Ignoring the angular momentum and spin states, the states of an electron in a hydrogen atom consist of the energy eigenstates $|\psi_n \rangle$ with quantum number $n$. Since there are (countably) infinitely many of such states $\Omega$ the electron can occupy, does this mean its entropy is infinite? Or does "the entropy" of this single electron not have physical meaning? Thanks!
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2$\begingroup$ With what probability does the electron occupy each of these states? All you've stated is that the maximum entropy is infinite, which isn't saying much. $\endgroup$– probably_someoneCommented Feb 25, 2018 at 3:04
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As in the comments, you need to specify the probability distribution for the electron eigenstates. This distribution can depend on the hydrogen atom's environment - if the hydrogen atom is in a resonant cavity filled with radiation, then the probability that the electron will be found in its higher energy eigenstates is different from when the hydrogen atom is cooled to a few microkelvins.
Given the probability distribution $p_i$ for the probability to find the electron in the $i^{th}$ energy eigenstate, the Shannon entropy of the electron in bits is:
$$S = - \sum\limits_i p_i\,\log_2 p_i$$
answered Feb 25, 2018 at 12:38
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1$\begingroup$ The subtlety of this problem is that there is a pileup of states when approaching the continuum, and an uncountable (in the technical mathematical sense) number of continuum states. The partition function will diverge unless you apply a cutoff related to the volume per atom. $\endgroup$ Commented Feb 25, 2018 at 22:39
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$\begingroup$ @BertBarrois The appropriate pdfs, like the MB distribution will do this. Also the sum is kind of an abstraction - when we consider unbound states, the sum becomes an integral and the appropriate measure, depending on the physical conditions modelled, will render the integral convergent. But i sense we may be saying the same thing in different words. $\endgroup$ Commented Sep 5, 2021 at 19:36