How can we derive the continuity equation from Schrodinger equation if the potential is a complex function of position? What I tried was the general $1-D$ derivation of the Continuity equation from Schrodinger equation and got the result: $$\frac{\partial \rho}{\partial t} = -\frac{\partial j}{\partial x} + \frac{\rho(x,t)}{i\hbar}\left[V(x) - V^*(x)\right]$$ where, $j$ is the probability current density. My doubt is what physics does the term $\frac{\rho(x,t)}{i\hbar}\left[V(x) - V^*(x)\right]$ convey?
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$\begingroup$ Note that $V(x) - V^{*}(x) = 2i \Im(V(x))$ $\endgroup$– noxNov 9, 2021 at 23:11
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A complex potential means the Hamiltonian is not Hermitian, which in turn implies probability is not be conserved. Your computation shows this explicitly, by introducing a term which breaks the continuity equation. In other words, the extra term conveys the fact that probability is not conserved in this model.
As a side comment, I've seen people modelling the decay of an unstable particle by exploiting this fact. The probability of finding the particle somewhere would diminish and tend to zero due to the non-hermitian terms. I do not know whether such a model could be used to make useful predictions, but from a theoretical point of view I regard it as an interesting approach.
answered Nov 9, 2021 at 19:28