Complex potential resulting in exponentially decreasing standing waves

[GarethB]

I am required to show that the potential V= -Vo(1+iε) in the schrodinger equation results in stationary waves that represent exponentially decreasing plane waves. I am also required to calculate the absorption co-efficient.

My (inept) attempt at a solution;

I know that for a comlpex energy E=Er +iEi (Ei being the imaginary energy) we have
ψ^*(x,t)ψ(x,t)=ψ(x)^*e^{-i/h(Er+iEi)t}ψ(x)e^i/h(Er+iEi)t
ψ^*(x,t)ψ(x,t)=ψ^*(x)ψ(x)e^2i/h(Ei)t
Sorry the h should be h cross (h/2π) but I left this out. The general idea that I have is that if there is Ei (a complex part to the energy) then ψ^*(x,t)ψ(x,t) is not equal to ψ^*(x)ψ(x) due to an exponential term involving Ei.
Thus the probability density has an exponential either increasing or decreasing factor. Is this what the problem was asking for? If so is Ei(the imaginary part of the energy) just the complex part of V (i.e -Voε)?

I have no idea where to start about calculating the absorption coefficient. I have seen papers that speak about the coefficients of the oncoming and reflected wave (due to a step in the potential), but no boundaries or other potentials were given in the question, so how can we have a reflected wave? I am totally confused by this.

 

[NateD]

Any help would be greatly appreciated.The potential V=-Vo(1+iε) in the Schrödinger equation results in a complex energy E=Er +iEi, where Er is the real part of the energy and Ei is the imaginary part. The time-dependent wavefunction is then given by ψ*(x,t)ψ(x,t)=ψ*(x)ψ(x)e2i/h(Ei)t, which represents an exponentially decreasing plane wave.The absorption coefficient can be calculated using the Beer-Lambert law, which states that the absorption coefficient α is proportional to the imaginary part of the energy Ei, according to the equation:α=C*Ei, where C is a constant. Thus, in this case, the absorption coefficient is proportional to Voε.