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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?
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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.

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  • $\begingroup$ Your expression for the canonical distribution doesn't have a normalization factor, and you say it's taken care of. What do you mean by this? It seems like we should divide your expression by the partition function, which is some constant. $\endgroup$ Apr 17 '20 at 13:17
  • $\begingroup$ The normalization factor for distribution $f(p,q,t)$ is $c = [\int dpdq f(p,q,t)]^{-1}$ - it is already present in your equation. If I take a normalized distribution, as you suggest, then this factor is 1 and need not be included in your first equation. You either use normalized distributions or write the normalization factors explicitly, but doing both is a waste of energy. $\endgroup$ Apr 17 '20 at 13:22

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