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==Let's do replica!==
==Let's do replica!==
To make progress in disordered systems, we need to analyze the moments of the partition function. For simplicity, we consider polymers starting at <math>0</math> and ending at <math>x</math>. We recall that:   
To make progress in disordered systems, we need to analyze the moments of the partition function. <strong>Goal</strong> 
 
From Valentina's lecture, recall that if 
 
<center> 
<math> 
\frac{\overline{Z(x,t)^2}}{ (\overline{Z(x,t)})^2}=1 
</math> 
</center> 
 
then the partition function is self-averaging, and 
 
<center> 
<math> 
\overline{\ln Z(x,t)} =\ln\overline{Z(x,t)} 
</math>. 
</center> 
 
The condition above is sufficient but not necessary. It is enough that 
 
<center> 
<math> 
\frac{\overline{Z(x,t)^2}}{ (\overline{Z(x,t)})^2} < \text{const} 
</math>, 
</center> 
 
when <math>t \to \infty</math>, to ensure the equivalence between annealed and quenched averages. 
 
In the following, we compute this quantity, which corresponds to a two-replica calculation. 
 
 
 
 
 
 
For simplicity, we consider polymers starting at <math>0</math> and ending at <math>x</math>. We recall that:   


* <math>V(x,\tau)</math> is a Gaussian field with   
* <math>V(x,\tau)</math> is a Gaussian field with   

Revision as of 10:37, 2 February 2025

Goal : final lecture on KPZ and directed polymers at finite dimension. We will show that for d>2 a "glass transition" takes place.


KPZ : from 1d to the Cayley tree

We know a lot about KPZ, but there is still much to understand:

  • In d=1, we have found θ=1/3 and a glassy regime present at all temperatures. The stationary solution of the KPZ equation describes, at long times, the fluctuations of quantities such as Emin[x]Emin[x]. However, it does not determine the actual distribution of Emin for a given x. In particular, we have no clear understanding of the origin of the Tracy-Widom distribution.
  • In d=, an exact solution exists for the Cayley tree, predicting a freezing transition to a 1RSB phase (θ=0).
  • In finite dimensions greater than one, no exact solutions are available. Numerical simulations indicate θ>0 in d=2. The case d>2 remains particularly intriguing.

Let's do replica!

To make progress in disordered systems, we need to analyze the moments of the partition function. Goal

From Valentina's lecture, recall that if

Z(x,t)2(Z(x,t))2=1

then the partition function is self-averaging, and

lnZ(x,t)=lnZ(x,t).

The condition above is sufficient but not necessary. It is enough that

Z(x,t)2(Z(x,t))2<const,

when t, to ensure the equivalence between annealed and quenched averages.

In the following, we compute this quantity, which corresponds to a two-replica calculation.




For simplicity, we consider polymers starting at 0 and ending at x. We recall that:

  • V(x,τ) is a Gaussian field with
V(x,τ)=0,V(x,τ)V(x,τ)=Dδd(xx)δ(ττ)
  • From Wick's theorem, for a generic Gaussian field W, we have
exp(W)=exp[W+12(W2W2)]

The first moment

The first moment of the partition function is straightforward to compute and corresponds to a single replica:

Z(x,t)=x(0)=0x(t)=x𝒟x(τ)exp[1T0tdτ12(τx)2]exp[1TdτV(x(τ),τ)]

Note that the term T2W2=dτ1dτ2V(x,τ1)V(x,τ2)=Dtδ0 exhibits a short-distance divergence due to the delta function. Hence, we can write:

Z(x,t)=1(2πtT)d/2exp[12x2tT]exp[Dtδ02T2]

The second moment

For the second moment, there are two replicas:

  • Step 1: The second moment is

Z(x,t)2=𝒟x1𝒟x2exp[0tdτ12T[(τx1)2+(τx2)2]]exp[1T0tdτ1V(x1(τ1),τ1)1T0tdτ2V(x2(τ2),τ2)]

  • Step 2: Using Wick's theorem, we obtain

Z(x,t)2=exp[Dtδ0T2]𝒟x1𝒟x2exp[0tdτ12T[(τx1)2+(τx2)2DT2δd[x1(τ)x2(τ)]]

  • Step 3: Changing coordinates** X=(x1+x2)/2;u=x1x2, we get

Z(x,t)2=(Z(x,t))2u(0)=0u(t)=0𝒟uexp[0tdτ14T(τu)2DT2δd[u(τ)]]u(0)=0u(t)=0𝒟uexp[0tdτ14T(τu)2]

Discussion

Hence, the quantity Z(x,t)2/(Z(x,t))2 can be computed.

  • The denominator u(0)=0u(t)=0𝒟uexp[0tdτ14T(τu)2] is the free propagator and gives a contribution (4Tt)d/2 .
  • Let us define the numerator
W(0,t)=u(0)=0u(t)=0𝒟uexp[0tdτ14T(τu)2DT2δd[u(τ)]]


Remark 2: From Feynman-Kac we can write the following equation

tW(x,t)=H^W(x,t)

Here the Hamiltonian reads:

H^=2T2DT2δd[u]

The single particle potential is time independent and actractive .

W(x,t)=x|exp(H^t)|0

At large times the behaviour is dominatated by the low energy part of the spectrum.

  • In d2 an actractive potential always gives a bound state. In particular the ground state has a negative energy E0<0. Hence at large times
W(x,t)=e|E0|t

grows exponentially. This means that at all temperature, when t

lnZ(x,t)lnZ(x,t)
  • For d>2 the low part of the spectrum is controlled by the strength of the prefactor DT2. At high temperature we have a continuum positive spectrum, at low temperature bound states exist. Hence, when t
{lnZ(x,t)=lnZ(x,t)forT>TclnZ(x,t)lnZ(x,t)forT<Tc

This transition, in d=3, is between a high temeprature, θ=0 phase and a low temeprature θ>0 no RSB phase.