 Transfer matrix of a delta potential

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 $\det M=1$.