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.

3 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.

    Reply

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