Time varying forces

1. Oct 3, 2006

quasar987

The SE is written in terms of a potential energy. It says, "given a particle in a region where the potential is V(,x,y,z), solve me if you want to know the probability density."

But not all forces can be represented by a potential energy. What does QM says, for exemple in the case of a particle in a time varying electric field?

2. Oct 4, 2006

dextercioby

Time varying potentials induce transitions among states on the time independent hamiltonian. That's what we have perturbation theory for.

Daniel.

3. Oct 4, 2006

StatMechGuy

In the case of a time-varying electric field, you would probably have to use the vector potential stuff as

$$\vec{E} = -\frac{1}{c} \frac{\partial \vec{A}}{\partial t} - \nabla \varphi$$
$$\vec{B} = \nabla \times \vec{A}$$

and then choose your vector potential accordingly.

edit: It's also interesting to note what happens if, say, your potential energy is not Hermitian. I recommend you explore that exercise a bit, as you get some interesting results regarding the normalization of the wave function. Also, it's important to note that most non-conservative forces (such as friction) are macroscopic, and, as far as I know, have no quantum analogs.

Last edited: Oct 4, 2006
4. Oct 4, 2006

quasar987

But the potential in the SE is a scalar potential. What would you do with $\vec{A}$?

5. Oct 4, 2006

Galileo

The S.E. in 'fundamental form' is $$i\hbar \frac{\partial}{\partial t}|\Psi\rangle = H|\Psi\rangle$$.
You should always get H from the classical Hamiltonian. The S.E. you wrote down cannot accomodate for all situations.

In EM, the conjugate momentum $\vec p$ of the position $\vec r$ is not $m\vec v$ but:
$$\vec p = m\vec v+q\vec A$$, and your Hamiltonian becomes:

$$H=\frac{1}{2m}\left(\vec p-q\vec A)^2+qU$$
where U and A are the potentials. The important thing is that they satisfy:

$$\vec E = -\vec \nabla U - \frac{\partial}{\partial t}\vec A$$
$$\vec B = \vec \nabla \times \vec A$$

(U and A are not unique. You have so-called gauge freedom. Different gauges will lead to the same physical results in EM and QM ofcourse. This is called gauge invariance).

By the way. At a fundamental level (microscopic scale), only conservative systems play a role anyway.

Last edited: Oct 4, 2006