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The delta-function well · Quantum Mechanics | Math Infinitum

Intuition

The thinnest possible well is a delta-function spike pulling downward. It is solvable in two lines, and it has exactly one bound state: a wavefunction decaying away from the spike on both sides, with a corner at the spike itself. It is the simplest model of anything that holds a particle over a very short range.

A single stake driven into soft ground holds a tent rope at one point: the rope sags away from the stake on both sides, with a sharp bend at the stake. The bound state of the delta well has that shape.

The one bound state of the delta well $V=-\alpha\delta(x)$ , drawn with $\kappa=1$ : $\psi\propto e^{-\kappa|x|}$ with $\kappa=m\alpha/\hbar^{2}$ . The arrow marks the spike pulling downward; the corner in $\psi$ is its signature.

One bound state

For $V=-\alpha\,\delta(x)$ with $\alpha>0$ , a bound state must be $e^{-\kappa|x|}$ on both sides. The jump condition at the origin fixes , and so the single energy. Every other energy gives a solution that grows at infinity.

\begin{aligned}&\psi(x)=\sqrt{\kappa}\,e^{-\kappa|x|},\\ &\kappa=\frac{m\alpha}{\hbar^{2}},\qquad E=-\frac{m\alpha^{2}}{2\hbar^{2}}\end{aligned}

Properties

There is exactly one bound state, whatever the strength $\alpha$ .
The state is even, peaked at the spike with a corner there.
A stronger spike binds more tightly: $\kappa$ grows as $\alpha$ and $|E|$ as $\alpha^{2}$ .

The bound state of a delta well

\begin{aligned}&V=-\alpha\,\delta(x)\ \Longrightarrow\ \kappa=\frac{m\alpha}{\hbar^{2}}\\ &E=-\frac{\hbar^{2}\kappa^{2}}{2m}=-\frac{m\alpha^{2}}{2\hbar^{2}}\end{aligned}

Away from the origin the potential is zero and a bound state decays: $Ae^{-\kappa|x|}$ , the same on both sides by continuity. Its slope drops from $+\kappa A$ to $-\kappa A$ across the origin, and the jump condition says it must drop by times the value .

Proof steps

\psi=Ae^{-\kappa|x|},\qquad E=-\frac{\hbar^{2}\kappa^{2}}{2m}

Away from the spike $\psi''=\kappa^{2}\psi$ , and decay on both sides with continuity at 0 gives this.

\psi'(0^{+})-\psi'(0^{-})=-\kappa A-\kappa A=-2\kappa A

The slope is $-\kappa A$ on the right and $+\kappa A$ on the left.

-2\kappa A=-\frac{2m\alpha}{\hbar^{2}}A

The jump condition for a spike of strength $-\alpha$ .

\kappa=\frac{m\alpha}{\hbar^{2}}

Divide by $-2A$ .

E=-\frac{m\alpha^{2}}{2\hbar^{2}}

Substitute into the energy; normalising gives $A=\sqrt{\kappa}$ .

Applications

Practice

Binding Energy

The bound state of a delta well of strength $\alpha$ has energy $-m\alpha^{2}/2\hbar^{2}$ .

E=-\frac{m\alpha^{2}}{2\hbar^{2}}

Try it

With $m=\hbar=1$ , what is the bound-state energy of $V=-2\,\delta(x)$ ?