# What is the physical meaning of exciting a single-electron transition in a many-electron system?

Let's say I have a gas of Li atoms in the $1s^2 2s^1$ ground state configuration and I want to excite the $2s \to 3s$ transition of the outermost electron with a laser pulse on resonance. (That is the mean photon energy is equal to the energy difference of the two levels.)

There are some things I don't understand about this basic situation, and I'd be happy if somebody could explain me that.

• The wavefunction of this interacting system is $$\Psi=\Psi(\textbf{r}_1,\sigma_1,\textbf{r}_2,\sigma_2,\textbf{r}_3,\sigma_3) \neq \Psi_{100}(\textbf{r}_1,\sigma_1) \Psi_{100}(\textbf{r}_2,\sigma_2)\Psi_{200}(\textbf{r}_3,\sigma_3)$$ that is it isn't separable into products of wavefunctions containing the spin and spatial coordinates of the individual electrons. Because of this configuration seems to have no physical or any kind of meaning to me, because it seems to assume this separability. What is the point of using it then?

• What does it mean to excite the outermost electron? There is only an entangled system of 3 electrons in which all of them are indistinguishable, thus it seems to me that there is no meaning in talking about "inner" or "outer" electrons or their transitions. The only meaningful thing is to talk about the 3-electron system as a whole and about the excitations of this whole system. Why is this language of "exciting the 2s electron to the 3s state" is used then? What is the meaning of it?

• If the usage of the language "exciting the outermost 2s electron to the 3s state" is correct, what physical measurement can I use to tell that this excitation has happened? Maybe somehow I could measure the energies of each of the electrons before and after the excitations and then I'd find that two energy values out of the 3 are the same before and after the laser pulse, but one energy I get has increased by the photon energy. But this method doesn't seem to help much, first because in an entangled system the individual particles are in mixed states and their energy measurement yield a statistical mixture of values (so at best all I could say is that the mean of one of the distributions has increased), and second, because I cannot label the electrons and associate the energy-value distributions to them (because they are indistinguishable) so again I can't say that there was an outermost electron in a certain state and now it is in another state. All I can say that the energy measurements on individual electrons previously yielded a distribution of values which was the sum of three probability distributions with certain means (e.g. $\overline{E}_1=\overline{E}_2<\overline{E}_3$) and deviations and now it's the sum of some other 3 distributions from which one maybe has a higher mean than before. [E.g. $\overline{E}'_1=\overline{E}'_2<\overline{E}'_3(>\overline{E}_3)$] But nothing says that the distribution with the higher mean comes from the electron which gave the highest mean before.

• What is the physical meaning of the electron configuration of an N-electron system?

• What is the meaning of exciting the outermost electron from one state to the other?

• What measurement tells me that the outermost electron has been excited? (Actually even better, if somebody can tell me how to measure if the outermost electron is in a mixed state or in a pure superposition state, the superposition being made up from the eigenstates of the 1-electron energy operator. The operator that corresponds to observing the 1-electron energy.)

This became a bit long, if something is not clear, please indicate, and I'll try to clear up. Thank you in advance.

• How do you know that your system is initially in the $1s^2 2s^1$ state? – probably_someone Feb 7 '18 at 15:40
• Let's say it's a gas at $0~K$. – user3237992 Feb 7 '18 at 17:35