I was studying the expectation value of the energy of a particle in the groud state of the infinite square well after its expansion in terms of width (from $a$ to $2a$), which is:

$$\langle H\rangle= \int \Psi^* \hat H \Psi dx = \frac{2}{a} \int_{0}^{a} \sin \left(x\frac{\pi}{a}\right) \hat H \sin \left(x\frac{\pi}{a}\right) dx = \frac{\pi^2 \hbar^2}{2ma^2}$$

Notice that

the expectation value of H is independent of time; this is a manifestation of conservation of energy in quantum mechanics (see Griffiths; second edition; page 37).

This means that the expectation value of the energy does not change after the expansion.

I have been thinking about why does this happen and I just can think of instantaneous expansion. However this argument goes against the statement that no-random information cannot travel faster than light. Therefore, should I regard a random expansion instead of determining one?

If I am wrong in my argument, what am I missing?


Due to the disagreement on the provided answer (which I claim it has to be time-independent) I will explain myself better and try to write down all the necessary steps.

Let's focus on what is the most probable result for the energy. We know that:

$$\sum_{n = 1}^{\infty}|c_n|^2 = 1$$

So the approach I will use is based on getting the constant $c_n$, plug into it different values of $n$ and take the square of the yielded number.

We can always write down the initial wave function as a linear combination of stationary states:

$$\Psi (x, 0) = \sum_{n = 1}^{\infty} c_n \psi_n (x)$$

Of course, this is always the case if $\psi_n$ is a complete set of functions.

Let's now multiply both sides by $\psi_m^*$ and integrate.

$$\int \psi_m^*\Psi (x, 0) dx= \sum_{n = 1}^{\infty} c_n \int \psi_m \psi_n^*(x) dx= \sum_{n = 1}^{\infty} c_n \delta_{mn}$$

If $m = n$ (i.e exploiting the orthonormality of the stationary states) we get:

$$c_n = \int \psi_n^*\Psi (x, 0) dx$$

This method for getting the coefficients is known as the Fourier's trick.

OK, once here we just have to use the Fourier's trick on our problem:

$$c_n = \frac{\sqrt{2}}{a} \int_0^{a} sin ( \frac{x\pi}{a}) sin ( \frac{n x\pi}{2a})dx$$


$$\Psi (x, 0) = \sqrt{\frac{2}{a}}sin ( \frac{x\pi}{a})$$

$$\psi_n^* = \sqrt{\frac{2}{2a}}sin ( \frac{n x\pi}{2a})$$

Note that the stationary states for the infinite square well are real so it would not be necessary to use *.

Good; now if we solve the integral we get:

If n is even (except for n = 2):

$$c_n = 0$$

If n is odd:

$$c_n = \pm \frac{4\sqrt{2}}{\pi(n^2-4)}$$

As pointed out, n = 2 is not included in this solution so we get it separately:

$$c_2 = \frac{\sqrt{2}}{a} \int_0^{a} sin ( \frac{x\pi}{a}) sin ( \frac{2 x\pi}{2a})dx = \frac{\sqrt{2}}{a} \int_0^{a} sin^2 ( \frac{x\pi}{a}) dx = \frac{1}{\sqrt{2}}$$

If we want to get the probabilities we just have to square the three results obtained.

So clearly we see that the most probable outcome is associated with n = 2, so the most probable value for the energy will be:

$$E_2 = \frac{(2)^2 \pi^2 \hbar^2}{8ma^2} = \frac{\pi^2 \hbar^2}{2ma^2}$$

Which is the same value than before the expansion of the well occurred.

Here comes the key of the problem:

I had the following idea before performing any calculations: if the most probable value for the energy after the expansion of the well is the same than before the expansion, the expectation value should yield the same result as well.

And that is indeed what happens:

$$\langle H\rangle= \int \Psi^* \hat H \Psi dx = \frac{2}{a} \int_{0}^{a} \sin \left(x\frac{\pi}{a}\right) \hat H \sin \left(x\frac{\pi}{a}\right) dx = \frac{\pi^2 \hbar^2}{2ma^2}$$

PS: My original question was totally conceptual. However, I can post another question focusing on that and let this one to discuss the solving method.


closed as unclear what you're asking by Aaron Stevens, Jon Custer, ZeroTheHero, John Rennie, Buzz Jan 11 at 17:12

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  • $\begingroup$ Why do you think that the expectation value does not depend on $a$? The formula for the expectation value you gave has an explicit dependency on it... $\endgroup$ – denklo Jan 10 at 12:18
  • $\begingroup$ I did not mean that it does not depend on a. I meant that it does not change its value after the expansion $\endgroup$ – JD_PM Jan 10 at 12:21
  • 3
    $\begingroup$ You are really overly worried when you try to make sense of the "instantaneous" expansion of the well. This is just a drummed up problem to identify the initial state of the bigger well. $\endgroup$ – ZeroTheHero Jan 10 at 13:38
  • $\begingroup$ @ZeroTheHero I am overthinking about how non-random instantaneous expansion of the well is even possible... $\endgroup$ – JD_PM Jan 10 at 14:08
  • 1
    $\begingroup$ @JD_PM If the instantaneous expansion is throwing you off, instead just think of an expansion that is "fast enough" $\endgroup$ – Aaron Stevens Jan 10 at 14:18