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Born-Oppenheimer approximation confusion

  1. Jan 21, 2013 #1
    Hi, I'm reading about the Born-Oppenheimer approximation for a solid and they're doing the formalism of it. They say that we can basically consider the ions stationary with respect to the electrons because they move so little and so slowly in comparison to them.

    They say that ##R_i## are the positions of the ions and ##r_j## are the positions of the electrons, ##P_i## are the momenta of the ions, ##p_j## are the momenta of the electrons (all vectors but I'm just writing them like this here). Then they say that we'll look at "core" electrons separately from "valence" electrons, because core ones just hang out by the nuclei while valence ones move around. Given all this, the hamiltonian is:

    ##H = \sum\limits_i \frac{P_i^2}{2M} + \sum\limits_{j = cond. elecs} \frac{p_j^2}{2m} + \sum\limits_{i,i'} V_{i,i'}(|R_i - R_{i'}|) + (e^2/2) \sum\limits_{j,j'=cond. elecs} \frac{1}{|r_j - r_{j'}|} + \sum\limits_{i,j} V_{ei}(|r_j - R_i|) + E_{core}##

    (where ##E_{core}## is the energy of the "core" electrons that are "attached" to the nuclei.)

    Then they rewrite this as:

    ##H = T_i + T_e + V_{ii} + V_{ee} + V_{ei} + E_{core}##

    Then they say that we can write the full wavefunction as a combination of two functions (here, ##r## and ##R## are the sets of the positions of all the electrons/ions, not single ones):

    ##\Psi(r,R) = \sum\limits_n \Phi_n(R) \Psi_{e,n}(r,R)##

    Then, they just do the eigenvalue equation, ##H\Psi = E\Psi##:

    ##(T_i + V_{ii} + E_{core})\Psi + \sum\limits_n \Phi_n (T_e + V_{ee} + V_{ei})\Psi_{e,n}(r,R) = E\Psi##

    In the second term, the part with the explicit sum, they put ##\Phi_n## out in front because the operators directly following it "only operate on the electron part of the product wavefunction", according to my book. But here's my confusion: doesn't ##V_{ei}## act on the ion part of the wave function? It was defined as ##\sum\limits_{i,j} V_{ei}(|r_j - R_i|)##, which has that ##R_i## in it. What am I missing?

    Thank you!
     
  2. jcsd
  3. Jan 21, 2013 #2

    Cthugha

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    You are more or less missing the Born-Oppenheimer approximation itself. Recall what you said in the beginning:

    So your assumption is basically that your electron distribution does not drag your ions around. The distribution of the ions will stay as it is. Therefore you can treat the position of the ions as a parameter instead of a variable. So the potential will only depend on these positions, but not act on them.

    If you want a more intuitive explanation, what you do is not solving the coupled system of ions and electrons, but getting a solution for the electron system for a fixed set of ion positions.
     
  4. Jan 21, 2013 #3

    DrDu

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    No, up to what VortexLattice has described, the whole wavefunction is still completely general and the BO approximation has not been invoked, yet.
    V_ei clearly acts also on the nuclear wavefunction, but as it is a multiplicative operator, it does not matter whether it appears in front or after ##\Phi_n##.
     
  5. Jan 21, 2013 #4

    Cthugha

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    Oh sorry. Maybe I misread and should stop posting after midnight.
     
    Last edited: Jan 21, 2013
  6. Jan 21, 2013 #5
    Hmmm, this doesn't seem consistent though. If ##V_{ei} = \sum\limits_{i,j} V_{ei}(|r_j - R_i|)##, I see what you're saying about the order of the operator ##R_i## not mattering (that's what you're saying, right?), but then what about ##V_{ii}##? That also just has multiplicative factors of ##R##: ##V_{ii} = \sum\limits_{i,i'} V_{i,i'}(|R_i - R_{i'}|)##.

    So why isn't that in the second term as well?

    Thank you!
     
  7. Jan 22, 2013 #6

    DrDu

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    Feel free to put it there!
    But the ordering chosen is not a matter of mathematical necessity but depends on the approximations which will probably introduced in the continuation of the argument.
     
    Last edited: Jan 22, 2013
  8. Jan 22, 2013 #7
    Hmmm, but that changes the final result! The final result they reach for this is separating ##\Phi_n## and ##\Psi_{e,n}## by separating this equation:

    ##(T_i + V_{ii} + E_{core})\Psi + \sum\limits_n \Phi_n (T_e + V_{ee} + V_{ei})\Psi_{e,n}(r,R) = E\Psi##

    and after doing many manipulations, get:

    ##(T_i + V_{ii} + E_{core} + E_{e,n})\Phi_n = E_n\Phi_n##

    and

    ##(T_e + V_{ee} + V_{ei})\Psi_{e,n}(r,R) = E_{e,n}\Psi_{e,n}(r,R)##

    So, the placement of ##V_{ei}## makes it end up in either the nuclear or electronic equation.
     
  9. Jan 22, 2013 #8

    DrDu

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    V_ei depends on both nuclear and electronic coordinates. You want an equation for ##\Phi_n## which does not contain electronic coordinates. Likewise you want an equation for ##\Psi_{e,n}## which does not contain derivatives with respect to R.
     
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