The one dimensional Ising Model

The one-dimensional Ising model consists of a row of spins, each of which can point in one of two directions. We label these two directions by and . The essential feature is that neighboring spins interact: when the coupling is ferromagnetic, neighboring spins prefer to agree. Thus two adjacent positive spins and two adjacent negative spins are energetically favorable, while a positive spin next to a negative spin costs energy. One may also apply an external magnetic field, which prefers one sign of spin over the other.

This is a very small-looking model, but it contains several ideas that later reappear throughout statistical mechanics, probability theory, dynamical systems, and even parts of analysis. There is a finite-volume probability measure; there is a partition function; there are thermodynamic limits; there are boundary conditions; there are correlation functions; there are fluctuations of the total magnetization; and there is the question of whether a small boundary bias or a small external field can produce macroscopic order. In one dimension with finite-range interactions, the answer is ultimately no at every positive temperature. But that conclusion is not obvious from the local energy alone. At low temperature, the spins really do form very long domains of almost constant sign. The important point is that “very long” is not the same thing as “infinite.”

We will derive everything from the finite model. The logical order will be as follows. First we solve the zero-field open chain by changing variables from spins to bonds. This already gives the partition function, energy, entropy, domain-wall distribution, and correlation function. Next we treat the periodic chain, where a parity constraint appears. Then we introduce the transfer matrix, which solves the model in an arbitrary field. Finally, we use the transfer-matrix spectrum to study boundary effects, infinite-volume Gibbs states, fluctuations, and the absence of a positive-temperature phase transition.

Throughout, the coupling constant is assumed positive: $J>0.$ The inverse temperature is $\beta=\frac{1}{k_B T}.$ It is convenient to use the dimensionless parameters $\displaystyle K=\beta J,\quad H=\beta h,$ where is the interaction strength and is the external magnetic field. Large means low temperature, while small means high temperature.

The finite open chain

Consider an open chain of spins $\sigma_1,\sigma_2,\dots,\sigma_N\in \{-1,+1\}.$ For free boundary conditions, the Hamiltonian is

$\displaystyle \mathcal H_N^{\mathrm{free}}(\sigma) =-J\sum_{i=1}^{N-1}\sigma_i\sigma_{i+1} -h\sum_{i=1}^{N}\sigma_i.$

The first sum is the interaction energy. If , then , and that bond contributes to the energy. If the spins disagree, then , and the bond contributes . Thus disagreement costs an extra energy . The second sum is the field contribution. When , a positive spin contributes and a negative spin contributes , so the field favors positive spins. When , the situation is reversed.

The finite-volume partition function is the sum of all Boltzmann weights:

$\displaystyle Z_N^{\mathrm{free}}(K,H) =\sum_{\sigma_1,\dots,\sigma_N=\pm1} \exp\big (-\beta \mathcal H_N^{\mathrm{free}}(\sigma) \big ) \\ =\sum_{\sigma_1,\dots,\sigma_N=\pm1} \exp\big( K\sum_{i=1}^{N-1}\sigma_i\sigma_{i+1} + H\sum_{i=1}^{N}\sigma_i \big).$

The Gibbs probability of a spin configuration is

$\displaystyle \mu_N(\sigma) =\frac{1}{Z_N^{\mathrm{free}}(K,H)} \exp\big( K\sum_{i=1}^{N-1}\sigma_i\sigma_{i+1} + H\sum_{i=1}^{N}\sigma_i \big).$

For any function of the spins, its expectation is

$\displaystyle \langle F\rangle_N = \sum_{\sigma}F(\sigma)\mu_N(\sigma).$

At finite , there is no difficulty in principle: the model is just a probability measure on a finite set of cardinality . The real problem begins when becomes large. We would like to know whether quantities per spin stabilize. The basic object is the finite-volume pressure

$\displaystyle p_N(K,H)=\frac{1}{N}\log Z_N(K,H).$

If the limit exists, we define

$\displaystyle p(K,H)=\lim_{N\to\infty}p_N(K,H).$

The free-energy density is then

$\displaystyle f(\beta,h) =-\frac{1}{\beta}p(\beta J,\beta h).$

The main task is to compute this limit(s) and then understand what it says about the spins themselves.

Zero External Field

We begin with the most transparent case: $H=0.$ Then the Hamiltonian is

$\displaystyle \mathcal H_N^{\mathrm{free}}(\sigma)=-J\sum_{i=1}^{N-1}\sigma_i\sigma_{i+1}.$

The energy depends only on whether neighboring spins agree. This suggests replacing the spin variables by bond variables

$\displaystyle b_i=\sigma_i\sigma_{i+1}, \quad 1\le i\le N-1.$

Each is either or . The value means that the bond is aligned, while means that there is a domain wall between sites and . At first sight, it may seem that the bond variables cannot be independent because they came from spins. For an open chain, however, they are independent once one remembers the first spin. Indeed, suppose that and the bonds are given. Then the other spins are determined recursively: $\sigma_{i+1}=\sigma_i b_i.$ Thus, for example, $\sigma_2=\sigma_1b_1,$ $\sigma_3=\sigma_1b_1b_2,$ and in general

$\displaystyle \sigma_j=\sigma_1\prod_{r=1}^{j-1}b_r.$

Conversely, every choice of one initial spin and every sequence of bond values determines a unique spin configuration. Therefore the map $(\sigma_1,\dots,\sigma_N) \longmapsto (\sigma_1,b_1,\dots,b_{N-1})$ is a bijection between the spin configurations and the set $\{-1,+1\}\times\{-1,+1\}^{N-1}.$ The Hamiltonian becomes

$\displaystyle \mathcal H_N^{\mathrm{free}} =-J\sum_{i=1}^{N-1}b_i.$

Hence the partition function is

$\displaystyle Z_N^{\mathrm{free}}(K,0) =2\sum_{b_1,\dots,b_{N-1}=\pm1} \exp\big(K\sum_{i=1}^{N-1}b_i\big).$

The factor comes from the two possible values of . Since the exponent is a sum of terms each depending on only one bond, the sum factors:

$\displaystyle Z_N^{\mathrm{free}}(K,0) =2\prod_{i=1}^{N-1}\big( \sum_{b_i=\pm1}e^{Kb_i} \big).$

For one bond, $\sum_{b=\pm1}e^{Kb} =e^K+e^{-K}=2\cosh K.$ Therefore

$\displaystyle Z_N^{\mathrm{free}}(K,0)=2(2\cosh K)^{N-1}.$

This formula is exact for every . It is worth pausing over what has happened. The original model looked interacting because the spins were coupled. But after passing to bonds, the zero-field open chain becomes a collection of independent random variables. The price is only that one must retain a single initial spin to reconstruct the entire configuration.

