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The dipole interaction and fermions

  1. Feb 5, 2012 #1
    Hi

    When looking at the interaction between an EM-wave/field E with an isolated atom, we know that they interact via the time-dependent dipole interaction V = dE. This can be derived by looking at an electron bound to an atom, and it is used in systems, where the atom can be rightfully approximated as a two-level atom.

    However, how can we always say that the atom only has a single electron attached to it? Who says that there can't be two/three/... electrons attached to our approximate two-level atom? In that case I don't believe the dipole interaction is valid.

    Any help is appreciated.

    Best wishes,
    Niles.
     
    Last edited: Feb 5, 2012
  2. jcsd
  3. Feb 5, 2012 #2

    Physics Monkey

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    Why would it not be valid? The dipole may be a more complicated operator in the many-electron case, but it still represents the first term in an expansion of the electrostatic energy of a compact system in a background field, right?

    In other words, suppose the atom is neutral. The electrostatic energy is
    [tex]
    \sum_{a} q_a V(r_a)
    [/tex]
    where a labels particles and V is an external potential or light field. If you assume V is slowly varying on the scale of the atom (the usual approximation), then all the [itex] r_a [/itex] are close and you can freely Taylor expand V. What does this give?
     
  4. Feb 5, 2012 #3
    Thanks for replying. I'm not quite sure I understand your reasoning. In almost every book on quantum optics I've seen, they start with the Hamiltonian of a single electron bound to an atom to find that the interaction between the atom and an EM-field is given by V = dE. What you are then doing is to say that if it is a many-electron atom, then the dipole d is the same to first order?

    To first order in V we would get
    [tex]
    \sum_{a} q_a V(R)
    [/tex]
    for R being the coordinate of the atom. How can I find the dipole from this?

    Best,
    Niles.
     
  5. Feb 8, 2012 #4

    Physics Monkey

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    The first term you wrote amounts to saying the total charge of the atom responds to the potential at the atom's location. Now suppose that the atom is neutral. The term you wrote vanishes since the total charge is zero. What is the next term in the series?
     
  6. Feb 8, 2012 #5
    The term compromising the electrostatic energy between a single electron of the atom and the atom itself. What would the term after that be?
     
  7. Feb 8, 2012 #6

    Physics Monkey

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    You need to do the expansion and look at the terms.

    The energy is
    [tex]
    \sum_a q_a V(r_a+R)
    [/tex]
    where I've assumed that the [itex] r_a [/itex] are measured from position [itex] R [/itex] of the nucleus (effectively the center of mass). Expand in small [itex] r_a [/itex]:
    [tex]
    \sum_a q_a V(r_a+R) \approx \sum_a q_a V(R) + \sum_a q_a r_a (\nabla V)(R) + ...
    [/tex]

    Can you interpret these terms?
     
  8. Feb 8, 2012 #7
    Ah, I see. So I get
    [tex]
    \sum_a q_a V(r_a+R) = \sum_a q_a V(R) + \sum_a q_a r_a (\nabla V)(R) + \frac{1}{2}\sum_a q_a r_a^2 (\nabla(\nabla V))(R) + \frac{1}{6}\sum_a q_a r_a^3 (\nabla(\nabla(\nabla V)))(R) + \ldots
    [/tex]
    I see that the first term (=zeroth order) corresponds to the energy of of the atom, if the charges respond to the potential at the atom's location, as you said. The higher-order terms I can't interpret; do they even have a physical interpretation?

    Best,
    Niles.
     
    Last edited: Feb 8, 2012
  9. Feb 8, 2012 #8

    Physics Monkey

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    Basically [itex] r_a [/itex] small means that the potential varies slowly over the scale of the atom so that we can keep only the first few terms in the series.

    Do you know what the dipole moment of a group of charges looks like?
     
  10. Feb 8, 2012 #9
  11. Feb 8, 2012 #10

    Physics Monkey

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    A lot of expressions in the "Expression (general case)" section are almost identical to what we've written here. So what is the meaning of [itex] \sum_a q_a r_a [/itex]? Also, what is the physical meaning of [itex] \nabla V [/itex]?
     
  12. Feb 8, 2012 #11
    Oh, our second term is just a sum of dipoles multiplied by the gradient of the potential. Ah, and the gradient of the potential is the (minus) electric field at the atom (= dipole approximation). Very nice, so we get just what I was looking for.

    Thanks for being patient with me, this is really good.

    Best,
    Niles.
     
  13. Feb 9, 2012 #12
    I have been thinking about our result, and there is a link I don't see. The external potential V stems from an electric field acting on our atom, which causes the displacements of the electrons (=creates the dipoles). But since that is the case, how can V(R)=0? Just because the atom is neutral doesn't necessarily imply that the *external* potential is zero at R.
     
    Last edited: Feb 9, 2012
  14. Feb 9, 2012 #13

    Physics Monkey

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    I think you might have misunderstood. I'm not claiming that V(R) is zero. What I am claiming is that the sum over charges is zero. This sum includes the protons as well as the electrons, but the protons are heavy so we usually treat them as fixed by the center of mass motion.

    Also, you should distinguish between the average value of the dipole and the dipole operator itself. In the hydrogen atom the dipole operator is
    [tex]
    d = q_{\text{elec}} r_{\text{elec}} + q_{\text{prot}} r_{\text{prot}}.
    [/tex]
    Because the system is neutral we can compute the dipole moment about any point we want and get the same answer (a result from E&M). Thus the dipole moment about the center of mass (roughly the proton position) is simple
    [tex]
    d = - e r
    [/tex]
    where r is the electron position relative to the proton i.e. exactly the coordinate we always use to solve the hydrogen atom.

    Once you've understood this operator, you can compute expectation values of it. It is true that the expectation value is zero in the ground states and that the expectation value is proportional to the applied external field (in weak fields).

    What we showed was that for many electron atoms the dipole operator has a certain form in terms of the electronic coordinates, again assuming the protons are all fixed at the origin. One may then compute expectation values of this operator in various states.
     
  15. Feb 9, 2012 #14
    Ah, I see. Thanks for that clarification.

    Best wishes,
    Niles.
     
  16. Feb 9, 2012 #15

    Physics Monkey

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    No problem.
     
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