0
$\begingroup$

This is a very basic doubt from Statistical Mechanics which I haven't been able to resolve yet.

Suppose we take into account only single-particle states for a system, and we wish to look at the various possible scenarios with the help of Maxwell-Boltzmann, Fermi-Dirac, and Bose-Einstein probability distributions. I've read that $f(E)$ is the probability distribution function, which also represents the average number of particles in a given state. While deriving this for the Bose-Einstein (B-E) or Fermi-Dirac (F-D) distributions, we've seen that $f(E)$ saturates at $1$ for F-D, while it can tend to $\infty$ for a B-E distribution. But is this true for a single-particle state as well(not single state!)? Should $f(E)$ lie in $[0,1]$ alike for distinguishable particles, Bosons and Fermions? Should $f(E)$ still represent an average number of particles for a single-particle state, as it's now a fraction? Can someone provide a detailed explanation to help me kill this doubt?

$\endgroup$
1
$\begingroup$

You should probably correct your question. BE distribution starts from $\infty$ and asymptotically approaches 0 while FD starts at 1. Here is the image of the three statistics under discussion.

Probability distribution(in Statistical Mechanics) represents the average occupancy of states and not the number of particles in a state. Distribution function times the density of states gives the number density of particles (can be loosely thought of as the number of particles in the particular state)

These distribution arise because of the fundamental difference between two types of particles namely Fermions and Bosons. This is equally true for single-particle states as well, as in if you fill single-particle states with fermions, since only two particles(spin multiplicity) can occupy a state it is bound to follow FD distribution.

$\endgroup$
18
  • $\begingroup$ I totally agree with you, and the bounds of the graphs are such due to the temperature dependence, and I'm aware of it. The fundamental doubt that's been bugging me is that for B-E, $f(E)$ starts from $\infty$ at very low temperatures, so $f(E)$ can't surely represent probability of occupation in this case. So how is $f(E)$ different from the probability of occupation of a state? Also, shouldn't $f(E)$ saturate at 1 alike for all 3 types of objects for a single-particle state? $\endgroup$ – AntMan Mar 5 at 10:25
  • 1
    $\begingroup$ ahh... in strict mathematical term it is called a Probability Density Function and it can start from $\infty$ provided the area under the graph is unity. Just think of it as a weight function that adds weightage to the different possible level. I didnt understand the last question. Saturate to 1 where? At high energy levels? And is there any chance you are confusing single particle state with single particle system? $\endgroup$ – YouFoundMe Mar 5 at 10:50
  • $\begingroup$ Single particle state is a state where only one particle is involved, and doesn't interact with other particles of the system, am I correct? It necessarily might not be a single-particle system. By saturation I mean that in general, for fermions, at very low temperatures, when the energy of the single-particle state is less than the fermi-energy at that temperature, then $f(E)$ for that distribution tends to $1$. This is obviously not the case for B-E, where at low temperatures, the graph clearly extends till $\infty$. You are correct in naming $f(E)$ as the Probability Density function. $\endgroup$ – AntMan Mar 5 at 12:05
  • $\begingroup$ But my doubt is that, on integrating the probability distribution function along with the degeneracies of the state(if any), I should get the total number of particles in that state, and not the probability of occupancy of that state. So, how should the normalized area be unity under the graph for $f(E)$? The question which I ask basically boils down to this- Is the Probability Distribution function same as the probability of occupancy for a single-particle state for Bosons, Fermions, and Distinguishable particles alike? $\endgroup$ – AntMan Mar 5 at 12:09
  • $\begingroup$ Oh okay, I think I get it now! Can you kindly just confirm if the probability density function actually becomes the probability of occupation(apparently it seems obvious to me!) for a single-particle state for all the 3 distributions as I've mentioned above? Thanks a lot, by the way! $\endgroup$ – AntMan Mar 5 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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