***Dualities in two-dimensional Ising models


Suggested background: should be digestable for a second or third year physics major. Some exposure to statistical mechanics would help.

1. Introduction
2. Lattices
3. Ising model on a square lattice
     3.1. High temperature expansion
     3.2. Low temperature expansion
     3.3. Kramers-Wannier duality and the critical temperature
4. Ising model on triangular and hexagonal lattices
     4.1. Low and high temperature expansions
     4.2. Star-triangle transformation and the critical temperature

1. Introduction

This fall, I took an excellent class on the statistical mechanics of phase transitions, taught by Professor Dam Son. On our final exam, we were asked a problem which requires a clever trick: compute (exactly!) the critical temperature of the Ising model on a 2-dimensional triangular lattice. Since I’m not assuming any stat mech knowledge, let me provide a sketch of the problem. Imagine we have a 2D crystalline lattice of electrons that looks something like this:

Each electron has a spin which can point either up or down, and each electron only interacts with its nearest neighbors on the lattice (for example, on the triangular lattice, each electron has six nearest neighbors). Let’s further assume that two neighboring electrons whose spins are oppositely aligned contribute an energy {J} to the overall energy, while two electrons whose spins point in the same direction contribute {-J}. Then the overall energy associated to a particular configuration of spins on the lattice is

\displaystyle H = -J\sum_{\langle ij\rangle}S_i S_j, \ \ \ J>0

where {S_i=\pm 1} is the spin of the electron at lattice site {i} and the sum is taken so that only electrons which are neighbors contribute. This is called the Ising model.

There are a few qualitative observations we can make at this stage. Configurations of electrons which `go with the flow’ cost less energy; it isn’t too hard to see that having all the spins pointing in the same direction minimizes the energy of the system. At low temperatures (say absolute zero) the electrons will organize themselves to do just this; this is called the ordered phase.

At higher temperatures there will be thermal fluctuations which will allow some electrons to incur the energy cost associated with going against the grain, i.e. there will be pairs of electrons whose spins are anti-aligned. If one chooses the temperature to be sufficiently high, these thermal fluctuations dominate, and the spins will point in essentially random directions. This is called the disordered phase, and can be represented by a `percolation diagram’ where regions with spin up electrons are colored e.g. purple and regions with spin down electrons are colored e.g. blue.

This whole discussion can be summarized in a famous graph which plots the average value of the spins against the temperature.

One naturally wonders about intermediate temperatures. Is there a temperature at which one transitions from the ordered phase to the disordered phase? The graph above suggests yes (at least for the Ising models we’re interested in), and the temperature at which the phase transition occurs is called the critical temperature of the model, {T_\mathrm{c}}. The physics of systems at criticality is incredibly rich, hence our motivation to figure out what exactly the critical temperature is. In general this is very hard to do analytically, but for certain Ising models there is a trick involving a duality between ordered and disordered systems that allows you to solve this problem exactly. So I propose the following plan:

0) We’ll set up a bit of machinery involving lattices.
1) We’ll work our way towards defining this duality for the Ising model on a square lattice and argue that the system at its critical temperature is self-dual; intuitively, since our duality transforms ordered systems to disordered systems and vice versa, the point at which the Ising model transitions between these two phases will be a system which is mapped to itself under this duality transformation. This will recover the critical temperature.
2) We’ll see that this duality on its own won’t be enough to compute the critical temperature for the Ising model on a triangular lattice. Enter the star-triangle transformation. Using this additional ingredient will get us what we want, as well as the solution to the hexagonal lattice for free.

Here we go.

2. Lattices

This is a physics post, and I want it to be as painless as possible for math-phobes so I’m going to basically say as little as I need to say about lattices (and only through pictures!) so that I can get to the thermodynamics as fast as possible. If any of the definitions are confusing, the pictures accompanying them should clear things up. Caution: I may make up my own terminology or repurpose terminology that already exists, so try not to take my language too seriously!

