial x^2}\psi(x) + 2 \frac{\partial V}{\partial x}\frac{\partial \psi}{\partial x} + V(x)\frac{\partial^2 \psi}{\partial x^2}\right)$
n/a (show that)
n/a (show that)
$\hat{V}' = 0$ if $V(x) =$ constant (in which case $\hat{V}'' = 0$ also). This does make sense, as if the potential is constant, the force vanishes, so the particle is free.

Chapter 5: Time evolution

Exercise 5.1

In this exercise we will show that the SE for the quantum state (5.1) implies the SE for wavefunctions (5.2).

Substitute (1.4) into (5.1).
Explain why we can take $\frac{d}{dt}$ inside the integral on the left-hand side, as long as we replace it by $\frac{\partial}{\partial t}$ .
Use (1.21) to express the right-hand side in terms of $\op{H}$ instead of $\hat{H}$ .
Take the scalar product on both sides with $\bra{x'}$ and use (1.13) to arrive at (5.2) (after finally changing variable from $x'$ to $x$ ).

Answers

$i\hbar \frac{d}{dt}\infint \psi(x,t)\ket{x} dx = \hat{H} \infint \psi(x,t)\ket{x} dx$ .
We are only interested in change in time; we write $\frac{d}{dt}$ since $\ket{\psi(t)}$ is only a function of $t$ . With wavefunction, there is also an $x$ dependence, and so we must write $\frac{\partial}{\partial t}$ to make it explicit that this is just the rate of change in time.
$\infint (\op{H}\psi(x,t))\ket{x} dx$ .
n/a (essentially a show that).

Exercise 5.2

In this exercise we will confirm that $\ket{\psi(t)}$ from (5.4) satisfies the SE (5.1) with $\ket{\psi_\init}$ from (5.3) as initial condition.

Calculate $i\hbar \frac{d}{dt}\ket{\psi(t)}$ .
Calculate $\hat{H} \ket{\psi(t)}$ and therefore confirm that $\ket{\psi(t)}$ satisfies the SE (5.1).
Confirm that $\ket{\psi(0)} = \ket{\psi_\init}$ .
Re-carry out the above calculations in terms of $\psi(x,t) = \inner{x}{\psi(t)}$ and the SE (5.2).

Answers

$E_k\sum_{k=1}^d \alpha_k e^{-iE_k t/\hbar}\ket{E_k}$ .
$E_k\sum_{k=1}^d \alpha_k e^{-iE_k t/\hbar}\ket{E_k}$ . LHS = RHS, so solution.
n/a (essentially show that)
Both sides equal $E_k\sum_{k=1}^d \alpha_k e^{-iE_k t/\hbar}u_{E_k}(x)$

Exercise 5.3

In this exercise, we will show that for a particle whose initial state is an energy eigenstate, $\ket{\psi_\init} = \ket{E_\ell}$ , the momentum of the particle does not change in time.

Using the definition of the momentum wavefunction (2.12), show that
$\inner{p}{E_\ell} = \tilde{u}_{E_\ell}(p),$
(5.21)
where $\tilde{u}_{E_\ell}(p)$ is the momentum wavefunction associated to $u_{E_\ell}(x)$ , and $u_{E_\ell}(x) = \inner{x}{E_\ell}$ is the (spatial) energy eigenfunction corresponding to the energy level of energy $E_\ell$ .
Using part 1 and (5.14), find the momentum wavefunction $\tilde{\psi}(p,t)$ of the particle at time $t$ .
Hence show that the probability density $\pd(p,t)$ is independent of time. Why does this mean that the momentum of the particle isn’t changing in time?

Answers

n/a (show that).
$e^{iE_\ell t/\hbar}\tilde{u}_{E_\ell}(p)$ .
$\pd(p,t) = |\tilde{u}_{E_\ell}(p)|^2$ . This is independent of time.

Exercise 5.4

In this exercise, we will consider a particle that is in a superposition of bound and non-bound states. Consider a particle with an initial state given by

\ket{\psi_\init} = \sum_{k=1}^d \alpha_k \ket{E_k} + \int_{E_B}^\infty \phi_\init(E) \ket{E}dE,

(5.22)

where $\ket{E_k}$ are energy eigenstates with $E_k < E_B$ , and $\ket{E}$ are eigenstates with $E > E_B$ . Note that these states are orthgonal, $\inner{E}{E_k} = 0$ .

Show that the normalisation condition for such superpositon states is

\sum_{k=1}^d |\alpha_k|^2 + \int_{E_B}^\infty |\phi_\init(E)|^2 dE = 1.

(5.23)

Write down the state $\ket{\psi(t)}$ that $\ket{\psi_\init}$ evolves into at time $t$ .
Substitute this into the SE (5.1) to confirm that it is a solution.
Verify that $\ket{\psi(t)}$ satisfies the initial condition $\ket{\psi(0)} = \ket{\psi_\init}$ .

