Consider a particle in an infinite potential well with length $L$: $\forall x \in (0,L): V(x) = 0$ and $V(x) = \infty$ elsewhere. The wave function at time $t = 0$ is given by $$ \psi(x,0) = \begin{cases} N(x - L/2) & 0 \leq x \leq L \\ 0 & \text{elsewhere} \end{cases}$$ I determined the normalization constant as $N$ has $N = \sqrt{\frac{12}{L^3}}$. The problem is also asking me to find $$ \psi(x,0) = \sum_{n}^{\infty} c_n \psi_n(x) $$ and to determine the expansion coefficients $c_n$. I did that by using the orthonormality conditions. For the infinite potential well we know that the wave functions are given as $$\psi_n(x) = \sqrt{\frac{2}{L}} \sin(\frac{n\pi x}{L}). $$ So I used $$ c_m = \int \psi_m^*(x) \psi(x,0) dx $$ and found the expansion coefficients as $$ c_n = \begin{cases} 0 & \text{when n is odd} \\ - \frac{\sqrt{24}}{n \pi} & \text{when n is even} \end{cases}$$ I also need to find the wave function $\Psi(x,t)$ at any later time. I wrote $$ \Psi(x,t) = \sum_n c_n \psi_n(x) \exp(-iE_nt / \hbar) $$ where $$E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2} $$ for the potential well. But the final question of this problem asks me to find the expectation value of the energy. So this means I have to find $$ \langle H \rangle = \int \Psi^*(x,t) H \Psi(x,t) dx = \int \big( \sum_m c_m \psi_m \exp(iE_mt / \hbar \big) H \big( \sum_n c_n \psi_n \exp(-iE_nt / \hbar) \big) dx $$ ? But how do I calculate this expression? Or is there some better way to find the expectation value of the energy? Griffiths textbook says (chapter two) that if $$ \Psi(x,t) = \sum_n c_n \psi_n(x) \exp(-iE_nt / \hbar) $$ then $$ \langle H \rangle = \sum_n^{\infty} E_n |c_n|^2. $$ But how do I know this series will converge?

  • $\begingroup$ Related: physics.stackexchange.com/q/38181/2451 $\endgroup$
    – Qmechanic
    Aug 7, 2016 at 21:09
  • $\begingroup$ @count_to_10. So for the infinite well potential we have $H = \frac{p^2}{2m}$ right. So I just use this in the integral above, and substitute for $p$ the operator and do the integral? But can I simplify the summations? $\endgroup$
    – Kamil
    Aug 7, 2016 at 21:45
  • $\begingroup$ I don't know. But I think in my case, convergence is irrelevant. If I assume that $\langle H \rangle \sum_n E_n | c_n|^2$ is correct, then in my case I would have $\sum_n \bigg( \frac{n^2 \pi^2 \hbar^2}{2mL^2} \bigg) \frac{24}{n^2 \pi^2} = \frac{12 \hbar^2}{mL^2}. $ The $n$'s cancelled, and I guess I can just drop the summation (not sure though). Still need to check if this has the dimensions of energy. $\endgroup$
    – Kamil
    Aug 7, 2016 at 21:57
  • $\begingroup$ Your initial wave finction doesn't satisfy the boundary conditions. In other words, it is too rapidly changing somewhere. In order to represent such a rapid change as a superposition of smooth wave functions, you need populate higher harmonics with huge energy. That is why the sum diverges. Use smoother initial wave function (zero at both ends) with a reasonable transition to zero (i.e., withing a finite thickness), and your higher harmonics will be more supressed. $\endgroup$ Jul 20, 2021 at 4:48
  • $\begingroup$ Note, physically there are mechanisms to get rid of high energy states in your initial state - via radiation, etc., so it will never stay the same physically. $\endgroup$ Jul 20, 2021 at 4:49

2 Answers 2


After making edits taking Valter Moretti's corrections into account I now feel quite confident in this answer.

I have followed your calculations and they seem correct. But you certainly can't "drop the summation" as you mention in the comments. The sum reduces (with $n = 2k$) as

$$ \left<E\right> = \sum_{k=1}^{\infty} E_n |c_n|^2 = \sum_{k=1}^{\infty} \frac{n^2 \pi^2 \hbar^2}{2mL^2} \frac{24}{n^2 \pi^2} = \sum_{k=1}^{\infty} \frac{12 \hbar^2}{mL^2} = \frac{12 \hbar^2}{mL^2} \sum_{k=1}^{\infty} 1 = \infty. $$

The fact is that the expectation value of the energy is actually infinite. Such a situation might seem bizarre, but as professor Moretti pointed out it is actually not an impossible situation; you will never measure an infinite value of the energy. The probability $|c_n|^2$ of measuring the energy $E_n$ still goes to zero as $n$ goes to infinity. An infinite expectation value simply means that if you take many measurements and average them, the average will increase without bound. This does not break any particular physical principles.

Actually, as detailed in the answers to the question that Qmechanic referenced, infinite energy expectation values are typical of discontinuous wavefunctions such as the one in your initial condition. In fact, the coefficients $c_n$ that you have derived show that $\psi$ is discontinuous, because it ts a theorem of Fourier analysis that if $\psi$ is continuous, then there is a constant $K$ such that $|c_n| \leq K/n^2$. In this case, $c_n$ goes to zero as $1/n$, not $1/n^2$.

  • 2
    $\begingroup$ I do not think that an infinite expectation value of an observable (like the energy) is non-physical necessarily. What is non-physical is an infinite value of a single outcome of the measurement of an observable. The expectation value is just a partially conventional measure of the quantity of energy of the system, which is not defined in this case since the state is a superposition of states with definite energy. The fact that "energy is conserved" means here that the probabilities to have certain (finite!) outcomes when measuring the energy are constant in time. $\endgroup$ Aug 8, 2016 at 12:42
  • 2
    $\begingroup$ Even continuity of the wavefunction cannot be considered natural or necessary. If all wavefunctions were assumed to be continuous we would renounce to the completeness of the space of the states, i.e., the Hilbert space structure. $\endgroup$ Aug 8, 2016 at 12:48
  • $\begingroup$ @Valter Moretti Thank you for the correction. If I understand you right, an infinite expectation value of the energy then simply expresses that, although the outcome of a single measurement is always finite, there is no sensible finite "average" (just like the set {0,1,...} has no mean). If we measure the energy many times, the mean value of our measurements will grow without bound. And of course there's nothing impossible about that, because the probability of measuring ridiculously high energies still goes to zero as ridiculousness increases, so to speak. $\endgroup$ Aug 8, 2016 at 13:42
  • $\begingroup$ Yes, that is my opinion! $\endgroup$ Aug 8, 2016 at 15:04
  • $\begingroup$ @ValterMoretti Rewrote based on your comments. Thanks for dispelling my misconception! $\endgroup$ Aug 9, 2016 at 21:06

I'm glad I found this post. I was thinking about Problem 2.8 in Griffiths-Schroeter: Introduction to QM, 3rd ed., where the $\psi(x,0)$ is the characteristic function of an interval. Even though the exercise didn't ask about $<H>$, I computed it anyway just for fun, and was surprised that it didn't converge. At first I thought I made a mistake, but this make sense now.

The problem can be read here, together with a solution, though it doesn't discuss $<H>$: https://stemjock.com/STEM%20Books/Griffiths%20QM%203e/Chapter%202/GriffithsQMCh2p8.pdf


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