- #1
physicsphreak2
- 13
- 0
I know that for hydrogenic wavefunctions, the parity of a given state is (-1)^l. But does this mean that the probability densities for any such wavefunction is ALWAYS even?
I'm trying to understand the Stark effect, and specifically why there is no first-order correction for he ground state of atoms whose valence electron is in a p-orbital. The only reason I can think of is that we must have: \Delta E^{(1)} = \langle \psi_0 | z | \psi_0 \rangle = 0 (for a field along z)... so we end up with the integral of |\psi_0 |^2 z dz... and if we want to argue that this is the product of an even and odd function (which is odd, therefore integrating to 0), it seems like the probability density (as opposed to the underlying state) must always be even. Is this true?
I'm trying to understand the Stark effect, and specifically why there is no first-order correction for he ground state of atoms whose valence electron is in a p-orbital. The only reason I can think of is that we must have: \Delta E^{(1)} = \langle \psi_0 | z | \psi_0 \rangle = 0 (for a field along z)... so we end up with the integral of |\psi_0 |^2 z dz... and if we want to argue that this is the product of an even and odd function (which is odd, therefore integrating to 0), it seems like the probability density (as opposed to the underlying state) must always be even. Is this true?
