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 $E_{\min }[x]-E_{\min }[x']$. 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} and ending in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} . We recall that

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{Z_t[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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T^2 \overline{W^2} = \int d \tau_1 d\tau_2 \overline{V(x,\tau_1)V(x,\tau_2)}= D t \delta_0} has a short distance divergence due to the delta-function. Hence we can write:

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

The second moment

Exercise: L-4

• Step 1: The second moment is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{Z[x_t,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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{Z[x_t,t]^2 } = \exp\left[ \frac{D t \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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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]} }

Discussion

Hence, the quantity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{Z[x_t,t]^2}/ (\overline{Z[x_t,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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sim =\sqrt{4 T t}. * Let us define the numerator <center> <math> 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{Z[x_t,t]^2}/ (\overline{Z[x_t,t]})^2 <\text{const} } , when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \partial_t W(x,t) =- \hat H W(x,t) }

Now the Hamiltonian reads:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H= -2 T \nabla^2 - \frac{D}{T^2} \delta^d[u] }

Now the Hamiltonian is time independent:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(x,t) = \langle x|\exp\left( - \hat H t\rangle) |0\rangle }

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