Infinite square well – uncertainty principle

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.4.

We can calculate the mean values of position and momentum and verify the uncertainty principle for the infinite square well. The Schrödinger equation for the square well is, between ${x=0}$ and ${x=a}$:

$\displaystyle \frac{d^{2}\psi}{dx^{2}}=-\frac{2m}{\hbar^{2}}E\psi \ \ \ \ \ (1)$

The stationary states of the infinite square well are given by

$\displaystyle \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)e^{-i(n^{2}\pi^{2}\hbar/2ma^{2})t} \ \ \ \ \ (2)$

for ${0\leq x\leq a}$.

For ${x}$ we have

$\displaystyle \left\langle x\right\rangle =\frac{2}{a}\int_{0}^{a}x\sin^{2}(n\pi x/a)dx=a/2 \ \ \ \ \ (3)$

$\displaystyle \left\langle x^{2}\right\rangle =\frac{2}{a}\int_{0}^{a}x^{2}\sin^{2}(n\pi x/a)dx=a^{2}\left(\frac{1}{3}-\frac{1}{2n^{2}\pi^{2}}\right) \ \ \ \ \ (4)$

$\displaystyle \sigma_{x}^{2}=\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}=a^{2}\left(\frac{1}{12}-\frac{1}{2n^{2}\pi^{2}}\right)=\frac{n^{2}\pi^{2}-6}{12n^{2}\pi^{2}}a^{2} \ \ \ \ \ (5)$

For the momentum ${p}$ we have

$\displaystyle \left\langle p\right\rangle =\frac{2\hbar}{ai}\int_{0}^{a}\sin(n\pi x/a)(n\pi/a)\cos(n\pi x/a)dx=0 \ \ \ \ \ (6)$

$\displaystyle \left\langle p^{2}\right\rangle =\frac{2\hbar^{2}}{a}\int_{0}^{a}\sin(n\pi x/a)(n\pi/a)^{2}\sin(n\pi x/a)dx=\frac{n^{2}\pi^{2}\hbar^{2}}{a^{2}} \ \ \ \ \ (7)$

$\displaystyle \sigma_{p}^{2}=\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}=\frac{n^{2}\pi^{2}\hbar^{2}}{a^{2}} \ \ \ \ \ (8)$

The uncertainty principle here is then:

$\displaystyle \sigma_{x}\sigma_{p}=\hbar\sqrt{\frac{\pi^{2}n^{2}-6}{12}} \ \ \ \ \ (9)$

The smallest uncertainty will be for the state ${n=1}$ and is approximately ${0.568\hbar}$, which satisfies the condition ${\sigma_{x}\sigma_{p}\ge\hbar/2}$.

8 thoughts on “Infinite square well – uncertainty principle”

1. rituparna ghosh

sorry it’s my mistake, actually I was confusing between the stationary state and linear combination of them! my mistake! sorry

2. alok

Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.4 . Could you help me with the solution of , i am not getting “a/2”?

1. Mike Randle

It took me around an hour of confusion to realize that the oscillating terms reduce. Specifically, sin(2*pi*n) and cos(2*pi*n) become 0 and 1 respectively because n is defined only by integers. Once the oscillating terms reduce, you recover Griffith’s solution. It is just one of those things that I had never encountered before. If n were defined on the reals instead of integers, this simplification would not happen.