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Hamiltonians, non-interacting hydrogenic systems, spin correction. Help!

  1. Dec 30, 2008 #1
    Eyoop gents, just looking for a wee bit of help with a correction term in the energy of a ground state hydrogenic Helium nucleus. I tried searching, but I didn't find what I needed.

    Non-interacting hamiltonian for two hydrogenic atom is
    [tex]\hat{H}_{0} = -\frac{\hbar^{2}}{2m}\left(\nabla^{2}_{1} + \nabla^{2}_{2}\right) - \frac{Ze^{2}}{4\pi\epsilon_{0}}\left(\frac{1}{r_{1}} + \frac{1}{r_{2}}\right)[/tex]

    The overall Hamiltonian is

    [tex]\hat{H} = \hat{H_{0}} + V[/tex]

    [tex]V = \frac{e^{2}}{4\pi\epsilon_{0}r_{1 2}}[/tex]

    Z = 2 for He nucleus.

    The ground state of [tex]\hat{H_{0}}[/tex] is the product wavefunction [tex]\Psi_{100}\left(\vec{r_{1}}\right)\Psi_{100}\left(\vec{r_{2}}\right)[/tex]

    [tex]E_{n} = -\frac{Z^{2}}{n^{2}}Ry[/tex]

    The first part asks what the corresponding energy is.

    That's given by [tex]\hat{H}\Psi_{100} = E_{1}}\Psi_{100} [/tex]

    For no interaction [tex]\hat{H_{0}}\Psi_{100}\left(\vec{r_{1}}\right)\Psi_{100}\left(\vec{r_{2}}\right) = \hat{H}\left(\vec{r_{1}}\right)}\Psi_{100}\left(\vec{r_{1}}\right)\Psi_{100}\left(\vec{r_{2}}\right) + \hat{H}\left(\vec{r_{2}}\right)}\Psi_{100}\left(\vec{r_{1}}\right)\Psi_{100}\left(\vec{r_{2}}\right)

    = 2E_{1}\Psi_{100}\left(\vec{r_{1}}\right)\Psi_{100}\left(\vec{r_{2}}\right)[/tex]

    [tex]2E_{1} = -2Z^{2}Ry = -8Ry[/tex]

    "What is the ground state when spin is included?"

    I know it should be of the form of [tex]2E_{1} = -2Z^{2}Ry + correction term= -8Ry + correction[/tex], but I can't for the life of me find the term. Help what is it???
  2. jcsd
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