Resonant states and transmission coefficient in Quantum mechanics

Question

A resonant state is defined as the solution of the Schrodinger equation (an eigenfunction of the Hamiltonian) under the boundary conditions that only the outgoing waves exist outside the scattering potential. Let An incident wave $A𝑒^{𝑖𝑘𝑥}$from the left (Re 𝑘>0) be scattered to result in a reflection wave $𝐵𝑒^{𝑖𝑘𝑥}$ and a transmission wave $C𝑒^{𝑖𝑘𝑥}$ The condition 𝐴=0 means that we have outgoing waves only. This boundary condition is referred to as the Siegert condition.

But if there is no Incident wave then how can there be a reflected/transmitted wave? This can also be seen by the fact that transmission probability which is defined by \begin{equation} \left|\frac{J_T}{J_I}\right| \end{equation} where $J_T \And J_I$ are probability current for transmitted and incident wave, is not defined when $A = 0$

Answer 1

The basic idea is that resonances are not true eigenstates of the hamiltonian, and they are not states for which the total norm is conserved. In essence:

if there is no Incident wave then how can there be a reflected/transmitted wave?

By continuously losing population, which "leaks out" at infinity. The amplitude of the eigenfunction at any given point (and, with it, the integral of the probability density over any finite region) decays exponentially with time.

The way this is achieved is that the resonance state is mostly an eigenstate of the hamiltonian, but the eigenvalue has a nonzero imaginary part, which is responsible for the decay, and which is inversely proportional to the half-life of the state. (I say "mostly" an eigenstate because it's not normalizable, and the precise relationship is rather tricky, but there's no need to go into those details on a first reading).