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Suppose two non interacting electrons, in a time independent potential, described by the equation:

\begin{equation} {H} \psi(r_1, r_2) = \frac{-\hbar^2}{2m} (\nabla^2_1 + \nabla_2^2) \psi(r_1, r_2) + V(r_1) \,\psi + V(r_2) \,\psi = E \end{equation} We can separate this equation supposing $\psi(r_1, r_2) = \psi_1(r_1) \psi_2(r_2)$. In the process of solving we get $E = E_1 + E_2 $, so:

\begin{equation} H \psi_1(r_1) \psi(r_2) = (E_1 + E_2)\psi_1(r_1) \psi(r_2) = E\psi_1(r_1) \psi(r_2) \end{equation}

Now if I say that the total wave function must be antisymmetric, I can combine both solutions:

$\psi_{TOTAL} = \frac{1}{\sqrt{2}}[\psi_1(r_1) \psi_2(r_1) - \psi_2(r_1) \psi_1(r_2)]$

But now, this wave function is not necessarily an eigenvector of $H$, so does writing the wave function like this change the energy of the total state? I know the first term of the total wave function is eigenvector of $H$, but the second one not, so what are the possibles energies for this system?

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In your particular case where you consider separable wave functions (Hamiltonian has your particles non-interacting), I don't get why you say that the form $\psi_{TOTAL} = \frac{1}{\sqrt{2}}[\psi_1(r_1) \psi_2(r_1) - \psi_2(r_1) \psi_1(r_2)]$ is not necessarily a solution.

Indeed, this is just a linear combination of Hartree products of one-electron wavefunctions. You were OK to accept those procuts individually as a solution, so why would a linear combination not be?

Plus, if you do understand that $<\psi_1(r_1) \psi(r_2)| H |\psi_1(r_1) \psi(r_2)> = E_1 + E_2 = E$, you can check very easily what is going on for your slater determinant: $$ H |\psi_1(r_1) \psi_2(r_2)-\psi_2(r_1) \psi_1(r_2)> = H |\psi_1(r_1) \psi_2(r_2)> -H |\psi_2(r_1) \psi_1(r_2)>$$ $$= (E1+E2)|\psi_1(r_1) \psi_2(r_2)> - (E1+E2)|\psi_2(r_1) \psi_1(r_2)> = E |\psi_1(r_1) \psi_2(r_2)-\psi_2(r_1) \psi_1(r_2)>$$

so that $<\psi_{TOTAL}|H|\psi_{TOTAL}>=E$, giving you the same Energy

EDIT: To see that the exchanged product yields the same value, consider that you can spearate your Hamiltonian in two: $H=H_1+H_2$ where $$H_i= \frac{-\hbar^2}{2m} \nabla^2_i + V(r_i) $$ each of those hamiltonian acts only on one particle so you have: $$H |\psi_1(r_1) \psi_2(r_2)> = (H_1 |\psi_1(r_1)>)\otimes|\psi_2(r_2)> + \psi_1(r_1)> \otimes (H_2 |\psi_2(r_2)>) $$ with $H_1 |\psi_1(r_1)>=E_1|\psi_1(r_1)>$ and $H_2 |\psi_2(r_2)>=E_2|\psi_2(r_2)>$

Now, you can realize that the two hamiltonians are the same, only they are acting on different particles. Right? So if you have $H_1 |\psi_1(r_1)>=E_1|\psi_1(r_1)>$, you MUST have $H_2 |\psi_1(r_2)>=E_1|\psi_1(r_2)>$ because H1 and H2 are those same hamiltonians (only acting on r1 or r2), and give the same eigenvalue for the same eigenvector. Therefore: $$H|\psi_2(r1)\psi_1(r2)>=(H_1 |\psi_2(r_1)>)\otimes|\psi_1(r_2)> + \psi_2(r_1)> \otimes (H_2 |\psi_1(r_2)>) $$ $$=E_2 |\psi_2(r_1)>\otimes|\psi_1(r_2)> + E_1\psi_2(r_1)> \otimes |\psi_1(r_2)>=(E_1+E_2)|\psi_2(r_1)\psi_1(r_2)>$$

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  • $\begingroup$ I know it is a solution, my question was if it was a solution with the same eigenvalue. The point of confusion is because I can't see why $ H \psi_2(r_1) \psi_1(r_2) = E_2 +E_1 $. If I only know the case with changed coordinates. $\endgroup$
    – Socrates
    Nov 1, 2019 at 4:04
  • $\begingroup$ Ask yourself why you accepted in the first place that $\psi_1(r_1) \psi_2(r_2)$ was an eigenvector of H with eigenvalue $E_1+E_2$. The same line of thought should lead you to accept the same for the exchanged wavefunction, since the Hamiltonain is the same for both particles. On a side note, you are writing your eigenvalue relationship wrong; the hamiltonian does not reduce a wavefunction to a real value $\endgroup$
    – Barbaud Julien
    Nov 1, 2019 at 4:14
  • $\begingroup$ I supposed that I solved the equation and found out, now if you solve for the other case why would you get the same $\endgroup$
    – Socrates
    Nov 1, 2019 at 4:15
  • $\begingroup$ In the process of separating variables, whe define two constans, in the end we get that $E$ is the sum of both. $\endgroup$
    – Socrates
    Nov 1, 2019 at 4:17
  • $\begingroup$ edited to answer by separating the hamiltonian $\endgroup$
    – Barbaud Julien
    Nov 1, 2019 at 4:39

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