Scattering from a Delta Potential

Scattering from a delta potential is the shortest exact example of a nontrivial one-dimensional scatterer. The particle is free everywhere except at one point, yet the point interaction changes the reflected and transmitted waves through a derivative jump condition.

Use

$$V(x)=\lambda\delta(x),$$

where $\lambda$ may be positive or negative. Positive $\lambda$ is a repulsive point barrier; negative $\lambda$ is an attractive point well. Define the inverse length scale

$$\gamma=\frac{m\lambda}{\hbar^2}.$$

The sign of $\gamma$ remembers whether the interaction is repulsive or attractive.

Matching Conditions

For $x\ne0$, the particle is free. The time-independent Schrödinger equation is

$$-\frac{\hbar^2}{2m}\psi''(x) +\lambda\delta(x)\psi(x) =E\psi(x).$$

The wavefunction is continuous,

$$\psi(0^-)=\psi(0^+)=\psi(0),$$

but its derivative jumps:

$$\psi'(0^+)-\psi'(0^-) = \frac{2m\lambda}{\hbar^2}\psi(0) =2\gamma\psi(0).$$

This jump condition is obtained by integrating the Schrödinger equation across a small interval containing the origin. It is the same rule used for the bound-state Delta-Function Potential.

Left-Incident Scattering State

Take $E\gt 0$ and define

$$k=\frac{\sqrt{2mE}}{\hbar}.$$

For a wave incident from the left,

$$\psi(x)= \begin{cases} e^{ikx}+r e^{-ikx}, & x\lt 0,\\ t e^{ikx}, & x\gt 0. \end{cases}$$

The continuity condition gives

$$1+r=t.$$

The derivative jump gives

$$ik t-ik(1-r)=2\gamma t.$$

Using $t=1+r$, this becomes

$$ik r=\gamma(1+r).$$

Solving,

$$r = \frac{\gamma}{ik-\gamma} = \frac{-i\gamma}{k+i\gamma},$$

and

$$t = \frac{ik}{ik-\gamma} = \frac{k}{k+i\gamma}.$$

These amplitudes contain both magnitude and phase information. The phase is physically relevant when point scatterers are combined or when wave packets interfere.

Reflection and Transmission

The left and right asymptotic potentials are equal, so the incident and transmitted wave numbers are both $k$. Therefore

$$R=\lvert r\rvert^2, \qquad T=\lvert t\rvert^2.$$

From the amplitudes,

$$R = \frac{\gamma^2}{k^2+\gamma^2}, \qquad T = \frac{k^2}{k^2+\gamma^2}.$$

Thus

$$R+T=1.$$

The result is a clean current-conservation check. A real delta potential redistributes the incident flux between reflected and transmitted channels but does not absorb probability.

Attractive Versus Repulsive Delta

The probabilities $R$ and $T$ depend on $\gamma^2$, so a repulsive delta barrier and an attractive delta well of the same integrated strength magnitude have the same reflection and transmission probabilities at a fixed energy.

The amplitudes themselves are not the same. Changing the sign of $\lambda$ changes the phases:

$$t(\gamma)=\frac{k}{k+i\gamma}, \qquad t(-\gamma)=\frac{k}{k-i\gamma}.$$

This phase distinction matters in interference problems. For example, two point interactions separated by a distance can produce energy-dependent transmission structure even when a single point interaction has the simple probability above.

Energy Limits

At high energy,

$$k\gg \lvert\gamma\rvert,$$

so

$$T\approx1, \qquad R\approx\frac{\gamma^2}{k^2}.$$

The particle has a short wavelength and is only weakly affected by the finite integrated strength.

At low energy,

$$k\ll \lvert\gamma\rvert,$$

so

$$T\approx\frac{k^2}{\gamma^2}, \qquad R\approx1.$$

In one dimension, even a point interaction reflects almost completely at sufficiently low energy, unless the interaction strength is zero.

Bound-State Pole Preview

The transmission amplitude is

$$t(k)=\frac{k}{k+i\gamma}.$$

Its pole occurs when

$$k+i\gamma=0, \qquad k=-i\gamma.$$

For an attractive delta potential, $\lambda\lt 0$, so $\gamma\lt 0$. Then

$$k=i\lvert\gamma\rvert$$

lies on the positive imaginary $k$ axis and corresponds to a normalizable bound state. The energy is

$$E = \frac{\hbar^2k^2}{2m} = -\frac{\hbar^2\gamma^2}{2m} = -\frac{m\lambda^2}{2\hbar^2}.$$

This is the same bound-state energy found directly on the Delta-Function Potential page. For a repulsive delta potential, $\gamma\gt 0$, the pole is on the negative imaginary $k$ axis and does not represent a square-integrable bound state.

