I Question about one electron hydrogen atom angular moment

1. Dec 8, 2018

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 Lz.

$$L_z = mħ$$
$$L^{2} = l(l+1)ħ^{2}$$

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?

2. Dec 8, 2018

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.

3. Dec 9, 2018

king vitamin

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).

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:
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).

4. Dec 9, 2018

PeroK

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 $\pm \frac{\hbar}{2}$.