Author Topic: Exchange symmetry (Read 3848 times) Tweet Share

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.

mark_alec · « **Reply #1 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.

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Author Topic: Exchange symmetry (Read 3848 times) Tweet Share

Cthulhu

Exchange symmetry

mark_alec

Re: Exchange symmetry

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