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

1. Dec 30, 2008

ScotchDave

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 $$\hat{H_{0}}\Psi_{100}\left(\vec{r_{1}}\right)\Psi_{100}\left(\vec{r_{2}}\right) = \hat{H}\left(\vec{r_{1}}\right)}\Psi_{100}\left(\vec{r_{1}}\right)\Psi_{100}\left(\vec{r_{2}}\right) + \hat{H}\left(\vec{r_{2}}\right)}\Psi_{100}\left(\vec{r_{1}}\right)\Psi_{100}\left(\vec{r_{2}}\right) = 2E_{1}\Psi_{100}\left(\vec{r_{1}}\right)\Psi_{100}\left(\vec{r_{2}}\right)$$

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

"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???