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The hydrogen atom has one electron with specific energy levels allowed,

enter image description here

If the electron is at the lowest energy level -13.59, then it takes a photon with energy of more than 10.19 but less than 12.08 to send this electron definitely to the next allowed energy level of -3.40, is this right? Excess energy from the precisely allowed level will be radiated away as photon.

If the hydrogen atom's single electron is excited to very high energies, for example n=100, it is called a "Rydberg atom".

enter image description here

The very high energy electron results in a much "larger" radius of the hydrogen atom.

Suppose E is the energy necessary to send the electron from n=100 to n=102 level.

Then, if we place two such atoms near enough,

enter image description here

(https://www.researchgate.net/figure/A-model-for-the-dipole-dipole-interaction-between-two-Rydberg-atoms-States-1-and-2-are_fig2_228072829)

In this case, after successfully sending one atom to n=102 using a photon of energy E, the other atom near enough to the first atom can no longer be excitable to same level n=102 using a photon of the same energy E. A higher energy photon is needed to push it to the next energy level. The most it can settle for given E is now only to get to n=101.

I don't quite understand why this is so, if the energy level is exactly predetermined for each level, why will a nearby electron with high energy raise the energy requirement for its neighboring electron?

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    $\begingroup$ If the electron is at the lowest energy level -13.59, then it takes a photon with energy of more than 10.19 but less than 12.08 to send this electron definitely to the next allowed energy level of -3.40, is this right? No. A photon of energy 10.19 eV is dramatically more likely to be absorbed than one of 11 eV. This is why there are quite sharp absorption lines in spectra. When you studied classical mechanics, did you learn about driven oscillators and resonance? $\endgroup$
    – Ghoster
    Commented Dec 5, 2022 at 20:46
  • $\begingroup$ @Ghoster thank you. so it's far more like to happen if we give it nearly the exact energy required to reach the target level. $\endgroup$
    – James
    Commented Dec 5, 2022 at 20:48
  • $\begingroup$ Yes, that’s right. $\endgroup$
    – Ghoster
    Commented Dec 5, 2022 at 20:49

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The energy levels that you have written down come from a particular Hamiltonian, in which the potential energy is given by the Coulomb interaction between the nucleus (a single proton) and an electron for a single atom. When there are two atoms involved, the Hamiltonian is different, so one should generally expect the energy level diagram to change.

In this model, specifically, one must write the potential energy of the full Hamiltonian including the effects of both protons and electrons, one proton and one electron from each of the two atoms. These interact via the Coulomb force and thus shift each other's energy levels around. Specifically, the chance of an interaction increases with the energy level in a Rydberg atom (the effective radius for which each of the electrons feels each other's presence is given by the large Rydberg radius) and that interaction typically follows a $1/r^6$ or $1/r^{12}$ dependence originating from Van-der-Waals-type interactions.

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  • $\begingroup$ thank you vm. There is one last part i am unclear about: will E in this case most certainly send the first electron to n=102, or is there a significant chance of E sending the electron only to n=101 instead (considering @ghoster 's comment). If the raison d'etre for the process is "Any time you can bring two systems together in such a way that the final state of one particle depends on the input state of the other, you can make an entangled state by making that input state a quantum superposition. This will necessarily lead to a pair of particles each of which is in an indeterminate state[...] $\endgroup$
    – James
    Commented Dec 6, 2022 at 0:26
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    $\begingroup$ @James not so fast, keep in mind that resonance (exact energy) is important if you want a good probability of a transition. The two atoms that have been brought together have a set of energy levels that can be computed exactly, so you'd want a laser with close to that transition frequency if you want to send electrons to a particular state. Otherwise, off-resonant interactions will cause things like phase shifts but not much population transfer. $\endgroup$
    – Quantum Mechanic
    Commented Dec 6, 2022 at 1:31
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    $\begingroup$ @James The only time resonance "doesn't matter" is when you give enough energy to ionize the electron - that's because there's a continuum of free states, so any excess energy will just go into the kinetic energy of the now-unbound electron. $\endgroup$
    – Quantum Mechanic
    Commented Dec 6, 2022 at 1:31
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    $\begingroup$ @James I don't know the goal but there are a few common procedures. One is to excite a particular electron to n=102, bring another atom close to it, and show that you need difference laser frequencies to excite the electron from the other atom. Another is to have a group (two or more...) of atoms close together and shine a laser with a particular frequency on the whole group. Then you won't know which atom has its electron get excited, so you get a superposition state where the excitation is "shared" among the different atoms $\endgroup$
    – Quantum Mechanic
    Commented Dec 6, 2022 at 16:31
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    $\begingroup$ @James that's correct! The collective state is a "Dicke state" that is a superposition of each of the electrons being in the excited state. If we call the ground state $|g\rangle$ and the excited (Rydberg) state $|r\rangle$, we have the superposition $(|r\rangle\otimes |g\rangle+|g\rangle\otimes |r\rangle)/\sqrt{2}$, and similarly if there are more than 2 atoms $\endgroup$
    – Quantum Mechanic
    Commented Dec 6, 2022 at 19:41

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