# The role of dimensionality

We have seen (although not really directly) that the dimensionality of the Ising model is crucial for the existence of phase transitions; in particular we have seen that for ${\displaystyle d=1}$ there are no phase transitions, while for ${\displaystyle d>1}$ they can occur. Sometimes the dimension of a model above which phase transitions occur is called lower critical dimension, so in our case we can say that the lower critical dimension of the Ising model is one.

We shall now use a heuristic argument in order to show that this is indeed the case for the Ising model, and that for ${\displaystyle d=1}$ there can be no long range order. This argument will also allow us to estimate (even if a bit roughly) the critical temperature of the two-dimensional Ising model.

## One dimension

Let us consider a one-dimensional Ising model at ${\displaystyle T=0}$; we know that in this case the system is completely ordered, namely all the spins point in the same direction, say upwards. If we now increase the temperature a little bit, then some spins will randomly flip due to thermal fluctuations; what we want to see is if the ordered state of the system is stable under this random spin flips[1]. Therefore, let us consider a one-dimensional Ising model of ${\displaystyle N}$ spins at ${\displaystyle T=0}$ (with all spins pointing upwards) with nearest neighbour interactions and without any external field, i.e. the Hamiltonian of the system is:

${\displaystyle {\mathcal {H}}=-J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}}$
and for the sake of convenience let us take periodic boundary conditions. The entropy of such a state is null (the system has only one possible configuration) so its free energy will be equal to its internal energy, and since there are ${\displaystyle N}$ parallel bonds we will have:
${\displaystyle F_{N}^{\text{ord.}}=U_{N}^{\text{ord.}}=-JN}$
Let us now flip a single spin:

One-dimensional Ising model at ${\displaystyle T=0}$${\displaystyle T=0}$
One-dimensional Ising model with a spin flipped

In this case, since two previously parallel bonds become antiparallel, the internal energy of the system becomes:

${\displaystyle U_{N}^{\text{flip}}=-J(N-2)+2J}$
and as of the entropy of the system:
${\displaystyle S_{N}^{\text{flip}}=k_{B}\ln N}$
since the flipped spin can be in any of the ${\displaystyle N}$ sites of the lattice (namely, the system can have ${\displaystyle N}$ different configurations). Therefore, from the thermodynamic definition of free energy (${\displaystyle F=U-TS}$):
${\displaystyle F_{N}^{\text{flip}}=-J(N-2)+2J-k_{B}T\ln N}$
and the variation of free energy due to the spin flipping is:
${\displaystyle \Delta F_{N}=F_{N}^{\text{flip}}-F_{N}^{\text{ord.}}=4J-k_{B}T\ln N}$
We can therefore see that for fixed ${\displaystyle T>0}$, ${\displaystyle \Delta F\to -\infty }$ if ${\displaystyle N\to \infty }$: in the thermodynamic limit it is energetically convenient for the system to flip spins. We can therefore continue with the operation of spin flipping until there are no other parallel spins left: the long range order of the system is unstable under thermal fluctuations, and so as soon as ${\displaystyle T>0}$ the one-dimensional Ising model can't exhibit a spontaneous magnetization (namely, there are no phase transitions). We could have obtained the same result with a slightly different approach, namely considering the system divided into two different magnetic domains, positively and negatively magnetized, respectivel:

One-dimensional Ising model with a domain flipped

In this case the internal energy of the system is:

${\displaystyle U_{N}^{\text{flip}}=-J(N-2)+2J=-JN+4J}$
On the other hand the entropy of the system is again the same. Therefore, we should have found ${\displaystyle \Delta F_{N}=F_{N}^{\text{flip}}-F_{N}^{\text{ord.}}=4J-k_{B}T\ln N}$, deducing the same conclusions.

## Two dimensions

We now consider a two-dimensional Ising model, with the same properties as before, defined on a lattice with coordination number ${\displaystyle z}$. Let us consider the system at ${\displaystyle T=0}$ with all the spins pointing upwards, and flip a domain of spins; the boundary of this domain will be a closed path, and let us suppose that it is made of ${\displaystyle n}$ bonds:

Two-dimensional Ising model with a domain flipped

The difference in internal energy between the ordered state and one with a flipped domain is therefore[2]:

${\displaystyle \Delta U=2Jn}$
The entropy is much trickier to compute exactly, so we will only give an estimate. If we consider the boundary of the domain as a random walk an upper limit to the number of possible configurations of this boundary is ${\displaystyle z^{n}}$; however, we are not taking into account the fact that the domain wall cannot intersect itself (otherwise there would be more than one domain), namely it must be a self-avoiding random walk. A first rough correction to this estimate could be supposing that at each step the domain wall can only go in ${\displaystyle z-1}$ directions since it must not immediately go back on itself, so a slightly better upper limit to the number of possible configurations of the system is ${\displaystyle (z-1)^{n}}$. Thus we can estimate the difference in entropy between the ordered and the domain-flipped state as:
${\displaystyle \Delta S\approx k_{B}Tn\ln(z-1)}$
However, our assumption still allows the domain boundary to intersect itself so we are surely overestimating ${\displaystyle \Delta S}$.

Therefore, the change in free energy due to the flipping of a domain of spins is:

${\displaystyle \Delta F=\left[2J-k_{B}T\ln(z-1)\right]n}$
We now see that the behaviour of ${\displaystyle \Delta F}$ is not well defined in the thermodynamic limit, namely when ${\displaystyle n\to \infty }$. In particular we will have ${\displaystyle \Delta F\to -\infty }$ only if:
${\displaystyle T>{\frac {2J}{k_{B}\ln(z-1)}}:=T_{c}}$
We therefore find something very interesting: in two dimensions the long range order of the Ising model is unstable under thermal fluctuations only when ${\displaystyle T>T_{c}}$, while for ${\displaystyle T it is stable, or in other words the system will always exhibit a spontaneous magnetization even in the absence of any external fields. We thus have found that for ${\displaystyle d=2}$ the Ising model undergoes a phase transition at ${\displaystyle T=T_{c}}$; let us note that this critical temperature depends on the coordination number ${\displaystyle z}$, so it is a characteristic of the lattice considered and not a universal property of the system[3]. If for example we consider a square lattice, ${\displaystyle z=4}$ and we get ${\textstyle T_{c}\approx 1.82{\frac {J}{k_{B}}}}$, while as we have seen previously the exact result gives ${\textstyle T_{c}\approx 2.27{\frac {J}{k_{B}}}}$: this discrepancy is due to the fact that we have overestimated the change in entropy of the system.

1. Of course, in the thermodynamic limit the flipping of a finite number of spins will not be sufficient, in general, in order to destroy the long range order; the only way to do so is flipping a non-zero fraction of spins.
2. This can be understood thinking about the one-dimensional case. For ${\displaystyle d=1}$ in fact we have ${\displaystyle \Delta U=4J}$ when we flip a single spin (and therefore there are two antiparallel bonds), so we can argue that if we flip ${\displaystyle N}$ spins the variation of internal energy is ${\displaystyle 2Jn}$ (if not exactly, at least of the same order).
3. But, as we have stressed many times, the behaviour of the thermodynamic properties of the system in the neighbourhood of ${\displaystyle T_{c}}$ is universal.