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 $E_{\min }[x]-E_{\min }[x']$. 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 $W$ field we have

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

The first moment

The first moment of the partition function is

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

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

${\overline {Z(x,t)}}={\frac {1}{(2\pi tT)^{d/2}}}\exp \left[-{\frac {1}{2}}{\frac {x^{2}}{tT}}\right]\exp \left[{\frac {Dt\delta _{0}}{2T^{2}}}\right]$

The second moment

Exercise: L-4

• Step 1: The second moment is

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

• Step 2: Use Wick and derive:

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

• Step 3: Now change coordinate $X=(x_{1}+x_{2})/2;\;u=x_{1}-x_{2}$ and get:

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

Discussion

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

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

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

Remark 1: From T-I, remember that if

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

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 $t\to \infty$, to have the equivalence between annealed and quenched averages.

Remark II: From L-3, we derive using Feynman-Kac, the following equation

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

Here the Hamiltonian reads:

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

The single particle potential is time independent and actractive .

$W(x,t)=\langle x|\exp \left(-{\hat {H}}t\right)|0\rangle$

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

• In $d\leq 2$ 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 $t\to \infty$

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

• 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 $t\to \infty$

${\begin{cases}{\overline {\ln Z(x,t)}}=\ln {\overline {Z(x,t)}}\quad {\text{for}}\;T>T_{c}\\\\{\overline {\ln Z(x,t)}}\ll \ln {\overline {Z(x,t)}}\quad {\text{for}}\;T<T_{c}\end{cases}}$

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