Three kinds of distributions. The states occupation rates:

F.D. $n_{i}=\frac{1}{e^{\beta (\varepsilon _{i}-\mu )}+1}$ B.E. $n_{i}=\frac{1}{e^{\beta (\varepsilon _{i}-\mu )}-1}$ Boltzmann $n_{i}=e^{-\beta (\varepsilon _{i}-\mu )}$

$(i$ labels different energies for a single particle, for the same $i$ there are still $G_{i}$ degeneracy.$)$

While, the density matrix for all three kinds of distribution is $\rho _{m}=\frac{e^{-\frac{E_{m}}{kT}}}{Z(T)}$

$(m$ labels different states of the $N$ particle system.$)$

It seems that Boltzmann distribution is more close to the density matrix expression. why is that?

$1$. If there is only one particle in the system, then

$\varepsilon _{i}=E_{i}$

$n_{i}=e^{-\beta (\varepsilon _{i}-\mu )}=\rho _{i}=\frac{e^{-\frac{E_{i}}{kT}}}{Z(T)}$

$Z(T)=e^{-\mu }$

$2$. If there are $N$ particles in the system, then

$\sum_{all\, particles} \varepsilon _{i}=E_{n}$

$\rho _{n}=\frac{e^{-\frac{E_{n}}{kT}}}{Z}= \frac{e^{-\frac{\sum \varepsilon _{i}n_{i}G_{i}}{kT}}}{Z}=\frac{\prod \left (e^{-\beta \varepsilon_{i} } \right )^{n_{i}G_{i}}}{Z}=\prod \left ( e^{-\beta (\varepsilon _{i}-\mu )} \right )^{n_{i}G_{i}}=\prod_{all\, particles}n_{i}$

$Z(T)=e^{-\mu N }$

$(i$ labels both different particle and its own energy number.$)$

Boltzmann distribution has more obvious analogy with density matrix distribution.

My question is: Is the analogy right? Can we still get the result

$Z(T)=e^{-\mu N }$

$\sum_{all\, particles} \varepsilon _{i}=E_{n}$

$\rho _{n}=\prod_{all\, particles}n_{i} $

for F.D. and B.E. distributions?

  • $\begingroup$ Why do you call them occupation rates, they don't have units of $s^{-1}$? $\endgroup$ – boyfarrell May 28 '14 at 11:18
  • $\begingroup$ @boyfarrell occupation numbers is a more conventional term, but I assume that's what OP means. The thought process could have been something like rate $\sim$ probability $\endgroup$ – Danu May 28 '14 at 12:51

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