202104142106 Notes on Schrödinger equation for the delta-function barrier

The Schrödinger equation for the delta-function barrier reads

$\displaystyle{-\frac{\hbar^2}{2m}}\dfrac{\mathrm{d}^2 \varphi}{\mathrm{d}x^2}+\alpha \delta (x)\varphi =E\varphi$

It yields both bound states ( $E<0$ ) and scattering states ( $E>0$ ). For the time being we consider only scattering states.

For $x<0$ , $V=\alpha \delta (x)=0$ .

The Schrödinger equation reads

$\displaystyle{\dfrac{\mathrm{d}^2\varphi}{\mathrm{d}x^2}=-\frac{2mE}{\hbar^2}\varphi=-k^2\varphi}$ ,

where $\displaystyle{k\stackrel{\textrm{def}}{=} \frac{\sqrt{2mE}}{\hbar}}$ is real and positive.

The general solution being

$\varphi (x)=Ae^{ikx}+Be^{-ikx}$ ,

and this time we cannot rule out either term, since neither of them blows up. Similarly, for $x>0$ ,

$\varphi (x)=Fe^{ikx}+Ge^{-ikx}$

The continuity of $\varphi (x)$ at $x=0$ requires that

$F+G=A+B$

The derivatives are

$\mathrm{d}\varphi /\mathrm{d}x=ik(Fe^{ikx}-Ge^{-ikx})$ for $x>0$ ;

$\mathrm{d}\varphi /\mathrm{d}x=ik(Ae^{ikx}-Be^{-ikx})$ for $x<0$ .

Here we define

$\displaystyle{\Delta (\mathrm{d}\varphi /\mathrm{d}x)=\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg|_{0^+}-\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg|_{0^-}}$

Hence, $\Delta (d\varphi /dx)=ik(F-G-A+B)$ .

By the second boundary condition, we have

$\displaystyle{\Delta \bigg(\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg)=\frac{2m}{\hbar^2}\lim_{\epsilon \to 0}\int_{-\epsilon}^{+\epsilon}V(x)\varphi (x)\,\mathrm{d}x}$ ,

which in turn gives

$\displaystyle{\Delta \bigg(\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg)=\frac{2m\alpha}{\hbar^2}\varphi (0)=\frac{2m\alpha}{\hbar^2}(A+B)}$ .

Combining the results of first derivative continuity and second boundary condition, we write

$\displaystyle{ik(F-G-A+B)=\frac{2m\alpha}{\hbar^2}(A+B)}$ ,

Calculations:

Suppose the wave is coming from the left so that there will be no wave scattering from the right, i.e., $G=0$ .

Recall that $F+G=F+0=F=A+B$ by the continuity condition.

Calculations:

$\begin{aligned} F& =A+B\\ & = A-\frac{i\beta}{1+i\beta}A\\ & = \frac{1}{1+i\beta}A\\ \end{aligned}$

Reflection coefficient $R$ :

$\begin{aligned} R & \equiv \frac{|B|^2}{|A|^2}\\ & = \bigg|-\frac{i\beta}{1+i\beta}\bigg|^2\\ & = \frac{|i\beta|^2}{|1+i\beta|^2}\\ & = \frac{(\sqrt{\beta^2})^2}{(\sqrt{1^2+\beta^2})^2}\\ & = \frac{\beta^2}{1+\beta^2}. \end{aligned}$

We could find the transmission coefficient $T$ in two ways:

(1) By the formula $\displaystyle{T\equiv \frac{|F|^2}{|A|^2}}$ ; or
(2) By the fact that the sum of reflection coefficient and transmission coefficient has to be unity,

i.e., $R+T=1$ .

Calculations:

By (2):

$\begin{aligned} T & = 1-R\\ & = 1-\frac{\beta^2}{1+\beta^2}\\ & = \frac{1}{1+\beta^2} \end{aligned}$

Notice that $R$ and $T$ are functions of $\beta$ , hence:

$\displaystyle{R=\frac{1}{1+(2\hbar E/m\alpha^2)}}$ ,

$\displaystyle{T=\frac{1}{1+(m\alpha^2/2\hbar^2E)}}$ .

Readers should verify this result.

Compare this delta-function barrier with the delta-function trap, we notice that the reflection coefficient and transmission coefficient are each identical respectively in two cases.

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