The general relationship between bound states and scattering poles is a major theme of scattering theory; see Bound States and Scattering Poles for the broader picture. Delta Potential Scattering carries this exact result into a convention-explicit $S$-matrix audit, including parity phases, poles, zeros, and the bound-state residue.

Narrow-Barrier Limit

The delta potential can be obtained from a narrow rectangular barrier or well. Let a barrier of width $a$ and height $V_0$ shrink while its area remains fixed:

$$a\to0, \qquad V_0\to\infty, \qquad V_0a=\lambda.$$

The detailed two-interface structure collapses to a single matching condition at the origin. The finite-barrier transmission formula reduces to

$$T = \frac{1}{1+\gamma^2/k^2} = \frac{k^2}{k^2+\gamma^2}.$$

This limiting relation is useful conceptually: the point interaction is not “nothing.” It is the zero-width limit of a family with nonzero integrated strength.

Parity View

Because $\delta(x)$ is even, the scattering can also be analyzed in parity channels. Odd wavefunctions vanish at the origin:

$$\psi_{\mathrm o}(0)=0.$$

They do not feel the delta interaction. Even wavefunctions generally have $\psi_{\mathrm e}(0)\ne0$, so their derivatives acquire the jump. In this language, the point interaction modifies only the even channel.

The left-incident amplitudes above recombine the even and odd parity channels into traveling waves. This is a useful bridge to more advanced one-dimensional scattering language, where phase shifts rather than $r$ and $t$ may be emphasized.

Common Mistakes

• Requiring $\psi'$ to be continuous at the delta interaction.
• Forgetting that $\psi$ itself remains continuous for this standard point interaction.
• Treating $T$ as an interior amplitude instead of an outgoing-current ratio.
• Assuming an attractive and repulsive delta have the same scattering phases because they have the same $R$ and $T$.
• Missing the bound-state pole for $\lambda\lt 0$.
• Thinking a zero-width potential must have zero effect even when its integrated strength is fixed.

Where This Is Used

• Delta-Function Potential gives the bound-state version and derives the jump condition.
• Reflection and Transmission Coefficients supplies the current-based definitions of $R$ and $T$.
• Finite Potential Barrier is the finite-width model whose narrow limit gives a delta interaction.
• Distributional Derivatives explains why derivative jumps create delta terms.
• Double Delta Potential applies the same matching rule at two points.
• Bound States and Scattering Poles develops the pole interpretation.
• Delta Potential Scattering checks the same amplitudes in lead and parity-channel conventions.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.
• S. Flügge, Practical Quantum Mechanics, Springer, 1999.

Exercises

1. Derive the amplitudes $r$ and $t$ for left incidence.

Solution

Continuity gives

$$1+r=t.$$

The derivative jump gives

$$ik t-ik(1-r)=2\gamma t.$$

Using $t=1+r$,

$$2ik r=2\gamma(1+r),$$

so

$$(ik-\gamma)r=\gamma.$$

Therefore

$$r=\frac{\gamma}{ik-\gamma} =\frac{-i\gamma}{k+i\gamma}.$$

Then

$$t=1+r =\frac{k}{k+i\gamma}.$$

2. Show explicitly that $R+T=1$.

Solution

The probabilities are

$$R=\frac{\gamma^2}{k^2+\gamma^2}, \qquad T=\frac{k^2}{k^2+\gamma^2}.$$

Adding them gives

$$R+T = \frac{\gamma^2+k^2}{k^2+\gamma^2} =1.$$

3. Explain why the attractive delta has a bound-state pole but the repulsive delta does not.

Solution

The pole of

$$t(k)=\frac{k}{k+i\gamma}$$

is at

$$k=-i\gamma.$$

For an attractive delta, $\lambda\lt 0$ and $\gamma\lt 0$, so $k=i\lvert\gamma\rvert$. This gives a decaying bound-state wavefunction and energy

$$E=-\frac{\hbar^2\gamma^2}{2m}.$$

For a repulsive delta, $\gamma\gt 0$, so the pole is at $k=-i\gamma$, on the negative imaginary axis. The corresponding exponential would grow rather than decay in the normalizable bound-state construction, so it is not a bound state.

4. What happens to the transmission probability as $k\to0$ for fixed nonzero $\lambda$?

Solution

For fixed nonzero $\lambda$, $\gamma$ is fixed and nonzero. The transmission probability is

$$T=\frac{k^2}{k^2+\gamma^2}.$$

As $k\to0$,

$$T\to0, \qquad R\to1.$$

Thus the point interaction becomes perfectly reflecting in the zero-energy limit.