Identical particles far apart

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.

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: $$\langle x_1, x_2 | \psi_{tot} \rangle =\frac{1}{2} \left(\langle x_1 | \psi_A \rangle \langle x_2 | \psi_B \rangle \pm \langle x_2 | \psi_A \rangle \langle x_1 | \psi_B \rangle \right ) = \frac{1}{2}\left( \psi_A(x_1)\psi_B(x_2) \pm \psi_A(x_2)\psi_B(x_1) \right )$$
• 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? Commented Nov 8, 2022 at 16:39