Schrodinger Wave Function Question

1. Sep 3, 2004

Ed Quanta

I have to show that if a wave function (Schrodinger) has a potential V(x) and the wave function's complex conjugate has a potential V'(x) and V(x) does not equal V'(x),
this contradicts the continuity equation dp/dt + div J =0

where p=charge density, and J=current density.

Can someone help me with this problem? I am unsure of how to represent dp/dt, even though I am pretty sure p= the modulus of the wave function squared. I am unsure of how to find J in terms of the Hamiltonian which I think is necessary to solve this thing.

2. Sep 3, 2004

shchr

I don't understand the meaning of the above sentence. Usually potentials are real functions.

3. Sep 3, 2004

nrqed

I assume you know the expression for the current density. You should start by proving the continuity equation, which is a useful exercise in any case. The first step is to write

dp/dt = d(Psi* Psi)/dt = (d Psi*/dt) Psi + Psi* (d Psi/dt).

The second step is to write explicitly Schrodinger equation (the time dependent one, of course) and then take its complex conjugate to get the equation for Psi. From the Sch eq, you get an equation for dPsi/dt and dPsi*/dt. You plug this into dp/dt, above. You will then obtain div J plus a term proportional to (Psi Psi* V - Psi Psi* V') which is zero for V=V'.

Pat

4. Sep 3, 2004

nrqed

Sometimes complex potentials are useful to model some effects. The most obvious is the case of an unstable particle. One cand model the effect of the decay by including a complex piece in the potential proportional to the half life. Then probabibility is not conserved, $d \rho/dt \neq 0$. More precisely, one finds that the integral of $\Psi \Psi^*$ goes like $e^{-t/\tau}$ instead of being equal to one for all t.

Of course, this is because the states to which the particle decays are not included in the theory. if they were, the time evolution would be unitary. But it's simpler to violate unitarity by simply making thepotential complex. There are other cases, even in quantum field theory, where this is a useful trick.

Pat

5. Sep 4, 2004

nrqed

Oops, I meant "to get the equation for Psi*", of course.

Pat

6. Sep 4, 2004

Ed Quanta

Where do we get this term from? and what is div of J? I know that
J=h/2mi(psi*(dpsi/dx)-psi(dpsi*/dx)). But unlike the derivatives of the wave function with respect to t which can be described in terms of the Hamiltonian, I do not see how the derivatives of the wave function with respect to x can be expressed in terms of the Hamiltonian. I am sorry, I am still pretty dam confused on what the equations should look like. I really appreciate the help.

By the way I currently have dp/dt=i/h(momentum^2/2m+V'(x))psi -i/h(momentum^2/2m+V'(x))psi*. Since dpsi/dt-i/hHpsi and dpsi*/dt=i/hHpsi* were H is the hamiltonian.

7. Sep 4, 2004

Tom Mattson

Staff Emeritus
You get it from subtracting the two equations he referred to. If V does not equal V', then that residual term is left over from the subtraction.

It's the divergence of the probability current.

Do exactly what he said. Start with the Schrodinger equation, and multiply it by Ψ*:

Ψ*[(h2/8&pi;2m)d2Ψ/dx2+VΨ]=(ih/2π)Ψ*(dΨ/dt)

Then take the complex conjugate of the Schrodinger equation, and multiply it by Ψ. That gives you 2 equations, which you subtract to get the continuity equation.

Last edited: Sep 4, 2004