Taking logarithms gives

$\displaystyle \log Z_N^{\mathrm{free}}(K,0) =\log 2 + (N-1)\log(2\cosh K).$

Dividing by and letting , we obtain

$\displaystyle p(K,0)=\log(2\cosh K).$

Thus the free energy per spin is

$\displaystyle f(\beta,0)=-\frac{1}{\beta} \log\big (2\cosh(\beta J)\big ).$

The finite endpoint correction disappears after division by . This is the simplest first example of a general thermodynamic principle: changes affecting only finitely many sites do not alter the bulk free energy.

Because the bond sum factorizes, the bond variables are independent under the zero-field open-chain Gibbs measure. For a single bond,

$\displaystyle \mathbb P(b_i=+1)=\frac{e^K}{e^K+e^{-K}},$

while

$\displaystyle \mathbb P(b_i=-1)=\frac{e^{-K}}{e^K+e^{-K}}.$

Thus the probability that a specified bond is a domain wall is

$\displaystyle \rho_{\mathrm{wall}} = \mathbb P(b_i=-1)=\frac{1}{1+e^{2K}}.$

This formula already gives a concrete low-temperature picture. When is large,

$\displaystyle \rho_{\mathrm{wall}} = \frac{e^{-2K}}{1+e^{-2K}}=e^{-2K}+O(e^{-4K}).$

Since , this becomes

$\displaystyle \rho_{\mathrm{wall}}=\exp\big( -\frac{2J}{k_BT} \big) + O\big( \exp\big( -\frac{4J}{k_BT} \big)\big).$

The energy cost of a wall is , and its probability is correspondingly suppressed by the Boltzmann factor . The important point is that this probability is never zero at positive temperature. It may be unimaginably small, but on an infinite chain there are infinitely many places where a wall may occur.

The expected value of a bond is

$\displaystyle \langle b_i\rangle=\frac{e^K-e^{-K}}{e^K+e^{-K}}=\tanh K.$

Since , this is also the nearest-neighbor correlation:

$\displaystyle \langle \sigma_i\sigma_{i+1}\rangle =\tanh K.$

The energy per bond is

$\displaystyle -J\langle \sigma_i\sigma_{i+1}\rangle =-J\tanh K.$

In the limit of a large chain, the distinction between energy per bond and energy per spin disappears, so

$\displaystyle u(\beta,0)=-J\tanh(\beta J).$

The specific heat per spin is the temperature derivative of the internal energy: $c(T) =\frac{du}{dT}.$ We have $u(T)=-J\tanh K,$ with $K=\frac{J}{k_BT}.$ Differentiating gives

$\displaystyle c(T)=\frac{du}{dT}= k_BK^2{\text{sech}}^2K.$

This is finite for every positive temperature. It tends to zero both as and as . There is a smooth maximum at an intermediate temperature, but there is no singularity. In particular, no derivative of the free energy becomes infinite or discontinuous at positive temperature.

The entropy per spin may be obtained from $f=u-Ts.$ Since , where is the pressure, this can be rearranged as $s=k_B\bigl(p+\beta u\bigr).$ Therefore the entropy per spin is

$\displaystyle s(K) =k_B \big ( \log(2\cosh K)-K\tanh K \big ).$

There is another direct probabilistic way to see this. The first spin is equally likely to be positive or negative, and once and all the bonds are known, the entire spin configuration is determined. Hence the total entropy is the entropy of the initial spin plus the sum of the entropies of the independent bonds. The initial spin contributes exactly , The entropy of one bond is

$-k_B\Big [ \mathbb P(b_i=+1)\log\mathbb P(b_i=+1) + \mathbb P(b_i=-1)\log\mathbb P(b_i=-1) \Big ].$

Thus, for the open chain,

$\displaystyle \frac{S_N}{k_B} = \log 2 + (N-1)\big [ \log(2\cosh K)-K\tanh K \big ].$

After dividing by and letting , the contribution of the single initial spin disappears, and one recovers the entropy density formula above.

At high temperature, , so the interaction between neighboring spins becomes weak. Hence $\frac{s(K)}{k_B} =\log 2-\frac{K^2}{2}+O(K^4).$ In particular, $\lim_{K\to0}\frac{s(K)}{k_B}=\log2.$ . This agrees with the exact picture at : all spin configurations of an -site chain have equal probability, so $S_N=k_B\log(2^N)=Nk_B\log2.$ At low temperature, , $\lim_{K\to\infty}s(K)=0.$ The reason is that almost every bond is aligned. here remain two ground states, all positive and all negative, but their contribution to the total entropy is only . After division by , this becomes , which vanishes as .

Correlations

The bond variables also give the full two-point function. Suppose that . Then

$\displaystyle \sigma_i\sigma_j= (\sigma_i\sigma_{i+1}) (\sigma_{i+1}\sigma_{i+2}) \cdots (\sigma_{j-1}\sigma_j).$

Every spin between and appears twice and cancels because . Thus

$\displaystyle \sigma_i\sigma_j= \prod_{r=i}^{j-1}b_r.$

Since the bonds are independent,

$\displaystyle \langle \sigma_i\sigma_j\rangle=\prod_{r=i}^{j-1}\langle b_r\rangle.$

Using , we obtain

$\displaystyle \langle \sigma_i\sigma_j\rangle=(\tanh K)^{j-i}.$

By symmetry, this can be written as

$\displaystyle \langle \sigma_i\sigma_j\rangle=(\tanh K)^{|i-j|}.$

This is one of the fundamental formulas of the model. It gives an exact answer, not merely an asymptotic one.

Because for finite positive , the correlation decays exponentially as the distance grows. We define the correlation length by writing the decay in the form $(\tanh K)^r =\exp\big( -\frac{r}{\xi(K)} \big).$ Taking logarithms gives

$\displaystyle \xi(K) =-\frac{1}{\log(\tanh K)}.$

At high temperature, is small and , so the correlation length is short. At low temperature, the correlation length becomes very large. To see the precise growth, write $\tanh K=\frac{1-e^{-2K}}{1+e^{-2K}}.$ Then

$\displaystyle \xi(K)= \frac{1}{2}e^{2K} \big ( 1+O(e^{-4K}) \big ).$

Thus the correlation length grows exponentially as the temperature tends to zero:

$\displaystyle \xi(T) \sim \frac{1}{2} \exp\big( \frac{2J}{k_BT} \big).$

This is why a finite low-temperature chain can look strongly ordered. If its length is much smaller than , it will often contain no domain wall at all. But if its length is much larger than , it breaks into many domains.

The periodic chain

Now impose periodic boundary conditions: $\sigma_{N+1}=\sigma_1.$ The Hamiltonian becomes

$\displaystyle \mathcal H_N^{\mathrm{per}}(\sigma)=-J\sum_{i=1}^{N}\sigma_i\sigma_{i+1}.$

We again define bonds $b_i=\sigma_i\sigma_{i+1}, \quad 1\le i\le N.$ However, the bonds are no longer completely free. Multiplying them all together gives

$\displaystyle \prod_{i=1}^{N}\sigma_i\sigma_{i+1}.$

Every spin appears twice in the product, so

$\displaystyle \prod_{i=1}^{N}b_i=\prod_{i=1}^{N}\sigma_i^2= 1$

Thus the number of bonds with value must be even. A ring cannot contain exactly one domain wall: after changing from positive to negative somewhere, the spin configuration must eventually change back in order to close the circle.

