What is the entropy of a hydrogen atom, a bound proton and electron?

First attempt: The standard molar entropy of hydrogen gas is 130.68 $J \, mol^{-1} K^{-1}$ at $298 K$. $1 \, mol = 6.02214076×10^{23}$.

Therefore, one hydrogen atom has an entropy of $2.17 \times 10^{-22} J/K$.

A thermodynamic definition of entropy is $S=-k_B \sum_{i} p_i \log{p_i}$ where $p_i$ is the probability of a given microstate and $k_B$ is Boltzmann’s constant. This closely resembles the formula for Shannon entropy, $H=-\sum_i p_i \log{p_i}$ where $p_i$ is the probability of a message $m_i$ taken from some message space $M$. See info-entropy relationship.

Dividing by $k_B$ “yields” the Shannon information, which appears to be 15.72 nats, roughly three bytes. This disagrees with the linked question, which yields a much larger information content.

Second attempt:

The probabilities $ P_i $ can be obtained from the Boltzmann distribution:

$ P_i = \frac{{e^{-E_i/(k_B T)}}}{{\sum_j e^{-E_j/(k_B T)}}} $

where $ E_i $ is the energy of the $ i $th eigenstate.

The energy levels of a hydrogen atom are given by the formula:

$ E_n = -\frac{{13.6 \, \text{eV}}}{{n^2}} $

where $ n $ is the principal quantum number (1, 2, 3, ...).

At $ T = 300 K$, using the formula above:

  1. Calculate the probabilities for each energy eigenstate:

    $ P_n = \frac{{e^{-(-13.6 \, \text{eV})/(n^2 k_B \cdot 300 \, \text{K})}}}{{\sum_{j=1}^{\infty} e^{-(-13.6 \, \text{eV})/(j^2 k_B \cdot 300 \, \text{K})}}} $

  2. Calculate the entropy using the formula: $ S = -k_B \sum_{n=1}^{\infty} P_n \log(P_n) $

The sum in the denominator of $P_n$ diverges.

In this paper, the “Shannon entropy” is computed and appears to be something like $3+\ln(\pi)$. Any help in understanding where I’m going wrong would be greatly appreciated!

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1 Answer 1


Not a direct answer but there is a discussion in Peierls "Surprises in Theoretical Physics" about the fact that the partition function of the hydrogen atom diverges, and what one should make of this. One passage:

"We have to conclude, with some surprise, that the chance of the atom remaining in the ground state or in any other finite state is zero. To make this a little more quantitative, consider an atom in a finite volume, say a sphere of radius $a$. Then states extending over a radius much less than $a$ will have the same energy as in free space, and states extending much beyond $a$ will not exist. Since the mean radius is $a_0 n^2$, with $a_0$ the Bohr radius, we can find the right order of magnitude by cutting off the sum at $n = (a/a_0)^{1/2}$, which, for large $a$, gives $\frac{2}{3}(a/a_0)^{3/2}$. On the other hand, it is easy to see that the continuous spectrum contributes, in the same limit, an amount proportional to the volume, i.e. to $a^3$. So in a really large volume the dominant state for the electron is to be in a state of positive energy; the atom is ionized.

"If we are really dealing with a single atom in an infinite volume, this is physically the correct answer, because the equilibrium for ionization depends on the available volume. This is evident from the consideration of the rate of ionization and recombination, which must balance in equilibrium. Whatever the mechanism, the rate of ionization is independent of the volume, whereas the rate of recombination depends on the chance of the electron meeting the nucleus again, which is inversely proportional to the volume."


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