We can myopically think of a 2-dimensional lattice as a repeating set of points in the plane, which will usually be drawn so that points which are nearest neighbors are connected by links. We will really only need to deal with three lattices in this post: the square lattice, the triangular lattice, and the hexagonal lattice. In these cases, the links bound squares, triangles, and hexagons which tile the plane, and we call each tile a cell. (Note: I’ve drawn things so that the vertices of the polygons are the lattice sites and their edges are the links. These lattices also extend infinitely in all directions, despite the fact that we have not yet invented infinitely large computer screens.)

Every lattice has a dual lattice; it’s obtained simply by putting a point at the center of each cell in the original lattice. Connecting the points of the dual lattice with perpendicular bisectors to the links of the original lattice gives you the links of the dual lattice. For example, the duality between the triangular and hexagonal lattices is depicted below.

Usually something only deserves to be called a `duality’ if the dual of the dual recovers the original object. And indeed, you can convince yourself by staring at the pictures above that the dual lattice of the dual lattice is just the original lattice you started with. Can you see why the square lattice is (conveniently) self-dual, i.e. is it’s own dual lattice?

By an Ising graph on a lattice, we mean a subset of the links of that lattice with the property that each point is touched by an even number number of links. We can draw an Ising graph by highlighting the links that belong to it.

The length of an Ising graph is just the number of links it has. Importantly, the smallest Ising graphs on the square lattice (aside from the unique graph of length zero) have length 4, and correspond to square loops. Similarly, the smallest Ising graphs on the triangular and hexagonal lattices have length 3 and 6 respectively (draw them).

Let’s say we’re given a configuration of spins. We can associate a domain wall drawing to this configuration by separating the regions which are spin up from the regions that are spin down. Here’s an example on the triangular lattice.

Staring at the last two pictures, one realizes that the Ising graph we drew on the hexagonal lattice is precisely the domain wall drawing we drew on the triangular lattice. This relationship is general: the domain wall drawings of a lattice are in one-to-one correspondence with the Ising graphs of its dual lattice.

One final remark: it will be convenient for us to `compactify’ our lattice. This simply amounts to picking some square region for the lattice to lie in and identifying the boundaries so that the lattice is really defined over a torus. If you haven’t seen ideas like this before, don’t worry; it’s not central to what follows. It just allows us to have a finite lattice without having to worry about boundary conditions (and without incurring serious physical consequences).

All the machinery is in place now. But why did we even bother? In the next sections, we will define the partition function of the Ising models and interpret its terms `diagrammatically’ in terms of Ising graphs and domain wall drawings.

3. Ising model on a square lattice

Let’s focus on the physics now. We will use units in which the Boltzmann constant is equal to one, {k_B = 1}. Recall that the partition function for a thermodynamic system is defined as

\displaystyle \mathcal{Z} = \sum_\alpha e^{- \beta E_\alpha}, \ \ \ \ \beta=1/T

where the sum is over all the possible states of the system, and {E_\alpha} is the energy of the {\alpha}th state. The partition function is a convenient way to encode all the thermodynamic information about the system that we care about. For example, the average energy of the system at temperature {T} can be computed from the partition function with the formula

\displaystyle \langle E \rangle = - \frac{\partial \log \mathcal{Z}}{\partial \beta}

and the free energy is defined through

\displaystyle F = - T \log \mathcal{Z}.

Let’s start by computing two different expansions of the partition function for the Ising model on the square lattice, one for low temperatures and one for high temperatures.

3.1. High temperature expansion

Define

\displaystyle \mathcal{H}(\mathfrak{S}) = K \sum_{\langle i j \rangle}S_iS_j, \ \ \ K = J/T.

Then by definition the partition function is

\displaystyle \mathcal{Z}_{\square}(K) = \sum_{\mathfrak{S}} \exp\big[\mathcal{H}(\mathfrak{S})\big]

where the sum is over all possible spin configurations on the square lattice. This is just a shorthand notation:

\displaystyle \sum_{\mathfrak{S}} = \sum_{S_1=\pm 1}\cdots\sum_{S_N=\pm 1}

where {N} is the total number of lattice sites. Writing this out a little more explicitly gives us

\displaystyle \mathcal{Z}_\square(K) = \sum_{\mathfrak{S}} \prod_{\langle ij \rangle}e^{KS_iS_j}.

