**Required math: calculus **

**Required physics: **Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.38.

Suppose a particle is in the ground state in an infinite square well in the interval . If the right wall suddenly moves to , what effect does this have on the allowable energies?

The ground state for an infinite square well of width is

The stationary states for a well of width are

The initial wave function can be expanded as a series using the new states:

where the are found using the orthogonal property of the stationary states:

For , this integral evaluates to

For , we have

The first few probabilities for stationary states are then

All higher coefficients are smaller, with all even coefficients from 4 upwards being zero.

The most probable energy is therefore

This happens to be the ground state of the original well, so, unsurprisingly, the most probable energy for the particle is the one it starts with.

The next most probable energy is

The average energy *could* be found if we could sum the infinite series

Clearly this isn’t an easy sum, but we can calculate the expectation value of the Hamiltonian directly as an integral to get the result

If you’re interested, a side effect of this result is that we can evaluate the sum above:

This series doesn’t converge all that quickly (it goes as ). The sum of the first 100 terms gives 1.991893832.

A slight rearrangement gives yet another way of calculating :

### Like this:

Like Loading...

*Related*

GHi, thanks for your explanation! But I have got a few questions regarding this, what if it is a symmetric well of -a/2<x<a/2 and it is expanded suddenly to -3a/2<x<3a/2 ? how will the probability of its energy state vary? Say, it is initially in the ground state.

growescienceIf all you’re interested in are the energy levels, the location of the well doesn’t matter, so you could analyze your problem with a well that starts out at and then expands to . You’d then just have to run through the same calculation replacing by . I don’t think that would give rise to any particular difficulties, but I haven’t checked it.

Brett JCan you tell me why the bouds on the integral on the Hamiltonian (expectation value for energy) and the bounds on the integrals for finding |Cn|^2 are 0 to a instead of 0 to 2a? Shouldn’t the bounds cover the whole well?

gwrowePost authorEquation 3 is writing the original wave function (eqn 1) as a series using the expanded well stationary states. The original wave function is non-zero only for (that is, it’s zero for the region of the expanded well), so when we work out the integral from 0 to reduces to an integral from 0 to .

For working out the average energy, again we have to use the actual wave function of the particle (which is equation 1), and that’s zero in the interval so again the integral is from 0 to .

In essence, the question is asking for the probability of measuring in a square well of width for a particle whose wave function is non-zero over only half the well at a particular time . The particle’s wave function is not a stationary state for the expanded well, so we need to first express it as a series over stationary states for that well. As increases, the wave function will spread out to fill the expanded well, and the general solution is

Since the complex exponential term in doesn’t affect the square modulus of the coefficient of , however, the probabilities of finding the particle in a given energy state don’t change with time.

Brett Jahh, very good. Thanks for the clarification!

Pingback: Infinite square well – expanding well | Physics pages