Delta-Function Potential as Self-Adjoint Extension

A delta potential is a self-adjoint realization of a differential expression with a matching condition, rather than multiplication by an ordinary function. Its resolvent differs from the free resolvent by a rank-one operator. That formula connects the derivative jump, the attractive bound state, and the limit of an impenetrable barrier. Delta-Function Potential develops the elementary model solution; this page develops its operator realization.

Required background. Quadratic Form Methods constructs the point interaction and derives its domain. Free-Particle Hamiltonian supplies the Green function used below.

A point interaction is encoded in a derivative jump

Let $\alpha=\hbar^2/(2m)>0$ and let $g\in\mathbb R$ have units of energy times length. On $L^2(\mathbb R)$ the closed form

$$q_g[\psi]=\alpha\|\psi'\|^2+g|\psi(0)|^2, \qquad Q=H^1(\mathbb R),$$

determines the self-adjoint operator $H_g$. The trace estimate and form construction are proved in the required background. Restoring $\alpha$, its domain is

$$\begin{aligned} D(H_g)=\{\psi\in H^1(\mathbb R):\;& \psi|_{\mathbb R_-},\psi|_{\mathbb R_+}\in H^2,\\ &\psi'(0+)-\psi'(0-)=\frac g\alpha\psi(0)\}. \end{aligned}$$

The action on each open half-line is $-\alpha\psi''$. Global $H^1$ membership imposes continuity of $\psi$. The derivative may jump.

This domain also explains the formal distributional notation. For a continuous piecewise $H^2$ function,

$$\psi''=\psi''_{\mathrm{pieces}}+ \bigl[\psi'(0+)-\psi'(0-)\bigr]\delta_0.$$

Thus the singular terms in $-\alpha\psi''+g\psi(0)\delta_0$ cancel exactly on $D(H_g)$, leaving an $L^2$ vector. Applying the formal expression to arbitrary $H^2$ vectors would not produce the interacting operator domain.

The punctured free differential operator admits other self-adjoint extensions, including ones allowing discontinuous functions. The continuous matching condition above selects the delta family; it is not a classification of all point interactions.

The resolvent differs from the free one by rank one

Write $R_g(z)=(H_g-z)^{-1}$. For $z\notin[0,\infty)$ choose $k$ with $z=\alpha k^2$ and $\operatorname{Im}k>0$. The free resolvent kernel is

$$G_z(x)=\frac{i}{2\alpha k}e^{ik|x|}.$$

The distributional equation for $\psi=R_g(z)f$ gives

$$\psi=R_0(z)f-g\psi(0)G_z.$$

Evaluation at zero is legitimate because $R_0(z)f\in H^2$ and $G_z\in H^1$. If $1+gG_z(0)\ne0$, it yields

$$\psi(0)=\frac{(R_0(z)f)(0)}{1+gG_z(0)}.$$

Consequently

$$R_g(z;x,y)=G_z(x-y)- \frac{gG_z(x)G_z(-y)}{1+gG_z(0)}.$$

The second term is bounded and rank one: $G_z$ is an $L^2$ function, and integration against $G_z(-y)$ is a bounded functional. No claim that the delta distribution itself is a bounded Hilbert-space perturbation is required. The formula can conversely be checked against the matching domain and the resolvent equation, establishing it without a formal operator sum.

The negative-energy pole gives the unique bound state

At $z=-\alpha\kappa^2$, $\kappa>0$, one has $G_z(0)=1/(2\alpha\kappa)$. The denominator vanishes precisely when

$$\kappa=-\frac{g}{2\alpha}>0.$$

Thus an attractive interaction $g<0$ has one bound state,

$$E_b=-\frac{g^2}{4\alpha},\qquad \psi_b(x)=\sqrt{\kappa}\,e^{-\kappa|x|}.$$

The normalization uses $\int e^{-2\kappa|x|}dx=1/\kappa$. The resolvent denominator has only this negative zero, so there are no additional negative eigenvalues. For $g\geq0$, the nonnegative form also excludes negative energies directly.

The finite-rank resolvent difference preserves the essential spectrum, which is $[0,\infty)$. Square-integrability excludes positive-energy piecewise oscillatory eigenfunctions and the zero-energy piecewise affine solution. The explicit one-dimensional spectral analysis gives absolutely continuous spectrum on $[0,\infty)$; there is no singular-continuous part.

For comparison, the standard left-incident scattering amplitudes are

$$r(k)=\frac{g}{2i\alpha k-g},\qquad t(k)=\frac{2i\alpha k}{2i\alpha k-g},\qquad k>0.$$

Their matching derivation belongs to Scattering from a Delta Potential. They satisfy $|r|^2+|t|^2=1$, and their upper-half-plane pole for $g<0$ is the same bound-state pole found from the resolvent.

Exercise: an infinite repulsive barrier separates the line

Use the resolvent formula to identify the limit $g\to+\infty$ at a fixed nonreal $z$.

Solution

The coefficient $g/(1+gG_z(0))$ tends to $1/G_z(0)$. Because the remaining rank-one operator is fixed and bounded, the convergence is in operator norm. The limiting kernel is

$$G_z(x-y)-\frac{G_z(x)G_z(-y)}{G_z(0)}.$$

For $x$ and $y$ on opposite sides of zero the two terms coincide and cancel. On either half-line the remaining expression is the method-of-images Dirichlet Green function. The limit is therefore the direct sum of the two half-line Dirichlet Hamiltonians. This is norm-resolvent convergence to decoupled dynamics, not convergence of $H_g$ as bounded operators.

References

• Albeverio, Sergio, Friedrich Gesztesy, Raphael Høegh-Krohn, and Helge Holden, Solvable Models in Quantum Mechanics, 2nd ed., AMS Chelsea Publishing, 2005, publisher edition — one-dimensional point interactions and their resolvents.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text, Chapters 6 and 9 — form perturbations and one-dimensional spectral theory.