Because the spins take on values {\pm 1}, we can use the identity {e^{KSS'} = \cosh K ( 1 + S S'\tanh K )} to further rewrite this as

\displaystyle \mathcal{Z}_\square(K) = (\cosh K)^{N_\ell}\sum_{\mathfrak{S}}\prod_{\langle ij \rangle}(1+S_iS_j\tanh K)

where {N_\ell} is the total number of nearest neighbor links on the square lattice. We would like to multiply out the product to obtain an expansion of the partition function in powers of {\tanh K}. To do this, let’s introduce a convenient shorthand: a link connecting two neighboring lattice sites {i} and {j} on the lattice will be denoted {\ell = \langle i j \rangle}. Then if we define {\pi(\ell) = S_i S_j}, I claim we can write the product

\displaystyle \prod_{\langle i j\rangle}(1+S_iS_j\tanh K) = \sum_n \sum_{\{\ell_1,\dots,\ell_n\}}^\ast \pi(\ell_1)\cdots\pi(\ell_n)(\tanh K)^n

where the sum {\sum_{\{\ell_1,\dots,\ell_n\}}^\ast} is over all subsets of links of size {n}. Implicit in the definition of subset is that no two links are equal, {\ell_i\neq \ell_j} and order does not matter, so that we do not count e.g. {\{\ell_1,\ell_2\}} and {\{\ell_2,\ell_1\}} twice in the sum! If it is hard to see where this formula comes from, try multiplying out e.g. {(1+x_1)\cdots(1+x_4)} and seeing that it agrees; this corresponds to the imaginary situation of having four total links in the lattice.

Consider a subset of links {\{\ell_1,\dots,\ell_n\}} so that some lattice site {i} is touched by an odd number of these links. The term {\pi(\ell_1)\cdots\pi(\ell_n) (\tanh K)^n} will not contribute in the partition function. To see this, notice that, since {S_i} appears an odd number of times in {\pi(\ell_1)\cdots \pi(\ell_n)}, we will have that

\displaystyle \pi(\ell_1)\cdots\pi(\ell_n)\big\vert_{S_i=1} = - \pi(\ell_1)\cdots\pi(\ell_n)\big\vert_{S_i=-1}

i.e. the coefficient evaluated with {S_i=1} will have the opposite sign compared to when it is evaluated with {S_i=-1}. Remember that in the partition function, the sum over spin configurations can be written as {\sum_{\mathfrak{S}} = \sum_{S_1=\pm 1} \cdots \sum_{S_{N} = \pm 1}} so that moving {\sum_{S_i=\pm 1}} all the way to the front forces the term to vanish.

The discussion above implies that a term involving {\{\ell_1,\dots,\ell_n\}} can only contribute if every lattice site is touched by an even number of the links {\ell_i} (it’s OK for a lattice site not to be touched by any of the links as well). Furthermore, for all such subsets, {\pi(\ell_1)\cdots\pi(\ell_n) = 1}. Now, we will see the full power of the machinery we developed in the previous section. By definition, the subsets that contribute are precisely the Ising graphs on the square lattice! We can combine this entire discussion to simplify things considerably. Let {G_\square^{(n)}(N)} be the number of distinct Ising graphs of length {n} on a square lattice with {N} sites. Evaluating the sum on spin configurations first gives us

