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. 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 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

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

• 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:

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

Exercise L4-A: 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:

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

• 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

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