I am confused about the definition of the partition function. From the class of statistical thermodynamics, I've learned that the partition function for a system with continuous energy can be expressed as $$ Q = \int \exp(\frac{-U(x)}{k_{B}T})dx \tag{1} $$ And from the partition function, we can calculate the Helmholtz free energy by using $F=-k_{B}T \ln Q$. Since the partition function can be regarded as the number of accessible states (in discrete systems) or the normalization constant of a probability distribution, I imagine that at a constant temperature, the partition function $ Q $ should also be a constant. Therefore, $F$ should be constant at a constant temperature.

So far it makes sense to me. However, if we consider, say, the dissociation of a single NaCl, we know that at different ion-pair distances, the values of free energy are also different. That is, the partition function should also vary with the ion-pair distance. However, I could not see any part in Eqn (1) accounts for other variables (i.e. order parameters or collective variables) such as ion-pair distance.

I guess the Eqn (1) is correct but not strict enough. Therefore, I'm wondering what the most strict definition of the partition is. Could somebody explain this to me? Many thanks in advance!


1 Answer 1


You should think about the meaning of $x$. For instance, in the casse of NaCl dissociation it would consist of the coordinates of all Na$^+$ and Cl$^-$ particles, i.e. it would be a highdimensional vector with about $3\times6\times10^{23}$ components. Notice that usually also the momenta are also included, but i assume these were integrated out...

  • $\begingroup$ Hi, thank you for your comment. What I meant was the dissociation of a single NaCl. Still, $Q$ above is a constant, right? $\endgroup$
    – Jack
    Jul 11, 2019 at 6:52
  • $\begingroup$ Well, it depends implicitely on the number of particles (one of each kind, in your case) and the volume (since you integrate over x). Aside from these dependencies, yes, it's constant... $\endgroup$
    – denklo
    Jul 11, 2019 at 6:59
  • $\begingroup$ Does the potential $U(x)$ not change discontinuously for each species between times before and after the dissociation? $\endgroup$
    – Lili FN
    Jul 20, 2021 at 11:59

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