4
$\begingroup$

We are all familiar with the Boltzmann-Gibbs-Shannon entropy formula:

$H_{\text{BGS}} = -\sum_{k}p_{k}\log{p_{k}}$

In information theory, this can be interpreted as the expectation value of the "surprise."

Using the maximum entropy principle, one can derive the micro, macro, and grand canonical ensembles from this expression. However, when it comes to Bose-Einstein and Fermi-Dirac statistics, these cannot be derived directly. Instead, we need to use the following entropy expressions:

$H_{\text{FD}} = -\sum_{k}(1-p_{k})\log{(1-p_{k})} - \sum_{k}p_{k}\log{p_{k}}$

$H_{\text{BE}} = \sum_{k}(p_{k}+1)\log{(p_{k}+1)} - \sum_{k}p_{k}\log{p_{k}}$

Interestingly, both of these expressions include an additional term. It caught my attention that we have a +1 and a -1, similar to the adjustments in the Bose-Einstein and Fermi-Dirac distributions. I would like to understand the significance and physical meaning behind these terms. Is there an intuitive way to interpret them?

$\endgroup$
9
  • $\begingroup$ There was a misprint in $H_{\text{FD}}$. $\endgroup$
    – Gec
    Commented May 21, 2023 at 8:00
  • $\begingroup$ Something along these lines was asked before here on PSE several times, namely whether or not the FD or BE are really probabilistic distributions... $\endgroup$ Commented May 21, 2023 at 8:16
  • $\begingroup$ @TobiasFünke maybe this one? Probability density functions for the Bose-Einstein and Fermi-Dirac statistics. It' s useful and related (not a duplicate IMO). Same for: What is the probability that a single-particle bosonic quantum state is occupied? $\endgroup$
    – Quillo
    Commented May 21, 2023 at 10:31
  • 1
    $\begingroup$ Well, but the as the answer correctly points out, the $p_k$ in the latter two cases are no probabilities, so this is not really the Shannon entropy (as a functional of probabilitity distributions). $\endgroup$ Commented May 22, 2023 at 14:07
  • 1
    $\begingroup$ @TobiasFünke yes I wrote my comment before actually reading the answer. I now see how this could be seen as a duplicate (maybe more of a $\textit{soft}$ duplicate). $\endgroup$
    – IchVerlore
    Commented May 22, 2023 at 14:25

1 Answer 1

5
$\begingroup$

The similarity between $H_{\text{BGS}}$ and $H_{\text{FD}}$, $H_{\text{BE}}$ is formal. $H_{\text{BGS}}$ represents the entropy of a multiparticle system. In this case, $k$ enumerates the states of a multiparticle system and $p_k$ are the probabilities of these states. $H_{\text{FD}}$ and $H_{\text{BE}}$ express the entropy of quantum ideal gases. In this case, $k$ enumerates single-particle states, and $p_k$ are the average numbers of particles in these states, not probabilities. To emphasize the fact that $p_k$ are not probabilities in the later case, let's mention that in an ideal Bose-Einstein gas, $p_k$ can be greater than $1$.

Quantum ideal gases become classical in the limit of $p_k\ll 1$. In this limit, $H_{\text{FD}}$, $H_{\text{BE}}$ are formally reduced to $H_{\text{BGS}}$, while $p_k$ are still not probabilities, but the average number of particles. Thus, the additional terms in $H_{\text{FD}}$, $H_{\text{BE}}$ are related to the quantum statistics of indistinguishable particles.

$\endgroup$
2
  • $\begingroup$ I like this answer a lot. And now I see why @TobiasFünke said that this might be a duplicate. Of course $p_{k}$ aren't really probabilities. $\endgroup$
    – IchVerlore
    Commented May 22, 2023 at 14:12
  • $\begingroup$ @IchVerlore I'm glad if my answer was helpful $\endgroup$
    – Gec
    Commented May 22, 2023 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.