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 1d 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 $d=\mathrm{\infty }$, 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

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

• 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

The first moment of the partition function is simple to compute and corresponds to a single replica

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

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:

$\overline{Z(x,t)}=\frac{1}{(2\pi tT{)}^{d/2}}\exp [-\frac{1}{2}\frac{x^2}{tT}]\exp \left[\frac{Dt{\delta }_0}{2T^2}\right]$

The second moment

For the second moment instead the are two replica

• Step 1: The second moment is

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

• Step 2: Use Wick and derive:

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

• Step 3: Now change coordinate $X=(x_1+x_2)/2;u=x_1-x_2$ and get:

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

Discussion

Hence, the quantity $\overline{Z(x,t{)}^2}/(\overline{Z(x,t)}{)}^2$ can be computed.

• The denominator $\underset{u(0)=0}{\overset{u(t)=0}{\int }}{𝒟}u\exp [-\underset{0}{\overset{t}{\int }}d\tau \frac{1}{4T}({\partial }_{\tau }u{)}^2]$ is the free propagator and gives a contribution $\sim (4Tt{)}^{d/2}$ .
• Let us define the numerator

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

Remark 1: From Valentina's lecture, remember that if

$\frac{\overline{Z(x,t{)}^2}}{(\overline{Z(x,t)}{)}^2}=1$

the partition function is self-averaging and $\overline{\ln Z(x,t)}=\ln \overline{Z(x,t)}$. The condition above is sufficient but not necessary. It is enough that $\overline{Z(x,t{)}^2}/(\overline{Z(x,t)}{)}^2<\text{const}$, when $t\to \mathrm{\infty }$, to have the equivalence between annealed and quenched averages.

Remark 2: From Feynman-Kac we can write the following equation

${\partial }_{t}W(x,t)=-\hat{H}W(x,t)$

Here the Hamiltonian reads:

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

The single particle potential is time independent and actractive .

$W(x,t)=⟨x|\exp (-\hat{H}t)|0⟩$

At large times the behaviour is dominatated by the low energy part of the spectrum.

• In $d\le 2$ an actractive potential always gives a bound state. In particular the ground state has a negative energy $E_0<0$. Hence at large times

$W(x,t)=e^{|E_0|t}$

grows exponentially. This means that at all temperature, when $t\to \mathrm{\infty }$

$\overline{\ln Z(x,t)}\ll \ln \overline{Z(x,t)}$

• For $d>2$ the low part of the spectrum is controlled by the strength of the prefactor $\frac{D}{T^2}$. At high temperature we have a continuum positive spectrum, at low temperature bound states exist. Hence, when $t\to \mathrm{\infty }$

$\{\begin{array}{l}\overline{\ln Z(x,t)}=\ln \overline{Z(x,t)}\text{for}T>T_c\\ \\ \overline{\ln Z(x,t)}\ll \ln \overline{Z(x,t)}\text{for}T<T_c\end{array}$

This transition, in $d=3$, is between a high temeprature, $\theta =0$ phase and a low temeprature $\theta >0$ no RSB phase.