In Thermodynamics, one often encounters the derivation of pressure as the generalised force that belongs to the extensive state-variable of the volume. Postulates: One looks just at a system of many classical particles, that are interacting by a given Hamilton function on the phase-space $H(\Gamma)$. In short, one postulates that the entroy

$S[\rho] = \int \rho(\Gamma) ln(\rho(\Gamma)) d\Gamma$

is maximal in the equilibrium of the state, and then adds boundaries, that state something about the mean values of observables, like

$V_0 = \int V(\Gamma) \rho(\Gamma) d\Gamma$

Solving this variatioen-problem with a boundary (entropy maximal) requires a lagrangian multiplicator, which is p, the pressure.

My problem here is: I can't imagine how I can define some function on the phase space that measures "Volume". How do I do that? Even if I coarse grain every particle, it still would not meet my expections of some kind of volume. Same in quantum mechanics: Is there some kind of volume-operator?

It get's even more wicked when I go to the grand-canonical ensemble with changing particlenumber: While before I had at least an Idea what the function V(\Gamma) should look like (some envelope of every particle), I now don't even know what I should imagine for this function.

Edit: It would maybe be a similar Question to ask for a "Volume-Operator" for a quantum-mechanical many-body-system. Does something like that exist? If not, how do you derive the definition of pressure in a statistical-mechanics-way?


I have been wondering the same question recently. There is a way to define the isobaric partition function that seems to work pretty well with large systems, but I'm not convinced that it's appropriate with small ones. The idea is not to define a measure of the volume; instead define volume indirectly via the canonical ensemble, as follows.

In the canonical ensemble you enclose your system in an impermeable box of some volume $V$. You then find that the allowed microstates $\{i\}$ have corresponding energies $\{E_i\}$. Now, maximising the entropy of the vector of probabilities $\boldsymbol{p} \equiv (p_1, p_2, \ldots)$ subject to the constraint of average energy $\bar{E} = \sum_i p_i E_i$ and normalisation is the same as unconditionally maximising the function $$ \Lambda(\boldsymbol{p}) \equiv -\sum_i p_i \ln{p_i} - \beta \Big( \sum_ip_i E_i - \bar{E}\Big) - \lambda \Big(\sum_i p_i - 1\Big). $$

Now, notice that because we enclosed the system in a box of volume $V$, the eigenvalues are really functions of the volume, i.e. $E_i = E_i(V)$; similarly the probabilities that maximise entropy also depend on the volume, i.e. $p_i = p_i(V)$. Imagine we have a set of allowed volumes $V_j$, and denote $p_i(V_j) \equiv p_{i,j}$. So if instead of a rigid box we had one that could stretch, but no constrain on the volume, we would be maximising the function $$ \Lambda(\boldsymbol{p}) \equiv -\sum_{i,j} p_{i,j} \ln{p_{i,j}} - \beta \Big( \sum_{i,j} p_{i,j} E_i(V_j) - \bar{E}\Big) - \lambda \Big(\sum_{i,j} p_{i,j} - 1\Big). $$

The last step is to constrain the volume: $\bar{V} = \sum_{i,j} p_{i,j} V_j$. Then we have $$ \Lambda(\boldsymbol{p}) \equiv -\sum_{i,j} p_{i,j} \ln{p_{i,j}} - \beta \Big( \sum_{i,j} p_{i,j} E_i(V_j) - \bar{E}\Big) - \lambda \Big(\sum_{i,j} p_{i,j} - 1\Big) - \beta P \Big( \sum_{i,j} p_{i,j} V_j - \bar{V}\Big), $$ where $\beta P$ is the new Lagrange multiplier, and maximising this function leads to the isothermal-isobaric distribution $$ p_{i,j} \propto \mathrm{e}^{-\beta[E_i (V_j) + P V_j]}. $$ The partition function becomes $$ \Delta(\beta, P) = \sum_{i,j} \mathrm{e}^{-\beta[E_i(V_j) + P V_j]}, $$ or allowing the volume to vary continuously, $\sum_j \to \int \mathrm{d}V$, and then $$ \Delta(\beta, P) = \int \mathrm{d}V \sum_{i} \mathrm{e}^{-\beta[E_i(V) + P V]} = \int \mathrm{d}V \: \mathrm{e}^{-\beta P V} \sum_{i} \mathrm{e}^{-\beta E_i(V)} = \int \mathrm{d}V \: \mathrm{e}^{-\beta P V} Z(\beta, V), $$ where $Z(\beta, V)$ is the canonical partition function at inverse temperature $\beta$ and volume $V$.

I think this is most easily understood with the example of the particle in the box. There is a set of eigenstates $\{n\}$ with energies $E_n = \pi^2 n^2 / 8 L^2$, where $L$ is the size of the box. We see that changing the size of the box changes the energies; the particle can be in any of these states at any box length; the equilibrium isothermal-isobaric distribution is achieved by maximising entropy such that there is a certain average energy and certain average box size.

There is a feature of this approach that I do not like, namely that, for a given box size, it assigns the same volume to each eigenstate no matter where the probability distribution $\lvert \psi \lvert^2$ is concentrated. I do not know how to resolve this.

  • $\begingroup$ So what you do is effectively: You don't look at the thermodynamic System "gas", but instead, you look at the System "gas + surrounding box". The Volume is then not an observable that is related to the state of the gas, but instead it related to the state of the box. That was also my approach to solve this, and if you think about it carefully, there is no other way to define volume well. $\endgroup$ Jun 21 '16 at 8:41

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