# A conceptual question related to statistical mechanics

Statistical mechanics allows us to consider an ensemble of systems, each of which consisting of only a single particle. Once we write the partition function for the system of one particle, we can easily derive all the thermodynamic quantities. One can accept that the internal energy computed from the partition function is the average energy of the system. But how to interpret pressure? What does pressure mean in the case of a system of one particle.

• In this case (and Pooya's great answer for a particle in a box) one can ascribe meaning to pressure, but I just wanted to point out that you can easily get into trouble applying thermodynamics or statistical mechanics to small systems. The assumptions we use and sweep under the rug can be violated and give really weird results. – psio Apr 30 '16 at 5:51
• I would also recommend that you have a look at the oldish but beautiful little book "Statistical Mechanics and the Foundations of Thermodynamics" by Anders Martin-Löf (Lecture Notes in Physics 101, 1979). The discussion of the pressure starts on page 23. – Yvan Velenik Apr 30 '16 at 7:49
• @EntropicallyDriven, can you elaborate your views. I too have a niggling doubt but am unable to express it well and tell what exactly am I suspecting when I derive pressure from partition function for a single particle system. – Amey Joshi Apr 30 '16 at 11:11
• Well, start with temperature: how do you define a thermodynamic temperature for a single particle? Temperature is usually introduced by considering two systems in equilibrium and finding their most probable configuration, resulting in the definition of $T^{-1} = dS/dE$. But we can safely find a "most probable" configuration only because we assume we have gigantic systems with loads of particles, and therefore fluctuations in energy around equilibrium are small (scale like $N^{1/2}$, while the energy scales like $N$. For one particle, none of those assumptions hold. – psio Apr 30 '16 at 15:28
• @EntropicallyDriven, I see your point. But can we not get over that problem if we assume our one particle system in a heat bath comprising of an enormous number of identical systems, weakly interacting with each other? Let me try proceeding along the same lines to check if pressure can be interpreted likewise. – Amey Joshi Apr 30 '16 at 17:52

Pressure is defined as the rate of increase in internal energy to rate of decrease in volume, i.e. $$P=-\frac{\partial U}{\partial V}$$

Assume a particle in a box, for example the classic infinite quantum potential well of width $L$. The quantized energy is $$E_n=\frac{n^2h^2}{8mL^2}$$

In a 3D box this becomes

$$E_{n_x,n_y,n_z}=\frac{(n_x^2+n_y^2+n_z^2)h^2}{8mL^2} = \frac{(n_x^2+n_y^2+n_z^2)h^2}{8mV^{2/3}}$$

where $V$ is the volume.

The ground state energy for this system is $$E_0 = \frac{3h^2}{8mV^{2/3}}$$ therefore the pressure is

$$P=-\frac{\partial E_0}{\partial V}=\frac{h^2}{4mV^{5/2}}$$

So you can see even a single particle can excert pressure on the boundaries of its bounding box!

• I understand your point. In fact, such a derivation can also be carried out for an ensemble with one particle. Mathematically, I understand what is happening. I am a little troubled by the doubt that we are probably pushing something under the rug - something that Entropydriven alluded in his comment. Let me check out the reference that Yvan has suggested. – Amey Joshi Apr 30 '16 at 11:09

A canonical ensemble is a collection of weakly interacting systems, in thermal equilibrium with each other. The individual systems can be a single particle/molecule if the energy of interaction between the particles is negligible in comparison with their own (kinetic) energy. The energy of ensemble is then just the sum of energies of individual particles, each particle considered as a 'system'. In such situations, one can consider a molecular partition function $q = \sum\exp(-\beta E_i)$, where $\beta = -1/(kT)$ and $E_i$ are the energy levels of the molecule. We can then write the total partition function as $Z = q^N/N!$. (We divide by $N!$ to account for indistinguishability of molecules.) Since the energy of interaction is negligible, all macroscopic effects are a sum of effects due to individual particles. For example, if we have just a thousand molecules in a container of $1$ $m^3$ then the pressure of the gas will be $1000$ times the 'pressure due to one molecule'.

An ideal gas is one example of an ensemble of 'loosely coupled systems'. Another one could be where we can write total energy as a sum of translational, vibrational and rotational energies when there is no coupling between them. One can then write the total pressure as being due to each of these components.

In general, if the partition function of an ensemble can be written as a product of partition functions of subsystems then all thermodynamic functions can be written as sums of contributions from the individual subsystems. (This is because of the logarithmic relation between Helmholtz free energy and partition function.)

• That is not a standard definition of "ensemble". The systems in a statistical ensemble are explicitly NOT interacting, because they are virtual copies/versions of a single system. Really, an ensemble is nothing more than a probability distribution over possible states. It's like if you didn't know where your cat was: you would say there are many possible places where your cat could be ('virtual copies' of your cat), not that there are a large number of weakly interacting cats. – Nanite May 18 '16 at 6:16
• Nanite, please refer to en.wikipedia.org/wiki/Canonical_ensemble for a definition of canonical ensemble. Further, that was not the main purpose of the question. – Amey Joshi May 18 '16 at 12:45