Problem Set 3 - Siavash Davani's Website

Exercise 1 (Infinite square well potential)

Assume we have a particle with mass $m$ in an infinite square well with length $a$ . The potential is therefore similar to the case discussed in the lecture

V(x) = \begin{cases} 0, & 0 \le x \le a \\ \infty, & \text{otherwise} \end{cases}.

(1)

We saw that

\psi_n(x) = \sqrt{\frac{2}{a}} \sin(\frac{n\pi}{a}x),

(2)

and the corresponding energies

E_n=\frac{\pi^2\hbar^2}{2ma^2}n^2

(3)

are solutions to the time-independent Schrodinger equation.

Now assume we have a particle in this potential with the initial wavefunction of

\Psi(x, t=0) = \frac{1}{2}\,\psi_1(x) + \frac{\sqrt{3}}{2}\,\psi_2(x)

(4)

Use the orthonormality property of the $\psi_n$ functions to show that the initial wavefunction $\Psi(x, 0)$ is normalized.
Solution
The orthonormality property is
$\int \psi_n^*(x) \psi_m(x) \, dx = \delta_{mn}.$
(5)
To check the normalization, we write
$\begin{align} \int \Psi^*(x, 0) \Psi(x, 0) \, dx &= \int \left( \frac{1}{2}\,\psi_1(x) + \frac{\sqrt{3}}{2}\,\psi_2(x) \right)^* \left( \frac{1}{2}\,\psi_1(x) + \frac{\sqrt{3}}{2}\,\psi_2(x) \right) \, dx \\ &= \int \left( \frac{1}{2}\,\psi_1^*(x) + \frac{\sqrt{3}}{2}\,\psi_2^*(x) \right) \left( \frac{1}{2}\,\psi_1(x) + \frac{\sqrt{3}}{2}\,\psi_2(x) \right) \, dx \\ &= \int \Bigg( \frac{1}{4}\,\psi_1^*(x)\psi_1(x) + \frac{\sqrt{3}}{4}\,\psi_2^*(x)\,\psi_1(x) \\ \nonumber &\quad\quad\quad + \frac{\sqrt{3}}{4}\,\psi_1^*(x)\psi_2(x) + \frac{3}{4}\,\psi_2^*(x)\psi_2(x) \Bigg) \, dx \\ &= \frac{1}{4}\,\delta_{11} + \frac{\sqrt{3}}{4}\,\delta_{21}+ \frac{\sqrt{3}}{4}\,\delta_{12} + \frac{3}{4}\,\delta_{22} \\ &= \frac{1}{4} + \frac{3}{4} = 1 \end{align}$
(6)
Write down the wavefunction after time $t$ evolves i.e. $\Psi(x, t)$ .
Hint: You do not need to solve any differential equation for that. You just need to know how each individual $\psi_n$ evolves.
Solution
Every $\psi_n(x)$ evolves as
$\psi_n(x,t) = \psi_n(x) \, e^{-\frac{i}{\hbar} E_n t}.$
(7)
And when time evolves, the general superposition state Ψ will evolve as if each if its $\psi_n$ components evolve independently
$\Psi(x, t) = \frac{1}{2}\,\psi_1(x)\,e^{-\frac{i}{\hbar}E_1t} + \frac{\sqrt{3}}{2}\,\psi_2(x)\,e^{-\frac{i}{\hbar}E_2t}.$
(8)
Show that the wavefunction $\Psi(x, t)$ from part $(b)$ stays normalized for all $t$ .
Solution
We know from general properties of the Schrodinger equation that a normalized wavefunction stays normalized when its dynamics is derived from the Schrodinger equation. But it is worth to see this once using direct calculations and to see how the terms actually cancel out. We take steps similar to part $(a)$ . In the calculations we drop the $x$ functionality of $\psi_n(x)$ functions to save space!
