Could the electron in the outer shell orbital be entangled with an electron within the inner S orbit of the atom? If so, how would this affect the properties of such an atom, such as emission and reactivity?
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They are entangled. The electrons are in a Slater determinant of all the atomic orbitals, so if there are $n$ electrons in states $1,2,3,4,\cdots n$, where the basis state is written $|1,2,3,\cdots n\rangle$ is the first electron in the first state and so on, the atomic state is:
$$|\psi\rangle = \frac 1 {\sqrt n}\sum_{\pi\in S_n}{\sigma(\pi)}|\pi(1,2,\cdots,n)\rangle$$
where $S_n$ is the symmetric group on $n$-letters, $\pi$ is a permutation operator and $\sigma(\pi)$ is it's parity.
Of course this is contract to the way we colloquially talk about filling states, e.g, "In lithium, the $1S$ shell is filled so the next electron goes into the $2S$ state"... but that's not an antisymmetric state.
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$\begingroup$ Thank you for your answer JEB. I am going to be absolutely honest, some parts of your answer go over my head so pardon my ignorance with this followup question. If these electrons are entangled, shouldn't, as an example, the EM energy emitted/absorbed by an outer shell electron also happen to one in the inner shell; which as far as I am aware, does not actually happen. How can this be? $\endgroup$ Commented Sep 14, 2021 at 16:04
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$\begingroup$ @C-Consciousness You should just write one down for lithium. I didn't have time because it's messy, but straightforward...and I forget atomic notation. Your comment is correct, all the electrons participate in photon exchange. Sometime our language simplifies QM, so don't forget we're taking linguistic short cuts. Also: electrons are indistinguishable: there is no such thing "the inner shell electrons", each of the 92 electron in U238 have $\sqrt{2/92}$ amplitude in the inner shell...you can never tell which $e-$ you are dealing with. $\endgroup$– JEBCommented Sep 15, 2021 at 13:00