---
title: "The Dirac-Delta Potential: A Single Bound State and Scattering"
module: Wave Mechanics in One Dimension
moduleNumber: 3
lessonNumber: 5
order: 305
summary: >
  A potential concentrated at a single point is solvable in closed form and isolates the
  physics of matching a wave function across a discontinuity. Integrating the Schrödinger
  equation across the spike gives a jump condition on the derivative; the attractive delta
  well then supports exactly one bound state, of energy set by the strength alone, while the
  same spike scatters an incoming beam with a transmission that rises from zero to one. The
  attractive well and the repulsive barrier scatter identically yet only the well binds.
topics: [Wave Mechanics in One Dimension]
draft: false
sources:
  - book: Griffiths & Schroeter
    ref: "Ch. 2 — Time-Independent Schrödinger Equation; §2.5 The Delta-Function Potential"
  - book: Shankar
    ref: "Ch. 5 — Simple Problems in One Dimension; §5.2 The Particle in a Box / singular potentials"
---

The [finite square well](/quantum-mechanics/wave-mechanics-1d/particle-in-infinite-and-finite-square-wells)
carries two length scales, its width and its depth, and the bound-state count depends on both
through the dimensionless product $R_0 = (a/\hbar)\sqrt{2mV_0}$. 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.[^gs-25]

## The delta well and the derivative-jump condition

Write the attractive potential as

$$
V(x) = -\alpha\,\delta(x),
\qquad \alpha > 0,
$$

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 $[-\epsilon, \epsilon]$ straddling the origin:

$$
-\frac{\hbar^2}{2m}\int_{-\epsilon}^{\epsilon}\frac{\d^2\psi}{\d x^2}\,\d x
\;-\;\alpha\int_{-\epsilon}^{\epsilon}\delta(x)\,\psi(x)\,\d x
\;=\; E\int_{-\epsilon}^{\epsilon}\psi(x)\,\d x.
$$

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

> **Theorem (Derivative-jump condition).** Across a potential $V(x) = -\alpha\,\delta(x)$ the
> wave function is continuous and its first derivative jumps by
> $$
> \Delta\psi' \equiv \psi'(0^+) - \psi'(0^-) = -\frac{2m\alpha}{\hbar^2}\,\psi(0).
> $$
> An attractive delta ($\alpha > 0$) bends the slope downward as $x$ crosses the origin,
> producing a downward-pointing cusp; a repulsive delta reverses the sign.

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.

$$
% caption: A finite well of shrinking width and growing depth at fixed area limits to a
% single spike; the bound wave function keeps its exponential tails but its rounded top
% sharpens into a cusp at the origin.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % axis
  \draw[black] (-3.4,0) -- (3.6,0) node[right, font=\scriptsize, text=black] {$x$};
  % delta spike drawn as downward arrow
  \draw[black, very thick, ->] (0,0) -- (0,-1.4);
  \node[black, font=\scriptsize, anchor=north east] at (0,-1.2) {well};
  % cusped bound state e^{-k|x|}
  \draw[acc, very thick] (-3.2,0.12) .. controls (-1.8,0.32) and (-0.8,1.2) .. (0,2.1);
  \draw[acc, very thick] (0,2.1) .. controls (0.8,1.2) and (1.8,0.32) .. (3.2,0.12);
  \node[acc, font=\scriptsize, anchor=south] at (0,2.1) {cusp};
  \node[acc, font=\scriptsize, anchor=west] at (2.2,0.35) {tail};
\end{tikzpicture}
$$

## The unique bound state

A bound state has $E < 0$. Set

$$
\kappa \equiv \frac{\sqrt{-2mE}}{\hbar} > 0,
$$

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

$$
\psi(x) =
\begin{cases}
B\,e^{\kappa x}, & x < 0, \\
B\,e^{-\kappa x}, & x > 0,
\end{cases}
$$

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

$$
-2\kappa B = -\frac{2m\alpha}{\hbar^2}\,B
\quad\Longrightarrow\quad
\kappa = \frac{m\alpha}{\hbar^2}.
$$

