# Infinite square well – average energy

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.9.

The average or expectation value of the energy of a particle in an infinite square well can be worked out either by using the series solution in the form

$\displaystyle \left\langle H\right\rangle =\sum_{n}\left|c_{n}\right|^{2}E_{n} \ \ \ \ \ (1)$

or directly using an integral, using ${H=p^{2}/2m}$ and ${p=\left(\hbar/i\right)\left(d/dx\right)}$:

$\displaystyle \left\langle H\right\rangle =-\frac{\hbar^{2}}{2m}\int_{0}^{a}\Psi^*(x,t)\frac{d^{2}}{dx^{2}}\Psi(x,t)dx \ \ \ \ \ (2)$

Since ${\Psi(x,t)}$ in the general case is a sum over stationary states, as in

$\displaystyle \Psi(x,t)=\sum_{n}c_{n}\psi_{n}(x)e^{-iE_{n}t} \ \ \ \ \ (3)$

and the ${\psi_{n}(x)}$ functions are orthogonal, all the terms of the form

$\displaystyle -\frac{\hbar^{2}}{2m}\int_{0}^{a}c_{n}^*\psi_{n}^*(x)e^{iE_{n}t}c_{m}\psi_{m}(x)e^{-iE_{m}t}dx \ \ \ \ \ (4)$

where ${m\ne n}$ in the integral evaluate to zero, and the remaining terms where ${n=m}$ are all independent of time since the complex exponentials cancel out, so ${\left\langle H\right\rangle }$ is independent of time. Thus if we know the wave function at any given time, we can use it to work out ${\left\langle H\right\rangle }$ at all times.

For example, if we have a parabolic wave function at ${t=0}$

$\displaystyle \Psi(x,0)=Ax(a-x) \ \ \ \ \ (5)$

for ${0\le x\le a}$, we can work out ${\left\langle H\right\rangle }$ directly from it. Applying the normalization condition we can find ${A}$.

 $\displaystyle \left|A\right|^{2}\int_{0}^{a}x^{2}(a-x)^{2}dx$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (6)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \sqrt{\frac{30}{a^{5}}} \ \ \ \ \ (7)$

We get for ${\left\langle H\right\rangle }$:

$\displaystyle \left\langle H\right\rangle =\frac{30}{a^{5}}\int_{0}^{a}x(a-x)\left(-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}x(a-x)\right)dx \ \ \ \ \ (8)$

Working out the second derivative and then the integral (use Maple or do by hand) we get ${\langle H\rangle=5\hbar^{2}/ma^{2}}$.