We next count the number of configurations with exactly walls. First choose the locations of the walls. Suppose there are domain walls. Then there are aligned bonds and misaligned bonds. Therefore

$\displaystyle \sum_{i=1}^{N}\sigma_i\sigma_{i+1} =(N-2m)-2m =N-4m.$

The Boltzmann weight is then $e^{K(N-4m)}.$ There are choices for the wall locations, and once the wall locations are fixed there are two choices for the sign of one reference spin. The complete spin configuration is then determined. Hence the number of configurations with exactly walls is $2\binom{N}{2m}.$ Therefore

$\displaystyle Z_N^{\mathrm{per}}(K,0) =2\sum_{m=0}^{\lfloor N/2\rfloor} \binom{N}{2m}e^{K(N-4m)}.$

To simplify this, factor out :

$\displaystyle Z_N^{\mathrm{per}}(K,0)=2e^{KN} \sum_{m=0}^{\lfloor N/2\rfloor} \binom{N}{2m}e^{-4Km}.$

Now use the identity

$\displaystyle \sum_{m=0}^{\lfloor N/2\rfloor} \binom{N}{2m}x^{2m}=\frac{(1+x)^N+(1-x)^N}{2}.$

Taking , we obtain

$\displaystyle Z_N^{\mathrm{per}}(K,0)=e^{KN} \big [ (1+e^{-2K})^N + (1-e^{-2K})^N \big ].$

So we find

$\displaystyle Z_N^{\mathrm{per}}(K,0)= (2\cosh K)^N + (2\sinh K)^N.$

Equivalently,

$\displaystyle Z_N^{\mathrm{per}}(K,0) =(2\cosh K)^N \big ( 1+\tanh^N K \big ).$

Dividing by after taking logarithms gives

$\displaystyle \frac{1}{N} \log Z_N^{\mathrm{per}}(K,0)=\log(2\cosh K) + \frac{1}{N} \log\big ( 1+\tanh^N K \big ).$

The second term tends to zero because . Hence

$\displaystyle \lim_{N\to\infty} \frac{1}{N} \log Z_N^{\mathrm{per}}(K,0)=\log(2\cosh K).$

Thus periodic and free boundary conditions have the same thermodynamic pressure. Their finite-volume partition functions differ, because a ring imposes the even-domain-wall constraint, but that constraint affects only terms that are negligible after taking the logarithm

For the periodic chain, the two-point function differs slightly from the open-chain formula because there are two routes around the circle between two sites. Let $t=\tanh K.$ For two sites at distance , the exact periodic correlation is

$\displaystyle \langle \sigma_0\sigma_r\rangle_N^{\mathrm{per}} =\frac{t^r+t^{N-r}}{1+t^N}.$

One can derive this from the transfer matrix, but it is also consistent with the geometry. The factor corresponds to moving directly from site to site ; the factor corresponds to going the other way around the ring.

For fixed , sending gives

$\displaystyle \langle \sigma_0\sigma_r\rangle = t^r.$

This agrees with the open-chain calculation. The limiting infinite-volume correlation does not remember whether the finite system was built with free or periodic endpoints.

Transfer Matrix Method

The bond-variable transformation becomes less useful when . The reason is that the field term depends on the individual spins: $H\sum_{i=1}^{N}\sigma_i.$ Knowing the bond variables tells us only the relative signs of spins. To reconstruct the field energy, one must also know the cumulative product of bonds from a reference point. Thus the measure is no longer a product measure in the bonds.

The correct replacement is the transfer matrix. The transfer matrix is not a mysterious physical construction. It is simply a way of organizing the finite sum defining the partition function. Since the Hamiltonian contains only nearest-neighbor interactions, one can sum over spins sequentially, carrying only the value of the previous spin. This is exactly what a two-state matrix does.

We split the field contribution equally between neighboring bonds. Define

$\displaystyle T_{\sigma,\tau} =\exp\big( K\sigma\tau + \frac{H}{2}(\sigma+\tau) \big),$

where . In the basis ordered as , this matrix is

$\displaystyle T =\begin{pmatrix} e^{K+H} & e^{-K}\\ e^{-K} & e^{K-H} \end{pmatrix}.$

For a periodic configuration , the product of the corresponding transfer-matrix entries is

$\displaystyle \prod_{i=1}^{N} T_{\sigma_i,\sigma_{i+1}} =\exp\big(K\sum_{i=1}^{N}\sigma_i\sigma_{i+1} + \frac{H}{2} \sum_{i=1}^{N}(\sigma_i+\sigma_{i+1}) \big ).$

Because the spins are periodic, every spin occurs exactly twice in the second sum:

$\displaystyle \sum_{i=1}^{N}(\sigma_i+\sigma_{i+1}) =2\sum_{i=1}^{N}\sigma_i.$

Therefore

$\displaystyle \prod_{i=1}^{N} T_{\sigma_i,\sigma_{i+1}}=\exp\big(K\sum_{i=1}^{N}\sigma_i\sigma_{i+1} + H\sum_{i=1}^{N}\sigma_i \big).$

Summing over all periodic spin sequences is exactly the trace of :

$\displaystyle Z_N^{\mathrm{per}}(K,H) ={\text{Tr}}(T^N).$

For free boundary conditions, define the endpoint vector

$\displaystyle v =\begin{pmatrix} e^{H/2}\\ e^{-H/2} \end{pmatrix}.$

Then one checks directly that

$\displaystyle Z_N^{\mathrm{free}}(K,H)= v^{\mathsf T}T^{N-1}v.$

The endpoint vector supplies the half-field factors missing from the two ends of the open chain.

The transfer matrix is real and symmetric, so it has two real eigenvalues and an orthonormal basis of eigenvectors. Its trace is

$\displaystyle {\text{Tr}}(T) =e^{K+H}+e^{K-H}=2e^K\cosh H.$

Its determinant is

$\displaystyle \det(T)=e^{2K}-e^{-2K}.$

Thus the eigenvalues are

$\displaystyle \lambda_{\pm} =\frac{\text{Tr}(T) \pm \sqrt{(\text{Tr}(T))^2-4\det(T)}}{2}.$

Therefore

$\displaystyle \lambda_{\pm} = e^K \big( \cosh H \pm \sqrt{\sinh^2H+e^{-4K}} \big).$

At zero field, this reduces to

$\displaystyle \lambda_+ =e^K+e^{-K}=2\cosh K,$

and

$\displaystyle \lambda_- = e^K-e^{-K} =2\sinh K.$

So we have

$\displaystyle Z_N^{\mathrm{per}}(K,H) ={\text{Tr}}(T^N) = (2\cosh K)^N + (2\sinh K)^N.$

Thus the transfer-matrix calculation contains the earlier periodic answer as a special case.

