Delta Potential Scattering

The one-dimensional delta interaction is exactly solvable, but it is not merely an elementary boundary-matching exercise. In one compact model it displays:

• reflection and transmission with nontrivial phases;
• a two-channel unitary scattering matrix;
• parity-channel diagonalization;
• analytic continuation in complex momentum;
• a bound-state pole for attraction and a non-normalizable lower-axis pole for repulsion.

Required background. Scattering from a delta potential derives the matching conditions and amplitudes. Here those amplitudes reveal parity phases, pole positions, and bound-state normalization; one-dimensional scattering explains the two-lead convention.

A Point Interaction

Consider

$$H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \lambda\delta(x),$$

with real $\lambda$. Define

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

The parameter $\gamma$ has dimensions of inverse length:

• $\gamma\gt0$ is a repulsive point barrier;
• $\gamma\lt0$ is an attractive point well.

For an incident energy $E\gt0$, let

$$k = \frac{\sqrt{2mE}}{\hbar} \gt0.$$

The goals are to derive the exact amplitudes, assemble the two-lead $S$ matrix, identify its parity phases, continue it to complex $k$, and recover the attractive bound state from its pole.

Matching Data

Away from the origin the particle is free. Integrating the Schrödinger equation across $x=0$ gives

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

and

$$\psi'(0^+)-\psi'(0^-) = 2\gamma\psi(0).$$

The wavefunction is continuous; its derivative is not. These two relations contain the entire point interaction.

For incidence from the left, these conditions give

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

The matching derivation explains each step. The same amplitudes apply for incidence from the right by parity symmetry.

The identity

$$t-r=1$$

is a useful algebraic check. It is also the statement that the odd parity channel, which vanishes at the origin, is unaffected by the interaction after the free-channel convention is removed.

Flux Check

The asymptotic wave number is $k$ on both sides, so the current ratios are

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

For real $k$ and $\gamma$,

$$\begin{aligned} R &= \frac{ \gamma^2 }{ k^2+\gamma^2 }, \\ T &= \frac{ k^2 }{ k^2+\gamma^2 }. \end{aligned}$$

Hence

$$R+T=1.$$

The stronger amplitude-level relation is

$$r^*t+t^*r=0.$$

Thus the reflected and transmitted amplitudes have the relative phase required by unitarity, not merely probabilities that happen to add to one.

Attraction and Repulsion at the Same Energy

Write

$$\kappa=\lvert\gamma\rvert,$$

and choose

$$k=2\kappa.$$

For the repulsive interaction, $\gamma=+\kappa$:

$$\begin{aligned} t_+ &= 0.8-0.4i, \\ r_+ &= -0.2-0.4i. \end{aligned}$$

For the attractive interaction, $\gamma=-\kappa$:

$$\begin{aligned} t_- &= 0.8+0.4i, \\ r_- &= -0.2+0.4i. \end{aligned}$$

Both signs give

$$T=0.8, \qquad R=0.2,$$

but their amplitudes are complex conjugates. Equal probabilities do not mean equal scattering data.

The transmission phases are

$$\arg t_\pm = \mp \arctan\left(\frac12\right) = \mp0.463648\ldots.$$

That sign can be observed through interference with a reference path or through composition with a second scatterer.

Top: the repulsive example at $k=2\gamma$ has $t=0.8-0.4i$ and $r=-0.2-0.4i$; the barrier label denotes the reduced potential $(m/\hbar^2)V=\gamma\delta(x)$. Middle: changing the sign of the interaction exchanges the pole and zero across the real $k$ axis; only the attractive upper-axis pole is a normalizable bound state. Bottom: the reduced even-channel eigenvalue stays on the unit circle, traversing conjugate arcs for equal-magnitude attraction and repulsion.

