# 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 there are no phase transitions, while for 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 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[edit | edit source]

Let us consider a one-dimensional Ising model at ; 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 spins at (with all spins pointing upwards) with nearest neighbour interactions and without any external field, i.e. the Hamiltonian of the system is:

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

In this case the internal energy of the system is:

## Two dimensions[edit | edit source]

We now consider a two-dimensional Ising model, with the same properties as before, defined on a lattice with coordination number . Let us consider the system at 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 bonds:

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

*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 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 . Thus we can estimate the difference in entropy between the ordered and the domain-flipped state as:

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

^{[3]}. If for example we consider a square lattice, and we get , while as we have seen previously the exact result gives : this discrepancy is due to the fact that we have overestimated the change in entropy of the system.

- ↑ 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. - ↑ This can be understood thinking about the one-dimensional case. For in fact we have when we flip a single spin (and therefore there are two antiparallel bonds), so we can argue that if we flip spins the variation of internal energy is (if not exactly, at least of the same order).
- ↑ But, as we have stressed many times, the behaviour of the thermodynamic properties of the system in the neighbourhood of is universal.