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 there is still much to understand:

• In $d=1$, we have found $\theta =1/3$ and a glassy regime present at all temperatures. The stationary solution of the KPZ equation describes, at long times, the fluctuations of quantities such as $E_{\min }[x]-E_{\min }[x']$. However, it does not determine the actual distribution of $E_{\min }$ for a given $x$. In particular, we have no clear understanding of the origin of the Tracy-Widom distribution.

• In $d=\infty$, an exact solution exists for the Cayley tree, predicting a freezing transition to a 1RSB phase ($\theta =0$).

• In finite dimensions greater than one, no exact solutions are available. Numerical simulations indicate $\theta >0$ in $d=2$. The case $d>2$ remains particularly intriguing.

Let's do replica!

To make progress in disordered systems, we need to analyze the moments of the partition function. Goal

From Valentina's lecture, recall that if

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

then 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

${\frac {\overline {Z(x,t)^{2}}}{({\overline {Z(x,t)}})^{2}}}<{\text{const}}$,

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

In the following, we compute this quantity, which corresponds to a two-replica calculation.

For simplicity, we consider polymers starting at $0$ and ending at $x$. 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 Wick's theorem, for a generic Gaussian field 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 } , we have

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

The first moment

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

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

Note that the term 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} exhibits a short-distance divergence due to the delta function. Hence, we can write:

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) } = \frac{1}{(2 \pi t T)^{d/2}}\exp\left[ -\frac{1}{2} \frac{ x^2}{t T} \right] \exp\left[ \frac{D t \delta_0}{2T^2} \right] }

The second moment

For the second moment, there are two replicas:

• 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: Using Wick's theorem, we obtain

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)^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: Changing coordinates** 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=(x_1+x_2)/2; \; u=x_1-x_2} , we 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 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)^2}/ (\overline{Z(x,t)})^2} can be computed.

• The denominator 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 \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

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(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 2: From Feynman-Kac we can write 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) }

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 .

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\right) |0\rangle }

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

• 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 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

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) = e^{ |E_0| t} }

grows exponentially. This means that at all temperature, 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}

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

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

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 \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 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 =3 } , is between a high temeprature, $\theta =0$ phase and a low temeprature 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 \theta>0} no RSB phase.