Themes > Science > Physics > Solid State Physics > Electrons in Metals > The Distribution Function

Let's say that our macroscopic system is in a particular state $ S^{(0)} = \left\{n^{(0)}_1,n^{(0)}_2,\ldots,n^{(0)}_s,\ldots\right\}$. The number of particles in a particular single-particle state $ s$ is $ n^{(0)}_s$.

What if we average over all possible macroscopic states $ S$? The occupation number of state $ s$ is likely to be different in different macroscopic states. The mean occupation number of quantum state $ s$ is

$\displaystyle \shadowbox{ $ \displaystyle \bar n_s \equiv \sum_{S = \{n_1,n_2,\ldots,n_s,\ldots\}} n_s P(S) $ }$ (26)

Substituting our result for $ P(S)$ gives
$\displaystyle \bar n_s = \frac{1}{Z} \sum_{S = \{n_1,n_2,\ldots,n_s,\ldots\}... ...\epsilon_1} e^{-\beta n_2 \epsilon_2} \cdots e^{-\beta n_s \epsilon_s} \cdots .$ (27)

This is the mean number of particles we would expect to see in single-particle state $ s$ at temperature $ T$.

Here's yet another way to look at this. Say the system has a single-particle eigenstate with eigenenergy $ \epsilon$. Then $ n(\epsilon)$, defined analogously to $ \bar n_s$ above, is the number of particles in that state. The quantity $ n(\epsilon)$ is called the Distribution Function. The distribution function turns out to be really important for understanding the properties of many-particle systems so we'll spend some time evaluating it for the different classes of quantum mechanical particles.

It's worth noting in passing that $ n(\epsilon)$ is not necessarily the number of particles with energy $ \epsilon$. There may be other degenerate single-particle states with the same eigenenergy. In this case, each of these degenerate states will be occupied by $ n(\epsilon)$ particles. We'll talk later about how to handle degenerate states.


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