$\begin{align} &\int \Psi^*(x, t) \Psi(x, t) \, dx \\ &= \int \left( \frac{1}{2}\,\psi_1\,e^{-\frac{i}{\hbar}E_1t} + \frac{\sqrt{3}}{2}\,\psi_2\,e^{-\frac{i}{\hbar}E_2t} \right)^* \left( \frac{1}{2}\,\psi_1\,e^{-\frac{i}{\hbar}E_1t} + \frac{\sqrt{3}}{2}\,\psi_2\,e^{-\frac{i}{\hbar}E_2t} \right) \, dx \\ &= \int \left( \frac{1}{2}\,\psi_1^*\,e^{\frac{i}{\hbar}E_1t} + \frac{\sqrt{3}}{2}\,\psi_2^*\,e^{\frac{i}{\hbar}E_2t} \right) \left( \frac{1}{2}\,\psi_1\,e^{-\frac{i}{\hbar}E_1t} + \frac{\sqrt{3}}{2}\,\psi_2\,e^{-\frac{i}{\hbar}E_2t} \right) \, dx \\ &= \int \Bigg( \frac{1}{4}\,\psi_1^*\psi_1 + \frac{\sqrt{3}}{4}\,\psi_2^*\,\psi_1\,e^{\frac{i}{\hbar}(E_2-E_1)t} + \frac{\sqrt{3}}{4}\,\psi_1^*\psi_2\,e^{\frac{i}{\hbar}(E_1-E_2)t} + \frac{3}{4}\,\psi_2^*\psi_2 \Bigg) \, dx \\ &= \frac{1}{4}\,\delta_{11} + \frac{\sqrt{3}}{4}\,\delta_{21}\,e^{\frac{i}{\hbar}(E_2-E_1)t}+ \frac{\sqrt{3}}{4}\,\delta_{12}\,e^{\frac{i}{\hbar}(E_1-E_2)t} + \frac{3}{4}\,\delta_{22} \\ &= \frac{1}{4} + \frac{3}{4} = 1. \end{align}$
(9)
As we can see, the phase terms corresponding to the time evolution do not contribute to the final normalization because they only appear where a Kronecker delta appears from orthogonal $\psi_n$ wavefunctions!
As time evolves, there will be a specific time $t_1$ when the wavefunction goes back to the initial state but takes an overall phase term as
$\Psi(x, t_1) = e^{i\theta} \, \Psi(x, 0).$
(10)
Derive the smallest non-zero time $t_1$ when this occurs and the corresponding θ.
Solution
We can rewrite the general solution from part $(b)$ as
$\begin{align} \Psi(x, t) &= \frac{1}{2}\,\psi_1(x)\,e^{-\frac{i}{\hbar}E_1t} + \frac{\sqrt{3}}{2}\,\psi_2(x)\,e^{-\frac{i}{\hbar}E_2t} \\ &= e^{-\frac{i}{\hbar}E_1t} \Bigg( \frac{1}{2}\,\psi_1(x) + \frac{\sqrt{3}}{2}\,\psi_2(x)\,e^{-\frac{i}{\hbar}(E_2-E_1)t} \Bigg). \end{align}$
(11)
If we want the state to be proportional to $\Psi(x, 0)$ , we must have
$e^{-\frac{i}{\hbar}(E_2-E_1)t} = 1$
(12)
which means
$\frac{1}{\hbar}(E_2-E_1)t = 2m\pi, \quad \text{for integer $m$} .$
(13)
The smallest $t$ will correspond to $m=1$ and we have
$t_1 = \frac{2\pi\hbar}{E_2-E_1} = \frac{2\pi\hbar}{\frac{\pi^2\hbar^2}{2ma^2}(2^2-1^2)} = \frac{4ma^2}{3\pi\hbar}.$
(14)
Calculate the expectation value of energy over time using
$\langle E(t) \rangle = \int \Psi^*(x, t) \, \hat{H} \, \Psi(x, t) \, dx,$
(15)