The Two-Lead Scattering Matrix

At fixed $k$, define incoming amplitudes by

$$\mathbf a_{\mathrm{in}} = \begin{pmatrix} a_L\\ a_R \end{pmatrix},$$

where $a_L$ enters from $x=-\infty$ and $a_R$ enters from $x=+\infty$. Define outgoing amplitudes by

$$\mathbf b_{\mathrm{out}} = \begin{pmatrix} b_L\\ b_R \end{pmatrix},$$

where $b_L$ leaves toward $x=-\infty$ and $b_R$ leaves toward $x=+\infty$.

Parity and time-reversal symmetry give

$$\mathbf b_{\mathrm{out}} = S_{\mathrm{lead}} \mathbf a_{\mathrm{in}},$$

with

$$S_{\mathrm{lead}}(k) = \begin{pmatrix} r&t\\ t&r \end{pmatrix}.$$

For zero interaction,

$$S_{\mathrm{lead}}^{(0)} = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}.$$

This is not the identity: a freely incoming wave from the left exits through the right lead. The lead basis labels spatial sides rather than propagation directions.

Unitarity

Direct multiplication gives

$$S_{\mathrm{lead}}^\dagger S_{\mathrm{lead}} = I$$

because

$$\lvert r\rvert^2+\lvert t\rvert^2=1, \qquad r^*t+t^*r=0.$$

The matrix is also symmetric,

$$S_{\mathrm{lead}}^T=S_{\mathrm{lead}},$$

as required by reciprocity for this real, time-reversal-invariant problem.

Remove the Free Lead Exchange

To compare phases with a convention in which free scattering is the identity, define

$$S_{\mathrm{red}} \equiv S_{\mathrm{lead}}^{(0)} S_{\mathrm{lead}}.$$

Since both matrices are symmetric under left–right exchange,

$$S_{\mathrm{red}} = \begin{pmatrix} t&r\\ r&t \end{pmatrix}.$$

Now

$$\lambda=0 \quad\Longrightarrow\quad S_{\mathrm{red}}=I.$$

This reduced matrix differs from the lead matrix only by a fixed free-channel convention. Probabilities are unchanged, but its eigenphases can be compared directly with the usual statement that a free phase shift vanishes.

Parity Channels

The normalized even and odd channel vectors are

$$\mathbf v_{\mathrm e} = \frac1{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad \mathbf v_{\mathrm o} = \frac1{\sqrt2} \begin{pmatrix} 1\\ -1 \end{pmatrix}.$$

They diagonalize $S_{\mathrm{red}}$:

$$\begin{aligned} S_{\mathrm e}(k) &= t+r = \frac{ k-i\gamma }{ k+i\gamma }, \\ S_{\mathrm o}(k) &= t-r = 1. \end{aligned}$$

The delta interaction changes only the even channel. For real $k$,

$$\lvert S_{\mathrm e}\rvert = \lvert S_{\mathrm o}\rvert = 1.$$

Write

$$S_{\mathrm e}=e^{2i\delta_{\mathrm e}}, \qquad S_{\mathrm o}=e^{2i\delta_{\mathrm o}}.$$

A continuous branch at positive $k$ can be chosen as

$$\delta_{\mathrm e}(k) = -\arctan \left( \frac{\gamma}{k} \right), \qquad \delta_{\mathrm o}(k)=0,$$

modulo integer multiples of $\pi$. At $k=2\kappa$,

$$S_{\mathrm e}^{(\pm)} = 0.6\mp0.8i,$$

where the upper sign denotes repulsion. These points lie on conjugate halves of the unit circle.

In the un-reduced lead matrix, the odd eigenvalue is $r-t=-1$. That minus sign is already present for a free particle and should not be mistaken for an interaction-induced phase.

Analytic Continuation in Momentum

For complex $k$, the amplitudes are meromorphic:

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

The interacting parity eigenvalue is

$$S_{\mathrm e}(k) = \frac{k-i\gamma}{k+i\gamma}.$$

Its pole is at

$$k_{\mathrm p} = -i\gamma,$$

and its zero is at the reflected point

$$k_{\mathrm z} = i\gamma.$$

For a real potential, this pole–zero pairing is the analytic continuation of real-axis unitarity.