\displaystyle  \begin{array}{rcl}  \mathcal{Z}_\square(K) &=& (\cosh K)^{N_\ell} \sum_n \sum_{\{\ell_1,\dots,\ell_n\}}^\ast\sum_{\mathfrak{S}} \pi(\ell_1)\cdots\pi(\ell_n)(\tanh K)^n = (\cosh K)^{N_\ell} \sum_n\sum_{ \substack{ \text{Ising}\\ \text{graphs}}} \sum_{\mathfrak{S}}(\tanh K)^n \\ &=&(\cosh K)^{N_\ell}2^N \sum_n \sum_{ \substack{ \text{Ising}\\ \text{graphs}}} (\tanh K)^n = (\cosh K)^{N_\ell}2^N \sum_n G_\square^{(n)}(N)(\tanh K)^n \\ &=& (\cosh K)^{N_\ell}2^N\left(1 + N(\tanh K)^4 + \cdots \right) \end{array}

where again {N_\ell} is the number of links and {N} is the number of lattice sites. So the partition function admits a diagrammatic interpretation in the same spirit as Feynman diagrams: for each Ising graph of length {n}, add {(\tanh K)^n}.

The first non-trivial term of the expansion has coefficient {N} because there are {N} squares on a square lattice with {N} sites.

3.2. Low temperature expansion

Let’s now develop an expansion that works for low temperatures. Recall we argued that when the temperature approaches zero, the only relevant configurations are the ones where every spin points in the same direction so as to minimize the energy. There are only two of these — every spin points up or every spin points down — and the energy is {-JN_\ell} in both cases, so the partition function limits to

\displaystyle \mathcal{Z}_\square(K)\sim 2 e^{KN_\ell} \text{ as } T\rightarrow 0.

As one increases the temperature, one can imagine configurations of spins in which one electron is oppositely aligned becoming more relevant. There are {2N} of these, {N} of them corresponding to one electron spin up and the rest spin down and the remaining {N} corresponding to one electron spin down and the rest spin up. The energy of such a configuration is larger by {4\cdot 2J}; and so we can obtain a better approximation to the partition function as

\displaystyle \mathcal{Z}_\square(K) \sim 2e^{KN_\ell}(1 + N(e^{-2K})^4)

We can continue in this way, throwing in spin configurations with larger and larger amounts of electrons with oppositely aligned spins. Recall: spin configurations are in one to one correspondence with domain wall drawings on the lattice. It is also true in general that the energy of a spin configuration is larger than the ground state configuration by precisely {n\cdot 2J} where {n} is the length of the domain wall drawing we associate to that spin configuration. We can therefore write the full expansion as

\displaystyle \mathcal{Z}_\square(K) = 2e^{KN_\ell}\sum_n D_\square^{(n)}(N)(e^{-2K})^n

where we have denoted the number of domain wall drawings of length {n} on a square lattice with {N} sites as {D^{(n)}_\square(N)}. But the domain wall drawings of a lattice are also in one to one correspondence with Ising graphs of the dual lattice. The square lattice is self-dual and has the same number of sites as the original lattice, and we get that

\displaystyle D^{(n)}_\square(N) = G^{(n)}_\square(N)

to get the following spectacular result:

\displaystyle \mathcal{Z}_\square(K) = 2e^{KN_\ell} \sum_n G^{(n)}_\square(N)(e^{-2K})^n = 2e^{KN_\ell}(1 + N(e^{-2K})^4 + \cdots).

This admits the same diagrammatic interpretation in terms of counting Ising graphs on the square lattice.

3.3. Kramers-Wannier duality and the critical temperature

To really bring out the similarity between the two expansions we’ve developed so far, define the series

\displaystyle f(X) = \sum_n G^{(n)}_\square(N)X^n.

Then

\displaystyle  \begin{array}{rcl}  \text{High-}T:& & \ \mathcal{Z}_\square(K) = (\cosh K)^{N_\ell}2^N f(\tanh K) = (\cosh K)^{N_\ell}2^N\left(1 + N(\tanh K)^4 + \cdots \right) \\ \text{Low-}T: & &\ \mathcal{Z}_\square(\tilde{K}) = 2\exp\left(\tilde{K}N_\ell\right)f\big(\exp(-2\tilde{K})\big) = 2\exp(\tilde{K}N_\ell)\left(1 + N\big(\exp(-2\tilde{K})\big)^4+\cdots\right). \end{array}

Up to a multiplicative constant out front, the high and low temperature expansions are just obtained by passing different arguments to the series defined by {f}. This suggests defining a duality between high and low temperature Ising models with the equation {e^{-2\tilde{K}} = \tanh K}, or

\displaystyle K\sim \tilde{K} = -\frac{1}{2}\log\left(\tanh K\right).

