The Dirac-Delta Potential: A Single Bound State and Scattering

The finite square well carries two length scales, its width and its depth, and the bound-state count depends on both through the dimensionless product $R_0=(a/\hslash )\sqrt{2m{V}_0}$. Shrinking the width while raising the depth so that the product $V_0\cdot (\text{width})$ stays fixed drives the well toward a Dirac-delta potential, a spike of zero width and infinite depth with a finite integrated strength. The limit is exactly solvable and strips the bound-state and scattering problems down to a single matching condition at one point.¹

The delta well and the derivative-jump condition

Write the attractive potential as

where $\alpha$ has units of energy times length and measures the strength. Away from the origin $V=0$, so the time-independent Schrödinger equation is free everywhere except at the single point $x=0$, and the only new physics is the condition joining the two half-line solutions there.

The wave function stays continuous at a delta, but its slope need not. Integrate the Schrödinger equation over a vanishing interval $[-ϵ,ϵ]$ straddling the origin:

The first integral is the change in slope, ${\psi }^{\prime }(ϵ)-{\psi }^{\prime }(-ϵ)$; the delta integral sifts out $\psi (0)$; the right side vanishes as $ϵ\to 0$ because $\psi$ is bounded and the interval shrinks. Taking the limit gives the jump condition.

The jump is proportional to the value of the wave function at the spike. Where $\psi (0)=0$ the delta is invisible (odd states never feel a central delta); where $\psi (0)\ne 0$ the slope kinks in proportion to the strength.

The unique bound state

A bound state has $E<0$. Set

so that away from the origin ${\psi }^{\prime \prime }={\kappa }^2\psi$. The normalizable solutions decay on each side,

with the two amplitudes forced equal by continuity at $x=0$. The slopes are ${\psi }^{\prime }(0^{-})=\kappa B$ and ${\psi }^{\prime }(0^{+})=-\kappa B$, so the jump is $\Delta {\psi }^{\prime }=-2\kappa B$. Inserting this and $\psi (0)=B$ into the jump condition,

This fixes a single value of $\kappa$, hence a single energy.

The normalization $B=\sqrt{\kappa }$ follows from ${\int }_{-\infty }^{\infty }|\psi {|}^2\mathbf{d}x=|B{|}^2{\int }_{-\infty }^{\infty }{e}^{-2\kappa |x|}\mathbf{d}x=|B{|}^2/\kappa =1$. Two features distinguish this from the finite well. First, the number of bound states does not grow with strength: a delta always binds exactly one state, because it has no width to fit additional half-wavelengths. Second, the wave function has a genuine kink at the origin, the cusp demanded by the derivative jump, whereas a smooth potential gives a smooth $\psi$.

Scattering off the delta

For $E>0$ the particle is unbound and the delta acts as a scatterer.³ With

send a beam in from the left. The general solution carries incident and reflected waves on the left and a transmitted wave on the right:

Continuity at $x=0$ gives $A+B=F$. The derivative jump, with ${\psi }^{\prime }(0^{+})=ikF$ and ${\psi }^{\prime }(0^{-})=ik(A-B)$, gives

Introduce the dimensionless strength

which compares the delta strength to the particle's wave number. The two matching equations solve to

The observable rates are the squared amplitude ratios; the wave number is the same on both sides, so no velocity weighting is needed:

Writing ${\beta }^2$ in terms of energies, with the bound-state depth $E_b\equiv m{\alpha }^2/2{\hslash }^2=|E|$ from the previous section,

so the transmission has the compact form

The single scale $E_b$ governs both sectors: it is the binding energy for $E<0$ and the crossover energy of the transmission curve for $E>0$. At $E=E_b$ the beam splits evenly, $T=R=\frac{1}{2}$.

The bound state as a pole of the transmission amplitude

The two sectors, bound ($E<0$) and scattering ($E>0$), are values of a single analytic function. The transmission amplitude is the ratio $t(k)\equiv F/A=1/(1-i\beta )$. Writing $\beta =m\alpha /{\hslash }^{2}k=\kappa /k$ with $\kappa =m\alpha /{\hslash }^2$,

Continued to complex wave number, $t(k)$ has one pole, at $k=i\kappa$ on the positive imaginary axis. The energy there is

exactly the bound-state energy found from the jump condition. The bound state and the scattering data are not separate calculations; the bound state is the pole of the same amplitude whose modulus on the real axis gives the transmission probability.

The statement is general. For any short-range one-dimensional potential the transmission amplitude is a meromorphic function of $k$ whose poles on the positive imaginary axis are the bound states, while poles just below the real axis are resonances. This analytic bookkeeping is the one-dimensional seed of the S-matrix viewpoint used throughout scattering theory.

The barrier and the well–barrier asymmetry

Reversing the sign gives a repulsive delta barrier,

The jump condition changes sign, $\Delta {\psi }^{\prime }=+\frac{2m\alpha }{{\hslash }^2}\psi (0)$, and the scattering calculation runs identically with $\beta \to -\beta$ in the transmitted amplitude, $F=A/(1+i\beta )$. Because $R$ and $T$ depend only on ${\beta }^2$, the reflection and transmission probabilities are exactly the same as for the attractive well of equal strength:

The scattering cannot tell a well from a barrier of the same $|\alpha |$; only the phase of the transmitted wave differs. The bound-state sector is where the sign matters. A repulsive delta has $V\ge 0$ everywhere, so no $E<0$ solution can be normalizable, and the barrier binds nothing. The well and the barrier are scattering-equivalent yet spectrally distinct: one holds a single bound state, the other holds none.

The delta potential is the coarsest model of a localized interaction, and its two results generalize. The single bound state is the prototype for shallow bound states in nuclear and molecular physics, where the binding energy scales as the square of the coupling. The scattering result is the one-dimensional version of a point interaction, whose energy dependence $T\sim E/(E+E_b)$ mirrors the low-energy behavior of a three-dimensional short-range scatterer. Softening the point into a smooth barrier of finite width restores the exponential tunneling factor of the rectangular barrier.