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Imagine we have a N-particle Hamiltonian without any interaction between the electron particles

$$ H = \sum_{i=1}^{N} \frac{p_i^2}{2m} + V(r_i)$$

then the solution to this equation $H\Psi = E \Psi$ will be a product wavefunction $\Psi = \psi_1..\psi_N.$

Now for this Hamiltonian it is true that all permutation operators $P$ commute with the Hamiltonian.

This means that there is a system of eigenfunctions to the Hamiltonian and the permutation operators and the eigenfunctions can be found by antisymmetrizing the single particle wavefunctions (Slater determinant). For this Hamiltonian we can apparently find a system of eigenfunctions that is completely antisymmetric.

Now assume that we have a Hamiltonian that does NOT commute with all permutation operators, because it acts differently on the electrons. For instance

$$H = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + V_1(r_1) + V_2(r_2),$$ where $V_1$ is very different from $V_2.$

What does second quantization tell us in this case, when the permutation operator does not commute with the Hamiltonian, are the "right" eigenstates of the Hamiltonian still antisymmetrized states?

Is there any physical difference in a symmetric in $N$-particles Hamiltonian vs a non-symmetric one?

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Now assume that we have a Hamiltonian that does NOT commute with all permutation operators, because it acts differently on the electrons. For instance H=K+V_1(r1)+V_2(r2), where V_1 is very different from V_2.

In the honest-to-god physical world that we live in you can not have $V_1$ being very different from $V_2$ because all electrons are identical, i.e., indistinguishable.

The anit-symmetrization postulate of quantum mechanics was derived from observations based on the honest-to-god physical world that we live in--a world in which all electron are indistinguishable. So this question is absurd. [Note to mods: I do not mean "absurd" in a rude way]

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  • $\begingroup$ well, you could think of two electrons being very far away from each other in different potentials, then this question is reasonable. I don't see your general problem. One could definitely think of constructing a model Hamiltonian this way, I would say. $\endgroup$ – Tokoyo Apr 10 '15 at 21:32
  • $\begingroup$ No, they would be subject to the same potential, but their positions would be different. E.g. evaluating 1/r at r=1 is "very different" from at r=10000. I.e., the values 1 an 1/10000 are "very different". The potential is the same. $\endgroup$ – hft Apr 10 '15 at 21:34
  • $\begingroup$ my idea is rather: Think about one electron being in a certain magnetic field that is restricted to the earth and the other one in a magnetic field restricted to mars. Do we need to use antisymmetrized states then or not? $\endgroup$ – Tokoyo Apr 10 '15 at 21:35
  • $\begingroup$ I feel like you are not hearing me... $\endgroup$ – hft Apr 10 '15 at 21:35
  • $\begingroup$ Anyways, the answer is yes, you need antisymmetrized states... $\endgroup$ – hft Apr 10 '15 at 21:36

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