What is the effect of a complex potential energy on particle absorption?

[Felicity]

Homework Statement

suppose V(x) is complex, obtain an expression for

∂/∂t P(x,t) and
d/dt ∫_-∞^∞dxP(x,t)

for absorption of particles the last quantity must be negative (since particles disappear, the probability of their being anywhere decreases). What does this tell us about the imaginary part of V(x)? (ch 2, problem 11 gasiorowicz)

Homework Equations

V(x) is the potential energy

schrodinger equation

∂/∂t P(x,t)= (∂ψ*)/∂t ψ+ψ*∂ψ/∂t

∂/∂t ∫_-∞^∞dxP(x,t)=-∫_-∞^∞dx ∂/∂x j(x,t)= 0

where j(x,t) is the probability current

but these may only be valid if V(x) is real

why?

The Attempt at a Solution

I see how one can calculate

∂/∂t P(x,t)= (∂ψ*)/∂t ψ+ψ*∂ψ/∂t

by plugging in the general schrodinger equation and its complex conjugate but in this situation V(x) must be real

Why does the potential energy V(x) have to be real though?

How would you find ∂/∂t P(x,t) if V(x) were complex?

I have not yet taken a complex analysis class so any recommendations of topics in complex analysis to look up would be appreciated

Any help would be greatly appreciated!

 

[Irid]

First, write down the Schrodinger equation:

$$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \Psi + V\Psi$$

and then it's complex conjugate (the same equation, only every imaginary unit must be added a minus sign). Then multiply each of them by $$\Psi$$ and $$\Psi^*$$ correspondingly and subtract. You'll see that if potential V has an imaginary part, it doesn't cancel and thus

$$\frac{\partial P}{\partial t} \neq 0$$

so the probability for finding the particle decreases in time.

 

[Felicity]

Thank you so much! I guess I just didn't see that making V(x) complex is as simple as turning it into (V(x) + some imaginary part)