3.1 Case 1: The Infinite Well

An infinite well. Image from Wikipedia

Inside the well, the potential is zero, so we have

$$ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)=E\psi(x),\text{ for }0\le x\le L. $$

General solution

$$ \psi(x)=A\sin(kx), $$

where we define $k\equiv\sqrt{{2mE}/{\hbar^2}}$. Applying boundary condition $\psi(L)=0$ gives

$$ A\sin(kL)=0. $$

Therefore

$$ kL=n\pi,\forall n\in\Bbb{Z}. $$

This result implies that

$$ \begin{cases}E=\frac{n^2\pi^2\hbar^2}{2mL^2},\\\\\psi(x)=A\sin(\frac{n\pi x}{L}).\end{cases} $$

Applying normalization condition gives

$$ \int_{\text{all space}}|\Psi(x,t)|^2dx=\int_0^L\left(A\sin\frac{n\pi x}{L}\right)^2dx=1. $$

The integral equals $A^2L/2$, so we have

$$ A=\sqrt{\frac{2}{L}}. $$

And here is the final result

$$ \bbox[silver]{\psi_n(x)=\begin{cases}\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L},&\text{ for }0\le x\le L\\\\0,&\text{else}\end{cases}} $$

and

$$ \bbox[silver]{E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}.} $$

Practice Problems for 3.1

  1. Suppose now the infinite well is at $|x|< a/2$, determine the allowed energies and corresponding wave functions.