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In 1 dimension what is the solution of the Schrödinger equation with potential

$$ V(x) = V_r + i V_i $$

Potentials are constant.

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1 Answer 1

The Hamiltonian $$H=T_{\text{kin}}+V_r+iV_i$$ will not be Hermitian as $$(iV_i)^*=-iV_i.$$

Technically, you can make an ansatz $$\Psi(x,t)=A\int\text{d}k\ \hat\Psi(k)\ \text e^{i(kx-\omega(k)t)}.$$

plug it into the differential equation and find

$$\omega(k)=\frac{\hbar^2 k^2}{2m}+V_r+iV_i,$$ or $$\hbar k=\pm\sqrt{2m(E-iV_i)},$$ where $E$ is some real number/numbers. You also want a boundary condition.

(As a vague remark, modelings of complex energies, which necessarily turn a phase like $\text e^{-i\omega}$ into something like the descending expression $\text e^{-\omega'}$, are associated with decay. But again, a particle that vanishes in time like that is probably not what you want to talk about.)

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Comment to the answer(v2): It should be stressed in the answer that the wavenumber $k$ is a manifestly real variable, not complex. Therefore the corresponding energies $E_k\equiv \hbar\omega_k$ are not real but complex. –  Qmechanic Oct 15 '12 at 13:59
    
as a comment i heard that a Schroedinguer equatio with the potential $ V(X) = ix^{3} $ may have ALL the energies real :) –  Jose Javier Garcia Oct 15 '12 at 18:30

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