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


1 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

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

  • 1
    $\begingroup$ Why the exchange operator is linear operator so that $\hat{\rho}\hat{\rho} \psi(A,B)=\psi(B,A) = (e^{\iota\theta})^2 \psi(A,B)$? $\endgroup$
    – yi li
    Jul 18, 2020 at 7:26
  • $\begingroup$ I still bit confused by the first question why can ignore phase in $\Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2) e^{i\phi}$ $\endgroup$
    – yi li
    Jul 18, 2020 at 7:30
  • 1
    $\begingroup$ since the particles are identical, under swapping operation the particle wavefunction must evolve with a pure phase, otherwise, if the wavefunction changes, the probability density of measuring them will change, and the particles will no longer remain identical $\endgroup$
    – sawan kt
    Jul 18, 2020 at 7:30
  • $\begingroup$ we did not include phase in $ \Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2) e^{i\phi}$, because the particles are identical, so we cannot relate the two indistinguishable entities, so they must be independent $\endgroup$
    – sawan kt
    Jul 18, 2020 at 7:34
  • $\begingroup$ Thanks for your explanation, since independent is the notion based on $|\psi(x)|^2$, if we just consider the probability density it's ok to ignore phase, but here is the wave function itself, can you elaborate more on "because the particles are identical, so we cannot relate the two indistinguishable entities"? $\endgroup$
    – yi li
    Jul 18, 2020 at 7:46

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