Solution for a Single Particle in an Infinite Quantum Well - Darin

Introduction

The particle in a box problem is a classic way of understanding yet another difference between the classical and quantum worlds. Imagine a box whose sides represent infinite potential, so no particle has enough energy to climb them and escape. Now say we put in an ordinary ball with some initial velocity and it bounces between the walls. The ball is no more likely to be in one place than another. But by solving the Schrödinger equation we'll see that a quantum particle isn't so egalitarian.

Mathematical Setup

[math]\displaystyle{ \frac{-\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi = E \Psi }[/math]
To describe the state of the quantum particle in the potential well we will find a function [math]\displaystyle{ \Psi }[/math] that satisfies the barrier conditions of the well and the Schrödinger equation (above), a differential equation.

Constructing [math]\displaystyle{ \Psi (x) }[/math]

For the region [math]\displaystyle{ x ≤ 0 }[/math] the wave function must be zero because the particle cannot overcome an infinite potential barrier. Similarly for [math]\displaystyle{ x ≥ L }[/math] .
Now, we just need to construct the solution/wave-function [math]\displaystyle{ \Psi }[/math] for the region [math]\displaystyle{ 0 ≤ x ≥ 0 }[/math] where the potential is zero. Let's rewrite the Schrödinger equation a bit to see what we're dealing with.
[math]\displaystyle{ \frac{\partial^2 \Psi}{\partial x^2} = - \frac{2m}{\hbar^2} E \Psi }[/math]
We can clearly see that a second derivative of [math]\displaystyle{ \Psi }[/math] is just a constant times the same function [math]\displaystyle{ \Psi }[/math]. This should remind you of sins and cosines or exponentials, such as
[math]\displaystyle{ sin(kx) }[/math] or [math]\displaystyle{ cos(kx) }[/math] where [math]\displaystyle{ k = \frac{2m}{\hbar^2} }[/math]. You could use complex exponentials instead of sins and cosines as well, but it turns out in this situation sins and cosines are slightly more convenient. A good exercise would be to use complex exponentials ([math]\displaystyle{ \Psi (x) = A e^{ikx} + B e^{-ikx} }[/math]) and verify that you get the right answer. Remember that differential equations may have several particular solutions and the general solution is a super position of them . So, [math]\displaystyle{ \Psi (x) = A sin(kx) + B cos(kx) }[/math], but this doesn't satisfy the boundary conditions. Plug in 0 and L and see that [math]\displaystyle{ \Psi }[/math] is not zero as it should be. This is because quantum particles have discrete energy levels so there is one more condition we must take into account, which is that for sin, [math]\displaystyle{ k = \frac{n \pi}{L} }[/math] and for cos, [math]\displaystyle{ k = \frac{n \pi}{2L} }[/math]. Plug in the boundary conditions to verify that [math]\displaystyle{ \Psi(0) = \Psi(L) = 0 }[/math]. Therefore, [math]\displaystyle{ \Psi (x) = A sin(\frac{n \pi x}{L}) + B cos(\frac{n \pi x}{2L}) }[/math].

We can find A and B by focusing on the region where [math]\displaystyle{ x = 0 }[/math]. [math]\displaystyle{ \Psi (0) = B = 0 }[/math], so [math]\displaystyle{ \Psi (x) =A sin(\frac{n \pi x}{L}) }[/math]. We also know that the wave function must be normalized.
[math]\displaystyle{ 1 = \int_{0}^L |\Psi|^2 dx = \int_{0}^L |A sin(\frac{n \pi x}{L})|^2 dx = \int_{0}^L A^2 sin^2(\frac{n \pi x}{L}) dx = A^2 [\frac{x}{2} - \frac{cos(\frac{2n \pi x}{L})}{4}]_{0}^L = A^2 (\frac{L}{2} - \frac{cos(2n \pi)}{4}) = A^2 (\frac{L}{2}) = 1 }[/math]. Therefore, [math]\displaystyle{ A = \sqrt{\frac{2}{L}} }[/math] and [math]\displaystyle{ \Psi = \sqrt{\frac{2}{L}} sin(\frac{n \pi x}{L}) }[/math] for n = 1, 2, 3, ...