Infinite square well – cubic sine initial state

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.37.

As an example of the infinite square well potential suppose the particle starts off with the wave function

$\displaystyle \Psi(x,0)=A\sin^{3}\left(\frac{\pi x}{a}\right) \ \ \ \ \ (1)$

for ${0\le x\le a}$ (and zero elsewhere).

First, we can find ${A}$ by normalizing (using software for the integral):

 $\displaystyle 1$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\int_{0}^{a}\sin^{6}\left(\frac{\pi x}{a}\right)dx\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\frac{5}{16}a\ \ \ \ \ (3)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{4}{\sqrt{5a}} \ \ \ \ \ (4)$

The general solution as a function of time is the series

$\displaystyle \Psi(x,t)=\sum_{n=1}^{\infty}c_{n}\Psi_{n}\left(x,t\right) \ \ \ \ \ (5)$

where ${\Psi_{n}}$ are the stationary states of the square well. We find the ${c_{n}}$ by considering the sum at ${t=0}$:

 $\displaystyle \Psi(x,0)$ $\displaystyle =$ $\displaystyle \sum_{n=1}^{\infty}c_{n}\Psi_{n}\left(x,0\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{a}}\sum_{n=1}^{\infty}c_{n}\sin\left(\frac{n\pi x}{a}\right) \ \ \ \ \ (7)$

Because the stationary states are orthogonal functions, we have

 $\displaystyle c_{n}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{a}}\int_{0}^{a}\sin\left(\frac{n\pi x}{a}\right)\Psi(x,0)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{a}}\frac{4}{\sqrt{5a}}\int_{0}^{a}\sin\left(\frac{n\pi x}{a}\right)\sin^{3}\left(\frac{\pi x}{a}\right)dx\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{24}{5\pi}\frac{\sqrt{10}\sin\left(n\pi\right)}{n^{4}-10n^{2}+9} \ \ \ \ \ (10)$

A quick glance at this result might make you think that ${c_{n}=0}$ for all ${n}$ because of the sine term. However, we need to be careful, since the denominator factors to

$\displaystyle n^{4}-10n^{2}+9=\left(n-1\right)\left(n+1\right)\left(n-3\right)\left(n+3\right) \ \ \ \ \ (11)$

and thus has zeroes at ${n=1,3}$. We can redo these integrals for these specific values of ${n}$ and we get

 $\displaystyle c_{1}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{a}}\frac{4}{\sqrt{5a}}\int_{0}^{a}\sin\left(\frac{\pi x}{a}\right)\sin^{3}\left(\frac{\pi x}{a}\right)dx\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{\sqrt{10}}\ \ \ \ \ (13)$ $\displaystyle c_{3}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{a}}\frac{4}{\sqrt{5a}}\int_{0}^{a}\sin\left(\frac{3\pi x}{a}\right)\sin^{3}\left(\frac{\pi x}{a}\right)dx\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{\sqrt{10}} \ \ \ \ \ (15)$

The full solution is therefore

$\displaystyle \Psi(x,t)=\frac{3}{\sqrt{5a}}\sin\left(\frac{\pi x}{a}\right)e^{-i\pi^{2}\hbar t/2ma^{2}}-\frac{1}{\sqrt{5a}}\sin\left(\frac{3\pi x}{a}\right)e^{-9i\pi^{2}\hbar t/2ma^{2}} \ \ \ \ \ (16)$

The particle will be found with energy ${E_{1}}$ with probability of ${\left|c_{1}\right|^{2}=0.9}$ and with energy ${E_{3}}$ with probability 0.1. Thus

 $\displaystyle \left\langle E\right\rangle$ $\displaystyle =$ $\displaystyle \left(0.9+0.1\times3^{2}\right)\frac{\pi^{2}\hbar^{2}}{2ma^{2}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.9\frac{\pi^{2}\hbar^{2}}{ma^{2}} \ \ \ \ \ (18)$

The mean position is found from

 $\displaystyle \left\langle x\right\rangle$ $\displaystyle =$ $\displaystyle \int_{0}^{a}x\left|\Psi(x,t)\right|^{2}dx \ \ \ \ \ (19)$

Working out the integrand, we get

 $\displaystyle \left|\Psi(x,t)\right|^{2}$ $\displaystyle =$ $\displaystyle -\frac{6}{5a}\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{3\pi x}{a}\right)\left(\sin\left(\frac{E_{1}t}{\hbar}\right)\sin\left(\frac{E_{3}t}{\hbar}\right)+\cos\left(\frac{E_{1}t}{\hbar}\right)\cos\left(\frac{E_{3}t}{\hbar}\right)\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle$ $\displaystyle +\frac{2}{a}-\frac{9}{5a}\cos^{2}\left(\frac{\pi x}{a}\right)-\frac{1}{5a}\cos^{2}\left(\frac{3\pi x}{a}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{6}{5a}\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{3\pi x}{a}\right)\cos\left(\frac{\left(E_{3}-E_{1}\right)t}{\hbar}\right)+\frac{2}{a}-\frac{9}{5a}\cos^{2}\left(\frac{\pi x}{a}\right)-\frac{1}{5a}\cos^{2}\left(\frac{3\pi x}{a}\right) \ \ \ \ \ (22)$

We can now do the integral using software with the result

$\displaystyle \int_{0}^{a}x\left|\Psi(x,t)\right|^{2}dx=\frac{a}{2} \ \ \ \ \ (23)$

The mean position is the midpoint of the well.

5 thoughts on “Infinite square well – cubic sine initial state”

1. Keith

A nicer way to do this would be to use the identity for sin^3(x), it spits out the coefficients directly but it would only work for an example like this.

1. Uzziel

Hi Keith, I was trying to do a similar thing but I got stuck (and that’s why I looked this up! )
Which identity worked for you?