Random energy model

The ${\displaystyle r}$-spin infinite-range model, in which all ${\displaystyle r}$-spin sets interact with a random, independent, identically distributed interaction constant, becomes the random energy model in a suitably defined ${\displaystyle r\to \infty }$ limit.[3]

More precisely, if the Hamiltonian of the model is defined by

${\displaystyle H({\boldsymbol {\sigma }})=\sum _{\{i_{1},\ldots ,i_{r}\}}J_{i_{1},\ldots i_{r}}\sigma _{i_{1}}\cdots \sigma _{i_{r}},}$

where the sum runs over all ${\displaystyle {N \choose r}}$ distinct sets of ${\displaystyle r}$ indices, and, for each such set, ${\displaystyle \{i_{1},\ldots ,i_{r}\}}$, ${\displaystyle J_{i_{1},\ldots ,i_{r}}}$ is an independent Gaussian variable of mean 0 and variance ${\displaystyle J^{2}r!/(2N^{r-1})}$, the Random-Energy model is recovered in the ${\displaystyle r\to \infty }$ limit.

As its name suggests, in the REM each microscopic state has an independent distribution of energy. For a particular realization of the disorder, ${\displaystyle P(E)=\delta (E-H(\sigma ))}$ where ${\displaystyle \sigma =(\sigma _{i})}$ refers to the individual spin configurations described by the state and ${\displaystyle H(\sigma )}$ is the energy associated with it. The final extensive variables like the free energy need to be averaged over all realizations of the disorder, just as in the case of the Edwards–Anderson model. Averaging ${\displaystyle P(E)}$ over all possible realizations, we find that the probability that a given configuration of the disordered system has an energy equal to ${\displaystyle E}$ is given by

${\displaystyle [P(E)]={\sqrt {\frac {1}{N\pi J^{2}}}}\exp \left(-{\dfrac {E^{2}}{J^{2}N}}\right),}$

where ${\displaystyle [\cdots ]}$ denotes the average over all realizations of the disorder. Moreover, the joint probability distribution of the energy values of two different microscopic configurations of the spins, ${\displaystyle \sigma }$ and ${\displaystyle \sigma '}$ factorizes:

${\displaystyle [P(E,E')]=[P(E)]\,[P(E')].}$

It can be seen that the probability of a given spin configuration only depends on the energy of that state and not on the individual spin configuration.[4]

The entropy of the REM is given by[5]

${\displaystyle S(E)=N\left[\log 2-\left({\frac {E}{NJ}}\right)^{2}\right]}$

for ${\displaystyle |E|<NJ{\sqrt {\log 2}}}$. However this expression only holds if the entropy per spin, ${\displaystyle \lim _{N\to \infty }S(E)/N}$ is finite, i.e., when ${\displaystyle |E|<-NJ{\sqrt {\log 2}}.}$ Since ${\displaystyle (1/T)=\partial S/\partial E}$, this corresponds to ${\displaystyle T>T_{c}=1/(2{\sqrt {\log 2}})}$. For ${\displaystyle T<T_{c}}$, the system remains "frozen" in a small number of configurations of energy ${\displaystyle E\simeq -NJ{\sqrt {\log 2}}}$ and the entropy per spin vanishes in the thermodynamic limit.