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If we take two electron system of non – interacting (hypothetical, not found in nature) electrons, than we can define Hartree product as: $$ \Psi(x_1,x_2) = \psi_1(x_1) \psi_2(x_2) $$

Since electrons don't interact we can write Schrödinger equation for each of them separately: $$ H_1\psi_1(x_1) = E_1\psi_1(x_1) $$ $$ H_2\psi_2(x_2) = E_2\psi_2(x_2) $$

Also, because electrons don't interact, both Hamiltonian and the energy of the system is the sum of Hamiltonians and energies for each electron: $$ H = H_1 + H_2 $$ $$ E = E_1 + E_2 $$

Schrödinger equation for the system is: $$ H\Psi = E\Psi $$

Plugging previous equations into the left hand side of equation and assuming that Hartree product is the solution to Schrödinger equation of the system, we get: $$ (H_1 + H_2)\psi_1(x_1)\psi_2(x_2) = E\psi_1(x_1)\psi_2(x_2) $$

Taking into account that both Hamiltonian operators are one-electron and linear, we can pull $ \psi_2(x_2) $ out of $ H_1 $ and $ \psi_1(x_1) $ out of $ H_2 $ which yields: $$ \psi_2(x_2)H_1\psi_1(x_1) + \psi_1(x_1)H_2\psi_2(x_2) = E\psi_1(x_1)\psi_2(x_2) $$

Taking into account Schrödinger equation for each of electrons separately: $$ \psi_2(x_2)E_1\psi_1(x_1) + \psi_1(x_1)E_2\psi_2(x_2) $$

Hartree product can be factored which gives: $$ \psi_2(x_2)\psi_1(x_1) (E_1 + E_2) = E\psi_1(x_1)\psi_2(x_2) $$

Taking into account that for non - interacting electrons, energy of the system is the sum of energies of each electron, we proved that Hartree product is the exact wavefunction for the system of non - interacting electrons since both sides of the last equation are equal.

In more general case we can write wavefunction for the system of $N$ non – interacting electrons as: $$ \Psi(x_1,x_2,...,x_N) = \prod_{n=1}^N \psi_n(x_n) $$

Wavefunction of the system isn't antisymmetric to electron exchange which means that in such a hypotethical system of non – interacting electrons it is possible to distinguish between them (it is known which electron is in which orbital).

I don't understand how is that possible since, according to my understanding, electron indistinguishability is a consequence of them being quantum particles which means that they lack exact position prior to measurement.

What does such a fact have to do with electron interaction since previous result shows that if interactions don't exist, we can suddenly distinguish between them?

If we measure positions of electrons in such a system, we should be able to tell which electron is which (which electron is in which orbital), but how is that possible if we don't know where both of electrons will be before the measurement is done?

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This is a question of linearity. You have constructed a differential equation where $\Psi_{12} = \psi_1(x_1)\psi_2(x_2)$ is a solution with a particular eigenvalue $E_1 + E_2$. Suppose that the exchanged wavefunction $\Psi_{21} = \psi_2(x_1)\psi_1(x_2)$ is also a solution with the same eigenvalue. (For instance, perhaps the non-interacting electrons are moving in the same external potential, so $H_i \psi_j(x_i) = E_j \psi_j(x_i)$, but $H_i f(x_j)=0$ for $i\neq j$.)

If two functions $\Psi_{12}$ and $\Psi_{21}$ are solutions to the same linear differential equation with the same eigenvalue, then linear combinations like $\Psi_{12} + e^{i\theta}\Psi_{21}$ are also solutions. To find the "particular solution," you have to account for external constraints on the problem, such as boundary conditions. The requirement for antisymmetry under exchange is such an external requirement, which arises in a natural way from relativistic quantum mechanics but which must be "bolted on" to the non-relativistic Schrödinger equation.

If you like, you can think of

$$ \hat{\text{exchange}}_{12}\Psi = -\Psi $$ as an additional eigenvalue equation which must be solved simultaneously with the Hamiltonian.

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  • $\begingroup$ How do we know that exchanged wavefunction $\Psi_{21}$ is a solution with the same eigenvalue? This needs a proof. $\endgroup$ Jun 12 at 18:34
  • $\begingroup$ I gave an example in the answer of some Hamiltonian behaviors which probably guarantee this symmetry, but the $i$ and $j$ are annoyingly small. I'll let you come up with your own proof for your own model. (Or, ask a follow-up question.) $\endgroup$
    – rob
    Jun 12 at 19:17
  • $\begingroup$ Fair enough. Let us see what I can do. $\endgroup$ Jun 12 at 19:49

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