L-4

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 $d=1$ to the Cayley tree

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

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

• In $d=\infty$, there is an exact solution for the Cayley tree that predicts a freezing transition to an 1RSB phase ($\theta =0$).

• In finite dimension, but larger than 1, there are no exact solutions. Numerical simulations find $\theta >0$ in $d=2$. The case $d>2$ 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 $0$ and ending in $x$. We recall that

• $V(x,\tau )$ is a Gaussian field with

${\overline {V(x,\tau )}}=0,\quad {\overline {V(x,\tau )V(x',\tau ')}}=D\delta ^{d}(x-x')\delta (\tau -\tau ')$

• From the Wick theorem, for a generic Gaussian 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 } field 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} (\overline{W^2}-\overline{W}^2)\right] }

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_t[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_t[x] } = \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

• Step 1:

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,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] }

Now you can change coordinate 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} 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,t]^2} = (\overline{Z[x_t,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]} }

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,t]^2}/ (\overline{Z[x_t,t]})^2} can be computed.

• Remark 1: From T-I, remember that if

${\frac {\overline {Z[x_{t},t]^{2}}}{({\overline {Z[x_{t},t]}})^{2}}}=1$

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,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,t]^2}/ (\overline{Z[x_t,t]})^2 <\text{const.} } when $t\to \infty$ 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 Z =- \hat H Z = }

Now the Hamiltonian reads:

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 H= -2 T \nabla^2 - \frac{D}{T^2} \delta^d[u] }

It is a