The purpose of this post is to prove a simple but somewhat unintuitive result in nonrelativistic \(1\)D scattering of quantum particles. Specifically, consider a potential energy landscape \(V(x)\) that decays asymptotically to \(\lim_{|x|\to\infty}V(x)=0\) at \(|x|\to\infty\) which may be asymmetric so that \(V(x)\) does not necessarily coincide with \(V(-x)\). As a result of such an asymmetry, one might expect that \(V(x)\) would scatter incident quantum particles differently depending on whether such quantum particles are incident on \(V(x)\) from the left \(x\to-\infty\) or from the right \(x\to\infty\). Nonetheless, it is an unintuitive but important result that actually the transmission amplitudes \(t_{\rightarrow}=t_{\leftarrow}\) are exactly the same while the reflection amplitudes \(r_{\rightarrow}=-r_{\leftarrow}^*t_{\rightarrow}/t^*_{\leftarrow}\) are congruent modulo a \(U(1)\) phase. In particular, one has the corollary that the left-incident and right-incident transmission and reflection probabilities are identical \(|t_{\rightarrow}|^2=|t_{\leftarrow}|^2\), \(|r_{\rightarrow}|^2=|r_{\leftarrow}|^2\) regardless of the precise nature of \(V(x)\), so long as it vanishes for sufficiently large \(|x|\).
To prove this, define the Wronskian state \(|W\rangle_{|\psi_1\rangle,|\psi_2\rangle}\in\mathcal H^{\otimes 2}\) of two states \(|\psi_1\rangle,|\psi_2\rangle\in\mathcal H\) by the antisymmetric Slater-like determinant \(|W\rangle_{|\psi_1\rangle,|\psi_2\rangle}:=|\psi_1\rangle\otimes P_{\mathcal H}|\psi_2\rangle-P_{\mathcal H}|\psi_1\rangle\otimes |\psi_2\rangle\); for instance one has the probability current state \(|J\rangle=|W\rangle_{K|\psi\rangle,|\psi\rangle}/2m\). One can explicitly calculate (thanks to \(H-V=T\propto P^2\)) that if \(|E\rangle_1,|E\rangle_2\) are any two degenerate \(H\)-eigenstates with the same energy \(E>0\) (e.g. a beam of particles of energy \(E\) thrown in from the left vs. a beam of the same energy \(E\) scattering in from the right), then \(|W\rangle_{|E_1\rangle,|E_2\rangle}\in\ker P_{\mathcal H^{\otimes 2}}\) has zero momentum, using the “product rule” \(P_{\mathcal H^{\otimes 2}}=P_{\mathcal H}\otimes 1_{\mathcal H}+1_{\mathcal H}\otimes P_{\mathcal H}\).
Thus, the Wronskian state \(|W\rangle_{|E_1\rangle,|E_2\rangle}\) is translationally invariant, meaning it is a constant across \(x\in\textbf R\). In particular, at \(x\to-\infty\), it is \(|W\rangle_{|k\rangle,|-k\rangle}=-2ikt_{\leftarrow}\) but at \(x\to\infty\) it is \(-2ikt_{\rightarrow}\) so we get \(t_{\rightarrow}=t_{\leftarrow}\). A similar argument replacing \(|k\rangle\mapsto K|k\rangle\) (but not touching the other one) leads to the condition \(\begin{pmatrix} t_{\leftarrow} & r_{\leftarrow}\end{pmatrix}^{\dagger}\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}t_{\rightarrow}\\r_{\rightarrow}\end{pmatrix}=0\) as claimed.