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Can we talk about the entropy for just one electron, such as the electron in hydrogen atom by the famous Boltzmann's formula?

$$S=k_B\ln\Omega$$

I guess the answer is OK.

But in thermodynamics, we also define the relation closely related to entropy

$$ dS = \dfrac{dQ}{T} $$

But it will be nonsense to talk about the temperature. So What's the paradox in both definitions?

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  • $\begingroup$ There is no paradox, but you are trying to apply thermodynamics in a regime where it is not applicable, i.e far from the thermodynamic limit. $\endgroup$
    – Nephente
    Jan 21, 2017 at 12:54
  • $\begingroup$ @Nephente You mean the thermodynamics will survive just in the limit of N/V finite. $\endgroup$
    – Jack
    Jan 21, 2017 at 13:00
  • $\begingroup$ Temperature is still well defined, the electron can absorb and emit energy. $\endgroup$
    – user126422
    Jan 21, 2017 at 18:22
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    $\begingroup$ But temperature is a macroscopic quantity in essence. $\endgroup$
    – Jack
    Jan 22, 2017 at 1:24
  • $\begingroup$ I would recommend reading this other post. It does not answer your question, but it's related. $\endgroup$
    – DanielSank
    Jan 22, 2017 at 8:14

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The bare Boltzmann formula is a bit basic to make sense of for a lone electron. You need to talk about a generalization: one thinks in terms of the Shannon entropy of the state.

We also need to be careful in that the entropy becomes conditional on knowledge that we already have.

An electron in a known pure quantum state has zero entropy: we know its state perfectly and need no information to define it.

If however, if it is in a mixed state, then there is indeed a nonzero entropy. Let's say the spin up / spin down is unknown to us; that is, it is in a classical probabilistic mixture of pure spin states. This situation might have arisen because we have sampled the electron from an ensemble, or it might have arisen through something like the Wigner's Friend thought experiment, where it is known that the is in a spin eigenstate but not which one by Wigner. Wigner's friend has all the measurement results, so right after the measurement, the entropy conditioned on being Wigner's friend is nought. If there are probabilities $p$ and $1-p$ of the electron's being spin up / down respectively, then the entropy conditioned on being Wigner is $-p\,\log p -(1-p)\,\log p$ (multiply by the Boltzmann constant if you want to give it the same dimensions as $Q/T$. More generally, if a quantum system is in a classical mixture of pure states described by Density Matrix $\rho$, then the above formula generalizes to the von Neumann entropy:

$$S=-\mathrm{tr}(\rho\,\log\rho)$$

Now to the temperature of an electron. Temperature is a parameter of a statistical distribution, namely it defines the Boltzmann distribution of an equilibrium ensemble of particles. As such, the notion is not directly applicable to a lone electron: we simply don't have a system of particles in thermodynamic equilibrium! However, we might have sampled the lone electron from an ensemble of electrons in thermodynamic equilibrium at temperature $T$. We can therefore think of its energy state as a classical mixture of energy eigenstates, and the von Neumann entropy of this mixture is (modulo multiplication by the Boltzmann constant) precisely the Gibbs entropy of the ensemble, calculated as a per-particle average entropy. One can then say that the electron is from a population at temperature $T$. If, for example, we add a small amount of heat to the electron population just before the sampling, then the population's entropy changes by $\mathrm{d}Q/T$, and this further entropy, divided per particle, is the change in the entropy attributable to the heat addition of the electron as a mixture of energy eigenstates.

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