# Helium two particle system of identical particles The equations [5.27]-[5.31] is a description of a two particle system (one electron, each in a hydrogenic atom). The assumption seems to be that these electrons are distinguishable (see equation [5.28]), but the discussion of these equations below, from "Because $\psi_0$ is a symmetric function, the spin state has to antisymemtric...", seems to assume that we are working with identical fermions, hence required to satisfy the antisymmetrization condition. Why is there this discrepancy?

Proposed Answer: Is the reasoning that even though the equations above assume that the fermions are distinguishable, if we assume that they are identical fermions then we get the same ground state anyway. This follows since the spatial part would be $$\psi_{0}(r_1,r_2) = A[\psi_{100}(r_1)\psi_{100}(r_2) + \psi_{100}(r_1)\psi_{100}(r_2)]$$ which reduces to $$\psi_{0}(r_{1}, r_{2}) = \frac{8}{\pi a^3}e^{\frac{}-2(r_1 +r_2){a}}$$ anyway after normalization?

Thanks.

As usual, in the first part of a "many-body" problem, one starts finding a non (anti)simmetrized wavefunction. This means that, one is temporarily neglecting the indistinguishability of the particles. Once this sort of "primitive" solution has been found, one re-introduces the fact that we are working with $\textit{indistinguishable}$ particles. How? Manually writing the (anti)symmetrized version of the primitive solution. In your "proposed answer" you've correctly written the simmetric spatial part of the wavefunction of your system (provided that both electrons are in the same single-particle state). As you can notice, in this example, after normalizing, nothing changes with respect to the "primitive" wavefunction.
• By "nothing changes with respect to the primitive wavefunction", you mean that if we consider the $\psi_0$ of my proposed answer then since $\int \int |\psi_0(r_1, r_2)|dr_1 dr_2 = 1$ we can show that $A = \frac{1}{2}$, thus $\psi_0$ is equivalent to equation [5.30], it follows that nothing changes? Also are you using "primitive" simply because the wave function $\psi_0$ is a product of stationary states? – user100411 Jan 28 '17 at 16:18