# Single-particle states in statistical distributions and the significance of Probability Distribution Function

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?

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.
• 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? – AntMan Mar 5 at 10:25
• 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? – YouFoundMe Mar 5 at 10:50
• 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. – AntMan Mar 5 at 12:05
• 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? – AntMan Mar 5 at 12:09