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

> **Theorem (The delta-well bound state).** The attractive delta $V = -\alpha\,\delta(x)$
> supports exactly one bound state, with
> $$
> \psi(x) = \sqrt{\kappa}\,e^{-\kappa|x|},
> \qquad \kappa = \frac{m\alpha}{\hbar^2},
> \qquad E = -\frac{\hbar^2\kappa^2}{2m} = -\frac{m\alpha^2}{2\hbar^2}.
> $$
> The energy is set by the strength $\alpha$ alone. There is one bound state for every
> $\alpha > 0$, no matter how weak, and never a second.

The normalization $B = \sqrt\kappa$ follows from
$\int_{-\infty}^{\infty}|\psi|^2\,\d x = |B|^2\int_{-\infty}^{\infty}e^{-2\kappa|x|}\,\d x
= |B|^2/\kappa = 1$. 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 $\psi$.

$$
% caption: The two exponential tails meet at the origin with slopes of equal magnitude and
% opposite sign; their difference is the derivative jump fixed by the delta strength.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (-3.2,0) -- (3.4,0) node[right, font=\scriptsize, text=black] {$x$};
  \draw[black, dashed] (0,0) -- (0,2.4);
  % left tail rising to peak
  \draw[acc, very thick] (-3.0,0.18) .. controls (-1.6,0.42) and (-0.7,1.35) .. (0,2.2);
  % right tail
  \draw[acc, very thick] (0,2.2) .. controls (0.7,1.35) and (1.6,0.42) .. (3.0,0.18);
  \fill[acc] (0,2.2) circle (2pt);
  % tangent slope indicators at origin
  \draw[black, thick] (-1.1,1.0) -- (0.2,2.65);
  \node[black, font=\scriptsize, anchor=east] at (-0.9,1.1) {slope up};
  \draw[black, thick] (1.1,1.0) -- (-0.2,2.65);
  \node[black, font=\scriptsize, anchor=west] at (0.9,1.1) {slope down};
\end{tikzpicture}
$$

> **Worked example.** Model a shallow surface trap as a delta well that binds an electron
> with decay length $1/\kappa = 1\ \text{nm}$, so $\kappa = 1\ \text{nm}^{-1}$. Using
> $\hbar c = 197.3\ \text{eV}\cdot\text{nm}$ and $m_e c^2 = 0.511\ \text{MeV}$, the binding
> energy is
> $$
> E = -\frac{\hbar^2\kappa^2}{2m_e} = -\frac{(\hbar c\,\kappa)^2}{2 m_e c^2}
> = -\frac{(197.3\ \text{eV})^2}{2(5.11\times10^5\ \text{eV})} = -0.038\ \text{eV},
> $$
> a shallow level of thermal order, produced by strength $\alpha = \hbar^2\kappa/m_e$. A beam
> sent in at the matching energy $E = E_b = 0.038\ \text{eV}$ transmits with
> $T = E/(E + E_b) = \tfrac12$: at its own binding energy the trap is exactly half-transparent.[^codata]

## Scattering off the delta

For $E > 0$ the particle is unbound and the delta acts as a scatterer.[^sh-54b] With

$$
k \equiv \frac{\sqrt{2mE}}{\hbar},
$$

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:

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

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

$$
ikF - ik(A - B) = -\frac{2m\alpha}{\hbar^2}\,F.
$$

Introduce the dimensionless strength

$$
\beta \equiv \frac{m\alpha}{\hbar^2 k},
$$

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

$$
F = \frac{A}{1 - i\beta},
\qquad
B = \frac{i\beta}{1 - i\beta}\,A.
$$

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

$$
R = \frac{|B|^2}{|A|^2} = \frac{\beta^2}{1 + \beta^2},
\qquad
T = \frac{|F|^2}{|A|^2} = \frac{1}{1 + \beta^2},
\qquad
R + T = 1.
$$

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

$$
\beta^2 = \frac{m^2\alpha^2}{\hbar^4 k^2} = \frac{m\alpha^2}{2\hbar^2 E} = \frac{E_b}{E},
$$

so the transmission has the compact form

$$
T(E) = \frac{1}{1 + E_b/E} = \frac{E}{E + E_b}.
$$

> **Theorem (Delta transmission).** A particle of energy $E > 0$ crossing a delta of strength
> $\alpha$ transmits with probability $T = E/(E + E_b)$, where $E_b = m\alpha^2/2\hbar^2$ is
> the magnitude of the well's bound-state energy. Transmission vanishes as $E \to 0$ and
> approaches unity as $E \to \infty$; a fast particle barely notices the spike.

