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</math></center>
</math></center>
It is crucial to assume <math>
It is crucial to assume <math>
  \bar{ \quad J\quad}=0 </math>, otherwise the model displays ferro/antiferro order. We sill discuss two distributions:
  \bar{ J}=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 σ=±1 and live on a lattice of N sitees i=1,2,,N. The enregy is writteen as a sum between the nearest neighbours <i,j>:

E=<i,j>Jijσiσj

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

J¯dJJπ(J)=0

It is crucial to assume J¯=0, otherwise the model displays ferro/antiferro order. We sill discuss two distributions:

  • Gaussian couplings: π(J)=exp(J2/2)/2π
  • Coin toss couplings, J=±1, selected with probability 1/2.

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

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