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

$E_{\min }[x]-E_{\min }[x^{\prime }]$. 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 $d>1$ the exponents are not known. There is an exact solution for the Cayley 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. For simplicity we consider polymers starting in $x_0=0$ and ending in $x_t$. We recall that

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

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

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

$\overline{\exp (W)}=\exp [\bar{W}+\frac{1}{2}(\overline{W^2}-\stackrel{2}{\stackrel{‾}{W}})]$

The first moment of the partition function is

$\overline{Z_t[x_t,t]}=\underset{x(0)=0}{\overset{x(t)=x_t}{\int }}{𝒟}{x}_1\exp [-\frac{1}{T}\underset{0}{\overset{t}{\int }}d\tau \frac{1}{2}({\partial }_{\tau }x{)}^2]\overline{\exp [-\frac{1}{T}\int d\tau V(x,\tau )]}$

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-function. Hence we can write:

$\overline{Z_t[x_1]}=\frac{1}{(2\pi tT{)}^{d/2}}\exp [-\frac{d}{2}\frac{x_t^2}{tT}]\exp \left[\frac{Dt{\delta }_0}{2T^2}\right]$

The second moment

• Step 1:

$\overline{Z[x_t,t{]}^2}=\exp \left[\frac{Dt{\delta }_0}{T^2}\right]\int {𝒟}{x}_1\int {𝒟}{x}_2\exp [-\underset{0}{\overset{t}{\int }}d\tau \frac{1}{2T}[({\partial }_{\tau }{x}_1{)}^2+({\partial }_{\tau }{x}_2{)}^2-\frac{D}{T^2}{\delta }^d[x_1(\tau )-x_2(\tau )]]$

Now you can change coordinate $X=(x_1+x_2)/2;u=x_1-x_2$ and get:

$\overline{Z[x_t,t{]}^2}=(\overline{Z[x_t,t]}{)}^2\frac{\underset{u(0)=0}{\overset{u(t)=0}{\int }}{𝒟}u\exp [-\underset{0}{\overset{t}{\int }}d\tau \frac{1}{4T}({\partial }_{\tau }u{)}^2-\frac{D}{T^2}{\delta }^d[u(\tau )]]}{\underset{u(0)=0}{\overset{u(t)=0}{\int }}{𝒟}u\exp [-\underset{0}{\overset{t}{\int }}d\tau \frac{1}{4T}({\partial }_{\tau }u{)}^2]}$

• Step 2: Hence, the quantity $\overline{Z[x_t,t{]}^2}/(\overline{Z[x_t,t]}{)}^2$ can be computed.

from the spectrum of the following Hamiltonian

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