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Hamiltonian for hydrogen atom?

  1. Dec 10, 2012 #1
    When I write down the Hamiltonian for the hydrogen atom why do we not include a radiation term or a radiation reaction term? If I had an electron moving in a B field it seems like I would need to have these terms included.
     
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  3. Dec 10, 2012 #2

    mfb

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    Where do you see a magnetic field? The electron "moves"*, but the nucleus does not (if you reduced the 2-body problem to a 1-body problem).
    "Classical" quantum mechanics (no quantum field theory) cannot include couplings to external radiation, or has to use effective models for that, so this is neglected in the derivation.

    Edit:
    *well, not really, but at least it has a wave function which has expressions similar to a velocity
     
    Last edited: Dec 10, 2012
  4. Dec 10, 2012 #3
    Generally, to begin with, the external magnetic field is ignored. You can add a magnetic field which interacts with the magnetic moment of the atom. This gives rise to the Zeeman effect, the splitting of energy levels based on the z-component of the total angular momentum (usually denoted m).

    http://en.wikipedia.org/wiki/Zeeman_effect
     
  5. Dec 11, 2012 #4
    I was just thinking that the electron was moving into its own B field that it created.
    Dont they have something like this in E&M?
     
  6. Dec 11, 2012 #5

    mfb

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    No, you don't get this.
    In quantum field theory, there is some sort of self-interaction, but that cannot be explained with a classical electromagnetic field.
     
  7. Dec 11, 2012 #6
  8. Dec 11, 2012 #7

    Jano L.

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    Only if the particle was extended in space. Then one part of the particle could move in the field of another part. However, there is not much evidence for such structured electron and it is difficult even to formulate such theory consistently, so most usually electrons are assumed as points, both in classical electrodynamics and in quantum theory.

    Because the Hamiltonian description is well suited for forces which are given by values of r and p. Radiation reaction force [itex]k\dot \mathbf{a}[/itex] does not fit into this framework - it contains second derivative of momentum.

    If it is external magnetic field (due to magnet), then one can include it via vector potential or terms like [itex]-\boldsymbol{\mu}\cdot\mathbf B[/itex] into the Hamiltonian. However, there is not much reason to include self-interaction of electron with its own field in the Hamiltonian. For example, most quantum-chemical calculations never use such terms and give quite good results (see Slater, Solid State and Molecular Theory: A Scientific Biography).
     
  9. Dec 11, 2012 #8
    ok thanks for all of your responses. If I had a relativistic electron moving in a B field would I then have a radiation term? The electron is a free particle moving through an external B field.
     
  10. Dec 11, 2012 #9

    Jano L.

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    In relativistic theory, for electron in external magnetic field, I would use

    [tex]
    H = \sqrt{(\mathbf p - \frac{q}{c}\mathbf A)^2c^2 + m^2c^4}
    [/tex]
    with [itex]\mathbf A[/itex] such that give the magnetic field in question.
     
  11. Dec 12, 2012 #10
    for hydrogen atom the only time parameter we can see is of order of10-10/107(bohr radius/velocity).velocity is only some approximate idea here.It is of order of 10-17,which is far from 10-24.
     
  12. Dec 12, 2012 #11

    mfb

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    This shows that radiative corrections are small.
     
  13. Dec 13, 2012 #12
    sure,it shows it.the parameter τ is the only parameter in classical electrodynamics which is relevant for considering whether radiative corrections should be included or not.
     
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