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= KPZ : from 1d to the Cayley tree=
= KPZ : from 1d to the Cayley tree=


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


* In <math>d=1</math> we found <math>\theta=1/3</math> and a glassy regime present at all temperatures. The stationary solution of the KPZ equation describes, at long times, the fluctions of quantities like  <math>E_{\min}[x] - E_{\min}[x']</math>. However it does not identify the actual distribution of <math> E_{\min}</math> for a given <math>x</math>. In particular we have no idea from where Tracy Widom comes from.
* In <math>d=1</math>, we have found <math>\theta=1/3</math> 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 <math>E_{\min}[x] - E_{\min}[x']</math>. However, it does not determine the actual distribution of <math>E_{\min}</math> for a given <math>x</math>. In particular, we have no clear understanding of the origin of the Tracy-Widom distribution.


* In <math>d=\infty</math>, there is an exact solution for the Cayley tree that predicts a freezing transition to an 1RSB phase (<math>\theta=0</math>).
* In <math>d=\infty</math>, an exact solution exists for the Cayley tree, predicting a freezing transition to a 1RSB phase (<math>\theta=0</math>).


* In finite dimension, but larger than 1, there are no exact solutions. Numerical simulations find <math>\theta >0</math> in <math>d=2</math>. The case <math>d>2</math> is very interesting.
* In finite dimensions greater than one, no exact solutions are available. Numerical simulations indicate <math>\theta > 0</math> in <math>d=2</math>. The case <math>d > 2</math> remains particularly intriguing.


==Let's do replica!==
==Let's do replica!==
To make progress in disordered systems we have to go through the moments of the  partition function. For simplicity we consider polymers starting in <math>0</math> and ending in  <math>x</math>. We recall that  
To make progress in disordered systems, we need to analyze the moments of the partition function.  From Valentina's lecture, recall that if 
* <math>V(x,\tau)</math> is a Gaussian field with
 
<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
<center> <math>
<center> <math>
\overline{V(x,\tau)}=0, \quad  \overline{V(x,\tau) V(x',\tau')} = D \delta^d(x-x') \delta(\tau-\tau')  
\overline{V(x,\tau)}=0, \quad  \overline{V(x,\tau) V(x',\tau')} = D \delta^d(x-x') \delta(\tau-\tau')  
</math></center>
</math></center>
* From the Wick theorem, for a generic Gaussian <math> W </math> field we have
 
* From Wick's theorem, for a generic Gaussian field <math> W </math>, we have
<center><math>
<center><math>
\overline{\exp(W)} = \exp\left[\overline{W} +\frac{1}{2} (\overline{W^2}-\overline{W}^2)\right] </math></center>
\overline{\exp(W)} = \exp\left[\overline{W} +\frac{1}{2} \left(\overline{W^2}-\overline{W}^2\right)\right]
</math></center>
 
===The first moment===
===The first moment===
The first moment of the partition function is simple to compute and corresponds to a single replica
The first moment of the partition function is straightforward to compute and corresponds to a single replica
<center> <math>
 
\overline{Z(x,t) } =\int_{x(0)=0}^{x(t)=x} {\cal D} x(\tau) \exp\left[- \frac{1}{T} \int_0^t d \tau \frac{1}2(\partial_\tau x)^2\right]  \overline{\exp\left[- \frac{1}{T} \int d \tau V(x(\tau),\tau ) \right]}
<center>
</math></center>
<math>
Note that the term <math> 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</math> has a short distance divergence due to the delta-function. Hence we can write:
\overline{Z(x,t) } =\int_{x(0)=0}^{x(t)=x} {\cal D} x(\tau) \exp\left[- \frac{1}{T} \int_0^t d \tau \frac{1}{2}(\partial_\tau x)^2\right]  \overline{\exp\left[- \frac{1}{T} \int d \tau V(x(\tau),\tau ) \right]}
<center> <math>
</math>
\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]
</center>
</math></center>
 
Note that the term <math> 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</math> exhibits a short-distance divergence due to the delta function. Hence, we can write:
 
<center>
<math>
\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] = Z_{\text{free}}(x,t,T)  \exp\left[ \frac{D  t \delta_0}{2T^2} \right]
</math> 
</center>
 
=== The second moment === 
 
For the second moment, there are two replicas: 
 
