Random energy model

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In the statistical physics of disordered systems, the random energy model is a toy model of a system with quenched disorder, such as a spin glass, having a first-order phase transition.[1][2] It concerns the statistics of a collection of N spins (i.e. degrees of freedom σ{σi}i=1N that can take one of two possible values σi=±1) so that the number of possible states for the system is 2N. The energies of such states are independent and identically distributed Gaussian random variables Ex𝒩(0,N/2) with zero mean and a variance of N/2. Many properties of this model can be computed exactly. Its simplicity makes this model suitable for pedagogical introduction of concepts like quenched disorder and replica symmetry.

Thermodynamic quantities

Critical energy per particle: hc=ln2.

Critical inverse temperature βc=2ln2.

Partition function Z(β)=seβH(s), which at large N becomes 2N𝔼E[eβE] when β<βc, that is, condensation does not occur. When this is true, we say that it has the self-averaging property.


Free entropy per particlef(β)=limN1NlnZ={ln2+14β2β<βc,βln2β>βc

Entropy per particles(h)=maxβ(f(β)βh)={ln2h2h[hc,+hc],0else 

Condensation

When β<βc, the Boltzmann distribution of the system is concentrated at energy-per-particle h=β/2, of which there are eN(ln2β2/4) states.

When β>βc, the Boltzmann distribution of the system is concentrated at h=hc, and since the entropy per particle at that point is zero, the Boltzmann distribution is concentrated on a sub-exponential number of states. This is a phase transition called condensation.

Participation

Define the participation ratio asY=EpE2=Ee2βE(EeβE)2The participation ratio measures the amount of condensation in the Boltzmann distribution. It can be interpreted as the probability that two randomly sampled states are exactly the same state. Indeed, it is precisely the Simpson index, a commonly used diversity index.

For each N,β, the participation ratio is a random variable determined by the energy levels.

When β<βc, the system is not in the condensed phase, and so by asymptotic equipartition, the Boltzmann distribution is asymptotically uniformly distributed over eN(ln2β2/4) states. The participation ratio is then eN(ln2β2/4)×(eN(ln2β2/4))2=eN(ln2β2/4)which decays exponentially to zero.

When β>βc, the participation ratio satisfieslimN𝔼[Y]=1βcβwhere the expectation is taken over all random energy levels.

Comparison with other disordered systems

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

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

H(σ)={i1,,ir}Ji1,irσi1σir,

where the sum runs over all (Nr) distinct sets of r indices, and, for each such set, {i1,,ir}, Ji1,,ir is an independent Gaussian variable of mean 0 and variance J2r!/(2Nr1), the Random-Energy model is recovered in the r limit.

Derivation of thermodynamical quantities

As its name suggests, in the REM each microscopic state has an independent distribution of energy. For a particular realization of the disorder, P(E)=δ(EH(σ)) where σ=(σi) refers to the individual spin configurations described by the state and H(σ) 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 P(E) over all possible realizations, we find that the probability that a given configuration of the disordered system has an energy equal to E is given by

[P(E)]=1NπJ2exp(E2J2N),

where [] 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, σ and σ factorizes:

[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]

S(E)=N[log2(ENJ)2]

for |E|<NJlog2. However this expression only holds if the entropy per spin, limNS(E)/N is finite, i.e., when |E|<NJlog2. Since (1/T)=S/E, this corresponds to T>Tc=1/(2log2). For T<Tc, the system remains "frozen" in a small number of configurations of energy ENJlog2 and the entropy per spin vanishes in the thermodynamic limit.

See also

References

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  1. Template:Cite book
  2. Michel Talagrand, Spin Glasses: A Challenge for Mathematicians (2003) Springer Template:ISBN
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  5. Template:Cite journal