Two-particle system wave function

Question

I have a similar question as Two particles system

That is:

why for two-particle without interaction will have wave function $\psi(x_1,x_2)=\psi_a(x_1)\psi_b(x_2)$

And when we exchange it will have the form $\psi(x_2,x_1)=\pm\psi(x_1,x_2)$.

and expression $\psi(x_1,x_2)=A[\psi_a(x_1)\psi_b(x_2)\pm\psi_a(x_2)\psi_b(x_1)]$

I was a bit confused by the first answer in the post above,why the last phase is irrelevant, so you get just the product of individual wavefunctions in $\Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2) e^{i\phi}$ . Since the point here is $\phi$ is not a constant,it depend on position $(x_1,x_2)$,even if it's constant,why we can ignore it?

And I get lost by the solution to the second question provided on the post,that is why $\Psi(x_1,x_2)=e^{i\phi}\Psi(x_2,x_1)$ implies $\Psi(x_2,x_1)=e^{i\phi}\Psi(x_1,x_2)$ ,since $e^{i\phi(x_1,x_2)}$ is function of ordered pair $(x_1,x_2)$,when we exchange $(x_1,x_2) \to (x_2,x_1)$why it has the same form?

I found another post it seems more reasonable solution

Answer 1

This is done for identical particles (really in QM we cannot distinguish between the two particles for ex. electrons or bosons)

consider some operator $\hat{\rho}$ which swaps two particles A and B.

$\hat{\rho} \psi(A,B)= e^{\iota\theta} \psi(A,B) $,

where $\psi(A,B)$ is the amplitude of wavefunction, which under swapping operation picks up a phase.

Now if we operate it twice we must get the same wavefunction,

$\hat{\rho}\hat{\rho} \psi(A,B)=\psi(B,A) = (e^{\iota\theta})^2 \psi(A,B) $

so $(e^{\iota\theta})^2 = 1$

so $e^{\iota\theta} = \pm 1$

Hence we get $\psi(B,A) = \pm \psi(A,B)$

Now suppose our particles are in states $\psi(A), \phi(B)$, to make then indistinguishable under swapping of A and B, we write them as superposition,

$\psi(A,B) = C[\psi(A)\phi(B)\pm\psi(B)\psi(A)]$,

now you try swapping them, you will get

$\psi(B,A) = \pm \psi(A,B)$