* Step 1: The second moment is 
 
<center> 
<math> 
\overline{Z(x,t)^2 } =\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] \right]  \overline{\exp\left[- \frac{1}{T} \int_0^t d \tau_1 V(x_1(\tau_1),\tau_1 ) - \frac{1}{T} \int_0^t d \tau_2 V(x_2(\tau_2),\tau_2 )\right]} 
</math> 
</center> 
 
* Step 2: Using Wick's theorem, we obtain
 
<center> 
<math> 
\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] 
</math> 
</center> 
and we can write:
<center> 
<math> 
\overline{Z(x,t)^2 } = (\frac{\overline{Z(x,t)}}{Z_{\text{free}}(x,t,T)})^2 \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] 
</math> 
</center>
* Step 3: Changing coordinates <math>X=(x_1+x_2)/2; \; u=x_1-x_2</math>, we get 
 
<center> 
<math> 
\overline{Z(x,t)^2} = (\overline{Z(x,t)})^2 \frac{\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]}{ Z_{\text{free}}(u=0,t, 2T)} 
</math>
</center>
where we used <math> Z_{\text{free}}^2(x,t,T)=Z_{\text{free}}(X=x,t,T/2)Z_{\text{free}}(u=0,t,2T) </math> with <math>Z_{\text{free}}(u=0,t,2T) = (4 \pi T t)^{d/2} </math>


=== The second moment ===
===The two replica propagator===


* Step 1: The second moment is
Let us define the propagator:
<center> <math>
<center> <math>
\overline{Z(x,t)^2 } =\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 \right]  \overline{\exp\left[- \frac{1}{T} \int_0^t d \tau_1 V(x_1(\tau_1),\tau_1 ) - \frac{1}{T} \int_0^t d \tau_2 V(x_2(\tau_2),\tau_2 )\right]}
  W(0,t)= \int_{u(0)=0}^{u(t)=0} {\cal D} \exp\left[-  \int_0^t d \tau  \frac{1}{4T}(\partial_\tau u)^2- \frac{D}{T^2} \delta^d[u(\tau)]\right] 
</math></center>
  </math> 
</center> 
Using the Feynman-Kac formula, we can write the following equation: 
<center> 
<math> 
\partial_t W(x,t) = - \hat H W(x,t)
</math>
</center>


* Step 2: Use Wick and derive:
Here, the Hamiltonian is given by:
<center> <math>
<center>
\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]
<math>
</math></center>
\hat H = - T \nabla^2 - \frac{D}{T^2} \delta^d[u]
</math>
</center>


* Step 3: Now  change coordinate <math>X=(x_1+x_2)/2; \; u=x_1-x_2</math> and get:
=== The Spectrum of the Two-Replica Hamiltonian ===
<center> <math>
\overline{Z(x,t)^2} = (\overline{Z(x,t)})^2 \frac{\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]}{\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]}
</math></center>


==Discussion==
The single-particle potential is '''time-independent and attractive'''. Since it is time-independent, we can use the spectral decomposition of the propagator. The long-time behavior is controlled by the low-energy part of the spectrum. In the presence of an attractive potential, we may have:  
Hence, the quantity <math>\overline{Z(x,t)^2}/ (\overline{Z(x,t)})^2</math> can be computed.  
* The denominator  <math>\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]  </math> is the free propagator and gives a contribution <math> \sim (4 T  t)^{d/2}</math> .
* Let us define  the numerator
<center> <math>
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]
</math></center>


<Strong>Remark 1:</Strong> From Valentina's lecture, remember that if
* A discrete set of eigenvalues corresponding to bound states, followed by a continuous spectrum.
<center> <math>
* Only a continuous spectrum.   
\frac{\overline{Z(x,t)^2}}{ (\overline{Z(x,t)})^2}=1
</math></center>
the partition function is self-averaging and <math> \overline{\ln Z(x,t)} =\ln\overline{Z(x,t)}
</math>.
The condition above is sufficient but not necessary. It is enough that <math>
\overline{Z(x,t)^2}/ (\overline{Z(x,t)})^2 <\text{const} </math>,  when <math> t\to \infty</math>, to have the equivalence between  annealed and quenched averages.


