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There are two identical particles, $a$ and $b$. The particle distance is large enough that interaction term, in the Hamiltonian, is negligible. The Hamiltonian of the system can be written as: $$\hat H=\hat H_a(x_a)+\hat H_b(x_b)$$ The eigestates are of the form $\psi(x_a,x_b)=\psi_a(x_a) \psi_b(x_b)$ where $\hat H_a\psi_a = E_a \psi_a$ and analogously, $\hat H_b\psi_b = E_b \psi_b$.

But there is a problem, if i switch the particle nothing change, because they are identical. So i have to write a symmetric, or an antisymmetric wave function. However, the classical simmetrization, or antisimmetrization does not work: $\frac 1 2 (\psi_(x_a) \psi_b(x_b) \pm \psi_a(x_b) \psi_b(x_a))$ is not an eigenstate of $\hat H$ because of the second term, indeed: $\hat H \psi_a(x_b) \psi_b(x_a) \neq E \psi_a(x_b) \psi_b(x_a)$.

So my question is, how can i create a symmetric, or antisymmetric state in this case? I thought that maybe, since particles are far apart, i should not worry with the symmetrization requirement. But on the the other hand, even if particles are very far, a switch between them does not change anything in the system.

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If you assume that you have identical particles then by definition the total Hamiltonian should be unchanged by exchange of the two particles.

Meaning that if you put particle $1$ in region of space $A$ its wave-function would be identical as if you were to put particle number $2$ in $A$ instead, the wave-function being $\psi_A$ in both case.

Maybe the confusion comes from the fact that you're using the same indexes for the region of space $A$, $B$ (or the parts of your total Hamiltonian) and the coordinates of the particles $x_1$, $x_2$.

So to answer your question you would have this for a symmetrized wave-function: $$ \langle x_1, x_2 | \psi_{tot} \rangle =\frac{1}{2} \left(\langle x_1 | \psi_A \rangle \langle x_2 | \psi_B \rangle \pm \langle x_2 | \psi_A \rangle \langle x_1 | \psi_B \rangle \right ) = \frac{1}{2}\left( \psi_A(x_1)\psi_B(x_2) \pm \psi_A(x_2)\psi_B(x_1) \right ) $$

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  • $\begingroup$ hi Mandr1, thanks for the answer. In the question the particles are $a$ and $b$. I don't understand the right hand term in your equation, are you sure it's correct? $\endgroup$
    – SimoBartz
    Nov 8, 2022 at 16:39

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