I Question about one electron hydrogen atom angular moment

[Rafael]

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

\frac{-ħ}{2 m}\nabla^{2} \Psi - \frac{e^{2}}{4 \pi \epsilon r} \Psi = E\Psi → \Psi_{nlm}

n measures energy, l measures L² and m measures L_z.

L_z = mħ
L^{2} = l(l+1)ħ^{2}

From what I understand, L_z is quantized, but L_x and L_y not. So if the electron is in state Ψ_nml or whatever superposition of the energy eigenstates, it has L_z 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 Ψ_n'm'l'which have L_z' quantized. But this is nosense because I can choose that the axis z' to be the old axis x, which have not L_x quantized.

What I am missing?
How does the two solutions (on S and S') compare?

 

[Paul Colby]

Well, the set S' are linear combinations of the S and visa versa. They form a $2l+1$ dimensional representation of the rotation group.

 

[king vitamin]

From what I understand, Lz is quantized, but Lx and Ly not.

This is not true, those operators are also quantized. However, if you choose a basis where the Lz operator is diagonal, the Lx and Ly will not be diagonal. Their eigenvalues are discrete, but the Lz eigenstates are superpositions of Lx/Ly eigenstates. Also, note that the different Lz states which share the same L^2 eigenvalue are all degenerate, and similarly, these non-diagonal Lx and Ly states will also be degenerate. You can always switch to a basis where these other states are diagonal (and then Lz will not be diagonal).

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

What you are looking for are the Wigner D-matrices: https://en.wikipedia.org/wiki/Wigner_D-matrix. This rotates the basis between different choices of which direction is diagonal.

One last comment:

n measures energy, l measures L2 and m measures Lz.

I want to mention that in a general problem with conserved angular momentum, the energy depends on both n and l, which also results in less degeneracy (but all states with the same l would still have the same energy). The Hydrogen atom is special in that the energy doesn't depend on l (this is related to some extra symmetry for 1/r potentials which is not present in general).

 

[PeroK]

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

\frac{-ħ}{2 m}\nabla^{2} \Psi - \frac{e^{2}}{4 \pi \epsilon r} \Psi = E\Psi → \Psi_{nlm}

n measures energy, l measures L² and m measures L_z.

L_z = mħ
L^{2} = l(l+1)ħ^{2}

From what I understand, L_z is quantized, but L_x and L_y not. So if the electron is in state Ψ_nml or whatever superposition of the energy eigenstates, it has L_z 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 Ψ_n'm'l'which have L_z' quantized. But this is nosense because I can choose that the axis z' to be the old axis x, which have not L_x quantized.

What I am missing?
How does the two solutions (on S and S') compare?

Note that these z-based eigenstates simply form a complete basis for all possible solutions. There is actually nothing special about the chosen z-axis. If you choose the x-axis instead of the z-axis, then you'd get a rotated set of basis eigenstates. And, any state of the hydrogen atom could be described as a combination of your $\psi_{nlm}$ states (using the z-axis) or your $\psi'_{nlm}$ states (using the x-axis).

These eigenstates do not represent "the only states that the hydrogen atom can have" - that would be a misinterpretation. Choosing the z-axis simply gives you a particular basis with which to describe all possible states.

Note also that the expected value of $L_x, L_y, L_z$ is not quantized. For a given state the expected value can be anything between $\pm \frac{\hbar}{2}$.