This section of Griffiths Introduction to Quantum Mechanics deals with Boltzmann, Fermi-Dirac, and Bose-Einstein distributions. I don't understand this line (highlighted in yellow):

enter image description here

Let's talk only of Maxwell-Boltzmann here to keep it simple. Originally, we had

$$N_n=d_ne^{-(\alpha+\beta E_n)}$$

This was explained in the book to be the equation for the most probable occupation number for distinguishable particles. Then, in the image above, the author divides by $d_n$ to result in "the number of particles in a particular state with that energy", but I don't quite understand this. Could someone explain this bit in simpler terms? Or with a simple example?

  • $\begingroup$ I think equation 5.103 is “a mean occupation of the states with energy ε”, or, in other words, it’s “a probability of occupation”. $\endgroup$
    – Orient
    Nov 16, 2019 at 6:33

1 Answer 1


The formulas in Griffiths are correct, but the explanation is pretty clumsy, because he's basically done the derivation 'in reverse'. For simplicity I'll just talk about the distinguishable particle case, but the others are similar.

The derivation in the forward direction looks like this: the Maxwell-Boltzmann distribution is the distribution that maximizes the entropy given fixed energy. Here, the entropy is defined as $$S \sim \sum p_i \log p_i$$ and the $p_i$ are the probabilities of occupancies of each state (not each energy level!). If you carry out the constrained optimization, using a similar method to Griffiths, you'll arrive at equation 5.103.

Now, the probability of occupancy of a state only depends on its energy. Let's say that the probability of occupancy of a state at some energy is $p_n = 1/2$, and the degeneracy is $d_n = 10^6$. Then by the law of large numbers, the total occupancy $N_n$ of this entire energy level will be very close to $p_n d_n = (1/2) 10^6$. The occupancy could certainly be more or less, but the probability distribution will be peaked about this central value.

The only problem with this approach is that the definition of $S$ is a little unintuitive. So instead, Griffiths works only with occupancy numbers $N_n$, so he can just "count the number of ways" to achieve those numbers instead of dealing with the probabilities $p_n$. Then, he implicitly takes the high $d_n$ limit, so that $N_n \approx p_n d_n$, and calculates $p_n = N_n / d_n$.

The high $d_n$ limit is necessary so that the probability estimated by this ratio is accurate. For example, if $p_n = 2/3$ but $d_n = 10$, the most likely occupancy number could be $N_n = 7$. Then dividing would give the approximation $p_n \approx 0.7$. For our calculated value of $p_n$ to be good, we must take $d_n$ to infinity.

A final muddy point is that Griffiths accidentally calls the probabilities $p_n$ "the most likely occupancy numbers of a state", even though this makes no sense because $p_n$ isn't even an integer, it's a probability between $0$ and $1$. This clumsy wording is because Griffiths has swept all of the probability language under the rug in favor of occupancy numbers, but it's just not right.

  • $\begingroup$ Honestly, I've been struggling with some of the wording in this book for 2 semesters now. Do you have any resources that ideally explain the entire book in other words or maybe just explains this section in the book more intuitively? $\endgroup$
    – DarthVoid
    Jan 24, 2017 at 18:17
  • 1
    $\begingroup$ @DarthVoid I had basically the same problem with Griffiths, and ended up having to relearn everything. I think Shankar is a good alternative reference. If you already know stat mech, you can just flip to the back of most books on it for a better derivation of these distributions. $\endgroup$
    – knzhou
    Jan 24, 2017 at 18:26
  • $\begingroup$ I'll check out Shankar's book, thanks for the tip. $\endgroup$
    – DarthVoid
    Jan 24, 2017 at 21:43

Your Answer

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

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