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View Full Version : Hamiltonians, non-interacting hydrogenic systems, spin correction. Help!


ScotchDave
Dec30-08, 02:20 PM
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
\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)

The overall Hamiltonian is

\hat{H} = \hat{H_{0}} + V

Where
V = \frac{e^{2}}{4\pi\epsilon_{0}r_{1 2}}

Z = 2 for He nucleus.

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

E_{n} = -\frac{Z^{2}}{n^{2}}Ry

The first part asks what the corresponding energy is.

That's given by \hat{H}\Psi_{100} = E_{1}}\Psi_{100}

For no interaction \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(\v ec{r_{1}}\right)\Psi_{100}\left(\vec{r_{2}}\right) + \hat{H}\left(\vec{r_{2}}\right)}\Psi_{100}\left(\v ec{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)

2E_{1} = -2Z^{2}Ry = -8Ry

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

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