The Hamiltonian of helium can be expressed as the sum of two hydrogen Hamiltonians and that of the Coulomb interaction of two electrons.
$$\hat H = \hat H_1 + \hat H_2 + \hat H_{1,2}.$$
The wave function for parahelium (spin = 0) is
$\psi(1,2) = \psi_S(r_1, r_2)\dot \xi_A(s_1, s_2)$ with the first being a symmetric spatial function and the second being an antisymmetric one.
We can separate this into the normalized function
$\psi_S(r_1,r_2) = \frac{1}{\sqrt{2}}[\psi_1(r_1)(\psi_2(r_2)+ \psi_1(r_2)(\psi_2(r_1)]=\psi_S(r_2,r_1)$
For orthohelium the functions look like this:
$\psi(1,2) = \psi_A(r_1, r_2)\dot \xi_S(s_1, s_2)$ $\psi_A(r_1,r_2) = \frac{1}{\sqrt{2}}[\psi_1(r_1)(\psi_2(r_2) - \psi_1(r_2)(\psi_2(r_1)]=-\psi_A(r_2,r_1)$
Show the ground state of helium is parahelium. The hint is what happens to the wavefunction.
OK, so I start with that the Hamiltonian of the given wave function(s)
$$H_1 = \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial^2 r_1^2} = E_1 \psi$$ $$H_2 = \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial^2 r_2^2} = E_2 \psi$$ $$H_{1,2} = -\frac{e^2}{4\pi\epsilon_0 r_{1,2}}$$
OK, I was trying to get a handle on how to get started with this. So I wanted to check if what I have above is "allowed" -- that is, is the second derivative (the nabla, really) of the psi functions treatable this way, since they all have two variables (really two position vectors) in them? Basically this is all about how to set up the initial differentials I would solve.
EDIT: One thing I thought of doing was this (for $H_1$):
$H_1 = -\frac{\hbar^2}{2m}(\frac{\partial \psi}{\partial^2 r_1^2}+\frac{\partial \psi}{\partial^2 r_2^2})=E_1(\psi_1(r_2)+\psi_1(r_1))$
but again I don't know if that's kosher.