3.2 Case 2 The Finite Well

This time, let's consider a potential

$$ U(x)=\begin{cases}0&0\le x\le L,\\ U_0&x\le0\ \text{or}\ x\ge L.\end{cases} $$

Now the time-independent Schrödinger equation can be expressed as

$$ \frac{d^2}{dx^2}\psi(x)=\frac{2m\left(U(x)-E\right)}{\hbar^2}\psi(x). $$

For simplicity, let us divide the whole space into three region: region (I) $x\le 0$, region (II) $0\le x\le L$ , and region (III) $x\ge L$, and also name the wave equations inside those three regions $\psi_\text{I}(x)$, $\psi_\text{II}(x)$, and $\psi_\text{III}(x)$ respectively.

Region II

As $U(x)=0$ for $0\le x\le L$, the equation becomes

$$ \frac{d^2}{dx^2}\psi(x)=-k^2\psi(x), $$

where $k=\sqrt{{2mE}/{\hbar^2}}$. As the differential equation is same as that of the infinite well, we have the same general solution

$$ \bbox[silver]{\begin{align} \psi_\text{II}=&A'\mathrm{e}^{ikx}+B'\mathrm{e}^{-ikx}\\ =&A\sin(kx)+B\cos(kx). \end{align}} $$

Region I and Region III

The differential equation can be rewritten as

$$ \frac{d^2}{dx^2}\psi(x)=\alpha^2\psi(x), $$

where $\alpha=\sqrt{2m(U_0-E)/\hbar^2}$. Consider the waves inside the well, i.e. $E

$$ \psi=C\mathrm{e}^{\alpha x}+D\mathrm{e}^{-\alpha x}. $$

The wave function should not diverge, i.e. $\psi(x)\rightarrow0$ as $x\rightarrow\pm\infty$. Therefore we have

$$ \bbox[silver]{\psi_\text{I}=C\mathrm{e}^{\alpha x},} $$

$$ \bbox[silver]{\psi_\text{III}=D\mathrm{e}^{-\alpha x}.} $$

Boundary Conditions

As stated before, wave equations should be $C^2$ functions. Therefore, wave functions and their derivatives should match at $x=0$ and $x=L$.

x=0

$\psi(x)$ continuous

$$ \psi_\text{I}(0)=\psi_\text{II}(0)\Rightarrow \bbox[silver]{C=B}. $$
$d\psi(x)/dx$ continuous

$$ \frac{d}{dx}\psi_\text{I}(0)=\frac{d}{dx}\psi_\text{II}(0)\Rightarrow\bbox[silver]{\alpha C=kA}. $$

x=L

$\psi(x)$ continuous

$$ \psi_\text{II}(L)=\psi_\text{III}(L)\Rightarrow \bbox[silver]{A\sin(kL)+B\cos(kL)=D\mathrm{e}^{-\alpha L}}. $$
$d\psi(x)/dx$ continuous

$$ \frac{d}{dx}\psi_\text{II}(L)=\frac{d}{dx}\psi_\text{III}(L)\Rightarrow\bbox[silver]{kA\cos(kL)-kB\sin(kL)=-\alpha D\mathrm{e}^{-\alpha L}}. $$

Pugging first two conditions into last two equations yields

$$ \begin{cases}\frac{\alpha}{k}C\sin(kL)+C\cos(kL)=D\mathrm{e}^{-\alpha L},\\ k\frac{\alpha}{k}C\cos(kL)-kC\sin(kL)=-\alpha D\mathrm{e}^{-\alpha L}, \end{cases} $$

which can be further simplified to

$$ \bbox[silver]{2\cot(kL)=\frac{k}{\alpha}-\frac{\alpha}{k}.} $$

The distance at which the amplitude of the wave become $1/\mathrm{e}$ of its original value (the amplitude at the boundary) is called the penetration depth. It is given by

$$ \delta=\frac{1}{\alpha}=\frac{\hbar}{\sqrt{2m(U_0-E)}}. $$

Practice Problem for 3.2

Study the case when potential energy is given by

$$ U(x)=\begin{cases}\infty&x\le0\\ 0&0<x<L\\ U_0&x\ge L. \end{cases} $$