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What happens if we re-excite an excited electron to a higher energy level? e.g. we excite hydrogen's electron from n=1 to n=3, then before it gets to return to the ground state, we hit it with another photon with the right frequency to move it to n=6:

1- is this even possible?

2- How does the electron releases it's energy and return to ground state?
A. will it release all of its energy as a single photon and return from n=6 to ground state in one step?
B. or it will first release a photon with the same frequency as the last photon it absorbed and return from n=6 to n=3, then release another photon and return from n=3 to n=1?
C. or it will return from n=6 to n=5 and release one photon, then from n=5 to n=4 and release another photon and so on?

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    $\begingroup$ This might help find references arxiv.org/abs/0904.2346 . It is all a matter of calculating probabilities for cascades.. $\endgroup$ – anna v Aug 19 '18 at 14:05
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  1. It is possible, but the probability to excite the electron to an even higher energy level is little, since the electron will stay at the higher (non-stable) energy level for a limited time as per QM.

  2. It is called relaxation, and it can go in multiple steps. Even though the electron was excited in one step or two steps as in your case, this does not mean that the relaxation is going to go in the same steps. Relaxation can occur in one step, when the electron returns to the ground state and releases one single photon with the corresponding energy. Or relaxation can go in multiple steps, with each step the electron releasing a photon with the corresponding energy levels.

A,B,C are all possible. It is all probabilities. The electron can return to ground state in one single step or multiple steps.

It is possible to excite the electron with multiple photons. Please see here: https://journals.aps.org/prb/abstract/10.1103/PhysRevB.97.134112

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    $\begingroup$ Your statement "the probability to excite the electron to an even higher energy level is little, since the electron will stay at the higher (non-stable) energy level for a limited time as per QM" doesn't always hold true, for example the electron could have been excited to a metastable state with a very long lifetime. $\endgroup$ – SabrinaChoice Aug 19 '18 at 20:59
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    $\begingroup$ Or, you can send both photons in at the same time. $\endgroup$ – SabrinaChoice Aug 19 '18 at 21:18
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To expand on Arpad's answer and offer a small correction:

It is indeed possible to excite an atom with a multiphoton process. You don't even need the intermediate state to be real, it can be 'virtual'. As long as the $n$ photons add up to the energy difference between states, it is possible to excite an electron in an atom to the higher energy state. For example: Say the energy difference between the ground and first excited state is $E=\hbar\omega = 2\pi\hbar*563THz$, then you can excite the transition with a single 532nm photon or two 1064nm photons.

You can also excite from one excited state to another, higher, excited state (like in your question). This is not necessarily 'unlikey' (..."the probability to excite the electron to an even higher energy level is little") as Arpad said; the intermediate state could be long-lived state (like a metastable state), or you can shine both lasers on your sample at the same time so that the necessary two photons arrive in quick succession.

If you take a path integral approach to the problem, the decay actually happens along all paths (in your example: A,B,C). The paths interfere constructively or destructively to give you the probability of a certain pathway. Some paths will dominate (be the most likely). You don't know which decay pathway is the most likely until you measure it over and over. With enough measurements, you will have a statistical distribution showing you the various likelihoods of different paths. In principle one could calculate this too.

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