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I am trying to solve the following equation (in 1D) and stuck in the middle of the way. Here's the equation:

$$i\frac{\partial\psi}{\partial t}=C\cdot\frac{\partial^{2}\psi}{\partial x^{2}}+iD\cdot\psi\,\,\,D,C\in\mathbb{R}$$

This means that the Hamiltonian $H=C\cdot P^{2}+iD$ (I think it's not hermitian)

Tried "brute forcing" it, I assumed separation of variables $\psi=T(t)S\left(x\right)$ and as usual denote a dot for time derivative and ' for space derivative.

$$i\dot{T}(t)\cdot S(x)=C\cdot T(t)S\left(x\right)''+iD\cdot T(t)S\left(x\right)$$

From here we can divide by $T\cdot S$ and get $$ i\frac{\dot{T}(t)}{T(t)}=C\cdot\frac{S\left(x\right)''}{S\left(x\right)}+iD$$ I don't know how to continue from here.

I know that for a constant potential $V=V_0$ I should get that solutions with $E>V_{0}$ are just free particles of with a defined momentum.

However here the potential is fully imaginary and I am kind of at a loss, Almost feels like it will be.

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Hint: what happens if you rephrase your Schrödinger equation using the transformation $$ \psi(x,t) = e^{D t}\varphi(x,t) $$ and re-express everything in terms of $\varphi$? This should make it clearer what role the (uniform!) complex potential plays.

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  • $\begingroup$ $$iDe^{Dt}\varphi\left(x,t\right)+ie^{Dt}\frac{\partial\varphi}{\partial t}=C\cdot e^{Dt}\frac{\partial^{2}\varphi}{\partial x^{2}}+iDe^{Dt}\varphi\left(x,t\right)$$ and I am left with a "free particle" behaviour for $\varphi$ $$\frac{\partial\varphi}{\partial t}=C\cdot\frac{\partial^{2}\varphi}{\partial x^{2}}$$ And in the end I have a solution of a free particle with growing amplitude in time, Am I correct? (It seems logical for me because the origin of my equation is dealing with nonlinear responses in media that includes gain were t is my "propogation" variable) $\endgroup$ – Pavel Penshin Mar 18 '19 at 19:21

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