L-9: Difference between revisions

From Disordered Systems Wiki
Jump to navigation Jump to search
Line 47: Line 47:
In your homewoork you solved a toy model for the interface:
In your homewoork you solved a toy model for the interface:
<center><math>
<center><math>
\partial_t h(r,t) =  \grad^2 r(r,t)  + F(r)
\partial_t h(r,t) =  \nable^2 r(r,t)  + F(r)
</math></center>
</math></center>
For simplicity, we assume Gaussian disorder
For simplicity, we assume Gaussian disorder
Line 53: Line 53:


You proved that:
You proved that:
* the roughness exponent of this model is  <math>\zeta_L=frac{4-d}{2}</math> below dimension 4
* the roughness exponent of this model is  <math>\zeta_L=\frac{4-d}{2}</math> below dimension 4
* The force per unit length acting on the center of the interface is <math> f= \sigma/\sqrt{L^d}</math>
* The force per unit length acting on the center of the interface is <math> f= \sigma/\sqrt{L^d}</math>
* at long times the  interface shape is  
* at long times the  interface shape is  

Revision as of 19:41, 24 March 2024

Multifractality

In the last lecture we discussed that the eigenstates of the Anderson model can be localized, delocalized or multifractal. The idea is to look at the (generalized) IPR

IPR(q)=n|ψn|2qLτq

The exponent τq is called multifractal exponent . Normalization imposes τ1=0 and the fact that the wave fuction is defined everywhere that τ0=d. In general τ0 is the fractal dimension of the object we are considering and it is simply a geometrical property.

  • Delocalized eigenstates

In this case, |ψn|2Ld for all the Ld sites. This gives

τqdeloc=d(q1)


  • Multifractal eigenstates.

This case correspond to more complex wave function for which we expect

|ψn|2LαforLf(α)sites

The exponent α is positive and f(α) is called multifractal spectrum . It is a convex function and its maximum is the fractal dimension of the object, in our case d. We can determine the relation between multifractal spectrum and exponent

IPR(q)=n|ψn|2qdαLαqLf(α)

for large L

τ(q)=minα(αqf(α))

This means that for α*(q) that verifies f(α*(q))=q we have

τ(q)=α*(q)qf(α*(q))


A metal has a simple spectrum. Indeed, all sites have α=d, hence f(α=d)=d and f(αd)=. Then α*(q)=d becomes q independent. A multifractal has a smooth spectrum with a maximum at α0 with f(α0)=d. At q=1, f(α1)=1 and f(α1)=α1.

Larkin model

In your homewoork you solved a toy model for the interface:

Failed to parse (unknown function "\nable"): {\displaystyle \partial_t h(r,t) = \nable^2 r(r,t) + F(r) }

For simplicity, we assume Gaussian disorder F(r)=0, F(r)F(r)=σ2δd(rr).

You proved that:

  • the roughness exponent of this model is ζL=4d2 below dimension 4
  • The force per unit length acting on the center of the interface is f=σ/Ld
  • at long times the interface shape is
h(q)h(q)=σ2qd+2ζL

In the real depinning model the disorder is however a non-linear function of h. The idea of Larkin is that this linearization is correct up, rf the length of correlation of the disorder along the h direction . This defines a Larkin length. Indeed from

(h(r)h(0))2=ddq(h(q)h(q)(1cos(qr)σ2r2ζL

You get

(h(L)h(0))2=rf2L=(rfσ)1/ζL

Above this scale, roguhness change and pinning starts with a crtical force

fc=σLd/(2ζL)

In d=1 we have L=(rfσ)2/3