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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 11:37, 2 February 2025

Goal : final lecture on KPZ and directed polymers at finite dimension. We will show that for 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 , we have found 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 . However, it does not determine the actual distribution of for a given . In particular, we have no clear understanding of the origin of the Tracy-Widom distribution.
  • In , an exact solution exists for the Cayley tree, predicting a freezing transition to a 1RSB phase ().
  • In finite dimensions greater than one, no exact solutions are available. Numerical simulations indicate in . The case 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

then the partition function is self-averaging, and

.

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

,

when , 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 and ending at . We recall that:

  • is a Gaussian field with
  • From Wick's theorem, for a generic Gaussian field Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W } , we have
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{\exp(W)} = \exp\left[\overline{W} +\frac{1}{2} \left(\overline{W^2}-\overline{W}^2\right)\right] }

The first moment

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

Note that the term Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T^2 \overline{W^2} = \int d \tau_1 d\tau_2 \overline{V(x,\tau_1)V(x,\tau_2)}= D t \delta_0} exhibits a short-distance divergence due to the delta function. Hence, we can write:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{Z(x,t) } = \frac{1}{(2 \pi t T)^{d/2}}\exp\left[ -\frac{1}{2} \frac{ x^2}{t T} \right] \exp\left[ \frac{D t \delta_0}{2T^2} \right] }

The second moment

For the second moment, there are two replicas:

  • Step 1: The second moment is

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{Z(x,t)^2 } = \exp\left[ \frac{D t \delta_0}{T^2} \right]\int {\cal D} x_1\int {\cal D} x_2 \exp\left[- \int_0^t d \tau \frac{1}{2T}[(\partial_\tau x_1)^2+ (\partial_\tau x_2)^2 - \frac{D}{T^2} \delta^d[x_1(\tau)-x_2(\tau)]\right] }

  • Step 3: Changing coordinates** Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X=(x_1+x_2)/2; \; u=x_1-x_2} , we get

Discussion

Hence, the quantity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{Z(x,t)^2}/ (\overline{Z(x,t)})^2} can be computed.

  • The denominator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{u(0)=0}^{u(t)=0} {\cal D} u \exp\left[- \int_0^t d \tau \frac{1}{4T}(\partial_\tau u)^2\right] } is the free propagator and gives a contribution .
  • Let us define the numerator
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(0,t)= \int_{u(0)=0}^{u(t)=0} {\cal D} u \exp\left[- \int_0^t d \tau \frac{1}{4T}(\partial_\tau u)^2- \frac{D}{T^2} \delta^d[u(\tau)]\right] }


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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \partial_t W(x,t) =- \hat H W(x,t) }

Here the Hamiltonian reads:

The single particle potential is time independent and actractive .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(x,t) = \langle x|\exp\left( - \hat H t\right) |0\rangle }

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

  • In Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d\le 2} an actractive potential always gives a bound state. In particular the ground state has a negative energy . Hence at large times
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(x,t) = e^{ |E_0| t} }

grows exponentially. This means that at all temperature, when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t\to \infty}

  • For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d > 2} the low part of the spectrum is controlled by the strength of the prefactor Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{D}{T^2} } . At high temperature we have a continuum positive spectrum, at low temperature bound states exist. Hence, when
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{cases} \overline{\ln Z(x,t)} = \ln\overline{Z(x,t)} \quad \text{for} \; T>T_c \\ \\ \overline{\ln Z(x,t)} \ll \ln\overline{Z(x,t)} \quad \text{for} \; T<T_c \end{cases} }

This transition, in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d =3 } , is between a high temeprature, phase and a low temeprature Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta>0} no RSB phase.