6.3: Lee-Yang Theory

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Analytic Properties of the Partition Function

How can statistical mechanics describe phase transitions? This question was addressed in some beautiful mathematical analysis by Lee and Yang⁷. Consider the grand partition function $\Xi$,

$$\Xi(T,V,z)=\sum_{N=0}^\infty z^N\,Q\ns_N(T,V)\,\lambda_T^{-dN}\ ,$$

where

$$Q\ns_N(T,V)={1\over N!}\int\!\!d^dx\ns_1\cdots\int\!\!d^dx\ns_N\>e^{-U(\Bx\ns_1\,,\,\ldots\,,\,\Bx\ns_N)/\kT}$$

is the contribution to the $N$-particle partition function from the potential energy $U$ (assuming no momentum-dependent potentials). For two-body central potentials, we have

$$U(\Bx\ns_1,\ldots,\Bx\ns_N)=\sum_{i<j}v\big(|\Bx\ns_i-\Bx\ns_j|\big).$$

Suppose further that these classical particles have hard cores. Then for any finite volume, there must be some maximum number $N\ns_V$ such that $Q\ns_N(T,V)$ vanishes for $N>N\ns_V$. This is because if $N>N\ns_V$ at least two spheres must overlap, in which case the potential energy is infinite. The theoretical maximum packing density for hard spheres is achieved for a hexagonal close packed (HCP) lattice⁸, for which $f\nd_{\ssr{HCP}}=\frac{\pi}{3\sqrt{2}}=0.74048$. If the spheres have radius $r\ns_0$, then $N\ns_V=V/4\sqrt{2}r_0^3$ is the maximum particle number.

Thus, if $V$ itself is finite, then $\Xi(T,V,z)$ is a finite degree polynomial in $z$, and may be factorized as

$$\Xi(T,V,z)=\sum_{N=0}^{N\ns_V} z^N\,Q\ns_N(T,V)\,\lambda_T^{-dN} = \prod_{k=1}^{N\ns_V} \bigg(1-{z\over z\ns_k}\bigg)\ ,$$

where $z\ns_k(T,V)$ is one of the $N\ns_V$ zeros of the grand partition function. Note that the $\CO(z^0)$ term is fixed to be unity. Note also that since the configuration integrals $Q\ns_N(T,V)$ are all positive, $\Xi(z)$ is an increasing function along the positive real $z$ axis. In addition, since the coefficients of $z^N$ in the polynomial $\Xi(z)$ are all real, then $\Xi(z)=0$ implies $\overline{\Xi(z)}= \Xi({\bar z})=0$, so the zeros of $\Xi(z)$ are either real and negative or else come in complex conjugate pairs.

Figure $\PageIndex{1}$: In the thermodynamic limit, the grand partition function can develop a singularity at positive real fugacity $z$. The set of discrete zeros fuses into a branch cut.

For finite $N\ns_V$, the situation is roughly as depicted in the left panel of Figure $\PageIndex{1}$, with a set of $N\ns_V$ zeros arranged in complex conjugate pairs (or negative real values). The zeros aren’t necessarily distributed along a circle as shown in the figure, though. They could be anywhere, so long as they are symmetrically distributed about the ${ Re}(z)$ axis, and no zeros occur for $z$ real and nonnegative.

Lee and Yang proved the existence of the limits

$$\begin{split} {p\over \kT}&=\lim_{V\to\infty} {1\over V}\,\ln\Xi(T,V,z)\\ n&=\lim_{V\to\infty} z\,{\pz\over\pz z}\bigg[{1\over V}\,\ln\Xi(T,V,z)\bigg]\ , \end{split}$$

and notably the result

$$n=z\,{\pz\over\pz z}\bigg({p\over\kT}\bigg)\ ,$$

which amounts to the commutativity of the thermodynamic limit $V\to\infty$ with the differential operator $z\,{\pz\over\pz z}$. In particular, $p(T,z)$ is a smooth function of $z$ in regions free of roots. If the roots do coalesce and pinch the positive real axis, then then density $n$ can be discontinuous, as in a first order phase transition, or a higher derivative $\pz^j p/\pz n^j$ can be discontinuous or divergent, as in a second order phase transition.

Electrostatic Analogy

There is a beautiful analogy to the theory of two-dimensional electrostatics. We write

$$\begin{split} {p\over\kT}&={1\over V}\sum_{k=1}^{N\ns_V}\ln\!\bigg(1-{z\over z\ns_k}\bigg)\\ &=-\sum_{k=1}^{N\ns_V}\Big[\phi(z-z\ns_k)-\phi(0-z\ns_k)\Big]\ , \end{split}$$

where

$$\phi(z)=-{1\over V}\,\ln (z)$$

is the complex potential due to a line charge of linear density $\lambda=V^{-1}$ located at origin. The number density is then

$$n=z\,{\pz\over\pz z }\bigg({p\over\kT}\bigg)=-z\,{\pz\over\pz z}\,\sum_{k=1}^{N\ns_V} \phi(z-z\ns_k)\ ,$$

to be evaluated for physical values of $z$, $z\in{\mathbb R}^+$. Since $\phi(z)$ is analytic,

$${\pz\phi\over\pz {\bar z}}={1\over 2}\,{\pz\phi\over\pz x} + {i\over 2}\,{\pz\phi\over\pz y}=0\ .$$

