The Question:

Consider a particle in a one-dimensional box, i.e. in a potential $$V(x)= \begin{cases} 0 & \text{ for } 0\leq x \leq L \\ \infty & \text{ for } x < 0 \text{ and } x > L . \end{cases} $$ Its wave function is a linear superposition of two stationary states and given by $$\psi(x,t)=\frac{1}{\sqrt{8}}\left[\sqrt{5}\psi_1(x,t)+i\sqrt{3}\psi_3(x,t)\right].$$ a) Use the stationary eigenstates in this potential (from your lecture notes or from a textbook), and write down $\psi(x,t)$.

Then calculate $|\psi(x,t)|^2.$ Make sure that your result for $|\psi(x,t)|^2.$ is manifestly real (no "$i$" left in there). Recall the relation $exp(i\theta)=cos\theta+isin\theta$.

What is the frequency of the oscillation in $|\psi(x,t)|^2$?

b) Calculate the expectation value of the Hamilton operator $\hat{H}$. Compare it with the energy eigenvalues $E_1$, $E_2$, $E_3$.

c) What is the probability for the particle to be in state $\psi_1$, and what are the probabilities for it to be in states $\psi_2$ and $\psi_3$, respectively?

d) Calculate the expectation values $\langle\hat{x}\rangle$ and $\langle\hat{p}\rangle$.

Is the particle's state stationary? Why, or why not?

My Attempt:
a) As given, in the question each superposed wave function are functions of time, thus the wave functions are of form: $$\psi_n(x,t)=\sqrt{\frac{2}{l}}sin\left(\frac{\pi x}{l}\right)e^{-\frac{iE_nt}{\hbar}},$$ Where the energy, $E_n$: $$E_n=\frac{\hbar^2}{2m}\left(\frac{n\pi}{L}\right)^2.$$ By substitution of states n= 1 and 3, and then back into the given superposition equation, we see: $$\psi(x,t)=\frac{1}{\sqrt{8}}\left[\sqrt{5}\left(\sqrt{\frac{2}{l}}sin\left(\frac{\pi x}{l}\right)e^{-\frac{i\hbar\pi ^2t}{2ml^2}}\right)+i\sqrt{3}\left(\sqrt{\frac{2}{l}}sin\left(\frac{3\pi x}{l}\right)e^{-9\frac{i\hbar\pi^2t}{2ml^2}}\right)\right].$$

Now, to find $|\psi(x,t)|^2,$ we can use the relation $|\psi(x,t)|^2=\psi^*(x,t)\cdot\psi(x,t).$

I am reasonably confident with the maths, here is my answer: $$\psi(x,t)\cdot\psi^*(x,t)=\frac{5}{4l}sin^2\left(\frac{\pi x}{l}\right)+\frac{3}{4l}sin^2\left(\frac{3\pi x}{l}\right)+\frac{\sqrt{15}}{2l}sin\left(\frac{\pi x}{l}\right) sin\left(\frac{3\pi x}{l}\right)sin\left(\frac{(E_3-E_1)t}{\hbar}\right)$$

Thus, it is clear that the frequency of oscillation is:$$\frac{(E_3-E_1)t}{\hbar}$$ My Problem-Part A-I am unsure on what the frequency of oscillation is actually referring to in this case, if anyone could provide some intuition for this, that'd be great.

b)Given the relation: $$<\hat{H}>=\sum_n |c_n|^2E_n,$$ $$<\hat{H}>=\lvert \frac{\sqrt{5}}{\sqrt{8}}\rvert^2\frac{\hbar^2\pi^2}{2ml^2}+\lvert\frac{i\sqrt{3}}{\sqrt{8}}\rvert^2\frac{9\hbar^2\pi^2}{2ml^2}$$ simplified, this gives: $$<\hat{H}>=\frac{32\hbar^2\pi^2}{ml^2}.$$ I have been asked to compare this with the eigenstates: $$E_1=\frac{\hbar^2\pi^2}{2ml^2},$$ $$E_2=\frac{4\hbar^2\pi^2}{2ml^2},$$ $$E_3=\frac{9\hbar^2\pi^2}{2ml^2}.$$ My problem- part b I can see the following $$64E_1=<\hat{H}>,$$ $$4E_2=<\hat{H}>,$$ $$\frac{64}{9}E_3=<\hat{H}>,$$ But I am struggling to see the relevancy, I think the explanation of this, may also to explain part d

c)I'm not sure where to start with this part

d)$$<x>=\int_0^L|\psi|^2x.dx$$ I have no problem with this integral, I have found the answer to be $\frac{1}{2}l$, which makes sense to me as it is the center point of the 'well'.

