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 $x_{0}=0$ and ending in $x_{t}$ . 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 of the partition function is

{\overline {Z_{t}[x_{t},t]}}=\int _{x(0)=0}^{x(t)=x_{t}}{\cal {D}}x_{1}\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 )\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_{t}[x_{1}]}}={\frac {1}{(2\pi tT)^{d/2}}}\exp \left[-{\frac {d}{2}}{\frac {x_{t}^{2}}{tT}}\right]\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 {\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 $X=(x_{1}+x_{2})/2;\;u=x_{1}-x_{2}$ and get:

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