Since , every entry of the transfer matrix is strictly positive. In particular, the larger eigenvalue is positive and strictly dominates the smaller one: $\lambda_+>\lambda_->0.$ Define $q(K,H) =\frac{\lambda_-}{\lambda_+}.$ Then $0<q(K,H)<1.$

Since $\displaystyle Z_N^{\mathrm{per}}(K,H) = \lambda_+^N+\lambda_-^N,$ we can factor out the dominant term:

$\displaystyle Z_N^{\mathrm{per}}(K,H) =\lambda_+^N \big ( 1+q(K,H)^N \big ).$

Therefore

$\displaystyle \frac{1}{N}\log Z_N^{\mathrm{per}}(K,H) = \log\lambda_+ + \frac{1}{N} \log\big( 1+q(K,H)^N \big).$

The second term tends to zero exponentially fast. Hence

$\displaystyle p(K,H)=\log\lambda_+.$

Substituting the formula for the larger eigenvalue gives

$\displaystyle p(K,H)= K +\log\big(\cosh H +\sqrt{\sinh^2H+e^{-4K}}\big).$

This is the exact infinite-volume pressure.

For free boundary conditions, the partition function has the form . Since the vector has strictly positive entries and the leading eigenvector of can also be chosen strictly positive, the leading eigenvalue contributes a nonzero coefficient. The lower eigenvalue contributes only a relative factor of order . Thus the free-boundary pressure has the same limit:

$\displaystyle \lim_{N\to\infty}\frac{1}{N}\log Z_N^{\mathrm{free}}(K,H)=\log\lambda_+.$

The bulk thermodynamics are therefore independent of whether the finite chain is closed into a ring or left with free endpoints.

Magnetization

The magnetization per spin is obtained by differentiating the pressure with respect to . It follows directly from differentiating the finite partition function.

Indeed,

$\displaystyle \frac{\partial Z_N}{\partial H} =\sum_{\sigma} \Big[\big( \sum_{i=1}^{N}\sigma_i\big) \exp\big( K\sum_i\sigma_i\sigma_{i+1} + H\sum_i\sigma_i \big) \Big ].$

Dividing by gives

$\displaystyle \frac{\partial}{\partial H}\log Z_N =\big \langle \sum_{i=1}^{N}\sigma_i \big \rangle_N.$

Hence

$\displaystyle \frac{\partial p_N}{\partial H} =\big \langle \frac{1}{N} \sum_{i=1}^{N}\sigma_i \big\rangle_N.$

After taking the thermodynamic limit, the magnetization is

$\displaystyle m(K,H) =\frac{\partial p}{\partial H}.$

Differentiating gives

$\displaystyle m(K,H) =\frac{\sinh H} {\sqrt{\sinh^2H+e^{-4K}}}.$

This formula gives the complete magnetization curve.

At zero field, $m(K,0)=0.$ More significantly, for every finite ,

$\displaystyle \lim_{H\downarrow0} m(K,H)=0.$

Indeed, as , the numerator behaves like , while the denominator tends to . Thus there is no spontaneous magnetization at positive temperature: $m_*(K) =\lim_{H\downarrow0}m(K,H)=0$

At zero field, the mean magnetization is zero, but the total magnetization fluctuates. Since correlations decay exponentially, the fluctuations remain of order $\sqrt N$ , although their variance is enlarged by the factor $e^{2K}$ . The cleanest way to see this is using the moment-generating function. For any real number $t$ ,

$\displaystyle \big \langle e^{tM_N/\sqrt N} \big \rangle_{K,0} = \frac{ Z_N(K,t/\sqrt N) }{ Z_N(K,0) }.$

For the periodic chain, the logarithm of the right side is

$\displaystyle \log Z_N(K,t/\sqrt N) - \log Z_N(K,0).$

The transfer-matrix formula gives

$\displaystyle \log Z_N(K,H) = Np(K,H) + O(1),$

uniformly for $H$ in a small fixed neighborhood of zero. The $O(1)$ term comes from the factor $1+q(K,H)^N$ , whose logarithm remains bounded. Now expand the analytic function $p(K,H)$ around $H=0$ . Since $p_H(K,0)=m(K,0)=0$ , we have

$\displaystyle p(K,H) = p(K,0) + \frac{1}{2}p_{HH}(K,0)H^2 + O(H^4).$

But $p_{HH}(K,0) = e^{2K}.$ Taking $H=t/\sqrt N$ , we get

$\displaystyle N \big(p(K,t/\sqrt N)-p(K,0) \big) =\frac{1}{2}e^{2K}t^2 + O(N^{-1}).$

Therefore

$\displaystyle \lim_{N\to\infty} \big \langle e^{tM_N/\sqrt N} \big \rangle =\exp\big (\frac{1}{2}e^{2K}t^2\big).$

The right side is the moment-generating function of a centered Gaussian random variable with variance $e^{2K}$ . Hence

$\displaystyle \frac{M_N}{\sqrt N} \Longrightarrow \mathcal N(0,e^{2K}).$

This says that the macroscopic magnetization is not of order $N$ ; it fluctuates only on the square-root scale. The variance becomes exponentially large at low temperature because spins are grouped into long correlated domains, but the fluctuations are still Gaussian at scales much larger than the correlation length.

Susceptibility

Differentiate the magnetization once more, we obtain

$\displaystyle \frac{\partial m}{\partial H}=\frac{e^{-4K}\cosh H}{\big (\sinh^2H+e^{-4K}\big )^{3/2}}.$

The physical susceptibility is differentiation with respect to . Since , we have

$\displaystyle \chi_h(K,H) =\frac{\partial m}{\partial h} =\beta\frac{\partial m}{\partial H}.$

At zero field, $\chi_h(K,0) =\beta e^{2K}.$ The susceptibility is therefore very large at low temperature:

$\displaystyle \chi_h(K,0)=\beta \exp\big(\frac{2J}{k_BT}\big).$

Again, this is not a phase transition. The quantity becomes large only as ; it remains finite at every fixed positive temperature.