<Strong>Remark 2:</Strong> From Feynman-Kac we can write the following equation
<center> <math>
\partial_t W(x,t) =-  \hat H W(x,t)
</math></center>
Here the Hamiltonian reads:
<center> <math>
\hat  H= -2 T \nabla^2 - \frac{D}{T^2} \delta^d[u]
</math></center>
The single particle  potential  is <Strong> time independent and actractive </Strong>.
<center> <math>
W(x,t) = \langle x|\exp\left( - \hat H t\right) |0\rangle
</math></center>
At large times the behaviour is dominatated by the low energy part of the spectrum.


* In <math> d\le 2</math> an actractive potential always gives a bound state. In particular the ground state has a negative energy <math> E_0 <0</math>. Hence at large times
As a funcion of the dimension we distiguish two cases:
<center> <math>
* For <math>d \leq 2</math>
W(x,t) = e^{ |E_0| t}  
An attractive potential always leads to the formation of a bound state.The ground state has a negative energy <math>E_0 < 0</math>. At long times, the propagator behaves as: 
</math></center>
<center>
grows exponentially. This means that at all temperature, when  <math> t\to \infty</math>
<math>
<center><math> \overline{\ln Z(x,t)}  \ll \ln\overline{Z(x,t)}
W(x,t) \sim e^{ |E_0| t}
</math></center>
</math>
</center>
This implies that at all temperatures, in the limit <math>t \to \infty</math>
<center>
<math>
\overline{\ln Z(x,t)}  \ll \ln\overline{Z(x,t)}
</math>
</center>


* For <math> d > 2</math> the low part of the spectrum is controlled by the strength of the prefactor <math>\frac{D}{T^2} </math>. At high temperature we have a continuum positive spectrum, at low temperature bound states exist. Hence,   when  <math> t\to \infty</math>  
* For <math>d > 2</math>
<center><math> \begin{cases}
The low-energy part of the spectrum is controlled by the prefactor <math>\frac{D}{T^2}</math>. At high temperatures, the spectrum remains continuous and positive.  At low temperatures, bound states appear. Thus, in the limit <math>t \to \infty</math>
\overline{\ln Z(x,t)} = \ln\overline{Z(x,t)} \quad \text{for} \; T>T_c \\
<center>
\\
<math>
\overline{\ln Z(x,t)} \ll \ln\overline{Z(x,t)} \quad \text{for} \; T<T_c
\begin{cases}
\end{cases}  
\overline{\ln Z(x,t)} = \ln\overline{Z(x,t)} \quad \text{for} \quad T > T_c \\
</math></center>
\\
This transition, in <math> d =3 </math>, is between a high temeprature, <math> \theta=0</math> phase and a low temeprature <math> \theta>0</math> <Strong> no RSB </Strong> phase.
\overline{\ln Z(x,t)} \ll \ln\overline{Z(x,t)} \quad \text{for} \quad T < T_c
\end{cases}
</math>
</center>
This transition, in <math>d = 3</math>, separates a high-temperature phase with <math>\theta = 0</math> and a low-temperature phase with <math>\theta > 0</math> and '''no RSB'''.

Latest revision as of 21:25, 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. 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 , we have

The first moment

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

Note that the term exhibits a short-distance divergence due to the delta function. Hence, we can write:

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

and we can write:

  • Step 3: Changing coordinates , we get

where we used with

The two replica propagator

Let us define the propagator:

Using the Feynman-Kac formula, we can write the following equation:

Here, the Hamiltonian is given by:

The Spectrum of the Two-Replica Hamiltonian

The single-particle potential is time-independent and attractive. Since it is time-independent, we can use the spectral decomposition of the propagator. The long-time behavior is controlled by the low-energy part of the spectrum. In the presence of an attractive potential, we may have:

  • A discrete set of eigenvalues corresponding to bound states, followed by a continuous spectrum.
  • Only a continuous spectrum.


As a funcion of the dimension we distiguish two cases:

  • For :

An attractive potential always leads to the formation of a bound state.The ground state has a negative energy . At long times, the propagator behaves as:

This implies that at all temperatures, in the limit :

  • For :

The low-energy part of the spectrum is controlled by the prefactor . At high temperatures, the spectrum remains continuous and positive. At low temperatures, bound states appear. Thus, in the limit :

This transition, in , separates a high-temperature phase with and a low-temperature phase with and no RSB.