Write L0ψ=−Uψ and use the outgoing inverse of L0. Adding the homogeneous incoming wave yieldsFor the attractive delta potential, let η=mα/(kℏ2). The equation reduces to ψ(x)=eikx+iηeik∣x∣ψ(0) and ψ(0)=1/(1−iη). Separating the two half-lines givesBoth sides have the same group velocity, soThe same amplitudes satisfy continuity and the exact delta-potential derivative jump ψ′(0+)−ψ′(0−)=−2mαψ(0)/ℏ2, providing an independent local check on the sign.
ψ=eikx−kℏ2im∫eik∣x−x′∣U(x′)ψ(x′)dx′.
t=1−iη1,r=1−iηiη,ψ={eikx+re−ikx,teikx,x<0,x>0.
T=1+η21,R=1+η2η2,T+R=1.
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