Are electrons in atoms always in eigenstates?

[snoopies622]

TL;DR Summary

Do the electron quantum numbers imply that those electrons are always in the corresponding eigenstates for those observables?

Going back to high school chemistry, i remember being taught that the electrons in an atom can each be identified with four quantum numbers - one for energy, two for angular momentum and one for spin. These numbers are integers except for the spin quantum numbers, which are either 1/2 or -1/2.

To me this seems to imply that at all times, every electron in an atom is in a state which is simultaneously an eigenstate for energy, total angular momentum, the angular momentum in the z direction, and spin.

Is this true, and if so why would that be the case?

 

[Vanadium 50]

I'm not sure what this even means.

What would a measurement be that wasn't in an eigenstate? That might help clarify,

 

[snoopies622]

I know that measuring something puts it in an eigenstate for what's being measured, but that's not what i'm asking about.

 

[PeterDonis]

the electrons in an atom can each be identified with four quantum numbers - one for energy, two for angular momentum and one for spin.

These numbers don't identify electrons, they identify particular states in the Hilbert space that form a useful basis for the space. More precisely, they form a useful basis of states in the Hilbert space for a single electron in the potential well of the atomic nucleus. But that does not mean that individual electrons in the atom are actually in one of these states. Electrons are indistinguishable, so all of the electrons in multi-electron atoms are entangled.

To see what this means, consider the following examples:

(1) A Helium atom in the ground state. There are two electrons, both in the 1s orbital. That means the two-electron system (if we neglect the nucleus--which only works in some circumstances) is in a joint state with $n = 1$, $l = 0$, $m = 0$. The spins $s$ of the electrons are opposite, but they are entangled, so neither one is in a definite spin state; the best you can say is that the total spin of the two-electron system is $0$ (at least, I believe that is the case for the He ground state).

(2) A Carbon atom in the ground state. Here the electron configuration is usually given as 1s2-2s2-2p2, but that does not mean you can pick out two electrons as the 1s electrons, two as the 2s electrons, and two as the 2p electrons. All 6 of the electrons are entangled, so you can't say that any particular one of the 6 is in any particular orbital. The most you can say is that the joint 6-electron state has two electrons with $n = 1$, $l = 0$, $m = 0$, two electrons with $n = 2$, $l = 0$, $m = 0$, and two electrons with $n = 2$, $l = 1$, and no definite value of $m$ (since you can't pick out any particular p orbital). (And none of the electrons will have a definite value of $s$.)

 

[snoopies622]

Thank you Peter, that's very helpful!

 

[Vanadium 50]

This is sounding a lot like "what are the electrons doing when we aren't measuring them?" which is not something that is easily addressed. (How would you test it?) Besides the other problems, you have the problem that what a pair of electrons or a filled shell of electrons is well-defined, the behavior of an individual electron is not.

You cannot, for example, point at an electron and say, "That one is in a 1s shell, and that other one is in a 2p".

 

[LittleSchwinger]

TL;DR Summary: Do the electron quantum numbers imply that those electrons are always in the corresponding eigenstates for those observables?

Is this true, and if so why would that be the case?

No it's not true. Electrons are usually not in an eigenstate corresponding to the quantum numbers you mentioned.

Typically they would be in what is called a mixed state. There's also the problem of indistinguishability that you can't consistently label which electron is which.

More detail would be above B level, so I'll leave it there.