Now, to find $<p>$, i can consider: $$<p>=m\frac{d}{dt}<x>,$$ clearly this leads to $<p>=0$,

Part d-My Problem This is what confuses me, surely its state should be stationary if the velocity is zero, which must be true assuming the particle is not massless. However I feel like I am mis-understanding the definition of its 'state' Also, I vaguely remember my lecturer saying something about this linking with the comparison in part b.

  • $\begingroup$ I think the expectation value of the energy should be between the lowest and highest states: $E_1 < \langle H \rangle < E_3 $, and given that $(5\cdot 1^2 + 3 \cdot 3^2)/8 = 1 \cdot 2^2$, you should find it equal to $E_2$. Check your math. Did you drop a $(1/\sqrt 8)^2$? $\endgroup$
    – JEB
    Feb 27, 2018 at 5:42
  • $\begingroup$ I did miss the factor of $(1/\sqrt{8})^2$, thanks :) $\endgroup$
    – George
    Feb 27, 2018 at 7:28
  • 1
    $\begingroup$ Part d) must have the time dependent terms included. The $\int \psi_1^* ~\psi_3 ~x~dx$ and vice-versa integrals are not zero. $\endgroup$
    – Bill N
    Feb 28, 2018 at 4:56

1 Answer 1


(a) When I read the question, I didn't know what was meant by oscillation frequency, but I think you figured it out. There is a time-dependence to the interference term.

(b) Check the math. Have you considered just factoring out $E_1$ in all the terms?

(c) Are you familiar with bra-ket notation? The probability to be in eigenstate $n$ is:

$$ \langle n | \psi \rangle $$


$$ |\psi \rangle = \sqrt{\frac 5 8}|1\rangle + i\sqrt{\frac 3 8}|3\rangle$$


$$ \langle n|m\rangle \equiv \int{\psi_n(x, t)^*\psi_m(x, t)dx} = \delta_{nm}.$$

It saves a lot MathJax in the long run, and is much clearer.

So you figured out the particle is on average in the middle of the box, but it is not in the middle of the box. Likewise with the momentum: it's moving in both directions, but if it had a nonzero mean--it would just leave the box.

  • $\begingroup$ Thanks for your help. However, I don't understand that notation (never seen it before). If I were looking to find the probability that the particle is in state $\psi_1$ would I just consider $\int_0^l ψ_1(x,t)∗ψ_1(x,t)dx$? $\endgroup$
    – George
    Feb 27, 2018 at 9:33
  • $\begingroup$ @George No, that expression is 1, since all eigenstates are normalized. In bra-key notation, the _amplitude is $\langle 1 | \psi \rangle = \frac{1} {\sqrt 5} \langle 1 | 1 \rangle + \frac{1}{\sqrt 3} \langle 1 | 3 \rangle $ $ = \frac{1}{\sqrt 5} $, so the probability is that squared (20%). Think of it a projecting $\psi$ onto an orthonormal basis. Note I edited the answer to make the a little clearer. $\endgroup$
    – JEB
    Feb 27, 2018 at 13:52
  • $\begingroup$ Thanks for your help, I read in a book however, that the probability of finding the particle in any of the eigenstates are the absolute square of the relevant coefficients. Surely this would make the answer: 5/8 for $E_1$ and 3/8 for $E_3$ and 0 for $E_2$? $\endgroup$
    – George
    Feb 27, 2018 at 19:38
  • 1
    $\begingroup$ @George You are correct. I had the coefficients upside down, and forgot the root 8. I was rushing, and flubbed it big time. $\endgroup$
    – JEB
    Feb 27, 2018 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.