Hi, I'm having trouble understanding angular moment of the one electron hydrogen atom.(adsbygoogle = window.adsbygoogle || []).push({});

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]

nmeasures energy,lmeasuresLand^{2}mmeasuresL._{z}

[tex]L_z = mħ[/tex]

[tex]L^{2} = l(l+1)ħ^{2}[/tex]

From what I understand,Lis quantized, but_{z}andL_{x}Lnot. So if the electron is in state_{y}Ψor whatever superposition of the energy eigenstates, it has_{nml}Lquantized._{z}

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 eigenstatesΨwhich have_{n'm'l'}Lquantized. But this is nosense because I can choose that the axis_{z'}z'to be the old axisx,which have notLquantized._{x}

What I am missing?

How does the two solutions (on S and S') compare?

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# I Question about one electron hydrogen atom angular moment

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