L-4: Difference between revisions

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* 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> 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>  
<center><math> \begin{cases}
<center><math> \begin{cases}
\overline{\ln Z(x,t)} = \ln\overline{Z(x,t)} \text{for} T>T_c \\
\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)} \text{for} T<T_c
\\
\overline{\ln Z(x,t)} \ll \ln\overline{Z(x,t)} \quad  \text{for} \; T<T_c
\end{cases}  
\end{cases}  
</math></center>
</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.
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.

Revision as of 10:51, 12 February 2024

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


KPZ : from to the Cayley tree

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

  • In we found and a glassy regime present at all temperatures. The stationary solution of the KPZ equation describes, at long times, the fluctions of quantities like . However it does not identify the actual distribution of for a given . In particular we have no idea from where Tracy Widom comes from.
  • In , there is an exact solution for the Cayley tree that predicts a freezing transition to an 1RSB phase ().
  • In finite dimension, but larger than 1, there are no exact solutions. Numerical simulations find in . The case is very interesting.

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 and ending in . We recall that

  • is a Gaussian field with
  • From the Wick theorem, for a generic Gaussian field we have

The first moment

The first moment of the partition function is

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) } =\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]} }

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} has 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

Exercise: L-4

  • Step 1: The second moment is
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 } =\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]} }
  • Step 2: Use Wick and derive:
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: Now change coordinate and get:
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 \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]} }

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 is the free propagator and gives a contribution 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 \sim (4 T t)^{d/2}} .
  • 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 1: From T-I, remember that if

the partition function is self-averaging and 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{\ln Z(x,t)} =\ln\overline{Z(x,t)} } . The condition above is sufficient but not necessary. It is enough that 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 <\text{const} } , when , to have the equivalence between annealed and quenched averages.

Remark II: From L-3, we derive using Feynman-Kac, 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 , is between a high 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} phase and a low temeprature ,math> \theta>0</math> no RSB phase.