The susceptibility can also be obtained directly from correlations. Let $M_N=\sum_{i=1}^{N}\sigma_i$ be the total magnetization. Since the external field appears in the Gibbs weight through the term $H M_N$ , differentiating the partition function with respect to $H$ inserts $M_N$ into the expectation. Thus

$\displaystyle \frac{\partial}{\partial H}\log Z_N=\langle M_N\rangle_N.$

Differentiating once more gives the standard fluctuation identity

$\displaystyle \frac{\partial^2}{\partial H^2}\log Z_N= \langle M_N^2\rangle_N-\langle M_N\rangle_N^2={\text{Var}}(M_N).$

Therefore the derivative of the magnetization per spin is the variance per spin:

$\displaystyle \frac{\partial}{\partial H}\left\langle \frac{M_N}{N}\right\rangle_N = \frac{1}{N}{\text{Var}}(M_N).$

At zero field, the measure is invariant under the global spin flip $\sigma_i\mapsto-\sigma_i$ , so $\langle M_N\rangle_N=0.$ Hence

$\displaystyle {\text{Var}}(M_N) = \langle M_N^2\rangle_N= \big \langle \left(\sum_{i=1}^{N}\sigma_i\right)^2 \big \rangle_N.$

Expanding the square gives

$\displaystyle {\text{Var}}(M_N) =\sum_{i=1}^{N}\sum_{j=1}^{N} \langle \sigma_i\sigma_j\rangle_N.$

Now use the zero-field correlation formula

$\displaystyle \langle \sigma_i\sigma_j\rangle=(\tanh K)^{|i-j|}.$

The terms with $i=j$ contribute $N$ , because $\sigma_i^2=1.$ For a fixed distance $r\ge1,$ there are $N-r$ pairs with $j=i+r$ and another $N-r$ pairs with $i=j+r.$ Thus there are $2(N-r)$ ordered pairs at distance $r.$ Therefore

$\displaystyle {\text{Var}}(M_N) = N+ 2\sum_{r=1}^{N-1}(N-r)(\tanh K)^r.$

Dividing by $N$ , we obtain

$\displaystyle \frac{{\text{Var}}(M_N)}{N}=1+ 2\sum_{r=1}^{N-1} \big (1-\frac{r}{N}\big)(\tanh K)^r.$

Since $0<\tanh K<1$ at every positive temperature, the sum has exponential decay. Hence, as $N\to\infty,$ the factor $1-r/N$ tends to $1$ for each fixed $r,$ and the limit becomes the infinite geometric sum:

$\displaystyle \lim_{N\to\infty}\frac{{\text{Var}}(M_N)}{N}= 1+2\sum_{r=1}^{\infty}(\tanh K)^r.$

Summing the series gives

$\displaystyle 1+2\sum_{r=1}^{\infty}(\tanh K)^r = 1+\frac{2\tanh K}{1-\tanh K} = \frac{1+\tanh K}{1-\tanh K}= e^{2K}.$

Thus

$\displaystyle \lim_{N\to\infty}\frac{\text{{Var}}(M_N)}{N} = e^{2K}.$

This agrees with the susceptibility computed earlier by differentiating the exact magnetization formula:

$\displaystyle \frac{\partial m}{\partial H}(K,0)=e^{2K}.$

So the zero-field susceptibility is exactly the integrated correlation function. The reason it grows at low temperature is now transparent: when $K$ is large, $\tanh K$ is close to $1,$ so correlations decay slowly. Many distant spins then contribute coherently to the fluctuation of $M_N,$ and therefore to the response to a small uniform field.

Correlations

The nearest-neighbor correlation is obtained by differentiating the pressure with respect to the coupling parameter . This is because multiplies the bond sum in the exponent. For the periodic chain,

$\displaystyle Z_N(K,H)= \sum_{\sigma}\exp \big (K\sum_{i=1}^{N} \sigma_i\sigma_{i+1} +H\sum_{i=1}^{N}\sigma_i \big).$

Differentiating gives

$\displaystyle \frac{\partial Z_N}{\partial K}=\sum_{\sigma} \big(\sum_{i=1}^{N}\sigma_i\sigma_{i+1} \big)\exp\big(K\sum_{i=1}^{N}\sigma_i\sigma_{i+1}+H\sum_{i=1}^{N}\sigma_i\big).$

After dividing by , one obtains

$\displaystyle \frac{\partial}{\partial K}\log Z_N=\big \langle \sum_{i=1}^{N}\sigma_i\sigma_{i+1} \big \rangle_N.$

The periodic Gibbs measure is translation-invariant, so every bond has the same expectation. Hence

$\displaystyle \frac{\partial}{\partial K} \big( \frac{1}{N}\log Z_N\big) =\langle \sigma_i\sigma_{i+1}\rangle_N.$

Thus the infinite-volume nearest-neighbor correlation is obtained by differentiating this expression with respect to

$\displaystyle \langle \sigma_i\sigma_{i+1}\rangle=\frac{\partial p}{\partial K}.$

In the thermodynamic limit, the pressure is

$\displaystyle p(K,H)=K+ \log\big(\cosh H+\sqrt{\sinh^2H+e^{-4K}}\big).$

The first term contributes For the logarithmic term, only the factor depends on

$\displaystyle \langle \sigma_i\sigma_{i+1}\rangle =1- \frac{2e^{-4K}}{\sqrt{\sinh^2H+e^{-4K}}\big (\cosh H+\sqrt{\sinh^2H+e^{-4K}}\big)}.$

The negative sign is natural. As the temperature rises, or equivalently as decreases, the spins become less aligned. Thus the correlation must lie below at every finite temperature.

At zero field, we have

$\displaystyle \langle \sigma_i\sigma_{i+1}\rangle =1-\frac{2e^{-4K}}{e^{-2K}(1+e^{-2K})} = \frac{e^K-e^{-K}}{e^K+e^{-K}}= \tanh K.$

This agrees exactly with the earlier bond-variable calculation. In that calculation, was itself the bond variable, whose mean was directly computed as The transfer-matrix calculation therefore recovers the same local correlation while also extending the formula to nonzero magnetic field.

The transfer matrix also gives the two-point correlation function at arbitrary external field. Let

$\displaystyle S=\begin{pmatrix} 1&0 \\ 0&-1 \end{pmatrix}.$

This matrix represents multiplication by the spin. Indeed, if the spin state is , then contributes the eigenvalue ; if the spin state is , it contributes the eigenvalue Thus inserting into a transfer-matrix product inserts a factor of the spin at that site. If we want to compute , we insert one copy of at site and another copy at site Between those two insertions there are transfer steps in one direction and transfer steps in the other direction. Therefore

$\displaystyle \langle \sigma_0\sigma_r\rangle_N=\frac{{\text{Tr}}\big (S T^r S T^{N-r}\big ) }{{\text{Tr}}(T^N)}.$

Now use the spectral decomposition of the transfer matrix. Since is real symmetric, it has an orthonormal eigenbasis. Let and be orthonormal eigenvectors corresponding to the eigenvalues and Let the corresponding projection operators be

$\displaystyle P_+=e_+e_+^{\mathsf T}, \quad P_-=e_-e_-^{\mathsf T}.$

Then we have the decomposition $\displaystyle T=\lambda_+P_+ + \lambda_-P_-.$ Consequently,

$\displaystyle T^r=\lambda_+^rP_+ + \lambda_-^rP_-.$

It is useful to factor out the dominant eigenvalue.