where $\hat{H}=-\frac{\hbar}{2m}\ppdv{}{x}+V(x)$ is the Hamiltonian operator. Does the expectation value change over time?
Hint: This calculation is not as difficult as it looks like. Just recall that $\psi_n(x)$ wavefunctions correspond to a well-known energy and have the property $\hat{H}\psi_n(x)=E_n\psi_n(x)$ . Then the orthonormality property will again simplify the integral.
Solution
With the help of the Hint, we first evaluate the term $\hat{H} \, \Psi(x, t)$ and then put it back to calculate the integral.
$\begin{align} \hat{H}\Psi(x, t) &= \hat{H}\Big( \frac{1}{2}\,\psi_1(x)\,e^{-\frac{i}{\hbar}E_1t} + \frac{\sqrt{3}}{2}\,\psi_2(x)\,e^{-\frac{i}{\hbar}E_2t} \Big) \\ &= \frac{1}{2}\,e^{-\frac{i}{\hbar}E_1t}\,\hat{H}\psi_1(x) + \frac{\sqrt{3}}{2}\,e^{-\frac{i}{\hbar}E_2t}\,\hat{H}\psi_2(x) \\ \intertext{Where in this step, the $\hat{H}$ passes the $e^{-\frac{i}{\hbar}E_nt}$ phase terms because the phase is only a function of time and has no $x$ dependency. Therefore, the operators in $\hat{H}=-\frac{\hbar}{2m}\ppdv{}{x}+V(x)$ have no effect on it. Next, we have} &= \frac{1}{2}\,e^{-\frac{i}{\hbar}E_1t}\,E_1\psi_1(x) + \frac{\sqrt{3}}{2}\,e^{-\frac{i}{\hbar}E_2t}\,E_2\psi_2(x), \end{align}$
(16)
where we used the $\hat{H}\psi_n(x)=E_n\psi_n(x)$ property. If we look closely, we see that the term $\hat{H}\Psi(x, t)$ is very similar to $\Psi(x, t)$ with the difference that each $\psi_n(x)$ component is multiplied by the corresponding energy $E_n$ . This means the whole integral can be evaluated very similar to the previous parts as
$\begin{align} &\int \Psi^*(x, t) \hat{H} \Psi(x, t) \, dx \\ &= \int \left( \frac{1}{2}\,\psi_1\,e^{-\frac{i}{\hbar}E_1t} + \frac{\sqrt{3}}{2}\,\psi_2\,e^{-\frac{i}{\hbar}E_2t} \right)^* \left( \frac{1}{2}\,E_1\psi_1\,e^{-\frac{i}{\hbar}E_1t} + \frac{\sqrt{3}}{2}\,E_2\psi_2\,e^{-\frac{i}{\hbar}E_2t} \right) \, dx \\ &= \int \left( \frac{1}{2}\,\psi_1^*\,e^{\frac{i}{\hbar}E_1t} + \frac{\sqrt{3}}{2}\,\psi_2^*\,e^{\frac{i}{\hbar}E_2t} \right) \left( \frac{1}{2}\,E_1\psi_1\,e^{-\frac{i}{\hbar}E_1t} + \frac{\sqrt{3}}{2}\,E_2\psi_2\,e^{-\frac{i}{\hbar}E_2t} \right) \, dx \\ &= \int \Bigg( \frac{1}{4}\,E_1\psi_1^*\psi_1 + \frac{\sqrt{3}}{4}\,E_1\psi_2^*\,\psi_1\,e^{\frac{i}{\hbar}(E_2-E_1)t} + \frac{\sqrt{3}}{4}\,E_2\psi_1^*\psi_2\,e^{\frac{i}{\hbar}(E_1-E_2)t} + \frac{3}{4}\,E_2\psi_2^*\psi_2 \Bigg) \, dx \\ &= \frac{1}{4}\,E_1\delta_{11} + \frac{\sqrt{3}}{4}\,E_1\delta_{21}\,e^{\frac{i}{\hbar}(E_2-E_1)t}+ \frac{\sqrt{3}}{4}\,E_2\delta_{12}\,e^{\frac{i}{\hbar}(E_1-E_2)t} + \frac{3}{4}\,E_2\delta_{22} \\ &= \frac{1}{4} E_1 + \frac{3}{4} E_2. \end{align}$