The $k$ plane is especially convenient because

$$E = \frac{\hbar^2k^2}{2m}$$

maps the upper and lower halves of the $k$ plane onto the two sheets of the energy plane. They meet at the threshold branch point $E=0$. A pole on the positive imaginary $k$ axis has negative real energy and a decaying spatial wavefunction.

Attractive Pole and Bound State

Let

$$\gamma=-\kappa, \qquad \kappa\gt0.$$

Then

$$t(k) = \frac{k}{k-i\kappa},$$

and the pole is at

$$k=i\kappa.$$

At this value, the outgoing wave on each side becomes a decaying exponential:

$$\psi_{\mathrm b}(x) = C e^{-\kappa\lvert x\rvert}.$$

The jump condition gives

$$-2\kappa C = 2\gamma C,$$

which is satisfied precisely because $\gamma=-\kappa$.

Normalization fixes

$$1 = \int_{-\infty}^{\infty} \lvert\psi_{\mathrm b}(x)\rvert^2\,dx = \frac{\lvert C\rvert^2}{\kappa},$$

so one may choose

$$C=\sqrt\kappa.$$

The pole energy is

$$\begin{aligned} E_{\mathrm b} &= \frac{\hbar^2(i\kappa)^2}{2m} \\ &= -\frac{\hbar^2\kappa^2}{2m} \\ &= -\frac{m\lambda^2}{2\hbar^2}. \end{aligned}$$

This is the unique bound state of the attractive delta potential.

Pole Residue and Tail Normalization

Near $k=i\kappa$,

$$\begin{aligned} t(k) &= \frac{k}{k-i\kappa}, \\ r(k) &= \frac{i\kappa}{k-i\kappa}. \end{aligned}$$

Their residues are

$$\operatorname*{Res}_{k=i\kappa}t = i\kappa, \qquad \operatorname*{Res}_{k=i\kappa}r = i\kappa.$$

Consequently,

$$\operatorname*{Res}_{k=i\kappa} S_{\mathrm e}(k) = 2i\kappa.$$

Since the normalized bound-state tail has $C^2=\kappa$, this model obeys

$$\operatorname*{Res}_{k=i\kappa} S_{\mathrm e}(k) = 2iC^2.$$

The exact numerical factor depends on channel and state normalization conventions, but the structural lesson is general: a bound-state pole carries information not only about its energy but also about the normalization of its asymptotic tail.

Repulsive Lower-Axis Pole

For $\gamma=+\kappa$, the pole is instead

$$k=-i\kappa.$$

The corresponding outgoing continuation grows as

$$e^{+\kappa\lvert x\rvert}$$

and is not square-integrable. It is a lower-imaginary-axis virtual or antibound continuation, not an eigenvector of the self-adjoint Hamiltonian.

The repulsive interaction has a zero at $k=i\kappa$, but an $S$-matrix zero is not a bound state. Poles and zeros exchange when the sign of $\lambda$ is reversed:

$$S_{\mathrm e}(k;-\gamma) = \frac{1}{ S_{\mathrm e}(k;\gamma) }.$$

This identity explains why the real-axis phases are conjugate while the probabilities agree.

No Resonance Pole

A resonance in one-dimensional short-range scattering would appear at complex $k$ with

$$\operatorname{Re}k\ne0, \qquad \operatorname{Im}k\lt0.$$

The single delta interaction has only the purely imaginary pole $k=-i\gamma$. It can support one bound state when attractive, but it has no finite-width quasi-bound region and therefore no off-axis resonance pair.

Two separated point interactions are different: propagation between them supplies an additional phase and can create resonance structure. That model belongs to Double Delta Potential.

High-Energy Expansion and Born Check

For

$$\frac{\lvert\gamma\rvert}{k}\ll1,$$

the exact amplitudes expand as

$$\begin{aligned} t(k) &= 1 - i\frac{\gamma}{k} - \frac{\gamma^2}{k^2} + O\left( \frac{\gamma^3}{k^3} \right), \\ r(k) &= -i\frac{\gamma}{k} - \frac{\gamma^2}{k^2} + O\left( \frac{\gamma^3}{k^3} \right). \end{aligned}$$

The leading reflected amplitude,

$$r^{(1)} = -i\frac{\gamma}{k},$$

is the one-dimensional first-Born result for a contact interaction in this normalization. The exact denominator $1+i\gamma/k$ resums repeated interactions at the point.

