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.

  • $\begingroup$ That said, I think you are missing the Coulombic interactions in $H_1$ and $H_2$. $\endgroup$
    – Cameron Williams
    Commented May 14, 2014 at 3:21
  • $\begingroup$ $\large\Psi_{\rm S}$ and $\large\Psi_{\rm ortho}$ are not eigenfunctions of the total hamiltonian. It's usually done to compare energies in both cases. It just amounts to take the 'average' of $\large H_{12}$ with both functions and yields an estimated for the difference of energies. Basically, it's first order perturbation theory. In Bethe & Salpeter book, there are several approaches to the energies of the Helium atom. $\endgroup$ Commented May 14, 2014 at 3:39
  • $\begingroup$ OK, so what does that mean for setting up the initial part of the problem? $\endgroup$
    – Jesse
    Commented May 14, 2014 at 3:42

1 Answer 1


I'm not sure how exact you want to be about this but from my point of view in order to show that parahelium has a lower energy you don't need to know much about the actual wavefunctions. In fact it's impossible to write an exact wavefunction for helium because of the interaction term. I'm not going to answer this for you because this looks like a homework question but I will try to expand on the hint to make it easier for you to solve yourself. Because electrons are fermions we know that the wavefunction of this system must be totally antisymmetric under exchange of particles. In the parahelium case the spin state is antisymmetric under this exchange so we know the spatial wavefunction has to be symmetric. Think about the states you can construct under this restriction and compare to the case of orthohelium, where the electrons of the same spin cannot occupy the same energy state.

I feel like I should also add that your expressions for the hamiltonian are missing the $1/r$ potential terms but aside from that what you've written does look "kosher".

  • $\begingroup$ That helps and no, I wasn't looking for an answer the way you mean -- I just wanted to make sure I was starting it right. I wasn't sure it was OK to break up the $\psi(r_1,r_2)$ functions within the Hamiltonian that way. So I should have (With the potential terms) $H_1 = -\frac{\hbar^2}{2m}(\frac{\partial^2 \psi_1}{\partial r_1^2}+\frac{\partial^2 \psi}{\partial r_2^2}-\frac{2e^2}{4\pi \epsilon_0 r}-\frac{2e^2}{4\pi \epsilon_0 r})=E_1(\psi_1(r_2)+\psi_1(r_1))$ $\endgroup$
    – Jesse
    Commented May 14, 2014 at 11:19

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