This is called the Kramers-Wannier duality. Now here comes the magic. Phase transitions go hand in hand with mathematical singularities, and the only possible source of these is the series {f(X)}. If we assume (correctly) that the Ising model on the square lattice only has one point at which it undergoes a phase transition, then {f(X)} only has one singularity. We have two different ways of writing the singular point of this series in terms of the critical parameter, prescribed by the low and high temperature expansions. These give us the formula {e^{-2K_{\mathrm{c}}} = \tanh K_{\mathrm{c}}} which can be solved to get

\displaystyle K_{\mathrm{c}} = \frac{\log(1+\sqrt{2})}{2}\implies T_{\mathrm{c}} = \frac{2J}{\log(1+\sqrt{2})}.

Ta-da!

4. Ising model on triangular and hexagonal lattices

It’s natural to feel invigorated by this result and think, “I can tackle any lattice in the world! Bring them on!” Unfortunately, the machinery we developed in the previous section won’t get us quite that far. In general, the duality transformation we defined will relate e.g. a high temperature expansion on a lattice to a low temperature expansion on its dual lattice. The key property we exploited in the previous section was that the square lattice is self-dual, which afforded us a relationship between two regimes of the same model. If we apply this same idea to the triangular lattice, we get a relationship between high temperatures on the triangular lattice and low temperatures on the hexagonal lattice (its dual lattice) and vice versa. But hope is not lost — we will define one more transformation called the star-triangle transformation, which takes us from the hexagon to the triangle. With this, we will generalize the square lattice solution

\displaystyle \text{High-}T \text{ on }\square\xrightarrow{\text{KW}} \text{Low-}T \text{ on } \square

to a solution for the triangular and hexagonal lattices,

\displaystyle \text{High-}T \text{ on }\triangle\xrightarrow{\text{KW}} \text{Low-}T\text{ on }{\mathrm{hex}}\xrightarrow{\text{ST}}\text{Low-}T\text{ on } \triangle

\displaystyle \text{High-}T \text{ on }{\mathrm{hex}}\xrightarrow{\text{ST}} \text{High-}T \text{ on }\triangle\xrightarrow{\text{KW}}\text{Low-}T \text{ on }{\mathrm{hex}}

where KW denotes the Kramers-Wannier duality transformation and ST denotes the star-triangle transformation. (Unfortunately I have to denote the hexagonal lattice with `hex’ since the hexagon symbol is resisting all attempts at being put into WordPress). What we obtain in the end is the desired relation between high temperatures and low temperatures on the same lattice.

4.1. High and low temperature expansions

I won’t rederive the expansions for the triangular and hexagonal lattices from scratch, since the situation is nearly identical to that of the square lattice and we won’t really need them. Instead, I’ll just write down the final result here, with slight notational changes. The interested reader may wish to test their understanding by obtaining these results themselves.

High temperature expansions:

\displaystyle \mathcal{Z}_\triangle(K,S_\triangle) = (\cosh K)^{L_\triangle}2^{S_\triangle}\sum_nG_{\triangle}^{(n)}(S_\triangle)(\tanh K)^n= (\cosh K)^{L_\triangle}2^{S_\triangle}\left(1+2S_\triangle(\tanh K)^3 +\cdots\right)

\displaystyle \mathcal{Z}_{{\mathrm{hex}}}(K,S_{{\mathrm{hex}}}) = (\cosh K)^{L_{{\mathrm{hex}}}}2^{S_{{\mathrm{hex}}}}\sum_n G_{{\mathrm{hex}}}^{(n)}(S_{{\mathrm{hex}}})(\tanh K)^n = (\cosh K)^{L_{{\mathrm{hex}}}}2^{S_{{\mathrm{hex}}}}\left(1 + \frac{S_{{\mathrm{hex}}}}{2}(\tanh K)^6+\cdots\right)

