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</math></center>
</math></center>
It is crucial to assume <math>
It is crucial to assume <math>
  \bar{ \quadJ\quad}=0 </math>, otherwise the model displays ferro/antiferro order. We sill discuss two distributions:
  \bar{ \quad J\quad}=0 </math>, otherwise the model displays ferro/antiferro order. We sill discuss two distributions:
* Gaussian couplings: <math> \pi(J) =\exp\left(-J^2/2\right)/\sqrt{2 \pi}</math>
* Gaussian couplings: <math> \pi(J) =\exp\left(-J^2/2\right)/\sqrt{2 \pi}</math>
* Coin toss couplings, <math>J= \pm 1 </math>, selected  with probability <math>1/2 </math>.
* Coin toss couplings, <math>J= \pm 1 </math>, selected  with probability <math>1/2 </math>.

Revision as of 16:22, 12 November 2023

Spin glass Transition

Experiments

Parlare dei campioni di rame dopati con il magnesio, marino o no: trovare due figure una di suscettivita e una di calore specifico, prova della transizione termodinamica.

Edwards Anderson model

We consider for simplicity the Ising version of this model.

Ising spins takes two values and live on a lattice of sitees . The enregy is writteen as a sum between the nearest neighbours <i,j>:

Edwards and Anderson proposed to study this model for couplings that are i.i.d. random variables with zero mean. We set the coupling distribution indicate the avergage over the couplings called disorder average, with an overline:

It is crucial to assume , otherwise the model displays ferro/antiferro order. We sill discuss two distributions:

  • Gaussian couplings:
  • Coin toss couplings, , selected with probability .

Edwards Anderson order parameter

The SK model

Random energy model

Derivation

Bibliography

Bibliography

  • Theory of spin glasses, S. F. Edwards and P. W. Anderson, J. Phys. F: Met. Phys. 5 965, 1975

=