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# 6.3: Lee-Yang Theory

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$$\newcommand\Vpi ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[22], line 1, column 1 at template() at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/06:_Classical_Interacting_Systems/6.03:_Lee-Yang_Theory), /content/body/p/span, line 1, column 23 $$
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$$\newcommand\Vrho ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[24], line 1, column 1 at template() at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/06:_Classical_Interacting_Systems/6.03:_Lee-Yang_Theory), /content/body/p/span, line 1, column 23 $$
$$\newcommand\Vvarrho ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[25], line 1, column 1 at template() at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/06:_Classical_Interacting_Systems/6.03:_Lee-Yang_Theory), /content/body/p/span, line 1, column 23 $$
$$\newcommand\Vsigma ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[26], line 1, column 1 at template() at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/06:_Classical_Interacting_Systems/6.03:_Lee-Yang_Theory), /content/body/p/span, line 1, column 23 $$
$$\newcommand\Vvarsigma ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[27], line 1, column 1 at template() at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/06:_Classical_Interacting_Systems/6.03:_Lee-Yang_Theory), /content/body/p/span, line 1, column 23 $$
$$\newcommand\Vtau ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[28], line 1, column 1 at template() at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/06:_Classical_Interacting_Systems/6.03:_Lee-Yang_Theory), /content/body/p/span, line 1, column 23 $$
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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.

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

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

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,

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