Low temperature expansions:

\displaystyle \mathcal{Z}_\triangle(K,S_\triangle) = 2\exp\left(L_\triangle K\right)\sum_nD_{\triangle}^{(n)}(S_\triangle)\left(\exp(-2K)\right)^n

\displaystyle \mathcal{Z}_{{\mathrm{hex}}}(K,S_{{\mathrm{hex}}}) = 2\exp\left(L_{{\mathrm{hex}}}K\right)\sum_nD_{{\mathrm{hex}}}^{(n)}(S_{{\mathrm{hex}}})\left(\exp(-2K)\right)^n

Notation:

\displaystyle  \begin{array}{rcl}  S_{\triangle,{\mathrm{hex}}} &=& \#\text{ sites on }\triangle\text{ or }{\mathrm{hex}} \text{ lattice} \\ L_{\triangle,{\mathrm{hex}}} &=& \#\text{ links on }\triangle\text{ or }{\mathrm{hex}} \text{ lattice} \\ G_{\triangle,{\mathrm{hex}}}^{(n)}(S_{\triangle,{\mathrm{hex}}}) &=& \#\text{ Ising graphs of length }n \text{ on } \triangle\text{ or }{\mathrm{hex}} \text{ lattice with } S_{\triangle,{\mathrm{hex}}} \text{ sites} \\ D_{\triangle,{\mathrm{hex}}}^{(n)}(S_{\triangle,{\mathrm{hex}}}) &=& \#\text{ domain wall drawings of length }n \text{ on } \triangle\text{ or }{\mathrm{hex}} \text{ lattice with } S_{\triangle,{\mathrm{hex}}} \text{ sites} \end{array}

In explicitly writing out the high temperature expansions, we used the fact that, for example, {G^{(3)}_{\triangle}(S_\triangle)=2S_\triangle}. To see this, just count the number of triangles on a triangular lattice. There are two types of triangles: the ones facing up and the ones facing down. The ones facing up are in one-to-one correspondence with the sites of the lattice, and since these are exactly half of the total number of triangles the result follows. Can you see why {G^{(6)}_{{\mathrm{hex}}}(S_{{\mathrm{hex}}}) = S_{{\mathrm{hex}}}/2}?

Let’s use the duality between triangular and hexagonal lattices to rewrite the low temperature expansions. We stated earlier that Ising graphs are in one-to-one correspondence with domain wall drawings on the dual lattice. Caution: to obtain the hexagonal lattice from the triangular one, we place a lattice site at the center of each triangle. Therefore, the dual of a triangular lattice with {S_\triangle} sites is a hexagonal lattice with {2S_\triangle} sites so the duality relationship reads

\displaystyle D^{(n)}_\triangle(S_\triangle) = G^{(n)}_{{\mathrm{hex}}}(2S_\triangle).

By the same token,

\displaystyle D^{(n)}_{{\mathrm{hex}}}(S_{{\mathrm{hex}}}) = G^{(n)}_\triangle(S_{{\mathrm{hex}}}/2)

so that we can now write

\displaystyle \mathcal{Z}_\triangle(K,S_\triangle) = 2\exp\left(L_\triangle K\right)\sum_nG_{{\mathrm{hex}}}^{(n)}(2S_\triangle)\left(\exp(-2K)\right)^n = 2\exp\left(L_\triangle K\right)\left(1+S_\triangle\exp(-2K)^6+\cdots\right)

\displaystyle \mathcal{Z}_{{\mathrm{hex}}}(K,S_{{\mathrm{hex}}}) = 2\exp\left(L_{{\mathrm{hex}}}K\right)\sum_nG_{\triangle}^{(n)}(S_{{\mathrm{hex}}}/2)\left(\exp(-2K)\right)^n = 2\exp\left(L_{{\mathrm{hex}}}K\right)\left(1+S_{{\mathrm{hex}}}\exp(-2K)^3+\cdots\right).

