Interacting systems

Let us now consider a real gas, formed by molecules that interact amongst themselves via a potential energy $U$. The total energy of the gas is the sum of the translational kinetic energies of the centre of mass of each single molecule, plus the molecular internal energies, plus the energy due to the interactions between the molecules and a possible interaction with an external potential, the last two all described by $U$. The internal energies include rotational kinetic energy, vibrational energies and so on. In Sec.  we arrived at expression  for the partition function of a molecular gas with no molecule-molecule interactions. Now we need to take into account also the interaction energy $U$. If we assume that $U$ does not affect the internal degrees of freedom, then we can write the total energy (note, now this is the instantaneous energy, not the average one) as

$$E = E_{tr} + E_{int} + U = \sum_{i=1}^N \left (\frac{p_i^2}{2m} + \epsilon_i^{int} \right ) + U,$$ (4.47)

where ${\bf p}_i = m {\bf v}_i$ is the momentum of particle $i$, with $m$ its mass and ${\bf v}_i$ its velocity, and $\epsilon_i^{int}$ the internal energy of particle $i$. In Sec.  we saw how the additivity of the energies results into a partition function that partitions into a product of partition functions, because of the rule of combination of probabilities. In particular, the partition function of the perfect gas, which we rename $Z_P$ here, was written as

$$Z_P = \frac{1}{N!}[Z_1^{tr}]^N [Z_{int}]^N$$ (4.48)

because the translational and the internal energies $E_{tr}$ and $E_{int}$ were written as a sum of $N$ identical single particle contributions, and the $N!$ terms was included to account for the de-localisation of the identical particles. Continuing that argument, we can show that the partition function of the interacting system can be written as

$$Z = Z_P Q,$$ (4.49)

where

$$Q = \frac{1}{V^N}\int_V d^3{\bf r}_1 \dots d^3{\bf r}_N e^{-\beta U}.$$ (4.50)

We should stress that the partitioning of the energy in a kinetic plus a potential part is only possible if the system behaves classically. In the quantum case this is not true (the reason being that the kinetic energy operator in the Hamiltonian does not commute with the potential energy operator) and the energy cannot be broken down in kinetic plus potential contributions. As a result, the partition function cannot be written as in Eq. .

To see how Eqs.  and  arise, consider the Boltzmann probability for the system to have energy $E$:

$$p(E) \propto e^{-\beta \left( U + \sum_{i=1}^N \epsilon_i^{int} +... ...\beta \sum_{i=1}^N\epsilon_i^{int} } e^{-\beta \sum_{i=1}^N \frac{p_i^2}{2m}} .$$ (4.51)

The constant of proportionality in  is equal to the number of states that have energy $E$, divided by the partition function. Rather than working with exact energies $E$, it is more convenient to work with ranges. In particular, for a free particle in a volume $V$ the number of states with translational energy $p^2/2m \le \epsilon_{tr} \le (p+dp)^2/2m$ is $4\pi V p^2 dp / h^3$, and so the number of states in an elementary element of volume of phase space $d^3{\bf r} d^3{\bf p}$ is $d^3{\bf r} d^3{\bf p}/h^3$ (see appendix ). The density of states is the same also for a particle in a potential energy (we shall not prove it). To obtain the partition function we need to sum over all possible states, which means integrating over positions and momenta, and summing over the internal degrees of freedom. We have:

$$Z = \prod_{i=1}^N\sum_j e^{-\beta \epsilon_i^{int}(j)} \frac{1}{h... ...e^{-\beta \sum_{i=1}^N \frac{p_i^2}{2m}} d^3{\bf p}_1 \dots d^3{\bf p}_N \times$$

$$\int_V d^3{\bf r}_1 \dots d^3{\bf r}_N e^{-\beta U},$$ (4.52)

where $\epsilon_i^{int}(j)$ is the $j$-th internal energy level of particle $i$. Since the particles are identical, the internal energy spectrum is the same for every one and so:

$$\prod_{i=1}^N\sum_j e^{-\beta \epsilon_i^{int}(j)} = \left [ \sum_j e^{-\beta \epsilon_1^{int}(j)} \right ]^N = [Z_{int}]^N.$$ (4.53)

The integrals over the momenta are also all identical. These integral extend from minus infinity to plus infinity for each component and each particle, and are more conveniently calculated in polar coordinates, giving:

$$\int e^{-\beta \frac{p^2}{2m}} d^3{\bf p} = 4\pi \int_0^\infty p^2 e^{-\beta \frac{p^2}{2m}} dp = \left(2\pi mk_{\rm B}T\right )^{3/2}.$$ (4.54)

We can therefore re-write  as:

$$Z = [Z_{int}]^N \left(\frac{2\pi mk_{\rm B}T}{h^2}\right )^{3N/2} V^N \frac{1}{V^N} \int_V d^3{\bf r}_1 \dots d^3{\bf r}_N e^{-\beta U} =$$

$$[Z_{int}]^N [Z_1^{tr}]^N Q.$$ (4.55)

Note that Eq.  does not contain the $N!$ at the denominator. This factor has to be added explicitly if one is dealing with identical particles able to share the whole volume and it arises because the integrals over the $N$ positions would contain $N!$ identical contributions due to all possible permutations of particles visiting the same portion of space. If the particles are instead localised this factor must not be included.