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,
$T = R = \tfrac12$.

$$
% caption: Transmission through the delta rises from zero at threshold to one at high energy,
% crossing one half at the energy equal to the well's binding energy; reflection is the
% complementary curve.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (5.8,0) node[right, font=\scriptsize, text=black] {$E/E_b$};
  \draw[->, black] (0,0) -- (0,3.4) node[above, font=\scriptsize, text=black] {$R,\,T$};
  \draw[black, dashed] (0,3.0) -- (5.6,3.0);
  \node[black, font=\scriptsize, anchor=east] at (-0.05,3.0) {$1$};
  \draw[black, dashed] (0,1.5) -- (5.6,1.5);
  \node[black, font=\scriptsize, anchor=east] at (-0.05,1.5) {$\frac{1}{2}$};
  % E=Eb marker at x=1.2
  \draw[black, dashed] (1.2,0) -- (1.2,1.5);
  \node[black, font=\scriptsize, anchor=north] at (1.2,-0.05) {$1$};
  % T = E/(E+Eb): rises 0 -> 1
  \draw[acc, very thick] (0,0) .. controls (0.6,1.2) and (1.2,1.5) .. (2.0,2.1)
    .. controls (3.2,2.6) and (4.4,2.85) .. (5.6,2.92);
  \node[acc, font=\scriptsize, anchor=north west] at (3.4,2.6) {$T$};
  % R = Eb/(E+Eb): falls 1 -> 0
  \draw[black, very thick, dashed] (0,3.0) .. controls (0.6,1.8) and (1.2,1.5) .. (2.0,0.9)
    .. controls (3.2,0.4) and (4.4,0.15) .. (5.6,0.08);
  \node[black, font=\scriptsize, anchor=south west] at (3.4,0.35) {$R$};
\end{tikzpicture}
$$

$$
% caption: An incoming beam meets the point scatterer at the origin and divides into a
% reflected wave returning left and a transmitted wave continuing right, with the same
% wavelength on both sides since the potential is zero away from the point.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (-4.0,0) -- (4.2,0) node[right, font=\scriptsize, text=black] {$x$};
  \draw[black, very thick, ->] (0,0.1) -- (0,1.6);
  \node[black, font=\scriptsize, anchor=south] at (0,1.6) {scatterer};
  % incident arrow
  \draw[->, black, thick] (-3.6,-0.9) -- (-2.0,-0.9);
  \node[black, font=\scriptsize, anchor=north] at (-2.8,-0.95) {incident};
  % reflected arrow
  \draw[->, black, thick] (-2.0,-1.6) -- (-3.6,-1.6);
  \node[black, font=\scriptsize, anchor=north] at (-2.8,-1.65) {returned};
  % transmitted arrow
  \draw[->, black, thick] (2.0,-0.9) -- (3.6,-0.9);
  \node[black, font=\scriptsize, anchor=north] at (2.8,-0.95) {transmitted};
\end{tikzpicture}
$$

## 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
$\beta = m\alpha/\hbar^2 k = \kappa/k$ with $\kappa = m\alpha/\hbar^2$,

$$
t(k) = \frac{1}{1 - i\kappa/k} = \frac{k}{k - i\kappa}.
$$

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

$$
E = \frac{\hbar^2 k^2}{2m}\bigg|_{k = i\kappa} = -\frac{\hbar^2\kappa^2}{2m} = -E_b,
$$

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.