At low energy, $\lvert\gamma\rvert/k\gg1$, this expansion fails even if $\lvert\lambda\rvert$ is numerically small in dimensional units. The correct dimensionless control parameter is $\lvert\gamma\rvert/k$, not $\lambda$ by itself.

Limiting Checks

Zero coupling

As $\gamma\to0$,

$$t\to1, \qquad r\to0, \qquad S_{\mathrm{red}}\to I.$$

High energy

As $k\to\infty$ at fixed $\gamma$,

$$T\to1, \qquad R\sim\frac{\gamma^2}{k^2}.$$

Low energy

As $k\to0^+$ at nonzero fixed $\gamma$,

$$t\to0, \qquad r\to-1.$$

The point interaction becomes perfectly reflecting at threshold. The approach to that limit still carries sign-dependent phase information.

Sign reversal

For real $k$,

$$\begin{aligned} t(k;-\gamma)&=t(k;\gamma)^*, \\ r(k;-\gamma)&=r(k;\gamma)^*. \end{aligned}$$

This proves equality of $R$ and $T$ and conjugacy of the full amplitudes.

Validity and Scope

The delta potential is not an approximation once it is defined as a self-adjoint point interaction with the stated matching condition. Its use as a model of a finite-range interaction is an approximation whose validity requires wavelengths long compared with the physical range and no sensitivity to omitted effective-range structure.

The formulas also assume:

• a real coupling and hence a Hermitian Hamiltonian;
• equal free thresholds and masses on both sides;
• one point interaction at the origin;
• the standard continuous-wavefunction delta interaction rather than a more general point-interaction boundary condition.

A complex $\lambda$ would model absorption or gain and destroy unitarity. A derivative-delta interaction or a general $U(2)$ point interaction can make the wavefunction itself discontinuous and has different matching data.

Exercises

1. Re-derive the amplitudes

Starting from continuity and the derivative jump, derive $r(k)$ and $t(k)$ without using the formulas above.

Solution

Continuity gives

$$t=1+r.$$

The jump condition is

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

Substitution gives

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

or

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

Then

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

2. Diagonalize both S-matrix conventions

Find the parity eigenvalues of $S_{\mathrm{lead}}$ and $S_{\mathrm{red}}$. Explain the odd-channel sign difference.

Solution

For the even vector $(1,1)^T/\sqrt2$,

$$S_{\mathrm{lead}} \mathbf v_{\mathrm e} = (r+t) \mathbf v_{\mathrm e}.$$

For the odd vector $(1,-1)^T/\sqrt2$,

$$S_{\mathrm{lead}} \mathbf v_{\mathrm o} = (r-t) \mathbf v_{\mathrm o} = -\mathbf v_{\mathrm o}.$$

Thus the lead eigenvalues are $r+t$ and $-1$. Multiplication by the free swap matrix leaves the even eigenvalue unchanged and reverses the odd one:

$$S_{\mathrm e}^{\mathrm{red}} = r+t, \qquad S_{\mathrm o}^{\mathrm{red}} = t-r = 1.$$

The lead-basis minus sign is the eigenvalue of free left–right exchange in the odd channel, not scattering from the delta interaction.

3. Verify the k = 2κ calculation

For $\gamma=\pm\kappa$ and $k=2\kappa$, the two signs give the same $T=0.8$ and $R=0.2$. Can a transmission experiment with a reference path distinguish them? Prepare equal coherent amplitudes $1/\sqrt2$ in two paths. The first contains the delta scatterer; the second acquires a controllable phase $e^{i\phi}$. Route the reflected component into a separate orthogonal output so that it never returns to the scatterer or recombiner. Recombine the two forward paths with

$$B=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}.$$

Find the two forward output probabilities and the separate reflection probability per original incoming particle at $\phi=\pi/2$, for both signs of $\gamma$. Verify normalization.

