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Q:Dirac delta function potential

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

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

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

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

With

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

this has solutions

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

For \( x > 0 \) we must have
\begin{equation}\label{eqn:diracPotential:100}
\psi = a e^{-\kappa x},
\end{equation}

and for \( x < 0 \)
\begin{equation}\label{eqn:diracPotential:120}
\psi = b e^{\kappa x}.
\end{equation}

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

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

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

\begin{equation}\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.
\end{equation}

The RHS is zero

\begin{equation}\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.
\end{equation}

That leaves
\begin{equation}\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}
\end{equation}

In the \( \epsilon \rightarrow 0 \) limit this gives

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

Equating relations for \( \kappa \) we have

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

or

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

with

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

The normalization requires

\begin{equation}\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},
\end{equation}

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

References

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