$\displaystyle \frac{T^r}{\lambda_+^r}=P_+ + q(K,H)^rP_-.$

Because for finite , one has Thus the term involving is exponentially smaller than the term involving when the power becomes large. Now take the thermodynamic limit with fixed. In the numerator ${\text{Tr}}\big (S T^r S T^{N-r}\big ),$ the factor is dominated by its part. The denominator is similarly dominated by Therefore

$\displaystyle \langle \sigma_0\sigma_r\rangle =e_+^{\mathsf T} S(P_+ + q^rP_-)Se_+.$

Now expand the two terms. The term gives

$\displaystyle e_+^{\mathsf T}SP_+Se_+ =\big(e_+^{\mathsf T}Se_+\big)^2.$

The quantity is the one-point function, so

$\displaystyle m(K,H)=e_+^{\mathsf T}Se_+.$

Therefore the first contribution is The term gives

$\displaystyle q^r e_+^{\mathsf T}SP_-Se_+= q^r\big(e_-^{\mathsf T}Se_+\big)^2.$

To identify the coefficient, use and Then To finish the calculation, we identify the coefficient of the decaying term. Since $S^2=I$ and $P_++P_-=I,$ we have

$\displaystyle 1 = e_+^{\mathsf T}S(P_++P_-)Se_+ = \big (e_+^{\mathsf T}Se_+\big)^2 + \big(e_-^{\mathsf T}Se_+\big)^2.$

But $e_+^{\mathsf T}Se_+$ is the magnetization $m(K,H).$ Therefore the remaining coefficient is $1-m(K,H)^2.$ Substituting this into the spectral expansion gives

$\displaystyle \langle \sigma_0\sigma_r\rangle = m(K,H)^2 + \big(1-m(K,H)^2\big)q(K,H)^r.$

Thus the connected correlation, obtained by subtracting the product of the one-point functions, is

$\displaystyle \langle \sigma_0\sigma_r\rangle- \langle\sigma_0\rangle\langle\sigma_r\rangle = \big(1-m(K,H)^2\big)q(K,H)^r.$

Here the decay factor is the ratio of the two transfer-matrix eigenvalues:

$\displaystyle q(K,H) = \frac{\lambda_-}{\lambda_+} = \frac{ \cosh H-\sqrt{\sinh^2H+e^{-4K}}}{\cosh H+\sqrt{\sinh^2H+e^{-4K}}}.$

Since $0<q(K,H)<1$ at positive temperature, the connected correlation decays exponentially with distance. In zero field this ratio becomes

$\displaystyle q(K,0) = \frac{1-e^{-2K}}{1+e^{-2K}} =\tanh K.$

Also $m(K,0)=0.$ Hence the general formula reduces to the familiar zero-field result $\langle \sigma_0\sigma_r\rangle =(\tanh K)^r.$ So the transfer-matrix computation recovers the bond-variable answer at zero field, while showing that in nonzero field the full correlation tends to $m(K,H)^2$ ; only the connected part decays to zero.

Boundary Conditions

A vanishing magnetization under symmetric boundary conditions does not, by itself, prove the absence of an ordered phase. In a finite box with no external field, the Gibbs measure is invariant under the global spin flip , so the average spin is automatically zero. But this can happen even in a system that has symmetry breaking in the infinite-volume limit. For example, in the higher-dimensional Ising model below the critical temperature, symmetric boundary conditions give zero magnetization in finite volume, while plus boundary conditions and minus boundary conditions lead to different infinite-volume states. Thus, to test whether the one-dimensional model really has no ordered phase, one should impose fixed boundary spins and see whether their influence survives deep in the bulk.

Consider the zero-field chain with two fixed endpoint spins

$\displaystyle \sigma_0=a,\quad \sigma_{N+1}=b, \qquad a,b\in \{-1,+1\}.$

The spins $\sigma_1,\dots,\sigma_N$ are summed over, but the two endpoint spins are held fixed. The zero-field transfer matrix is

$\displaystyle T_0= \begin{pmatrix} e^K&e^{-K}\\ e^{-K}&e^K \end{pmatrix}.$

This matrix has the normalized eigenvectors

$\displaystyle u_+ = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}, \quad u_-= \frac{1}{\sqrt2} \begin{pmatrix} 1\\ -1 \end{pmatrix}$

with eigenvalues $2\cosh K$ and $2\sinh K,$ respectively. Hence

$\displaystyle T_0^n = (2\cosh K)^n u_+u_+^{\mathsf T} + (2\sinh K)^n u_-u_-^{\mathsf T}.$

Writing this formula in spin indices, where the row and column labels are $a,b\in\{-1,+1\},$ gives

$\displaystyle (T_0^n)_{a,b} = \frac{(2\cosh K)^n}{2} \big(1+ab(\tanh K)^n\big).$

Indeed, the $u_+$ contribution gives $1/2$ for every pair of boundary spins, while the $u_-$ contribution gives $ab/2$ ; the ratio of the two eigenvalues is $\tanh K.$

The partition function with fixed boundary spins $a$ and $b$ is just the transfer amplitude from $a$ to $b$ across $N+1$ bonds:

$\displaystyle Z_N^{a,b} =(T_0^{N+1})_{a,b}.$

To compute the expected spin at an interior site $i,$ split the chain at that site. If $\sigma_i=s,$ then the left part contributes $(T_0^i)_{a,s}$ and the right part contributes $(T_0^{N+1-i})_{s,b}.$ Therefore

$\displaystyle \langle \sigma_i\rangle_{a,b} = \frac{ \sum_{s=\pm1} s(T_0^i)_{a,s}(T_0^{N+1-i})_{s,b}}{(T_0^{N+1})_{a,b}}.$

Substituting the explicit formula for $(T_0^n)_{a,b}$ and simplifying gives the exact expression

$\displaystyle \langle \sigma_i\rangle_{a,b} = \frac{ a(\tanh K)^i + b(\tanh K)^{N+1-i}}{1+ab(\tanh K)^{N+1}}.$

This formula is the precise statement of how boundary influence decays. The left boundary contributes the term $a(\tanh K)^i$ , which decreases exponentially with the distance from the left endpoint. The right boundary contributes $b(\tanh K)^{N+1-i}$ , which decreases exponentially with the distance from the right endpoint. The denominator is a finite-size correction coming from the interaction between the two boundaries through the whole chain; it tends to $1$ when the endpoints are far apart, because $0<\tanh K<1$ for every finite positive temperature. Thus, if the observation site is taken deep into the bulk, meaning $\min \{i,\ N+1-i\}\to\infty,$ then both boundary terms vanish, and

$\displaystyle \langle \sigma_i\rangle_{a,b}\to0.$

In particular, even plus boundary conditions on both sides do not create a nonzero bulk magnetization. If $a=b=+1,$ then

$\displaystyle \langle \sigma_i\rangle_{+,+} = \frac{ (\tanh K)^i +(\tanh K)^{N+1-i} }{ 1+(\tanh K)^{N+1} },$

which tends to zero whenever the site is far from both boundaries. The boundary can polarize nearby spins, but its effect decays exponentially with distance. Therefore a plus boundary does not produce a genuine plus phase in the one-dimensional nearest-neighbor Ising model at positive temperature.