> **Theorem (Bound states as scattering poles).** The transmission amplitude
> $t(k) = k/(k - i\kappa)$, continued to complex $k$, has a pole at $k = i\kappa$ on the
> positive imaginary axis, sitting at $E = -\hbar^2\kappa^2/2m$. Every bound state of a
> short-range potential appears as such a pole of the scattering amplitude; reading off the
> poles recovers the bound spectrum from scattering alone.

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.

$$
% caption: Real wave numbers give scattering states with energy above zero; the single pole
% of the transmission amplitude sits on the positive imaginary axis and marks the one bound
% state below zero.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-3.2,0) -- (3.4,0) node[right, font=\scriptsize, text=black] {Re $k$};
  \draw[->, black] (0,-1.4) -- (0,3.0) node[above, font=\scriptsize, text=black] {Im $k$};
  % real axis scattering states (marks)
  \foreach \xx in {-2.6,-2.0,-1.4,1.4,2.0,2.6} {
    \fill[black] (\xx,0) circle (1.6pt);
  }
  \node[black, font=\scriptsize, anchor=north] at (2.0,-0.2) {scattering};
  % pole on positive imaginary axis
  \fill[acc] (0,1.8) circle (2.6pt);
  \draw[acc] (0,1.8) circle (4.5pt);
  \node[acc, font=\scriptsize, anchor=west] at (0.25,1.8) {pole: bound state};
\end{tikzpicture}
$$

## The barrier and the well–barrier asymmetry

Reversing the sign gives a repulsive **delta barrier**,

$$
V(x) = +\alpha\,\delta(x),
\qquad \alpha > 0.
$$

The jump condition changes sign, $\Delta\psi' = +\frac{2m\alpha}{\hbar^2}\psi(0)$, 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:

$$
T_{\text{barrier}}(E) = \frac{1}{1 + \beta^2} = T_{\text{well}}(E).
$$

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.

$$
% caption: The attractive well and the repulsive barrier of equal strength scatter with the
% same reflection and transmission, but only the downward spike supports a bound level below
% zero energy.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % well panel
  \draw[black] (-2.2,0) -- (2.2,0) node[right, font=\scriptsize, text=black] {$x$};
  \draw[black, very thick, ->] (0,0) -- (0,-1.5);
  \draw[acc, thick, dashed] (-1.6,-0.9) -- (1.6,-0.9);
  \node[acc, font=\scriptsize, anchor=south] at (-1.0,-0.85) {bound level};
  \node[align=center, font=\scriptsize, text=black] at (0,-2.1) {well: one bound state};
  % barrier panel
  \begin{scope}[xshift=6.0cm]
    \draw[black] (-2.2,0) -- (2.2,0) node[right, font=\scriptsize, text=black] {$x$};
    \draw[black, very thick, ->] (0,0) -- (0,1.5);
    \node[black, font=\scriptsize, anchor=west] at (0.1,0.9) {no bound state};
    \node[align=center, font=\scriptsize, text=black] at (0,-2.1) {barrier: none};
  \end{scope}
\end{tikzpicture}
$$

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](/quantum-mechanics/wave-mechanics-1d/barrier-penetration-and-quantum-tunneling).

[^gs-25]: **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_ 3rd ed., §2.5 — the delta-function well and barrier, the derivative-jump condition $\Delta\psi' = -(2m\alpha/\hbar^2)\psi(0)$, the single bound state $E = -m\alpha^2/2\hbar^2$, and the transmission $T = 1/(1+\beta^2)$ with the well–barrier scattering equivalence.
[^sh-54b]: **Shankar**, _Principles of Quantum Mechanics_ 2nd ed., Ch. 5 — matching solutions across singular one-dimensional potentials and the bound-state / scattering decomposition of the spectrum.
[^codata]: Constants from CODATA/NIST: $\hbar c = 197.3\ \text{eV}\cdot\text{nm}$ and $m_e c^2 = 0.5110\ \text{MeV}$. NIST, _CODATA Recommended Values of the Fundamental Physical Constants_, https://physics.nist.gov/cuu/Constants/.