# Is the mean two-spin correlation proportional to the internal energy of an Ising System?

The Ising model has this partition function

$$Z= \sum_{states}e^{-\beta E}= \sum_{\{\sigma \}}e^{\beta J \sum_{<i,j>}\sigma_i\sigma_j}$$

The internal energy can be calculated as:

$$U=\frac{1}{Z}\sum_{states}Ee^{-\beta E} = -\frac{\partial}{\partial \beta} ln(Z)$$

If I want to calculate the mean two-spin correlation $<\sigma_i\sigma_j>$ where $i$ and $j$ are neibouring sites I would do this:

$$<\sigma_i\sigma_j>=\frac{1}{Z}\sum_{states}\sigma_i\sigma_j\ e^{-\beta E} = \frac{1}{Z}\sum_{\{\sigma \}}\sigma_i\sigma_j\ e^{\beta J \sum_{<i,j>}\sigma_i\sigma_j}$$

Isn't this the same as this?

$$<\sigma_i\sigma_j>=\frac{1}{N^2}\frac{1}{J}\frac{\partial}{\partial \beta} ln(Z)$$

where the $N^2$ accounts for the summation over $<i,j>$ that drops with the derivative. If this is true, then:

$$U= -J N^2 <\sigma_i\sigma_j>$$

Is this true?

• Yes, if you are considering the 2d model with periodic boundary condition. There is just a factor of $2$ missing (there are twice as many edges as vertices). Commented Jan 2, 2018 at 9:38
• Thanks! I have a question, though. If U is an extensive quantity, that means that the correlation scales as 1/N... that means that in the thermodynamic limit it goes to zero... this can't be right. Commented Jan 2, 2018 at 17:13
• I understood your $N^2$ as denoting the number of vertices of the system (and this is the correct understanding for your formula to be correct). In that case, $U/N^2$ converges to the energy density, which is a finite, positive quantity. Commented Jan 2, 2018 at 17:58
• I thought N was the number of sites in the lattice so the total number of possible pairs <i,j> is 2N^2 (with the 2 factor as you pointed out). I always thought that the internal energy scales with the number of "particles" or, in this case, sites... is that not correct? it scales with the number or pairs? I'm not sure what you mean by vertices... sorry, not an expert on Ising exactly here. Thanks! Commented Jan 2, 2018 at 18:25
• The number of pairs of nearest neighboring spins on your lattice is equal to twice the number of spins (remember that each spin only interacts with its $4$ neighbors, not with all other spins!). The number of terms in the sum is twice the number of spins (rather than $4$ times), because you don't want to count each pair twice. So, if $N$ is the number of spins in your system, then your formula should have $2N$ rather than $N^2$. Commented Jan 2, 2018 at 19:10

$$U= -\frac{q}{2}J N <\sigma_i\sigma_j>$$
where $q$ is the number of neighbours in the lattice.