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 $d>2$ 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 $0$ and ending in $x$. 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 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

The first moment of the partition function is

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

Note that the term $T^{2}{\overline {W^{2}}}=\int d\tau _{1}d\tau _{2}{\overline {V(x,\tau _{1})V(x,\tau _{2})}}=Dt\delta _{0}$ has 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

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)^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:

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

• Step 3: Now change coordinate 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} 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)^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 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 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 (4 T t)^{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 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)^2}}{ (\overline{Z(x,t)})^2}=1 }

the partition function is self-averaging and 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)} =\ln\overline{Z(x,t)} } . The condition above is sufficient but not necessary. It is enough that ${\overline {Z(x,t)^{2}}}/({\overline {Z(x,t)}})^{2}<{\text{const}}$, 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 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) }

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, 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.