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Separating Hamiltonian functions. Helium atom.

  1. Jun 26, 2008 #1
    I'm using McQuarrie's "Quantum Chemistry" book for a little bit of light reading. He included a proof of a theorem that states that if a Hamiltonian function is separable, then the eigenfunctions of Schrodinger's equation are the products of the eigenfunctions of the simpler "separated" equations... Anyway...

    I just started studying the helium atom. Schrodinger's equation is written with two different laplacian operators, one that acts "only on the first electron", and one that acts "only on the second electron". However, the laplacian for both electrons is just a differential operator which contains a mixture of partial derivatives with respect to all three coordinates. I don't understand why you can assume that the first laplacian only acts "on the first electron". After all, a differential operator (D) acting on two functions (F and G):
    D(F*G) is not F*D(G) if D contains derivatives which pertain to F.

  2. jcsd
  3. Jun 26, 2008 #2
    That Hamiltonian operator is looking at each interaction separately and making a statement that all of the interactions taken together describe the system completely. Omitting the inter-nueclon terms (which have little effect on the motion of the electrons), and assuming a stationary nucleus (which we can do since the mass of the nucleus is much larger than the mass of the electrons) there are 5 terms: a laplacian term that deals with the momentum of one of the electrons relative to the nucleus, another laplacian term that deals with the momentum of the other electron relative to the nucleus, and three columb terms: one for each of the electrons interacting with the nucleus and one for the interaction between the two electrons (this one is sometimes omitted to simply the problem).

    So there are two momentum terms because each of the electron's momentum is independent of the others, so you need one term that deals with each.


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    Last edited by a moderator: May 3, 2017
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