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At the delta potential, the bound equation becomes $\psi (x)=\beta \psi (0)e^{-\kappa |x|}/\kappa$. A nonzero solution therefore requires $\kappa =\beta$, givingThere is one bound level. The transmission amplitude $t(k)$ has its physical-sheet pole at $k=i\beta$, which produces exactly this decaying state. The real-axis probability's rational continuation $T(k)=k^2/(k^2+{\beta }^2)$ has both $+i\beta$ and $-i\beta$ poles; the lower pole is not an additional bound state. It is essential to distinguish the analytic amplitude from an even continuation of its modulus squared.

Away from the delta potential, the decaying bound state obeys $E_0=-{\hslash }^2{\kappa }^2/(2m)$. Integrating the Schrödinger equation across the attractive interaction givesFor an even exponential, the left side is $-2\kappa \phi (0)$. ThereforeThe normalized bound state in a delta potential is ${\phi }_0(x)=\sqrt{\kappa }{e}^{-\kappa |x|}$, since $2\kappa {\int }_0^{\infty }{e}^{-2\kappa x}dx=1$. The single positive solution for $\kappa$ gives one bound state; an odd decaying wave cannot be both continuous and nonzero with this interaction.

Choose the even-channel scattering phase shift convention ${\phi }_{k,e}(x)={\pi }^{-1/2}\cos (k|x|+{\delta }_e)$ for $k>0$. The delta potential derivative jump givesso, choosing the branch that tends to zero as $k\to \infty$,An odd scattering eigenfunction is ${\phi }_{k,o}(x)={\pi }^{-1/2}\sin kx$: it vanishes at the interaction and its derivative is continuous, so it acquires no scattering phase shift. These parity channels, with the bound state, give the continuum part of the spectral resolution of a quantum propagator.

A point interaction concentrates a coefficient of a linear differential equation at a single interior point. On an interval $0<x<\ell$, with $0<c<\ell$, consider $L_{\alpha }y=y^{\prime \prime }+\alpha y(c){\delta }_c$. The domain consists of continuous functions that are twice differentiable away from $c$, with the required one-sided derivatives. Multiplication of the Dirac delta by $y$ means the unambiguous point value $y(c){\delta }_c$. It does not mean an undefined product of a delta with an arbitrary discontinuous function.

The equation is interpreted using distributional derivatives. Smooth forcing gives an ordinary equation on each side, with transmission conditions for a delta coefficient at the interaction. A delta potential in the one-dimensional Schrödinger equation is a physical example after choosing the corresponding sign and constants.

For $V(x)=g\delta (x-x_0)$, put $q=\sqrt{2mE}/\hslash >0$ and $\gamma =mg/({\hslash }^{2}q)$. Relate the coefficients of $A{e}^{iqx}+B{e}^{-iqx}$ immediately to the right and left by $v_R=D(x_0)v_L$. Continuity of the wavefunction and integration of the Schrödinger equation across the delta potential give ${\psi }^{\prime }(x_0^{+})-{\psi }^{\prime }(x_0^{-})=2mg\psi (x_0)/{\hslash }^2$. Solving these two matching equations givesFor scatterers at $x_1<\cdots <x_N$, the full transfer matrix is $M=D(x_N)\cdots D(x_1)$: the leftmost scatterer acts first on the coefficient column. With these globally phased plane waves, the propagation phases are already in each $D(x_j)$. For incidence only from the left, $t=1/M_{22}$ and $r=-M_{21}/M_{22}$ follow from $(t,0{)}^T=M(1,r{)}^T$ and $\det M=1$.

For a finite jump in a one-dimensional potential energy, with constant mass and no delta potential, both the wavefunction and its first spatial derivative are continuous. Integrating the time-independent Schrödinger equation across a vanishing interval makes the jump in ${\psi }^{\prime }$ vanish, since the integral of $(V-E)\psi$ vanishes. A jump in $\psi$ itself would produce a derivative of a Dirac delta function in ${\psi }^{\prime \prime }$, with no matching term in the equation. The second derivative can jump. This matching allows transmission and quantum tunnelling; a finite step is not a hard wall.