L-4

Goal 1: final lecture on KPZ and directed polymers at finite dimension. We will show that for $d>2$ a "glass transition" takes place.

Goal 2: We will mention some ideas related to glass transition in true glasses.

Part 1: KPZ in finite dimension

• In $d=1$ we found $\theta =1/3$ and a glassy regime present at all temperatures. Moreover, the stationary solution tell us that $E_{\min }[x]$ is a Brownian motion in $x$. However this solution 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 $d>1$ the exponents are not known. There is an exact solution for the Caley tree (infinite dimension) that predicts a freezing transition to an 1RSB phase ($\theta =0$).

Let's do replica!

To make progress in disordered systems we have to go through the moments of the partition function. We recall that

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

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

• From the Wick theorem, for a generic Gaussian $W$ field we have

${\overline {\exp(W)}}=\exp \left[{\overline {W}}+{\frac {1}{2}}({\overline {W^{2}}}-{\overline {W}}^{2})\right]$

The first moment of the partition function is

${\overline {Z_{t}[x_{1}]}}=\int {\cal {D}}x_{1}\exp \left[-{\frac {1}{T}}\int _{0}^{t}d\tau (\partial _{\tau }x)^{2}\right]{\overline {\exp \left[-{\frac {1}{T}}\int d\tau V(x,\tau )\right]}}$

Note that the term $T^{2}{\overline {W^{2}}}=\int d\tau _{1}d\tau _{2}{\overline {V(x,\tau _{1})V(x,\tau _{2})}}=Dt\delta (0)$ has a short distance divergence due to the delta-functiton. Hence we can write:

${\overline {Z_{t}[x_{1}]}}={\frac {1}{\sqrt {2\pi tT}}}\exp \left[-{\frac {x^{2}}{2tT}}\right]\exp \left[{\frac {Dt\delta (0)}{2T^{2}}}\right]$

Exercise L4-A: the second moment

Show:

${\overline {Z_{t}[x_{1}]Z_{t}[x_{2}]}}=\exp \left[{\frac {Dt\delta (0)}{T^{2}}}\right]$

\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{1}{T^2} \delta[x_1(\tau)-x_2(\tau)]\right] </math>