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 (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} ).

• In finite dimension, but larger than 1, there are no exact solutions. Numerical simulations find 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} 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 $x$. We recall 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 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 $W$ field 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} (\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(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} a small actractive potential leads to a continuum positive spectrum where, 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}

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

Only below a critical temperature, 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} } is large enough to have a bound state and a glassy phase. 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 math> \theta>0</math> no RSB phase.