4.2. Star-triangle transformation and the critical temperature

The only thing we have left to do is define the star-triangle transformation, which takes us from a model on the hexagonal lattice to a model on the triangular lattice. Notice that the hexagonal lattice is composed of two triangular sublattices, which we’ll denote by {A} and {B}. We’re going to `decimate’ the spins which lie on the {A} sublattice (depicted in purple below).

The partition function involves a sum over spin configurations which can be split up into sums over spin configurations on the {A} and {B} sublattices,

\displaystyle \sum_{\mathfrak{S}} = \sum_{\mathfrak{S}_B}\sum_{\mathfrak{S}_A}.

The decimation procedure mathematically corresponds to actually evaluating the sum over the {A} sublattice. What we’ll be left with is a sum over configurations on the {B} sublattice (which is triangular) and if we’ve done everything correctly, the resulting partition function will have the same form as the partition function for a triangular Ising model.

Before we do this, let’s cite a purely algebraic result. We want to find solutions for {C} and {K'} in the equation

\displaystyle C \exp\left(K'(S_1S_2+S_2S_3+S_3S_1)\right) = 2\cosh\big(K(S_1+S_2+S_3)\big)

in terms of {K} that hold for all {S_i = \pm 1}. Choosing {S_1=S_2=S_3=1} and {S_1=S_2=1}, {S_3=-1} gives two equations (these are the only independent ones) which one can solve to obtain

\displaystyle K' = \frac{1}{4}\log\left(\frac{\cosh(3K)}{\cosh(K)}\right), \ \ \ C = 2(\cosh(K))^{3/4}(\cosh(3K))^{1/4}.

Now, note that every link on the hexagonal lattice is attached to a unique site on the {A} sublattice, and moreover each site on the {A} sublattice has 3 such links attached to it. So define {N_a = \left\{S_1^{(a)},S_2^{(a)},S_3^{(a)}\right\}} to be the three spins (on the {B} sublattice) which neighbor the site {a} on {A}. Notice further that the sum over nearest neighbors on the hexagonal lattice can be rewritten by summing over the nearest neighbors of every site on the {A} sublattice

\displaystyle \sum_{\langle ij\rangle\in{\mathrm{hex}}} = \sum_{a\in A}\sum_{b\in N_a}.

Then, taking {C} and {K'} defined as above, we can rewrite the partition function of the hexagonal model after decimation as

\displaystyle  \begin{array}{rcl}  \mathcal{Z}_{{\mathrm{hex}}}(K,S_{{\mathrm{hex}}}) &=& \sum_{\mathfrak{S}}\exp\left[\mathcal{H}(\mathfrak{S})\right] = \sum_{\mathfrak{S}}\exp\left(\sum_{\langle ij\rangle}KS_iS_j\right) =\sum_{\mathfrak{S}}\prod_{\langle ij\rangle}\exp(KS_iS_j) \\ &=& \sum_{\mathfrak{S}_A}\sum_{\mathfrak{S}_B}\prod_{a\in A}\prod_{b\in N_a}\exp(KS_aS_b) = \sum_{\mathfrak{S}_B}\prod_{a\in A}\sum_{S_a=\pm 1}\exp\left[KS_a\left(S_1^{(a)}+S_2^{(a)}+S_3^{(a)}\right)\right] \\ &=& \sum_{\mathfrak{S}_B}\prod_{a\in A}C\exp\left[K'\left(S_1^{(a)}S_2^{(a)} + S_2^{(a)}S_3^{(a)} + S_3^{(a)}S_1^{(a)}\right)\right] = C^{S_{{\mathrm{hex}}}/2}\sum_{\mathfrak{S}_B}\exp\left(K'\sum_{\langle i j\rangle\in\triangle}S_iS_j\right) \\ &=& C^{S_{{\mathrm{hex}}}/2}\mathcal{Z}_\triangle(K',S_{{\mathrm{hex}}}/2) \end{array}

Here’s the conclusion: the star-triangle transformation relates a model on the hexagonal lattice with {S_{{\mathrm{hex}}}} sites and parameter {K} to a triangular model with {S_{{\mathrm{hex}}}/2} sites and parameter {K'},

\displaystyle K\sim K' = \frac{1}{4}\log\left(\frac{\cosh(3K)}{\cosh(K)}\right).

