Regarding the average occupation number for a Bose/Fermi gas we have:

$$\bar n_\epsilon=\frac 1 {e^{\beta(\epsilon_p - \mu)} \pm 1}$$.

Now the problem I am having has to do with the nomenclature of the above expression.

Some people call it : the average occupation number of an energy state with energy value $\epsilon_p$. But since we are observing a 1-Particle system, I don't think calling it like this makes sence.

Another way is (as I read in a pdf) : the distribution function $f(\epsilon_\nu)$ for the average occupation of a single-particle state $\nu$ with energy $\epsilon_\nu$ can be derived:

$$\bar n_\epsilon=\frac 1 {e^{\beta(\epsilon_p - \mu)} \pm 1}$$.

Is this correct? This would mean that the above expression is a Comulutative distribution function (CDF). When we want to know the average energy of the system (either a bose gas or a fermion gas) we calculate it in the following way:

$E=\int_0^\infty \epsilon g(\epsilon)\bar n_\epsilon d\epsilon $ .

If we observe a continoues variable X and it's average we have:

$\bar X=\int_a^b X \rho(x)dx$ where $\rho(x)$ is the PDF.

My questions are the following two:

  1. What is the proper nomenclature for what I called as "average occupation number of an energy state" ?

  2. If we compare the last 2 integrals does this mean that the product between the CDF ($\bar n_\epsilon$) and the density of states ($g(\epsilon)$) is a PDF?

  • $\begingroup$ Why do you think you're 'observing a 1-Particle system'? $\endgroup$ Commented Jul 10, 2021 at 19:52
  • $\begingroup$ Because that is what was said in the pdf that I was reading + it kind of makes sence, when you consider how we find the average number of particles for the system and the average energy as I showed above. I personally imagined $n_\epsilon$ expressing the aver. number of particles per energy state, when we observe all the microstates in which a system can be found. In each microstate for a particular energy state (value) that we are observing, the number of particles with this energy value varies from microstate to microstate. That's how I understand $n_\epsilon$ $\endgroup$
    – imbAF
    Commented Jul 10, 2021 at 19:57
  • $\begingroup$ What is CDF and PDF? $\endgroup$
    – ytlu
    Commented Jul 10, 2021 at 19:59
  • $\begingroup$ @ytlu cdf = Cumulative distribution function, pdf= probability distribution function $\endgroup$
    – imbAF
    Commented Jul 10, 2021 at 20:02
  • $\begingroup$ @Jakob well for example you can find the probability density of a particle being in height h in atmosphere, (you are observing a 1 particle system) and then by multiplying this PDF with the particle density at h=0, you find the number of particles at height h. You start with a single particle system and jump to a system of N particles. $\endgroup$
    – imbAF
    Commented Jul 10, 2021 at 20:04

1 Answer 1

  1. The term $\bar n$

$$\bar n(\epsilon) =\frac 1 {e^{\beta(\epsilon - \mu)} \pm 1}$$.

is known as the Fermi-Dirac distribution (the $+$ sign), and Bose-Einstein distribution for $-$ sign. The average occupation number of a state with energy $\epsilon$.

The chemical potential $\mu$ is to be determined from the following:

$$ N = \int_0^\infty g(\epsilon) \bar n(\epsilon) d\epsilon. $$ Where $g(\epsilon)$ is the density of states available, the number of states at energy $\epsilon$. The chemical potential is chosen to make this equality satisfied.

  1. In the area of solid state ( where these two distribution functions are mainly applied), the terms CDF and PDF are rarely heard.
  • $\begingroup$ And the fact that we are integrating from 0 to infinity, that means we are taking into consideration all the states (the state in which the one particle system can be found) and all the different values of energy that the states are having? $\endgroup$
    – imbAF
    Commented Jul 10, 2021 at 20:20
  • $\begingroup$ "The average occupation number of a state with energy " what it is meant with this? The probability that this state, with this particular energy value is being occupied by a particle? Or is the average number of particles which occupy the system? (meaning we are taking some sort of average over all the microstates in which the system ca be found)? Which of the two? $\endgroup$
    – imbAF
    Commented Jul 10, 2021 at 20:23
  • $\begingroup$ Yes. There must a lower bound for the available states. Otherwise, all particle will be pushed to the $-\infty$. We define the lower bound of state, $\epsilon = 0$. $\endgroup$
    – ytlu
    Commented Jul 10, 2021 at 20:24
  • $\begingroup$ It is a average occupation of a single state. The average occupation number can be ranging from $0$ to $\infty$. For Fermi-Dirac, the occupation of a state is either $0$ or $1$, rendering an average $0<\bar n < 1$. For boson, $0<\bar n < \infty$. $\endgroup$
    – ytlu
    Commented Jul 10, 2021 at 20:29
  • $\begingroup$ I am confused. If that's the case then $\bar n$ is not a CDF, because CDF can take values from 0 to 1. Which means that what I read was a mistake and that $\bar n$ is not a distribution (short for Commutative distribution function). So something ain't right. I don't understand but thank you for the help $\endgroup$
    – imbAF
    Commented Jul 10, 2021 at 20:31

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