Gibbs States

The preceding boundary calculation proves more than the statement that the average spin vanishes in the middle of a long chain. It proves uniqueness of the infinite-volume Gibbs state at every positive temperature. To explain what this means, we should distinguish two different questions. The first question is about one particular observable, for example $\sigma_i$ . The second question is about the entire probability law of any fixed finite block of spins. A system has more than one infinite-volume Gibbs state if different boundary conditions at infinity can produce genuinely different limiting probability laws in finite regions. Thus it is not enough to show only that one expectation vanishes; one wants to show that every local observable has the same limiting expectation, regardless of the boundary condition.

Fix a finite set of sites, say $A\subset\mathbb Z$ , and let $F$ be an observable depending only on the spins in $A.$ For example, $F$ could be $\sigma_0$ , or $\sigma_0\sigma_3$ , or the indicator of the event that a whole finite block has a specified spin pattern. Now put this set inside a much larger interval $[L,R]$ , impose boundary spins at $L-1$ and $R+1$ , and form the finite-volume Gibbs measure. The question is whether

$\displaystyle \langle F\rangle_{[L,R]}^{a,b}$

has a limit as $L\to-\infty$ and $R\to+\infty,$ and whether that limit depends on the boundary spins $a,b\in\{-1,+1\}.$ In general, we want to understand if the infinite limit depends on the details of how the limit is taken.

In the one-dimensional nearest-neighbor model, the only way the outside world can affect the interval is through the two endpoint spins. This is the feature of nearest-neighbor interactions. At zero field, the explicit formula

$\displaystyle \langle \sigma_i\rangle_{a,b} = \frac{ a(\tanh K)^i + b(\tanh K)^{N+1-i} }{1+ab(\tanh K)^{N+1}}$

already shows the essential mechanism: the influence of a boundary spin is multiplied by a power of $\tanh K.$ Since $0<\tanh K<1$ at every finite positive temperature, this influence decays exponentially with the distance from the boundary. The same idea applies not only to the single spin , but to every local observable. Suppose depends only on the spins in a fixed finite set Then changing the boundary spins at and can affect only through the transfer-matrix factors connecting the boundary to the set Since boundary influence is carried by the subleading eigenvalue, it is exponentially small in the distance from to the boundary. More precisely, for some constant , depending on but not on the size of the interval,

$\displaystyle \Big | \langle F\rangle_{[L,R]}^{a,b} -\langle F\rangle_{[L,R]}^{a',b'} \Big | \le C_F (\tanh K)^{{\text{dist}}(A,{L-1,R+1})}.$

Here denotes the distance from the observed set to the nearer boundary. Thus, as the endpoints are sent farther away, the difference between any two choices of boundary condition tends to zero. This is the precise sense in which the bulk forgets the boundary. The same conclusion can be expressed through the transfer matrix. A local expectation in a finite interval is built from a product of transfer matrices: one long product from the left boundary to the observed block, a finite product through the block, and another long product from the block to the right boundary. The spectral decomposition is

$\displaystyle T^n = \lambda_+^nP_+ + \lambda_-^nP_-.$

After factoring out , this becomes

$\displaystyle \frac{T^n}{\lambda_+^n} =P_{+} + \big(\frac{\lambda_-}{\lambda_+}\big)^nP_{-}.$

At positive temperature, the transfer matrix has strictly positive entries, so is strictly dominant. We see that the boundary-dependent part is suppressed by a factor of order Therefore, if the observed block is far from the boundary, all boundary dependence is exponentially small. Consequently, for every local observable the infinite-volume limit

$\displaystyle \lim_{L\to-\infty,\ R\to+\infty} \langle F \rangle_{[L,R]}^{a,b}$

exists and is independent of the boundary spins and This is the local meaning of uniqueness of the infinite-volume Gibbs state. An infinite-volume Gibbs state is a probability measure on configurations $\{\pm1\}^{\mathbb Z}$ whose conditional distribution inside every finite interval agrees with the finite-volume Ising Gibbs distribution determined by the two exterior endpoint spins. Uniqueness means that there is only one probability measure with this property. There is no hidden plus state and minus state at positive temperature. This is also the mathematical meaning of “no phase coexistence” here. In a system with phase coexistence, different boundary conditions at infinity select different infinite-volume Gibbs states. For example, plus boundary conditions might lead to a state with positive magnetization, while minus boundary conditions might lead to a state with negative magnetization. In the one-dimensional nearest-neighbor Ising model at positive temperature, this does not happen. Plus and minus boundary conditions influence only a finite boundary layer, and that influence decays exponentially into the chain. Deep in the bulk, all boundary conditions give the same local statistics. Therefore the model has a single positive-temperature equilibrium phase.

Markov-chain

The zero-field infinite chain can be described exactly as a stationary two-state Markov chain. Given the current spin $\sigma_i=s$ , the next spin has transition probabilities

$\displaystyle \mathbb P(\sigma_{i+1}=s'\mid\sigma_i=s)=\frac{e^{Kss'}}{2\cosh K}.$

Thus

$\displaystyle \mathbb P(\sigma_{i+1}=\sigma_i) =\frac{e^K}{2\cosh K}, \quad \mathbb P(\sigma_{i+1}=-\sigma_i) =\frac{e^{-K}}{2\cosh K}.$

The stationary distribution is uniform:

$\displaystyle \mathbb P(\sigma_i=+1) =\mathbb P(\sigma_i=-1) =\frac{1}{2}.$

Indeed, the probability of a spin sequence under this Markov chain is

$\displaystyle \frac{1}{2} \prod_{i=1}^{N-1} \frac{e^{K\sigma_i\sigma_{i+1}}}{2\cosh K}.$

This equals

$\displaystyle \frac{ \exp\big( K\sum_{i=1}^{N-1}\sigma_i\sigma_{i+1} \big)}{2(2\cosh K)^{N-1}},$

which is exactly the zero-field free-boundary Gibbs measure.

The transition matrix is

$\displaystyle P =\begin{pmatrix} \frac{e^K}{2\cosh K} & \frac{e^{-K}}{2\cosh K} \\ \frac{e^{-K}}{2\cosh K} & \frac{e^K}{2\cosh K} \end{pmatrix}.$

Its eigenvalues are $1$ and $\tanh K.$ The second eigenvalue measures how rapidly the chain forgets its starting state. This is exactly the same number that appears in the spin correlation function. Thus the correlation length is also the mixing length of the Markov chain.

Phase Transition:

There are several closely related ways to state the final conclusion about phase transition.

First, the infinite-volume pressure is explicitly

$\displaystyle p(K,H) =K + \log\big ( \cosh H+ \sqrt{\sinh^2H+e^{-4K}}\big ).$

For every finite , the quantity inside the square root is strictly positive: $\sinh^2H+e^{-4K}>0.$ Therefore the pressure is a real analytic function of and for every positive temperature. There is no nonanalytic point in the free energy at finite .