Before, we showed that composing KW then ST transformations (or vice versa) relates the high and low temperature regimes of the triangular and hexagonal models respectively. Therefore, we can combine the above with Kramers-Wannier duality,

\displaystyle K\sim \tilde{K} = -\frac{1}{2}\log\left(\tanh K\right)

to get the desired relationship,

\displaystyle K\sim \hat{K} = \frac{1}{4}\log\frac{\cosh\left(-\frac{3}{2}\log\tanh K\right)}{\cosh\left(-\frac{1}{2}\log\tanh K\right)}.

Using the same arguments as we used for the square lattice, we obtain the critical temperature of the triangular model as the fixed point of this combined duality transformation, which can be computed using Mathematica as

\displaystyle K^{(\triangle)}_{\mathrm{c}} = \log(3)/4 \implies T^{(\triangle)}_{\mathrm{c}} = 4J/\log(3).

Composing the other way around gives us the critical temperature of the hexagonal model,

\displaystyle K_{\mathrm{c}}^{({\mathrm{hex}})} = \frac{1}{2}\log(2+\sqrt{3}) \implies T_{\mathrm{c}}^{({\mathrm{hex}})}=\frac{2J}{\log(2+\sqrt{3})}.

So concludes our solution.

9 Comments for “***Dualities in two-dimensional Ising models”

Stanley

says:

Is this a close book exam?
It would be really hard if you don’t have reference at hand.
I wrote something similar to this when I was in third year. The high-low temperature expansion relates to something very interesting in lattice gauge theory.

brandonrayhaun

says:

My bad, I was unclear about this. The exam was a take-home and Dam Son had already done the square lattice case in class. So, much easier than I made it seem. 😛

Not your business, really.

says:

A square has 4 sites, there is one distinct square in a square, therefore 4=1 according to your G_square argument. Nope, I can’t see how lattice of N spins has N squares. A lattice of 4 spins has 1 square and 4 is not 1. A lattice of 9 spins has 4 unit squares too. Not 9.

The dual of a square is one point at the center
So D is different from G

So what’s the point ? Is it that N^2-(N-1)^2=2N-1 is negligible compared to N^2 as N grows to infinity ? So we say it’s about equal ?

I got scammed into thinking I understood until you didn’t carry out the calculations later. Obviously I can’t do it because it’s wrong
I hate physics, it’s always obviously wrong. I wish I didn’t spend 5 years on it. Such a big waste of time. But yeah I live in France unfortunately and France sucks at physics

thanks anyway I guess

brandonrayhaun

says:

Sorry for the late reply.

Note that we are working with periodic boundary conditions. So you should imagine that the lattice is on the surface of a donut if you like, sort of like in this picture — https://i.stack.imgur.com/g8wdG.png — and if you go through it carefully, you’ll find that a lattice with N sites has N squares. The situation you described of a lattice with 4 sites is not a lattice with periodic boundary conditions.

Alternatively, you can work without periodic boundary conditions, in which case the formula becomes better as N becomes large.

Sorry for the lack of clarity. But I do think that all of your confusions will be resolved if you take into account the periodic boundary conditions. Please let me know if you have any more questions.

Matt

says:

Hi. Thanks for your thorough article. I have a question. Is f(X) not a finite sum of analytic functions? If so, can it have a singular point?

brandonrayhaun

says:

Thanks very much for your question. You’re right – the singularities of the partition function appear in the thermodynamic limit, which is the limit that N goes to infinity. I should have been more clear about this!

astro

says:

Wow! We had the hexagonal lattice question in the exam and I got a lot more confused. Thank you your post clears up so many of my doubts. I know it must be tiring to write such a post, explaining everything and accompanying it with diagrams and all…I am really thankful.

brandonrayhaun

says:

I’m glad it helped! When you say “the exam”, do you mean that you took a course with Dam Son as well?

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