(17)
As we can see, the expectation value of energy does not change over time thus it is conserved!
Calculate the expectation value of position
$\langle x(t) \rangle = \int \Psi^*(x, t) \, x \, \Psi(x, t) \, dx.$
(18)
How does it change over time? How does the change relate to the time $t_1$ from part $(d)$ ?
Hint: This one can be a bit more difficult to solve! You need some integration by parts. Feel free to also use integral solvers like WolframAlpha.
Solution
Let us do the labor! The integral is
$\langle x(t) \rangle = \int \Psi^*(x, t) \, x \, \Psi(x, t) \, dx = \int x \, \abs{\Psi(x, t)}^2 \, dx.$
(19)
In part $(b)$ , we already calculated $\abs{\Psi(x, t)}^2$ as
$\begin{align} \abs{\Psi(x, t)}^2 &= \frac{1}{4}\,\psi_1^*\psi_1 + \frac{\sqrt{3}}{4}\,\psi_2^*\,\psi_1\,e^{\frac{i}{\hbar}(E_2-E_1)t} + \frac{\sqrt{3}}{4}\,\psi_1^*\psi_2\,e^{\frac{i}{\hbar}(E_1-E_2)t} + \frac{3}{4}\,\psi_2^*\psi_2 \\ \intertext{Because $\psi_1$ and $\psi_2$ are real, we have $\psi_1=\psi_1^*$ and $\psi_2^*=\psi_2$, then} &= \frac{1}{4}\,\abs{\psi_1}^2 + \frac{3}{4}\,\abs{\psi_2}^2 + \frac{\sqrt{3}}{4}\,\psi_1\psi_2\,\Big[ e^{\frac{i}{\hbar}(E_2-E_1)t} + e^{-\frac{i}{\hbar}(E_2-E_1)t} \Big] \\ &= \frac{1}{4}\,\abs{\psi_1}^2 + \frac{3}{4}\,\abs{\psi_2}^2 + \frac{\sqrt{3}}{4}\,\psi_1\psi_2\,\Big[ 2\cos(\frac{E_2-E_1}{\hbar}t) \Big] \\ &= \frac{1}{2a} \sin^2(\frac{\pi}{a}x) + \frac{3}{2a} \sin^2(\frac{2\pi}{a}x) + \frac{\sqrt{3}}{a}\sin(\frac{\pi}{a}x)\sin(\frac{2\pi}{a}x)\cos(\frac{E_2-E_1}{\hbar}t) \end{align}$
(20)
Therefore, we have
$\begin{align} \langle x(t) \rangle &= \frac{1}{a}\int_0^a \frac{1}{2} x\sin^2(\frac{\pi}{a}x) + \frac{3}{2} x\sin^2(\frac{2\pi}{a}x) + \sqrt{3}\,x\sin(\frac{\pi}{a}x)\sin(\frac{2\pi}{a}x)\cos(\frac{E_2-E_1}{\hbar}t) \, dx \\ &= \frac{1}{a} \left( \frac{1}{2} \frac{a^2}{4} + \frac{3}{2} \frac{a^2}{4} + \sqrt{3}\frac{-8a^2}{9\pi^2}\cos(\frac{E_2-E_1}{\hbar}t) \right) \\ &= \frac{a}{2} - \frac{8\sqrt{3}}{9\pi^2} \, a \cos(\frac{E_2-E_1}{\hbar}t) \end{align}$
(21)
For calculating the integral, we can use integration by parts as well as trigonometric identities like $\cos(A-B)-\cos(A+B)=2\sin(A)\sin(B)$ .
If we approximate the final answer
$\langle x(t) \rangle = \frac{a}{2} - \frac{8\sqrt{3}}{9\pi^2} \, a \cos(\frac{E_2-E_1}{\hbar}t) \approx \frac{a}{2} - 0.156 \, a \cos(\frac{E_2-E_1}{\hbar}t),$
(22)
which means the particle oscillates around the midpoint $\frac{a}{2}$ over time. And we can see that the period of oscillation is
$T = \frac{2\pi\hbar}{E_2-E_1},$
(23)
which is exactly the time $t_1$ calculated in part $(d)$ .

