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}}.

Leave a Reply

Your email address will not be published. Required fields are marked *