I Does putting a hydrogen atom in a box mix angular momentum states?

[Twigg]

If you put a hydrogen atom in a box ($\psi=0$ on the walls of the box), spherical symmetry will be broken so $n$,$l$,$m_l$ are no longer guaranteed to be good quantum numbers. In general, the new solutions will be a linear combination of all the $|n,l,m_l\rangle$ states. I know that $l$ and $m_l$ are good quantum numbers whenever the potential or boundary conditions are spherically symmetric (only depend on the radial coordinate). I know the box boundary condition depends on both $\theta$ and $\phi$ in spherical coordinates. Is that enough to conclude that $l$ and $m_l$ no longer good quantum numbers?

The reason I'm unsure is that the act of putting a box around the hydrogen atom doesn't feel like it would torque the wavefunction (maybe shear it, but not torque it), so it seems weird that the expectation value of the angular momentum should change. So, I would expect $\langle L^2 \rangle$ to still be $l(l+1)$ and $\langle L_z \rangle$ to still be $\hbar m_l$. Is this even possible, if the wavefunctions are a linear combination of different $|l,m_l \rangle$ states? Am I talking nonsense? Sorry!

 

[PeterDonis]

The reason I'm unsure is that the act of putting a box around the hydrogen atom doesn't feel like it would torque the wavefunction (maybe shear it, but not torque it), so it seems weird that the expectation value of the angular momentum should change

Change from what? There is no single expectation value of angular momentum that must always be the case if spherical symmetry is not broken. Different states can have different expectation values of angular momentum. That will still be true with your different boundary condition.

What will change is which states are eigenstates of the new Hamiltonian that you have induced by making the boundary a box instead of a sphere.

the act of putting a box around the hydrogen atom doesn't feel like it would torque the wavefunction

Are you envisioning actually taking a hydrogen atom that was in a spherical cavity and putting it in a box-shaped cavity? Of course you could do this, and you could do it without applying any torque to the atom (at least in principle), but that doesn't seem to be what you're envisioning in the first part of your post. There it just seems like you're asking what the effect of the different boundary conditions will be on the space of solutions. That's a different question.

 

[Twigg]

What will change is which states are eigenstates of the new Hamiltonian that you have induced by making the boundary a box instead of a sphere.

Are you envisioning actually taking a hydrogen atom that was in a spherical cavity and putting it in a box-shaped cavity? Of course you could do this, and you could do it without applying any torque to the atom (at least in principle), but that doesn't seem to be what you're envisioning in the first part of your post. There it just seems like you're asking what the effect of the different boundary conditions will be on the space of solutions. That's a different question.

Oh, yeah that makes sense. I confused the eigenstates of the atom-in-a-box with the non-stationary states you would get by starting with a hydrogen atom in free space in an eigenstate and "turning on" the box boundary condition. The one has no bearing on the other. I think that settles it. Thanks!

Edit: To clarify, I'm now feeling confident that the atom-in-a-box eigenstates would be linear combinations of $|l,m_l\rangle$'s for different $l$'s and $m_l$'s.