Ensemble average in microcanonical and canonical ensembles

Question

In Huang's statistical mechanics, the general ensemble average of a dynamical variable $O(p,q)$ is defined as:

$$ \langle O \rangle=\frac{\int O(p,q)f(p,q,t)dpdq}{\int f(p,q,t)dpdq} $$

where $f(p,q,t)$ is the distribution function obeying Liouville's theorem. How does this expression simplify for the microcanonical and canonical ensembles?

For the microcanonical ensemble, the probability of a system with Hamiltonian $H$ being in an energy range $[E,E+\delta E]$ is equal for all possible microstates (i.e., combinations of $p$ and $q$). So the distribution function $f$ cancels and we have:

$$ \langle O \rangle=\frac{\int_{E<H<E+\delta E} O(p,q)dpdq}{\int_{E<H<E+\delta E} dpdq} $$

For the canonical ensemble, the distribution function $f=ce^{-\beta H}$, where $c$ is a constant obtained by the normalization condition of probability distributions. Substituting this into our general ensemble average definition, you get:

$$ \langle O \rangle=\frac{\int O(p,q)e^{-\beta H}dpdq}{\int e^{-\beta H}dpdq} $$

where the integrals are over all possible values of $(p,q)$.

• Is my reasoning behind these simplifications correct?
• Is the probability distribution $f$ that I used for each ensemble the same one that is in Liouville's theorem?

Answer 1

The first equation is an expression forbthe average in respect to an arbitrary probability density $f(p,q,t)$! Note that it does not have to he normalized - the denominator is the normalization factor.

In the microcanonical ensemble the probability density is constant everywhere in the energy shell $[E, E+\delta E]$, and zero everywhere else, which is indicated here by the integration limits. For the canonical ensemble $$f(p,q,t) = e^{-\beta H(p,q,t)}.$$ As I have mentioned, the normalization factor is already taken care of, and indeed it disappears from your calculation.