Exercise 2 (Quantum harmonic oscillator)

In the lectures, we discussed and solved the harmonic oscillator problem. The Schrodinger equation in this case was

-\frac{\hbar^2}{2m}\ppdv{\Psi}{x} + \frac{1}{2}m\omega^2 x^2 \Psi = E\Psi,

(24)

for the angular frequency ω and a particle with mass $m$ . The wavefunctions which solve the equation are

\psi_n(x)= \left( \frac{m\omega}{\pi\hbar} \right)^{1/4} \frac{1}{\sqrt{2^n n!}} \, \mathcal{H}_n(\xi) e^{-\xi^2/2}, \quad\quad \xi=\sqrt{\frac{m\omega}{\hbar}}x

(25)

and they correspond to energies

E_n = \left( n+\frac{1}{2} \right) \hbar\omega.

(26)

As we can see, although we have the closed-form expression of the wavefunctions, most often the wavefunctions seem complicated to work with. The ladder operators

a_+ = \frac{1}{\sqrt{2m}}\left( \frac{\hbar}{i}\pdv{}{x} + im\omega x \right)

(27)

and

a_- = \frac{1}{\sqrt{2m}}\left( \frac{\hbar}{i}\pdv{}{x} - im\omega x \right)

(28)

helped a lot in solving the equation. We will see in this problem, that the ladder operators are also very useful for doing calculations involving the wavefunctions and deriving useful properties of them. For solving the tasks in this problem, you do not need to substitute the exact form of the $\psi_n$ functions into the integrals.

