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Wave Equation for the Helium Atom/No of terms = 2 or 3?

  1. Jun 8, 2004 #1
    Hi friends...

    Sometime back, I encountered the Self Consistent Field Method in Quantum Mechanics, which is used to compute wave functions in complex atoms. The book I read this from is "Practical Inorganic Chemistry" by Clyde and Day. The method is explained through an argument about the potential energy function (V) of the Helium atom (a sum of three terms due to the three particle system).

    According to it, in the case of Helium which has two electrons (and one nucleus with a +2e charge), the wave equation is

    [tex]
    \frac{1}{m_1}\nabla^2_{1} \psi_{T} +
    \frac{1}{m_2}\nabla^2_{2} \psi_{T} +
    \frac{8\pi^{2}}{h^{2}} (E-V)\psi_{T} = 0
    [/tex]

    where [tex]m_{1}[/tex] and [tex]m_{2}[/tex] are electron masses. Now, I think this is incomplete and should include a third term for the mass of the nucleus with charge +2e. So in my opinion, the wave equation should be,

    [tex]
    \frac{1}{m_1}\nabla^2_{1} \psi_{T} +
    \frac{1}{m_2}\nabla^2_{2} \psi_{T} +
    \frac{1}{m_3}\nabla^2_{3} \psi_{T} +
    \frac{8\pi^{2}}{h^{2}} (E-V)\psi_{T} = 0
    [/tex]

    where [tex]m_{3}[/tex] is the mass of the nucleus. The book adopts a convention I have not come across elsewhere (so far): there is a laplacian operator for each particle (hence the subscripts). This of course, leads to an intitally tedious looking set of expressions while separation of variables.

    Please tell me which of the abovementioned equations is correct, as I am wondering why such an advanced book should fail to explain it.


    Cheers
    Vivek
     
  2. jcsd
  3. Jun 13, 2004 #2
    maverick,

    Think about how big m3 is compared to m1 and m2 (nearly 4 orders of magnitued!).
     
  4. Jun 14, 2004 #3

    turin

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    Homework Helper

    In an exagerated sense, you can leave out the mass of the nucleas for the same basic 2 reasons that you can leave out the mass of the earth in a classical first year projectile problem. The Earth doesn't move so much and we don't care about the motion of the Earth in such a problem anyway.

    The V in your Hamiltonian takes the nucleas into consideration. Essentially, the nucleas is just the point at the bottom of the r-1 potential well.
     
    Last edited: Jun 14, 2004
  5. Jun 25, 2004 #4
    Hi turin/jdavel

    Thanks (sorry for this late acknowledgement..I was out).

    Cheers
    Vivek
     
  6. Jun 25, 2004 #5

    Njorl

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    Science Advisor

    On an undergrad quantum exam, I read a question as, "Please express and find solutions for the wave equation of potassium."

    I nearly feinted. I thought there might be a trick, and pondered it for a while. I finished the rest of the exam, which was trivial by comparrisson, and came back to it. The professor saw me agonizing over it and asked if there was a problem. I asked how in the world we could solve the potassium atom in an exam period.

    He looked at me and said, "Positronium, not potassium."

    Njorl
     
  7. Jun 27, 2004 #6
    :rofl:

    Interesting...
     
  8. Jul 3, 2004 #7

    vanesch

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    Staff Emeritus
    Science Advisor
    Gold Member

    Hello,

    What you write is exact, however, as others pointed out, you can separate the dynamics of the nucleae and of the electrons because of the big mass difference. This is formalised under the name of Born-Oppenheimer approximation. It goes as follows (for molecules):

    1) Consider the positions of the nucleae fixed, and solve the quantum problem (in one way or another) of the electrons in the electrostatic potential of the nucleae (and the electrodynamic interactions amongst the electrons).
    This gives us the stationary states of the n-electron system, parametrised in the positions of the nucleae.

    2) for the ground state of the electron system, this gives you an energy level which is also parametrized in the positions of the nucleae: consider this as the POTENTIAL ENERGY of the nucleae.

    3) consider the quantum system of the nucleae (forget about the electrons) with as potential energy the function mentioned in 2). Solve this quantum problem: you have the quantized motion of the nucleae.

    Feynman proved mathematically that this algorithm gives the correct solution to the full quantum problem in the limit of the mass ratio nucleae/electrons large.

    Of course for an atom, 2 and 3 make no sense, so you can limit yourself to step 1)

    Furthermore, even 1) is usually too complicated to be solved as such. So other approximation techniques are used in that case, and the first one is the self-consistent field method (Hartree Fock), where the electron-electron individual interactions are replaced by a simple common potential, averaged over the wave functions squared. This wipes electron-electron position correlations under the carpet.

    cheers,
    Patrick.
     
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