User:Pavithransiyer/Sherrington Kirkpatrick Model

The Hamiltonian or total energy for the model are respectively given by:

$$ H = \sum_{ij}J_{ij}S_{i}S_{j} - h\sum_{i}S_{i} $$

$$ Z = \text{Tr} e^{-\beta H} = \text{Tr}_{S}\exp\left\{\beta\sum_{ij}J_{ij}S_{i}S_{j} + \beta h \sum_{i}S_{i}\right\} $$

Having obtained the partition function, the next step is to obtain an expression for the free energy for the system, from which we can obtain the thermodynamics of the system. Notice that the free energy of the system cannot depend on the specific nature of the individual links. Hence the Free energy must be averaged with respect to the distribution of links. The free energy of this model is therefore given by averaging the free energy obtained for a particular realization of the distribution (denoted by $$F[J_{ij}]$$), over the entire distribution. Hence we write $$F = \overline{F[J_{ij}]} = -\beta\overline{\ln Z[J_{ij}]}$$. In general, one would rather expect to take the logarithm of the $$\overline{Z_{ij}}$$, rather that $$\overline{\ln Z[J_{ij}]}$$, where the latter case is a lot difficult than the other. The distinction between the two methods comes from the underlying assumption that the disorder is quenched. Since the disorder is frozen, the timescale at which the spin system assumes a different distribution of the links is much larger compared to that at which the spin fluctuates at the lattice sites. Therefore averaging the partition function directly: $$F = \ln\int P[J_{ij}]Z[J_{ij}]$$ is not correct as it treats the fluctuations in the links happening at the same time scale as the spin fluctuations. Therefore, we adopt the procedure of averaging over the free energy of all the realizations of the disorder.

If it was $$Z$$ and not $$\ln Z$$ which had to be averaged over, then this can be achieved by means of a single Gaussian integral. On the contrary, averaging over $$\ln Z[J_{ij}]$$ is difficult and requires the following identity, which is the replica trick:

$$\ln Z = \lim_{n\to 0}\dfrac{Z^{n} - 1}{n}$$

Now averaging over the logarithm reduces to averaging over an arbitrary power of $$Z$$ which is not much difficult from averaging over $$Z[J_{ij}]$$.

$$ F = \int dP[J_{ij}] F[J_{ij}] = -\dfrac{1}{\beta}\lim_{n\to 0}\int dP[J_{ij}] \dfrac{Z^{n}[J_{ij}] - 1}{n} $$

Taking the $$n^{th}$$ power of the partition function gives:

$$ Z^{n}[J_{ij}] = \left[\text{Tr} e^{-\beta H}\right]^{n} = \left[\text{Tr}_{S}\exp\left\{\beta\sum_{ij}J_{ij}S_{i}S_{j} + \beta h \sum_{i}S_{i}\right\}\right]^{n} $$

$$ Z^{n}[J_{ij}] = \Prod_{\alpha} $$

The variables $$J_{ij}$$ can take values $$\pm 1$$ corresponding to Ferromagnetic and Anti-Ferromagnetic nature of links, but their distribution is assumed to be Gaussian (one can assume a binomial distribution as well, or any distribution for that matter) with a mean value of $$\dfrac{J_{0}}{N}$$ and a variance of $$\dfrac{J^{2}}{N}$$, described by the density function:

$$ P[J_{ij}] = \dfrac{1}{J}\sqrt{\dfrac{1}{N}}\exp\left\{-\dfrac{N}{2J^{2}}\left(J_{ij} - \dfrac{J_{0}}{N}\right)^{2}\right\} $$