Use the orthonormality of the $\psi_n$ functions
$\int \psi_n^*(x) \psi_m(x) \, dx = \delta_{mn},$
(29)
and the fact that the ladder operators transform the wavefunctions as
$a_+ \psi_n(x) \propto \psi_{n+1}(x) \quad\text{and}\quad a_- \psi_n(x) \propto \psi_{n-1}(x),$
(30)
to show that the two integrals below hold for any $n$ .
$\int \psi_n^*(x) \, a_+ \, \psi_n(x) \, dx = 0$
(31)
$\int \psi_n^*(x) \, a_- \, \psi_n(x) \, dx = 0$
(32)
Solution
This task is very straightforward. For any $n$ , we have
$\begin{align} \int \psi_n^*(x) \, a_+ \, \psi_n(x) \, dx &\propto \int \psi_n^*(x) \, \psi_{n+1}(x) \, dx \\ &\propto \delta_{n,n+1} = 0, \end{align}$
(33)
and similarly for $a_-$ .
The position $x$ can be written in terms of the ladder operators as
$x = A ( a_+ - a_- )$
(34)
for some constant $A$ . Find $A$ .
Solution
We can see that
$a_+ - a_- = \frac{2i}{\sqrt{2m}} m\omega x,$
(35)
which means
$x = \frac{-i}{\omega\sqrt{2m}} ( a_+ - a_- ).$
(36)
Now using the results of parts $(a)$ and $(b)$ , show that the expectation value of position is zero for all wavefunctions $\psi_n$ (for any $n$ ).
$\langle x \rangle = \int \psi_n^*(x) \, x \, \psi_n(x) \, dx = 0.$
(37)
Hint: Pay attention to the place of $x$ inside the integral! It is important to keep it where it is before substituting the ladder operators in its place. You will later on know more about this systematically in the lectures.
Solution
If we substitute $x$ from part $(b)$ into the integral, we have
$\begin{align} \langle x \rangle &= \int \psi_n^*(x) \, x \, \psi_n(x) \, dx \\ &= \frac{-i}{\omega\sqrt{2m}} \int \psi_n^*(x) \, ( a_+ - a_- ) \, \psi_n(x) \, dx \\ &= \frac{-i}{\omega\sqrt{2m}} \left[ \underbrace{\int \psi_n^*(x) \, a_+ \, \psi_n(x) \, dx}_{0} - \underbrace{\int \psi_n^*(x) \, a_- \, \psi_n(x) \, dx}_{0} \right] \\ &= 0. \end{align}$
(38)
Use a method similar to part $(c)$ to calculate
$\langle x^2 \rangle_n = \int \psi_n^*(x) \, x^2 \, \psi_n(x) \, dx,$
(39)
and then the uncertainty in position $\sigma_x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}$ as a function of $n$ .
Hint: You can find the properties $[a_-, a_+]=\hbar\omega$ and $a_+ a_- + \frac{1}{2}\hbar\omega = E$ useful!
Solution
We can write
$\begin{align} \langle x^2 \rangle &= \int \psi_n^*(x) \, x^2 \, \psi_n(x) \, dx \\ &= \left(\frac{-i}{\omega\sqrt{2m}}\right)^2 \int \psi_n^*(x) \, (a_+ - a_-)(a_+ - a_-) \, \psi_n(x) \, dx \\ \intertext{Pay attention to the order of the ladder operators in the next step!} &= \frac{-1}{2m\omega^2} \int \psi_n^*(x) \, ( a_+a_+ + a_-a_- - a_+a_- - a_-a_+ ) \, \psi_n(x) \, dx \\ \intertext{For a reason similar to part $(a)$, we have $\int \psi_n^*(x) \, a_+^2 \, \psi_n(x) \, dx = \int \psi_n^*(x) \, a_-^2 \, \psi_n(x) \, dx = 0$, which means the integral simplifies to} &= \frac{-1}{2m\omega^2} \int \psi_n^*(x) \, (- a_+a_- - a_-a_+ ) \, \psi_n(x) \, dx \\ &= \frac{1}{2m\omega^2} \int \psi_n^*(x) \, (a_+a_- + a_-a_+ ) \, \psi_n(x) \, dx \\ \intertext{Now we use $a_- a_+ - a_+ a_- =\hbar\omega$ to write} &= \frac{1}{2m\omega^2} \int \psi_n^*(x) \, (2 a_+a_- + \hbar\omega ) \, \psi_n(x) \, dx \\ &= \frac{1}{m\omega^2} \int \psi_n^*(x) \, (a_+a_- + \frac{1}{2}\hbar\omega ) \, \psi_n(x) \, dx \\ &= \frac{1}{m\omega^2} \int \psi_n^*(x) \, E \, \psi_n(x) \, dx \\ \intertext{And now, we know that the wavefunctions $\psi_n$ have the special property that $E \psi_n(x) = E_n \psi_n(x)$. This means we have} &= \frac{1}{m\omega^2} E_n \int \psi_n^*(x) \psi_n(x) \, dx \\ &= \frac{1}{m\omega^2} E_n \int \abs{\psi_n(x)}^2 \, dx \\ &= \frac{1}{m\omega^2} E_n = \frac{1}{m\omega^2} \, \hbar\omega \left( n+\frac{1}{2} \right) = \frac{\hbar}{m\omega} \left( n+\frac{1}{2} \right) \end{align}$
(40)
Although it took a lot of steps to get there, we just needed to use the general information we had regarding the ladder and corresponding position and energy operators to calculate the result of integrals which otherwise needed explicit calculations involving Hermite polynomials and Gaussian functions!
And finally, the uncertainty of position is
$\sigma_x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} = \sqrt{\frac{\hbar}{m\omega} \left( n+\frac{1}{2} \right)}.$
(41)