Note: there’s an error below (and in the associated PDF).  10 points to anybody that finds it (I’ve fixed it in my working version of phy1520.pdf)

### Q:Dirac delta function potential

Problem 2.24/2.25 [1] introduces a Dirac delta function potential

\label{eqn:diracPotential:20}
H = \frac{p^2}{2m} – V_0 \delta(x),

which vanishes after $$t = 0$$. Solve for the bound state for $$t < 0$$ and then the time evolution after that.

### A:

The first part of this problem was assigned back in phy356, where we solved this for a rectangular potential that had the limiting form of a delta function potential. However, this problem can be solved directly by considering the $$\Abs{x} > 0$$ and $$x = 0$$ regions separately.

For $$\Abs{x} > 0$$ Schrodinger’s equation takes the form

\label{eqn:diracPotential:40}
E \psi = -\frac{\Hbar^2}{2m} \frac{d^2 \psi}{dx^2}.

With

\label{eqn:diracPotential:60}
\kappa =
\frac{\sqrt{-2 m E}}{\Hbar},

this has solutions

\label{eqn:diracPotential:80}
\psi = e^{\pm \kappa x}.

For $$x > 0$$ we must have
\label{eqn:diracPotential:100}
\psi = a e^{-\kappa x},

and for $$x < 0$$
\label{eqn:diracPotential:120}
\psi = b e^{\kappa x}.

requiring that $$\psi$$ is continuous at $$x = 0$$ means $$a = b$$, or

\label{eqn:diracPotential:140}
\psi = \psi(0) e^{-\kappa \Abs{x}}.

For the $$x = 0$$ region, consider an interval $$[-\epsilon, \epsilon]$$ region around the origin. We must have

\label{eqn:diracPotential:160}
E \int_{-\epsilon}^\epsilon \psi(x) dx = \frac{-\Hbar^2}{2m} \int_{-\epsilon}^\epsilon \frac{d^2 \psi}{dx^2} dx – V_0 \int_{-\epsilon}^\epsilon \delta(x) \psi(x) dx.

The RHS is zero

\label{eqn:diracPotential:180}
E \int_{-\epsilon}^\epsilon \psi(x) dx
=
E \frac{ e^{-\kappa (\epsilon)} – 1}{-\kappa}
-E \frac{ 1 – e^{\kappa (-\epsilon)}}{\kappa}
\rightarrow
0.

That leaves
\label{eqn:diracPotential:200}
\begin{aligned}
V_0 \int_{-\epsilon}^\epsilon \delta(x) \psi(x) dx
&=
\frac{-\Hbar^2}{2m} \int_{-\epsilon}^\epsilon \frac{d^2 \psi}{dx^2} dx \\
&=
\frac{-\Hbar^2}{2m} \evalrange{\frac{d \psi}{dx}}{-\epsilon}{\epsilon} \\
&=
\frac{-\Hbar^2}{2m}
\psi(0)
\lr
{
-\kappa e^{-\kappa (\epsilon)}

\kappa e^{\kappa (-\epsilon)}
}.
\end{aligned}

In the $$\epsilon \rightarrow 0$$ limit this gives

\label{eqn:diracPotential:220}
V_0 = \frac{\Hbar^2 \kappa}{m}.

Equating relations for $$\kappa$$ we have

\label{eqn:diracPotential:240}
\kappa = \frac{m V_0}{\Hbar^2} = \frac{\sqrt{-2 m E}}{\Hbar},

or

\label{eqn:diracPotential:260}
E = -\inv{2 m} \lr{ \frac{m V_0}{\Hbar} }^2,

with

\label{eqn:diracPotential:280}
\psi(x, t < 0) = C \exp\lr{ -i E t/\hbar – \kappa \Abs{x}}.

The normalization requires

\label{eqn:diracPotential:300}
1
= 2 \Abs{C}^2 \int_0^\infty e^{- 2 \kappa x} dx
= 2 \Abs{C}^2 \evalrange{\frac{e^{- 2 \kappa x}}{-2 \kappa}}{0}{\infty}
= \frac{\Abs{C}^2}{\kappa},

so
\label{eqn:diracPotential:320}
\boxed{
\psi(x, t < 0) = \inv{\sqrt{\kappa}} \exp\lr{ -i E t/\hbar – \kappa \Abs{x}}. } There is only one bound state for such a potential. After turning off the potential, any plane wave $$\label{eqn:diracPotential:360} \psi(x, t) = e^{i k x – i E(k) t/\Hbar},$$ where $$\label{eqn:diracPotential:380} k = \frac{\sqrt{2 m E}}{\Hbar},$$ is a solution. In particular, at $$t = 0$$, the wave packet $$\label{eqn:diracPotential:400} \psi(x,0) = \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x} A(k) dk,$$ is a solution. To solve for $$A(k)$$, we require $$\label{eqn:diracPotential:420} \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x} A(k) dk = \inv{\sqrt{\kappa}} e^{ – \kappa \Abs{x} },$$ or $$\label{eqn:diracPotential:440} \boxed{ A(k) = \inv{\sqrt{2\pi \kappa}} \int_{-\infty}^\infty e^{-i k x} e^{ – m V_0 \Abs{x}/\Hbar^2 } dx. }$$ The initial time state established by the delta function potential evolves as \label{eqn:diracPotential:480} \boxed{ \psi(x, t > 0) = \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x – i \Hbar k^2 t/2m} A(k) dk.
}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.