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The relationship between occupation number (which is the number of particles at a certain energy level) and the density of states is as follows:

$$n(E) = D(E)F(E)$$

where $D(E)$ is the DOS and $F(E)$ is the Fermi function.

But intuitively this formula seems to have a problem. As I understand the fermi function, it is a probability density function that gives the likelihood for say an electron to possess a certain amount of energy among various values of energies. So if there's 400 electrons, and f(E1) = 0.5, then the energy level E1 will be occupied by 200 electrons. There is no DOS information required for this calculation. If E1 had 400 states, it will be half filled, if it had 200 it will be fully filled. But the above equation says that no matter what the occupation of E1 will be always half as n(E) would be half of D(E). Where am I mistaken?

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As I understand the fermi function, it is a probability density function that gives the likelihood for say an electron to possess a certain amount of energy among various values of energies.

No, that's not right. The Fermi function $$f(E) = \frac{1}{e^{(E-\mu)/kT}+1}$$ gives the probability that a single-particle state which has energy $E$ is occupied at temperature $T$, given that the system has chemical potential $\mu$. More specifically, if $x$ is a single-particle state and $E[x]$ is the energy of that state, then $$\mathrm{Prob}(x\text{ is occupied}) = f\big(E[x]\big) = \frac{1}{e^{(E[x]-\mu)/kT}+1}$$In particular, it is not a probability density but rather a genuine probability.

So if there's 400 electrons, and f(E1) = 0.5, then the energy level E1 will be occupied by 200 electrons.

No. If $f(E_1)=0.5$, then there is a 50/50 chance that any given single-particle state with energy $E_1$ is occupied. If $1000$ single-particle states each have energy $E_1$, then at any given time we would expect $500$ of them to be occupied and $500$ of them to be empty.

If $D(E)$ is the number of single-particle states per unit volume with energy between $E$ and $E+\mathrm dE$ (i.e. the density of states) and $f(E)$ is the probability that each of these states is occupied, then it follows that $n(E)=D(E)f(E)$ is the expected number of occupied single-particle states per unit volume with energy between $E$ and $E+\mathrm dE$.

In the case of a finite system with discrete energies, we would also have that if $g(E)$ is the number of single-particle states with energy $E$, then $N(E)=g(E)f(E)$ is the expected number of occupied single-particle states with energy $E$, which can be obtained from the above by integrating over a single energy level and multiplying by the volume of the system.

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  • $\begingroup$ This means that all microstates with the the same energy are equally probable. But shouldn't the probability of occupation over all microstates add up to 1 ? I can see that integrating f(E) over the entire range of energy gives 1, but will Probability(x is occupied) integrated over all x give 1 ? $\endgroup$ Aug 31, 2022 at 19:04
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    $\begingroup$ @Examination12345 We're not talking about microstates, we're talking about single-particle states. A microstate would be a specification of the occupancy of each single-particle state. The integral of $f(E)$ over all energies is not equal to $1$, which can be seen by noting that $\int f(E) \mathrm dE$ has dimensions of energy (since $f$ itself is dimensionless). $\endgroup$
    – J. Murray
    Aug 31, 2022 at 19:10
  • $\begingroup$ i think i have got it except for one last thing. The fermi function just gives the probability that a state is filled with some electron, it does not talk about the choice a single electron takes. This is possible when there are enough no of electrons and one state being filled does not limit the supply of electrons for other states to be filled. But what if the total electron count is limited such that even if a state has a high f(E), there aren't enough electrons to fill it ? Would a high f(E) mean that an electron spend more time in there ? Thanks !! $\endgroup$ Sep 1, 2022 at 6:57
  • $\begingroup$ @J. Murray, ""The integral of f(E) over all energies is not equal to 1, which can be seen by noting that ∫f(E)dE has dimensions of energy (since f itself is dimensionless)"" $\sum\limits_{i}^{} f(E_i) =<N>$ - average number of particles in the system, so $f(E_i)=<n(E_i)>$ average number of particles on level $E_i$. The introduction of $D(E)$ is connected with the transition from summing to integration $\endgroup$ Sep 1, 2022 at 12:23

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