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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?



1 Answer 1


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.

In order to explain the discrepancy between expected theoretical result and the experimental one, you should take into account the repulsive interaction between the electron, which makes the hamiltonian non-separable and hence the wavefunction non-factorizable. A more complicated approach is need to takle the interacting system.

  • $\begingroup$ 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? $\endgroup$
    – user100411
    Jan 28, 2017 at 16:18
  • $\begingroup$ Yes, that's why I said that nothing changes. Concerning the second part of your comment, the word "primitive" is not official. Actually, to my knowledge, a standard definition of not-yet-(anti)symmetrized "wavefunction" doesn't exist. I use it to underline the fact that it cannot be interpreted as a wavefunction yet, but it's the seed to build the actual wavefunction. And yes, it's simply the product of stationary states. $\endgroup$
    – AndreaPaco
    Jan 29, 2017 at 0:51
  • $\begingroup$ @JohnDoe, you should consider accepting the answer if you're happy with it. Otherwise, please give the community more details about your question. $\endgroup$
    – AndreaPaco
    Jan 31, 2017 at 9:31

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