If we decompose the complex potential $\phi=\phi\ns_1+i\phi\ns_2$ into real and imaginary parts, the condition of analyticity is recast as the Cauchy-Riemann equations,

$${\pz\phi\ns_1\over\pz x}={\pz\phi\ns_2\over\pz y}\qquad,\qquad {\pz\phi\ns_1\over\pz y}=-{\pz\phi\ns_2\over\pz x}\ .$$

Thus,

$$\begin{split} -{\pz\phi\over\pz z}&=-{1\over 2}\,{\pz\phi\over\pz x}+{i\over 2}\,{\pz\phi\over\pz y}\\ &=-{1\over 2}\bigg({\pz\phi\ns_1\over\pz x} + {\pz\phi\ns_2\over\pz y}\bigg) + {i\over 2}\bigg({\pz\phi\ns_1\over\pz y} - {\pz\phi\ns_2\over\pz x}\bigg)\\ &=-{\pz\phi\ns_1\over\pz x}+i\,{\pz\phi\ns_1\over\pz y}=E\ns_x-i E\ns_y\ , \end{split}$$

where $\BE=-\bnabla\phi\ns_1$ is the electric field. Suppose, then, that as $V\to\infty$ a continuous charge distribution develops, which crosses the positive real $z$ axis at a point $x\in{\mathbb R}^+$. Then

$${n\ns_+-n\ns_-\over x}=E\ns_x(x^+)-E\ns_x(x^-)=4\pi\sigma(x)\ ,$$

where $\sigma$ is the linear charge density (assuming logarithmic two-dimensional potentials), or the two-dimensional charge density (if we extend the distribution along a third axis).

Example

As an example, consider the function

$$\begin{split} \Xi(z)&={(1+z)^M\,(1-z^M)\over 1-z} \\ &=(1+z)^M\,\big(1+z+z^2+\ldots+z^{M-1}\big)\ . \end{split}$$

The $(2M-1)$ degree polynomial has an $M^{\ssr{th}}$ order zero at $z=-1$ and $(M-1)$ simple zeros at $z=e^{2\pi i k/M}$, where $k\in\{1,\ldots,M\!-\!1\}$. Since $M$ serves as the maximum particle number $N\ns_V$, we may assume that $V=M v\ns_0$, and the $V\to\infty$ limit may be taken as $M\to\infty$. We then have

$$\begin{split} {p\over\kT}&=\lim_{V\to\infty} {1\over V}\,\ln\Xi(z)\\ &={1\over v\ns_0}\,\lim_{M\to\infty}{1\over M}\ln\Xi(z)\\ &={1\over v\ns_0}\,\lim_{M\to\infty}{1\over M}\bigg[ M\ln(1+z) + \ln\!\big(1-z^M\big) - \ln(1-z)\bigg]\ . \end{split}$$

The limit depends on whether $|z|>1$ or $|z|<1$, and we obtain

$${p\,v\ns_0\over\kT}=\begin{cases} \ln(1+z) & \hbox{ if $|z|<1$} \\ &\\ \Big[\ln(1+z)+\ln z\Big] & \hbox{ if $|z|>1$}\ . \end{cases}$$

Figure $\PageIndex{2}$: Fugacity $z$ and $pv_0/k_{\ssr{B}}T$ versus dimensionless specific volume $v/v_0$ for the example problem discussed in the text.

Thus,

$$n=z\,{\pz\over\pz z}\bigg({p\over\kT}\bigg)=\begin{cases} {1\over v\ns_0}\cdot{z\over 1+z} & \hbox{ if $|z|<1$} \\ &\\ {1\over v\ns_0}\cdot\Big[{z\over 1+z}+1\Big] & \hbox{ if $|z|>1$}\ . \end{cases}$$

If we solve for $z(v)$, where $v=n^{-1}$, we find

$$z=\begin{cases} {v\ns_0\over v-v\ns_0} & \hbox{ if $v>2v\ns_0$} \\ &\\ {v\ns_0-v\over 2v-v\ns_0} & \hbox{ if ${1\over 2}v\ns_0 < v < {2\over 3} v\ns_0$} \ . \end{cases}$$

We then obtain the equation of state,

$${p\, v\ns_0\over\kT}=\begin{cases} \ln\!\Big({v\over v-v\ns_0}\Big) & \hbox{ if $v>2v\ns_0$} \\ &\\ \ln 2 & \hbox{ if ${2\over 3} v\ns_0 <v<2v\ns_0$} \\ &\\ \ln\!\Big({v(v\ns_0-v)\over (2v-v\ns_0)^2}\Big) & \hbox{ if ${1\over 2}v\ns_0 < v < {2\over 3} v\ns_0$}\ . \end{cases}$$