construction of the scattering states, $\psi_p(x)$ incident from the left, $\psi'_p(x)$ incident from the right (for $p>0$) \begin{eqnarray} \psi_p(x)=\left\{\begin{array}{cc} e^{ipx}+r(p)e^{-ipx},&x\rightarrow-\infty\\ te^{ipx},&x\rightarrow+\infty \end{array}\right. \end{eqnarray} \begin{eqnarray} \psi'_p(x)=\left\{\begin{array}{cc} e^{-ipx}+r'(p)e^{ipx},&x\rightarrow+\infty\\ t'(p)e^{-ipx},&x\rightarrow -\infty \end{array}\right. \end{eqnarray} <sub>I write $x\rightarrow\pm\infty$, since if the line has a nonzero extension in the transverse $y$-direction it is not enough to take $x$ outside of the support of $V$, because of evanescent waves: waves that decay exponentially into the region where $V=0$, but have not yet decayed to zero.</sub> the reflection coefficients $r(p),r'(p)$ and transmission coefficients $t(p),t'(p)$ define the scattering matrix: $$S(p)=\begin{pmatrix} r(p)&t'(p) \\ t(p)&r'(p) \end{pmatrix}$$ unitarity: $S(p)S^\dagger(p)=\mathbb{1}$, time-reversal symmetry: $S^t(p)=S(p)$, so $t(p)=t'(p)$. These constraints may be incorporated in the polar decomposition $$S=\begin{pmatrix} e^{2i\phi}\sqrt{1-T}&e^{i\phi+i\phi'}\sqrt{T} \\ e^{i\phi+i\phi'}\sqrt{T}&-e^{2i\phi'}\sqrt{1-T} \end{pmatrix},$$ with $\phi(p),\phi'(p)\in[0,2\pi)$ and $T(p)\in[0,1]$. There is no simple closed-form expression for $\phi,\phi',T$ for arbitrary $V(x)$, this will typically require a numerical solution.