ATAR Notes: Forum

Uni Stuff => Science => Faculties => Physics => Topic started by: Cthulhu on April 26, 2010, 11:04:01 pm

Title: Exchange symmetry
Post by: Cthulhu on April 26, 2010, 11:04:01 pm: I have a wavefunction $\Psi(x_1,x_2)$ that when acted upon an operator $\mathcal{P}\Psi(x_1,x_2) = \Psi(x_2,x_1)$
I've found the eigenvalues $\lambda = \pm 1$ Now I need to show that the probability density is unaffected by particle interchange I know its got something to do with the eigenvalues but ... I don't know whats going on.
Title: Re: Exchange symmetry
Post by: mark_alec on May 18, 2010, 10:20:15 pm: You know that P[Psi] = +/-1*Psi. But the probability density is |Psi|^2 which is unaffected by a negative sign. Hence an exchange operation does not change the probability density.