Delta potential

A delta potential is a point interaction imposing continuity of the wavefunction and a jump in its derivative.

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 $H=p^2/(2m)-g\delta (x)$ with $g>0$, put $\kappa =mg/{\hslash }^2$ and $T>0$. The bound state in a delta potential has squared amplitude $\kappa$ at the origin. The even normalized continuum states have squared amplitude $k^2/[\pi (k^2+{\kappa }^2)]$ for $k>0$, and the odd states vanish there. The spectral resolution of a quantum propagator gives the displayed result. At long times, the continuum stationary point at $k=0$ is suppressed by its quadratic spectral weight: expanding that weight gives a continuum term $e^{-3i\pi /4}(2m/\hslash T{)}^{3/2}/(4\sqrt{\pi }{\kappa }^2)+O(T^{-5/2})$. The bound state term persists with constant modulus. This stationary phase expansion requires $\hslash {\kappa }^{2}T/(2m)\gg 1$.

Inside walls at $\pm a$, a central delta potential of strength $\kappa$ leaves odd parity operator eigenstates unchanged because their wavefunctions vanish at zero. For even states, continuity and the derivative jump ${\psi }^{\prime }(0+)-{\psi }^{\prime }(0-)=2m\kappa \psi (0)/{\hslash }^2$ give the displayed quantization equation. For $\kappa >0$, one root lies between $(n+1/2)\pi /a$ and $(n+1)\pi /a$ for every $n\ge 0$.

Put $c=m\kappa /{\hslash }^2$, $k_n=(n+1/2)\pi /a$, and ${\delta }_n=a({\lambda }_n-k_n)$. The quantization equation gives ${\delta }_n=\arctan (c/{\lambda }_n)$, so ${\lambda }_n{\delta }_n\to c$ and ${\delta }_n\to 0$. Therefore ${\lambda }_n^2-k_n^2\to 2c/a$, proving the displayed absolute energy shift. Its ratio to the unperturbed energy tends to zero. The normalized unperturbed even state has probability density $1/a$ at the origin, so its first-order delta-barrier expectation is also $\kappa /a$.

For equal delta potentials at $\pm a$, odd parity makes the interior wave proportional to $\sin (kx)$. The derivative jump gives the odd-channel denominator $k\cot (ka)+U_0-ik$. At $k=i\kappa$, its zero is the displayed equation. The function $2\kappa /(1-e^{-2\kappa a})$ increases strictly from $1/a$, proving exactly one odd bound state when $U_{0}a<-1$. For $U_{0}a=u\gg 1$, the outgoing pole near $ka=\pi$ is $\pi -\pi /u+(\pi -i{\pi }^2)/u^2+O(u^{-3})$, with positive resonance width.

An attractive one-dimensional delta potential has one even exponentially decaying bound state.

For delta interactions at points $x_j$, a Green-function solution reduces the Schrodinger equation to a finite linear system for the values $\psi (x_j)$.

The symmetric double-delta potentialseparates into even and odd parity channels. At either interaction, the wavefunction is continuous and its derivative has the jump

For $V(x)=g[\delta (x-a)+\delta (x+a)]$ with $a>0$ and $\nu =-mg/{\hslash }^2>0$, an even bound state obeys $\kappa =\nu (1+e^{-2a\kappa })$, and an odd bound state obeys $\kappa =\nu (1-e^{-2a\kappa })$, with energy $E=-{\hslash }^2{\kappa }^2/(2m)$ and $\kappa >0$. The even right-hand side is strictly decreasing from $2\nu$, so it meets $\kappa$ once. For the odd equation, $h(\kappa )=\nu (1-e^{-2a\kappa })-\kappa$ is strictly concave, starts at zero with slope $2a\nu -1$, and tends to negative infinity. It has one positive equation root precisely when $2a\nu >1$. Thus there is one bound state for $-{\hslash }^2/(2ma)\le g<0$, and two for $g<-{\hslash }^2/(2ma)$. For $g\ge 0$ positivity of the Hamiltonian operator excludes negative energy, and square integrability excludes nonnegative-energy exterior solutions. At $g=g_c$ the odd zero-energy solution is bounded but not square integrable: it is a zero-energy scattering resonance, not a second bound state.

For positive wavenumber $k$, the odd channel of the symmetric double-delta potential hasFor real $k$ its numerator and denominator are complex conjugates, making the channel unitary. Its poles are the zeros of $k{e}^{-ika}+U_0\sin (ka)$.

For $U_0<0$, an odd bound state has $k=i\kappa$ with $\kappa >0$ and obeysA nonzero positive solution exists exactly when $-U_{0}a>1$; equality is the zero-energy threshold.

For $U_{0}a\gg 1$, the lowest odd scattering resonance pole lies near the first odd mode trapped between the barriers:in its real part and to leading nonzero order in its imaginary part. Writing $E={\hslash }^2{k}^2/(2m)=E_R-i\Gamma /2$, the survival probability decays as $e^{-\Gamma t/\hslash }$.

For $V(x)=V_0\delta (x)$, $E={\hslash }^2{k}^2/(2m)$, and $\gamma =m{V}_0/({\hslash }^{2}k)$, continuity and the derivative jump giveThey obey $|r{|}^2+|t{|}^2=1$ and $r^{\ast }t+t^{\ast }r=0$.