[Physics] Conversion between continuous and discrete partition functions

lattice-modelpartition functionphase-spacestatistical mechanics

I am a bit confused about the two notations for partition functions, that is the continuous form

$$Z=\frac{1}{h^3}\int\mathrm{d}\boldsymbol{p}\mathrm{d}\boldsymbol{r}\mathrm{e}^{-\beta H(\boldsymbol{p},\boldsymbol{r})}$$

and the discrete form

$$Z=\sum\limits_i \mathrm{e}^{-\beta H_i} \, .$$

First of all, how is it possible to come from one of these expressions to the respective other one?

Suppose we have a single particle on a discrete lattice (each lattice cell with a site length of $a$) with no kinetic energy. We can then use the 2nd expression and get the partition sum as

$$Z=\sum\limits_i \mathrm{e}^{-\beta V_i} \, .$$

with $V_i$ the potential energy ($T_i=0$ assumed) at the site $i$ and the sum going over all lattice sites. How would the continuous expression look like?

$$Z=\frac{1}{h^3}\left(\int\mathrm{d}\boldsymbol{p}\right)\left(\int\mathrm{d}\boldsymbol{r}\mathrm{e}^{-\beta V(\boldsymbol{r})}\right)=\frac{1}{h^3}\left(\int\mathrm{d}\boldsymbol{p}\right)\left(\sum\limits_i\mathrm{e}^{-\beta V_i}a^3\right)\ldots$$

and can we convert this expression again into a sum to come back to the discrete expression?

Best Answer

To a physicist, an integral is often nothing other than a "continuous sum". The instruction $\sum_x f(x)$ often means an actual sum when $x$ is drawn from a countable set, but an integral when $x$ is drawn from an uncountable set.

Therefore, there is no "conversion" between your expressions - they all are valid ways of writing down the partition function for different types of systems:

  • Classical statistical system, finitely many microstates $i\in\{1,\dots,n\}$ with energy $E_i$: $Z = \sum_i \mathrm{e}^{-\beta E_i}$

  • Classical statistical system, infinitely many continuous microstates given by points $(q,p) \in \mathbb{R}^{6N}$ in phase space for $N$ particles: $Z = h^{-3}\int \mathrm{e}^{-\beta H(q,p)}\mathrm{d}^{3N}x\mathrm{d}^{3N}p$.

The integral is simply the only way to get something that is close to a "sum over all $(x,p)$" that is mathematically well-defined and behaves in the desired way.

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