Second, the spontaneous magnetization vanishes:

$\displaystyle \lim_{H\downarrow0}m(K,H)=0.$

Third, the correlation function decays exponentially:

$\displaystyle \langle \sigma_0\sigma_r\rangle = (\tanh K)^r.$

Fourth, fixed boundary conditions lose their influence exponentially fast as one moves into the bulk.

Fifth, the chain has a positive density of domain walls at every positive temperature:

$\displaystyle \rho_{\mathrm{wall}}=\frac{1}{1+e^{2K}}>0.$

The domain-wall explanation is especially instructive. A wall in one dimension is a point defect. It costs only the fixed energy , independent of the length of the system. At positive temperature, it therefore has a nonzero probability of occurring. Since the system has infinitely many possible locations for walls, an infinite chain contains infinitely many of them. The chain may spend an extremely long interval in the positive state and then an extremely long interval in the negative state, but it cannot sustain one sign forever.

All of these are manifestations of the same fact: the one-dimensional nearest-neighbor Ising model has a unique Gibbs state at positive temperature and no ordered phase.

The fact that there is no phase transition at positive temperature does not mean that the model is entirely nonsingular. The singularity occurs at zero temperature.

At zero field, the two ground states are

$\displaystyle \sigma_i=+1 \quad\text{for all }i,$

and

$\displaystyle \sigma_i=-1 \quad\text{for all }i.$

Both have energy per spin . If a field is present, it selects one of them. The zero-temperature energy density is

$\displaystyle e_0(h)= -J-|h|.$

This has a cusp at . The zero-temperature magnetization is

$\displaystyle m_{T=0}(h) = \begin{cases} +1,& h>0,\\ -1,& h<0. \end{cases}$

Thus the magnetization jumps at zero field only after the temperature has been taken to zero.

The order of limits matters. At fixed positive temperature,

$\displaystyle \lim_{h\to 0} m(T,h)=0.$

But if one first lets and then sends a positive field to zero, one obtains

$\displaystyle \lim_{h\downarrow0} \lim_{T\to0} m(T,h)= 1$

The zero-temperature point is therefore singular even though every positive-temperature point is analytic.

A useful low-temperature scaling observation comes from

$\displaystyle m(K,H)= \frac{\sinh H}{\sqrt{\sinh^2H+e^{-4K}}}.$

For small , this is approximately $\frac{H}{\sqrt{H^2+e^{-4K}}}.$ Thus the crossover from nearly zero magnetization to nearly full magnetization occurs on the field scale $H\asymp e^{-2K}.$ This is the same exponential scale that governs the domain-wall density and the inverse correlation length.

Renormalization

The one-dimensional Ising model also permits an exact coarse-graining transformation. Work at zero field and sum over every second spin. Consider three spins with interactions between and , and between and . Summing over the middle spin gives

$\displaystyle \sum_{\tau=\pm1} e^{K\sigma\tau+K\tau\rho}=\sum_{\tau=\pm1} e^{K\tau(\sigma+\rho)}.$

There are two cases. If , then , and the sum is $2\cosh(2K).$ If , then , and the sum is $2.$

We seek constants and such that

$\displaystyle \sum_{\tau=\pm1} e^{K\sigma\tau+K\tau\rho} = Ae^{K'\sigma\rho}.$

When , this requires

$\displaystyle Ae^{K'} =2\cosh(2K).$

When , it requires

$\displaystyle Ae^{-K'}=2$

Dividing the two equations gives $e^{2K'}=\cosh(2K).$ Thus

$\displaystyle K'=\frac{1}{2}\log(\cosh(2K)).$

The transformation is simpler if we use $t=\tanh K.$ We get

$\displaystyle \tanh K'=(\tanh K)^2.$

Thus we have $\displaystyle t\longmapsto t^2.$ under this transformation. For every finite positive temperature, $0\le t<1.$ Repeated transformation therefore sends $t,\ t^2,\ t^4,\ t^8,\dots$ to zero. The only nonzero fixed point is , corresponding to , or zero temperature. This coarse-graining picture gives another explanation for why the one-dimensional nearest-neighbor model has no positive-temperature critical point.

Conclusion

The exact solution of the one-dimensional Ising model is not just a computation of the partition function. It gives a complete picture of how order, fluctuations, boundary effects, and temperature fit together in one dimension.

At zero field, the essential simplification is that the model becomes independent in the bond variables ` $b_i=\sigma_i\sigma_{i+1}.$ ` A bond with ` $b_i=-1$ ` is a domain wall, and its probability is

$\displaystyle \rho_{\mathrm{wall}} =\frac{1}{1+e^{2K}}.$

Thus domain walls are rare at low temperature, but they are not absent at any finite temperature. The nearest-neighbor correlation is

$\displaystyle \langle \sigma_i\sigma_{i+1}\rangle = \tanh K,$

and the full zero-field two-point function is

$\displaystyle \langle \sigma_i\sigma_j\rangle = (\tanh K)^{|i-j|}.$

Therefore correlations decay exponentially. The correlation length is

$\displaystyle \xi(K) = -\frac{1}{\log(\tanh K)}.$

At low temperature, $K\to\infty,$ ` this becomes

$\displaystyle \xi(K)\sim \frac{1}{2}e^{2K}.$

So the low-temperature chain has very long ordered regions. But the typical domain size is still finite at every finite $K.$ `

The transfer-matrix calculation extends the solution to nonzero external field. The pressure is

$\displaystyle p(K,H) = K+ \log\big(\cosh H+ \sqrt{\sinh^2H+e^{-4K}} \big).$

Differentiating with respect to the field gives the magnetization

$\displaystyle m(K,H) = \frac{\sinh H} {\sqrt{\sinh^2H+e^{-4K}}}.$

For every finite $K,$ ` this magnetization is continuous at ` $H=0,$ ` and

$\displaystyle \lim_{H\downarrow 0} m(K,H)=0.$

Thus there is no spontaneous magnetization at positive temperature. Plus boundary conditions can polarize spins near the boundary, but their influence decays exponentially into the bulk. Consequently the infinite-volume Gibbs state is unique: plus, minus, free, or mixed boundary conditions all give the same local limiting state.

This is the central lesson of the one-dimensional model. Low temperature produces strong local alignment, long domains, large susceptibility, and a large correlation length. But none of these is the same as true long-range order. At every positive temperature, domain walls have finite energy cost and therefore positive density. An infinite chain contains infinitely many such walls, so the system repeatedly switches between positive and negative domains. The chain may look ordered on large finite scales, especially when $K$ is large, but it has no genuine ordered phase at any positive temperature.

The only singular limit is the zero-temperature limit. As $T \to 0,$ the domain-wall density tends to zero and the correlation length diverges. At exactly zero temperature, the two ground states, all positive and all negative, become distinct. Thus the one-dimensional nearest-neighbor Ising model has no positive-temperature phase transition, but it does have a zero-temperature ordering singularity.

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