Delta potentials and derivative discontinuities

Write $V(x)=-g\delta(x)$ with $g>0$. Since a delta distribution has inverse-length units, $g$ has units of energy times length. Integrate the stationary Schrödinger equation over a symmetric interval around zero. The integral of the bounded energy term tends to zero as the interval shrinks, while the contact term gives $-g\psi(0)$. The kinetic second derivative contributes the difference between the two one-sided slopes, fixing a finite jump with a definite sign.

The wavefunction itself remains continuous in this delta model. Allowing an arbitrary value jump would generate a derivative of a delta distribution in its second derivative, which the chosen interaction does not contain. This statement is a property of this particular point interaction, not every conceivable singular Hamiltonian. One may define other contact models by other boundary conditions, but they must be specified explicitly instead of being silently inserted into the delta-potential calculation.

For $E<0$, normalizability requires a decaying exponential on either side. Continuity makes the two amplitudes equal, giving $\psi=Ae^{-\kappa|x|}$ with $\kappa>0$. Its right slope is $-\kappa A$ and left slope is $+\kappa A$. Inserting their difference into the jump condition determines one positive decay constant, hence one bound energy. The attractive delta therefore has exactly one bound state for every positive strength, without a minimum attraction threshold.

An odd bound state cannot be hidden in this solution. Odd continuity would make its value at the origin zero, so the delta would have no effect there. A decaying free solution on either half-line whose endpoint value vanishes must then have zero amplitude. A repulsive delta reverses the jump sign and would require a negative decay constant for this ansatz, contradicting normalizability. It has scattering states but no negative-energy bound state on the full line.

The cusp does not mean probability leaks at the interaction point. For this real stationary bound state the probability current is zero on both sides, and the real matching strength is compatible with unitary evolution. A naive calculation using only the free differential expression would miss the contact energy at the cusp. One must apply the distributional equation with its matching domain, or use the corresponding quadratic energy form that includes the term $-g|\psi(0)|^2$.

A narrow square well approaches this contact model when its width tends to zero while its integrated attraction stays fixed. Holding depth fixed instead makes its integrated attraction vanish, a different limit. The contact model suppresses information about finite interaction range. It can describe long-wavelength phenomena relative to that range, but it is not evidence that microscopic forces truly have infinite height or that all short-range interactions share the same spectrum at arbitrary energies.