Confusion with superposition of states

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ntt
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Summary:

Can an electron be in a superposition of energy eigenstates during relaxation after excited?
Since my major is not physics. My QM knowledge is not pretty good (Mostly self study). I am sorry if this question was asked multiple times in the forum.

I've learned that wavefunction can be written as a linear combination of eigenfunctions due to completeness property.
If an electron is excited. When it is relaxing back to ground state. Can the state of the electron during relaxation a superposition of two discrete energy eigenstates?

If so, during relaxation, is there dynamical law that describes the change of coefficient of each eigenstate with time? (I've looked up the spontaneous emission but still don't have a clear idea what governs the time of relaxation like the design of laser which utilizes a metastable level)

From my understanding, when we observed photon emitted from relaxation, this means we already observe the system which the wave function then has already collapsed to the ground state and the relaxation time is very short. Is my understanding correct?

Any interpretation and correction of my concept and additional reading resources would be greatly appreciated.

Thank everyone in advance for the replies.
 

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vanhees71
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You have to carefully distinguish about which eigenstates you talk.

Take a hydrogen atom as a concrete example. You first treat it just as an electron in the Coulomb potential of the nucleus. Then you can evaluate the energy eigenstates of this Hamiltonian. If this were the full story, and your atom is in some energy eigenstate at ##t=0##, then it would stay in this state forever.

Now, however, there's also the electromagnetic field itself, and it's a quantum field, i.e., there are quantum fluctuations, leading to a coupling of the electron to the electromagnetic field. Taking this additional part of the Hamiltonian into account, which can be done perturbatively, you'll get a finite probability for a spontaneous transition from an excited atomic state to a lower atomic state and the emission of a photon.
 
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  • #3
ntt
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You have to carefully distinguish about which eigenstates you talk.

Take a hydrogen atom as a concrete example. You first treat it just as an electron in the Coulomb potential of the nucleus. Then you can evaluate the energy eigenstates of this Hamiltonian. If this were the full story, and your atom is in some energy eigenstate at ##t=0##, then it would stay in this state forever.

Now, however, there's also the electromagnetic field itself, and it's a quantum field, i.e., there are quantum fluctuations, leading to a coupling of the electron to the electromagnetic field. Taking this additional part of the Hamiltonian into account, which can be done perturbatively, you'll get a finite probability for a spontaneous transition from an excited atomic state to a lower atomic state and the emission of a photon.
Thank you very much for your reply.

Your reply lead my reading to Fermi's Golden rule which is what I was looking for.
 
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