# 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