Delta Potential

$U(x)=-\alpha \delta (x)$

• $\alpha$ has units of energy times length
• $\delta (x)$ has units of $\frac{1}{L}$

Bounded: $E<0$

$$\psi (x)=\{\begin{array}{ll}B{e}^{kx},& x\le 0\\ B{e}^{-kx},& x\ge 0\end{array}$$

• $B=\sqrt{k}=\frac{\sqrt{m\alpha }}{\hslash }$
  • Obtained by normalization
• $k\equiv \frac{\sqrt{-2mE}}{\hslash }=\frac{m\alpha }{{\hslash }^2}$
  • Unrelated to wavenumber
  • Second equation is obtained by integrating the TISE over a small area around $x=0$

$E=-\frac{m{\alpha }^2}{2{\hslash }^2}$

• Based on unit analysis and normalization
• Video Explanation

Scattering: $E>0$

$$\psi (x)=\{\begin{array}{ll}A{e}^{ikx}+B{e}^{-ikx},& x<0\\ F{e}^{ikx}+G^{-ikx},& x>0\end{array}$$

• $\frac{k\equiv \sqrt{2mE}}{\hslash }$
• For wave traveling rightward:
  • $G=0$ if the wave travels rightward since there would be no incident wave on the right of the delta potential traveling leftward
  • $F=A(1+2i\beta )-B(1-2i\beta )$
  • $\beta \equiv \frac{m\alpha }{{\hslash }^{2}k}=\frac{m\alpha }{\hslash \sqrt{2mE}}$
  • $B=\frac{i\beta }{1-i\beta }A$
  • $F=\frac{1}{1-i\beta }A$
  • $R=\frac{|B{|}^2}{|A{|}^2}=\frac{{\beta }^2}{1+{\beta }^2}=\frac{1}{1+\frac{2{\hslash }^{2}E}{m{\alpha }^2}}$
  • $T=\frac{|F{|}^2}{|A{|}^2}=\frac{1}{1+{\beta }^2}=\frac{1}{1+\frac{m{\alpha }^2}{2{\hslash }^{2}E}}$
  • Changing delta well to delta barrier doesn’t change $R$ and $T$