- #1
Rafael
- 5
- 0
Hi, I'm having trouble understanding angular moment of the one electron hydrogen atom.
Solving Schrodinger equation on a referece system (say S) I get the energy eigenstates. They depend on three quantum numbers, n, l, m
[tex] \frac{-ħ}{2 m}\nabla^{2} \Psi - \frac{e^{2}}{4 \pi \epsilon r} \Psi = E\Psi → \Psi_{nlm} [/tex]
n measures energy, l measures L2 and m measures Lz.
[tex]L_z = mħ[/tex]
[tex]L^{2} = l(l+1)ħ^{2}[/tex]
From what I understand, Lz is quantized, but Lx and Ly not. So if the electron is in state Ψnml or whatever superposition of the energy eigenstates, it has Lz quantized.
This is when my question arises, if I use another referece system (say S') which is rotated with respect S, and I solve Schrodinger equation again, I get another set of energy eigenstates Ψn'm'l'which have Lz' quantized. But this is nosense because I can choose that the axis z' to be the old axis x, which have not Lx quantized.
What I am missing?
How does the two solutions (on S and S') compare?
Solving Schrodinger equation on a referece system (say S) I get the energy eigenstates. They depend on three quantum numbers, n, l, m
[tex] \frac{-ħ}{2 m}\nabla^{2} \Psi - \frac{e^{2}}{4 \pi \epsilon r} \Psi = E\Psi → \Psi_{nlm} [/tex]
n measures energy, l measures L2 and m measures Lz.
[tex]L_z = mħ[/tex]
[tex]L^{2} = l(l+1)ħ^{2}[/tex]
From what I understand, Lz is quantized, but Lx and Ly not. So if the electron is in state Ψnml or whatever superposition of the energy eigenstates, it has Lz quantized.
This is when my question arises, if I use another referece system (say S') which is rotated with respect S, and I solve Schrodinger equation again, I get another set of energy eigenstates Ψn'm'l'which have Lz' quantized. But this is nosense because I can choose that the axis z' to be the old axis x, which have not Lx quantized.
What I am missing?
How does the two solutions (on S and S') compare?