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Time varying forces

  1. Oct 3, 2006 #1

    quasar987

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    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?
     
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  3. Oct 4, 2006 #2

    dextercioby

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

    Daniel.
     
  4. Oct 4, 2006 #3
    In the case of a time-varying electric field, you would probably have to use the vector potential stuff as

    [tex]
    \vec{E} = -\frac{1}{c} \frac{\partial \vec{A}}{\partial t} - \nabla \varphi
    [/tex]
    [tex]
    \vec{B} = \nabla \times \vec{A}
    [/tex]

    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
  5. Oct 4, 2006 #4

    quasar987

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    But the potential in the SE is a scalar potential. What would you do with [itex]\vec{A}[/itex]?
     
  6. Oct 4, 2006 #5

    Galileo

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    The S.E. in 'fundamental form' is [tex]i\hbar \frac{\partial}{\partial t}|\Psi\rangle = H|\Psi\rangle[/tex].
    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 [itex]\vec p[/itex] of the position [itex]\vec r[/itex] is not [itex]m\vec v[/itex] but:
    [tex]\vec p = m\vec v+q\vec A[/tex], and your Hamiltonian becomes:

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

    [tex]\vec E = -\vec \nabla U - \frac{\partial}{\partial t}\vec A[/tex]
    [tex]\vec B = \vec \nabla \times \vec A[/tex]

    (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
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