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. 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 $W$, we have

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

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 $T^{2}{\overline {W^{2}}}=\int d\tau _{1}d\tau _{2}{\overline {V(x,\tau _{1})V(x,\tau _{2})}}=Dt\delta _{0}$ exhibits 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

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

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

• Step 3: Changing coordinates** $X=(x_{1}+x_{2})/2;\;u=x_{1}-x_{2}$, 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 ${\overline {Z(x,t)^{2}}}/({\overline {Z(x,t)}})^{2}$ 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 Valentina's lecture, 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 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 <\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 2: From Feynman-Kac we can write the following equation

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

Here 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 \hat H= -2 T \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 $d\leq 2$ an actractive potential always gives a bound state. In particular the ground state has a negative energy 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 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 $t\to \infty$

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 \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 ${\frac {D}{T^{2}}}$. At high temperature we have a continuum positive spectrum, at low temperature bound states exist. Hence, 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}

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 $d=3$, is between a high 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} 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.