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The ground state of helium - showing that parhelium is ground state

  1. May 11, 2014 #1
    1. The problem statement, all variables and given/known data

    The Hamiltonian of helium can be expressed as the sum of two hydrogen Hamiltonians and that of the Coulomb interaction of two electrons.

    [itex]\hat H = \hat H_1 + \hat H_2 + \hat H_{1,2}[/itex]

    The wave function for parahelium (spin = 0) is

    [itex]\psi(1,2) = \psi_S(r_1, r_2)\dot \xi_A(s_1, s_2)[/itex] with the first being a symmetric spatial function and the second being an antisymmetric one.

    We can separate this into the normalized function

    [itex]\psi_S(r_1,r_2) = \frac{1}{\sqrt{2}}[\psi_1(r_1)(\psi_2(r_2)+ \psi_1(r_2)(\psi_2(r_1)]=\psi_S(r_2,r_1)[/itex]

    For orthohelium the functions look like this:

    [itex]\psi(1,2) = \psi_A(r_1, r_2)\dot \xi_S(s_1, s_2)[/itex]
    [itex]\psi_A(r_1,r_2) = \frac{1}{\sqrt{2}}[\psi_1(r_1)(\psi_2(r_2) - \psi_1(r_2)(\psi_2(r_1)]=\psi_A(r_2,r_1)[/itex]

    Show the ground state of helium is parahelium. The hint is what happens to the wavefunction.


    2. Relevant equations

    OK, so I start with that the Hamiltonian of the given wave function(s)

    $$H_1 = \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial^2 r_1^2} = E_1 \psi, H_2 = \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial^2 r_1^2} = E_2 \psi,
    H_{1,2} = -\frac{e^2}{4\pi\epsilon_0 r_{1,2}}$$


    3. The attempt at a solution

    OK, I was trying to get a handle on how to get started with this. My first thought was to treat each Hamiltonian separately and just get a wavefunction the way I would solving any Schrodinger, though I am not sure of the boundary conditions. Then I can get a value for energy. But my sense is there is a simpler way to do it. I don't need a whole walk-through I don't think, I am just trying to understand some of the slutions I do see out there.
     
  2. jcsd
  3. May 13, 2014 #2
    and looking into it some more I wanted to check if this was an OK (mathematically speaking) thing to do:
    One thing I thought of doing was this (for [itex]H_1[/itex]):

    $$H_1 = -\frac{\hbar^2}{2m}(\frac{\partial^2 \psi}{\partial r_1^2}+\frac{\partial^2 \psi}{\partial r_2^2})=E_1(\psi_1(r_2)+\psi_1(r_1))$$

    but again I don't know if that's kosher

    Thanks.
     
  4. May 14, 2014 #3

    DrClaude

    User Avatar

    Staff: Mentor

    You mean
    $$
    \psi_A(r_1,r_2) = \frac{1}{\sqrt{2}}[\psi_1(r_1) \psi_2(r_2) - \psi_1(r_2)\psi_2(r_1)]= - \psi_A(r_2,r_1)
    $$

    You're missing the Coulomb interaction between the electrons and the nucleus. Both ##H_1## and ##H_2## look like the single-electron (hydrogenic atom) Hamiltonian.

    That part should be trivial: it is easy to find the solution for a hydrogenic atom.

    You have the right approach. Look at the energy of the two-electron system by first neglecting the electron-electron interaction (treat it as a perturbation, even though it is not really one).
     
  5. May 14, 2014 #4
    OK, adding in the potential terms I wanted to make sure this was o to do:
    For the [itex]\psi(r_1,r_2)[/itex] function I should have (With the potential terms)
    $$H_1 = -\frac{\hbar^2}{2m}(\frac{\partial^2 \psi_1}{\partial r_1^2}+\frac{\partial^2 \psi}{\partial r_2^2}-\frac{2e^2}{4\pi \epsilon_0 r}-\frac{2e^2}{4\pi \epsilon_0 r})=E_1(\psi_1(r_2)+\psi_1(r_1))$$.

    To solve that I can further break that up as follows:

    $$-\frac{\hbar^2}{2m}\frac{\partial^2 \psi_1}{\partial r_1^2}-\frac{2e^2}{4\pi \epsilon_0 r_1}=E\psi_1(r_1)$$
    $$-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial r_2^2}-\frac{2e^2}{4\pi \epsilon_0 r}=E_1\psi_1(r_2)$$.

    Right? Add these energies up and I should have a different result for the assymetric and symmetric cases. (One of the functions is negative, for one thing).
     
  6. May 25, 2014 #5
    Just consider what happens to the spatial part of the wavefunction in the ground state. Is it spatially symmetric or anti-symmetric?

    Now, what must the spin part be then?
     
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