Answers

n/a (show that).
$\ket{\psi(t)} = \sum_{k=1}^d \alpha_k e^{-iE_kt/\hbar}\ket{E_k} + \int_{E_B}^\infty \phi_\init(E) e^{-iEt/\hbar}\ket{E}dE$
Both sides evaluate to 2. $\ket{\psi(t)} = \sum_{k=1}^d E_k \alpha_k e^{-iE_kt/\hbar}\ket{E_k} + \int_{E_B}^\infty E \phi_\init(E) e^{-iEt/\hbar}\ket{E}dE$
n/a (essentially a show that).

Exercise 5.5

In this exercise, we will show that the superposition principle holds. Consider the initial wavefunction

\psi_\init''(x) = \beta \psi_\init(x) + \gamma \psi_\init'(x),

(5.19)

where

\psi_\init(x)

and

\psi_\init'(x)

are two separate initial conditions, and β and γ are complex numbers, chosen so that

\ket{\psi''_\init}

is a normalised quantum state.

Furthermore, assume that $\psi(x,t)$ and $\psi'(x,t)$ are, respectively, solutions to the SE (5.2) with initial wavefunctions at $t=0$ equal to $\psi_\init(x)$ and $\psi'_\init(x)$ .

Show that
$\psi''(x,t) = \beta \psi(x,t) + \gamma \psi'(x,t).$
(5.20)
satisfies the SE (5.2) and has (5.19) as initial wavefunction at $t=0$ .

Answers

n/a (show that).

Chapter 6: Free particle

Exercise 6.2

Consider two particles, the first of which has wavefunction $\psi(x)$ , and the second of which has wavefunction $\psi'(x)$ , related to $\psi(x)$ via

\psi'(x) = \psi(x)e^{ip_0x/\hbar},

(6.15)

where $p_0$ is a real constant.

Show that the probability density of the two particles, $\pd(x)$ and $\pd'(x)$ are equal to each other, $\pd(x) = \pd'(x)$ . Explain why this shows that the two particles are in the same place.

The momentum wavefunction of the first particle is

\tilde\psi(p) = \frac{1}{\sqrt{2\pi\hbar}} \infint \psi(x) e^{-ipx/\hbar}dx.

(6.16)

Show that the momentum wavefunction of the second particle is

\tilde{\psi}'(p) = \tilde{\psi}(p-p_0).

(6.17)

If the first particle has average momentum $\av{p} = \overline{p}$ , show that the second particle has average momentum $\av{p} = \overline{p} + p_0$ .
If the first particle has standard deviation in momentum $\Delta p = \overline \Delta$ , find the standard deviation of the second particle.
Explain why the second particle has momentum $p_0$ more than the first particle.

Answers

n/a (show that).
n/a (show that).
n/a (show that).
n/a (show that).
$\pd'(p) = \pd(p-p_0)$ . The pdf is shifted by $p_0$ (with the density otherwise left unchanged in form). This is what it means to increase the momentum by $p_0$ .

Exercise 6.3

Consider the gaussian wavefunction from the main text,

\psi_\init(x) = \left(\frac{1}{a\sqrt{2\pi}}\right)^{1/2} e^{-x^2/4a^2}e^{ip_0 x/\hbar},

(6.11)

Using Exercise 6.2 and (3.3), or otherwise, find the momentum wavefunction $\tilde{\psi}_\init(p)$ of the particle.
Using part 1, write down the momentum wavefunction of the particle at time $t$ .

Answers

$\tilde{\psi}_\init(p) = \sqrt{\frac{\sqrt{2}a}{\sqrt{\pi}\hbar}}e^{-a^2(p-p_0)^2/\hbar^2}$
$\tilde{\psi}(p,t) = \sqrt{\frac{\sqrt{2}a}{\sqrt{\pi}\hbar}}e^{-a^2(p-p_0)^2/\hbar^2}e^{-ip^2t/2M\hbar}$

Exercise 6.4

Consider a quantum particle with an initial wavefunction given by

\psi_\init(x) = \begin{cases} N\left(1+\frac{x}{\ell}\right) &\text{ if } -\ell \leq x \leq 0, \\ N\left(1-\frac{x}{\ell}\right) &\text{ if } 0 \leq x \leq \ell, \\ 0 &\text{ otherwise,} \end{cases}

(6.18)

where $N > 0$ is a normalisation constant.

Show that in order for the wavefunction to be normalised
$N = \sqrt{\frac{3}{2\ell}}.$
(6.19)
Sketch the wavefunction.
Show that the initial momentum wavefunction of the particle is

\vph\tilde{\psi}_\init(p) = 2\sqrt{\frac{3\ell}{\pi\hbar}}\left(\frac{\sin (p\ell/\hbar)}{p\ell/\hbar}\right)^2.

(6.20)

Write down the wavefunction $\psi(x,t)$ . This should be in the form of an integral over $p$ (you are not expected to evaluate this integral!)
(optional) Numerically evalute the integral in part 4, to plot the probability density $\pd(x,t)$ for $t = 1, 2, \ldots, 5$ (assuming $\ell = \hbar = M = 1$ ).

Answers

n/a (show that).
n/a (sketch).
n/a (show that).
$\psi(x,t) = \frac{\sqrt{6 \ell}}{\pi \hbar}\infint \left(\frac{\sin (p\ell/\hbar)}{p\ell/\hbar}\right)^2 e^{ipx/\hbar} e^{-ip^2 t/2M\hbar} dp$ .
n/a (plot).