Solution

The amplitudes are $t=0.8\mp0.4i$ and $r=-0.2\mp0.4i$, with the upper sign for repulsion. The forward outputs are

$$\begin{pmatrix}A_+\\A_-\end{pmatrix} =B\frac1{\sqrt2}\begin{pmatrix}t\\e^{i\phi}\end{pmatrix} =\frac12\begin{pmatrix}t+e^{i\phi}\\t-e^{i\phi}\end{pmatrix}, \qquad P_\pm=\frac14|t\pm e^{i\phi}|^2.$$

At $\phi=\pi/2$ this gives

$$\begin{aligned} \gamma=+\kappa:&\quad (P_+,P_-)=(0.25,0.65),\\ \gamma=-\kappa:&\quad (P_+,P_-)=(0.65,0.25). \end{aligned}$$

Only half the initial probability enters the scattering path, so the separate reflection probability is $R/2=0.10$. Hence

$$P_++P_-+\frac R2 =\frac{1+T}{2}+\frac R2=1.$$

The output imbalance detects the sign-dependent phase of $t$ even though $T$ is unchanged. Probabilities conditioned on reaching a forward output would instead be divided by $P_++P_-=0.90$; they are different from the absolute probabilities calculated here.

4. Recover the bound state from the pole

For $\lambda\lt0$, show that the pole condition gives the normalized bound state and its energy.

Solution

Write $\gamma=-\kappa$ with $\kappa\gt0$. The denominator vanishes at

$$k=i\kappa.$$

The decaying solution is

$$\psi(x)=C e^{-\kappa\lvert x\rvert}.$$

Its derivative jump is $-2\kappa C$, equal to $2\gamma C$ because $\gamma=-\kappa$. Normalization gives

$$1 = 2\lvert C\rvert^2 \int_0^\infty e^{-2\kappa x}\,dx = \frac{\lvert C\rvert^2}{\kappa},$$

so $C=\sqrt\kappa$ up to a phase. The energy is

$$E_{\mathrm b} = -\frac{\hbar^2\kappa^2}{2m} = -\frac{m\lambda^2}{2\hbar^2}.$$

5. Relate the pole residue to the tail

Compute the residue of $S_{\mathrm e}$ at the attractive pole and express it in terms of the bound-state tail coefficient $C$.

Solution

For $\gamma=-\kappa$,

$$S_{\mathrm e}(k) = \frac{k+i\kappa}{k-i\kappa}.$$

Therefore

$$\operatorname*{Res}_{k=i\kappa} S_{\mathrm e} = \left. \left( k+i\kappa \right) \right|_{k=i\kappa} = 2i\kappa.$$

Since $C=\sqrt\kappa$,

$$\operatorname*{Res}_{k=i\kappa} S_{\mathrm e} = 2iC^2.$$

6. Identify the Born regime

Expand the exact reflected amplitude through second order in $\gamma/k$. At what energy scale is the first term reliable?

Solution

Using

$$\frac1{1+iz} = 1-iz-z^2+O(z^3),$$

with $z=\gamma/k$,

$$r = -i\frac{\gamma}{k} - \frac{\gamma^2}{k^2} + O\left( \frac{\gamma^3}{k^3} \right).$$

The first term is reliable when

$$\frac{\lvert\gamma\rvert}{k}\ll1,$$

or

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

The same scale is the magnitude of the attractive bound-state energy, which is a useful independent dimensional check.

References

• S. Albeverio, F. Gesztesy, R. Høegh-Krohn, and H. Holden, Solvable Models in Quantum Mechanics, 2nd ed., AMS Chelsea (2005).
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press (2018).
• C. J. Joachain, Quantum Collision Theory, 3rd ed., North-Holland (1983).
• R. G. Newton, Scattering Theory of Waves and Particles, 2nd ed., Dover (2002).
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover (2006).