# Infinite square well – change in well size

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 ${x\in[0,a]}$. If the right wall suddenly moves to ${x=2a}$, what effect does this have on the allowable energies?

The ground state for an infinite square well of width ${a}$ is

$\displaystyle \psi_{1}=\sqrt{\frac{2}{a}}\sin\left(\frac{\pi x}{a}\right) \ \ \ \ \ (1)$

The stationary states for a well of width ${2a}$ are

$\displaystyle \psi_{n}=\frac{1}{\sqrt{a}}\sin\left(\frac{n\pi x}{2a}\right) \ \ \ \ \ (2)$

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

$\displaystyle \psi_{1}=\sum_{n=1}^{\infty}c_{n}\psi_{n} \ \ \ \ \ (3)$

where the ${c_{n}}$ are found using the orthogonal property of the stationary states:

$\displaystyle c_{n}=\frac{1}{\sqrt{a}}\sqrt{\frac{2}{a}}\int_{0}^{a}\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{n\pi x}{2a}\right)dx \ \ \ \ \ (4)$

For ${n\ne2}$, this integral evaluates to

$\displaystyle c_{n}=\frac{4\sqrt{2}}{\pi\left(4-n^{2}\right)}\sin\left(\frac{n\pi}{2}\right) \ \ \ \ \ (5)$

For ${n=2}$, we have

 $\displaystyle c_{2}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{a}}\sqrt{\frac{2}{a}}\int_{0}^{a}\sin^{2}\left(\frac{\pi x}{a}\right)dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}} \ \ \ \ \ (7)$

The first few probabilities for stationary states are then

 $\displaystyle \left|c_{1}\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{32}{9\pi^{2}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.36\ \ \ \ \ (9)$ $\displaystyle \left|c_{2}\right|^{2}$ $\displaystyle =$ $\displaystyle 0.5\ \ \ \ \ (10)$ $\displaystyle \left|c_{3}\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{32}{25\pi^{2}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.13 \ \ \ \ \ (12)$

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

The most probable energy is therefore

 $\displaystyle E_{2}$ $\displaystyle =$ $\displaystyle \frac{2^{2}\pi^{2}\hbar^{2}}{2m\left(2a\right)^{2}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\pi^{2}\hbar^{2}}{2ma^{2}} \ \ \ \ \ (14)$

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

$\displaystyle E_{1}=\frac{\pi^{2}\hbar^{2}}{8ma^{2}} \ \ \ \ \ (15)$

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

 $\displaystyle \left\langle E\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}E_{2}+\sum_{\mathrm{odd}\; n}\frac{32}{\pi^{2}\left(4-n^{2}\right)^{2}}E_{n}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\pi^{2}\hbar^{2}}{8ma^{2}}\left[2+\sum_{\mathrm{odd}\; n}\frac{32n^{2}}{\pi^{2}\left(4-n^{2}\right)^{2}}\right] \ \ \ \ \ (17)$

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

 $\displaystyle \left\langle E\right\rangle$ $\displaystyle =$ $\displaystyle \int_{0}^{a}\psi_{1}(x)\left(-\frac{\hbar^{2}}{2m}\right)\frac{d^{2}}{dx^{2}}\psi_{1}(x)dx\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{ma}\int_{0}^{a}\sin\left(\frac{\pi x}{a}\right)\frac{d^{2}}{dx^{2}}\sin\left(\frac{\pi x}{a}\right)dx\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}\pi^{2}}{ma^{3}}\int_{0}^{a}\sin^{2}\left(\frac{\pi x}{a}\right)dx\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}\pi^{2}}{2ma^{2}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{2} \ \ \ \ \ (22)$

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

$\displaystyle \sum_{\mathrm{odd}\; n}\frac{32n^{2}}{\pi^{2}\left(4-n^{2}\right)^{2}}=2 \ \ \ \ \ (23)$

This series doesn’t converge all that quickly (it goes as ${1/n^{2}}$). The sum of the first 100 terms gives 1.991893832.

A slight rearrangement gives yet another way of calculating ${\pi}$:

$\displaystyle \sum_{\mathrm{odd}\; n}\frac{16n^{2}}{\left(4-n^{2}\right)^{2}}=\pi^{2} \ \ \ \ \ (24)$

## 6 thoughts on “Infinite square well – change in well size”

1. G

Hi, 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.

1. growescience

If 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 ${x\in\left[0,a\right]}$ and then expands to ${x\in\left[0,3a\right]}$. You’d then just have to run through the same calculation replacing ${2a}$ by ${3a}$. I don’t think that would give rise to any particular difficulties, but I haven’t checked it.

2. Brett J

Can 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?

1. gwrowe Post author

Equation 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 ${0 (that is, it’s zero for the region ${\left[a,2a\right]}$ of the expanded well), so when we work out ${c_{n}}$ the integral from 0 to ${2a}$ reduces to an integral from 0 to ${a}$.

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 ${\left[a,2a\right]}$ so again the integral is from 0 to ${a}$.

In essence, the question is asking for the probability of measuring ${E_{n}}$ in a square well of width ${2a}$ for a particle whose wave function is non-zero over only half the well at a particular time ${t=0}$. 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 ${t}$ increases, the wave function will spread out to fill the expanded well, and the general solution is

$\displaystyle \Psi\left(x,t\right)=\sum_{i=1}^{\infty}c_{n}\psi_{n}\left(x\right)e^{-iE_{n}t/\hbar} \ \ \ \ \ (1)$

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