L-4: Difference between revisions

From Disordered Systems Wiki
Jump to navigation Jump to search
 
(22 intermediate revisions by the same user not shown)
Line 13: Line 13:


==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.   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   
Line 38: Line 64:
<center>   
<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]   
\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>   
</math>   
</center>
</center>


=== The second moment ===   
=== The second moment ===   
Line 46: Line 72:
For the second moment, there are two replicas:   
For the second moment, there are two replicas:   


* **Step 1: The second moment is**  
* Step 1: The second moment is   


<center>   
<center>   
Line 54: Line 80:
</center>   
</center>   


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


<center>   
<center>   
Line 61: Line 87:
</math>   
</math>   
</center>   
</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>


* **Step 3: Changing coordinates** <math>X=(x_1+x_2)/2; \; u=x_1-x_2</math>, we get 
===The two replica propagator===


Let us define the propagator: 
<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> 
Using the Feynman-Kac formula, we can write the following equation: 
<center>   
<center>   
<math>   
<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]}  
\partial_t W(x,t) = - \hat H W(x,t)   
</math>   
</math>   
</center>   
</center>   


==Discussion==
Here, the Hamiltonian is given by:  
Hence, the quantity <math>\overline{Z(x,t)^2}/ (\overline{Z(x,t)})^2</math> can be computed.
<center>   
* 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> .
<math>
* Let us define the numerator
\hat H = - T \nabla^2 - \frac{D}{T^2} \delta^d[u]
<center> <math>
</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]
</center>
</math></center>
 
=== 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: 


<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.