Perturbation Theory

Suppose energy is the sum of two terms:

U=U₀+U₁

and we know how to calculate

$$Z_0 = \int \exp(-\beta U_0)\,d\tilde r$$

How can we calculate partition function Z?

Idea:
$$\begin{multline*} Z = \int \exp\bigl(-\beta (U_0+U_1)\bigr)\,d\tilde r =\ Z_0 \frac{1}{Z_0}\int\exp\bigl(-\beta (U_0+U_1)\bigr)\,d\tilde r \end{multline*}$$

But this is exactly  

$$Z= Z_0\left\langle \exp\bigl(-\beta U_1\bigr)\right\rangle_0$$ (25)

with $\left\langle \ldots\right\rangle_0$--averaging over reference system with energy U₀:

$$\left\langle \ldots\right\rangle_0 = \frac{1}{Z_0}\int\ldots\exp(-\beta U_0)\,d\tilde r$$

For the free energy:  

$$\begin{aligned} -\beta A =& \log \bigl(\frac{Z_0}{N!\Lambda... ...langle \exp(-\beta U_1)\right\rangle_0 = A_0 + A_1\end{aligned}$$ (26)

A₁ can be expanded in $\beta$, by expanding $\log\left\langle exp(-\beta U_1)\right\rangle_0$. 

$$A_1 = \left\langle U_1\right\rangle_0 + \sum_{n=2}^\infty\frac{ \omega_n}{n!}(-\beta)^n$$ (27)

This makes a series in powers of the perturbation energy U₁, averaged over the configurations of the unperturbed system. (They are contained in the $\omega$'s).

Van der Waals Equation of state:

Let's consider the hard-sphere system as a reference, and the attractive part of the molecular interaction v(r) as a perturbation.

$$\left\langle U_1\right\rangle_0 = \frac{\rho^2V}{2}\int_0^\infty v(r) g_{HS}(r) 4\pi r^2 dr$$

Assuming at low densities:

$$g_{HS}(r) = \begin{cases} 0\quad r<\sigma \\ 1\quad r\gt\sigma\end{cases}$$

$$\left\langle U_1\right\rangle_0= -aN\rho \quad \mathrm{where} a=-\int_\sigma^\infty v(r)r^2dr$$

For the hard spheres system:

$$Z_0=(V-Nb)^N, \quad F_0=-NkT\log(V-Nb)$$

Substituting this into (26,27), and differentiating with respect to volume, obtain the Van der Waals equation of state:

$$\frac{p}{kT}=\frac{\rho}{1-\rho b}-\frac{a\rho^2}{kT}$$