LBan-II

Introduction: Interfaces and Directed Polymers

The physical properties of many materials are governed by manifolds embedded in them. Examples include: dislocations in crystals, domain walls in ferromagnets or vortex lines in superconductors. We fix the following notation: - $D$: spatial dimension of the embedding medium – $d$: internal dimension of the manifold – $N$: dimension of the displacement (or height) field

These satisfy the relation:

$D=d+N$

We focus on two important cases:

Directed Polymers (d = 1)

The configuration is described by a vector function: ${\vec {x}}(t)$, where $t$ is the internal coordinate. The polymer lives in $D=1+N$ dimensions.

Examples: vortex lines, DNA strands, fronts.

Although polymers may form loops, we restrict to directed polymers, i.e., configurations without overhangs or backward turns.

Interfaces (N = 1)

The interface is described by a scalar height field: $h({\vec {r}},t)$, where ${\vec {r}}\in \mathbb {R} ^{d}$ is the internal coordinate and $t$ represents time.

Examples: domain walls and propagating fronts

Again we neglect overhangs or pinch-off: $h({\vec {r}},t)$ is single-valued

Note that using our notation the 1D front is both an interface and a directed polymer

Thermal Interfaces

The dynamics is overdamped, so that we can neglect the inertial term. The Langevin equation of motion is

$\partial _{t}h(r,t)=-\mu {\frac {\delta E_{pot}}{\delta h(r,t)}}+\eta (r,t)$

The first term $-\delta E_{pot}/\delta h(r,t)$ is the elastic force trying to smooth the interface, the mobility $\mu$ is the inverse of the viscosity. The second term is the Langevin noise. It is Guassian and defined by

$\langle \eta (r,t)\rangle =0,\;\langle \eta (r',t')\eta (r,t)\rangle =2dD\delta ^{d}(r-r')\delta (t-t')$

The symbol $\langle \ldots \rangle$ indicates the average over the thermal noise and the diffusion constant is fixed by the Einstein relation $D=\mu K_{B}T$. We set $\mu =K_{B}=1$

The potential energy of surface tension ($\nu$ is the stiffness) can be expanded at the lowest order in the gradient:

$E_{pot}=\nu \int d^{d}r{\sqrt {1+(\nabla h)^{2}}}\sim {\text{const.}}+{\frac {\nu }{2}}\int d^{d}r(\nabla h)^{2}$

Hence, we have the Edwards Wilkinson equation:

$\partial _{t}h(r,t)=\nu \nabla ^{2}h(r,t)+\eta (r,t)$

Scaling Invariance

The equation enjoys of a continuous symmetry because $h(r,t)$ and $h(r,t)+c$ cannot be distinguished. This is a condition of scale invariance:

$h(br,b^{z}t){\overset {inlaw}{\sim }}b^{\alpha }h(r,t)$

Here $z,\alpha$ are the dynamic and the roughness exponent respectively. From dimensional analysis

$b^{\alpha -z}\partial _{t}h(r,t)=b^{\alpha -2}\nabla ^{2}h(r,t)+b^{-d/2-z/2}\eta (r,t)$

From which you get $z=2$ in any dimension and a rough interface below $d=2$ with $\alpha =(2-d)/2$.

Width of the interface

Consider a 1-dimensional line of size L with periodic boundary conditions. We consider the width square of the interface

$w_{2}(t)=\left[\int _{0}^{L}{\frac {dr}{L}}\left(h(r,t)-\int _{0}^{L}{\frac {dr}{L}}h(r,t)\right)\right]^{2}$

It is useful to introduce the Fourier modes:

${\hat {h}}_{q}(t)={\frac {1}{L}}\int _{0}^{L}e^{iqr}h(r,t),\quad h(r,t)=\sum _{q}e^{-iqr}{\hat {h}}_{q}(t)$

Here $q=2\pi n/L,n=\ldots ,-1,0,1,\ldots$ and recall $\int _{0}^{L}dre^{iqr}=L\delta _{q,0}$. using de Parseval theorem for the Fourier series

$w_{2}(t)=\sum _{q\neq 0}|{\hat {h}}_{q}(t)|^{2}=\sum _{q\neq 0}\left({\hat {h}}_{q}(t){\hat {h}}_{-q}(t)\right)^{2}$

In the last step we used that ${\hat {h}}_{q}^{*}(t)={\hat {h}}_{-q}(t)$.

Solution in the Fourier space

show that the EW equation writes

$\partial _{t}{\hat {h}}_{q}(t)=-\nu q^{2}{\hat {h}}_{q}(t)+\eta _{q}(t),\quad {\text{with}}\;\langle \eta _{q_{1}}(t')\eta _{q_{2}}(t)\rangle ={\frac {2T}{L}}\delta _{q_{1},-q_{2}}\delta (t-t')$

The solution of this first order linear equation writes

${\hat {h}}_{q}(t)={\hat {h}}_{q}(0)e^{-\nu q^{2}t}+\int _{0}^{t}dse^{-\nu q^{2}(t-s)}\eta _{q}(s)$

• Assume that the interface is initially flat, namely ${\hat {h}}_{q}(0)=0$. Show that

$\langle {\hat {h}}_{q}(t){\hat {h}}_{-q}(t)\rangle ={\begin{cases}{\dfrac {T(1-e^{-2\nu q^{2}t})}{L\nu q^{2}}},&q\neq 0,\\[1.2em]{\frac {2T}{L}}t,&q=0.\end{cases}}$

• The mean width square grows at short times and saturates at long times:

$\langle w_{2}(t)\rangle ={\dfrac {T}{L\nu }}\sum _{q\neq 0}{\dfrac {1-e^{-2\nu q^{2}t}}{q^{2}}}={\begin{cases}T{\sqrt {\frac {2t}{\pi \nu }}},&t\ll L^{2},\\[1.2em]{\frac {T}{\nu }}{\frac {L}{12}},&t\gg L^{2}.\end{cases}}$

Directed polymers in random media

Let us consider polymers $x(\tau )$ of length $t$, starting in $0$ and ending in $x$ and at thermal equlibrium at temperature $T$. The partition function of the model writes as

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

For simplicity, we assume a white noise, ${\overline {V(x,\tau )V(x',\tau ')}}=D\delta (x-x')\delta (\tau -\tau ')$. Here, the partition function is written as a sum over all possible paths, corresponding to all possible polymer configurations that start at $0$ and end at $x$, weighted by the appropriate Boltzmann factor.

Polymer partition function and propagator of a quantum particle

Let's perform the following change of variables: $\tau =it'$. We also identifies $T$ with $\hbar$ and ${\tilde {t}}=-it$ as the time.

$Z(x,{\tilde {t}})=\int _{x(0)=0}^{x({\tilde {t}})=x}{\cal {D}}x(t')\exp \left[{\frac {i}{\hbar }}\int _{0}^{\tilde {t}}dt'{\frac {1}{2}}(\partial _{t'}x)^{2}-V(x(t'),t')\right]$

Note that $S[x]=\int _{0}^{\tilde {t}}dt'{\frac {1}{2}}(\partial _{t'}x)^{2}-V(x(t'),t')$ is the classical action of a particle with kinetic energy ${\frac {1}{2}}(\partial _{\tau }x)^{2}$ and time dependent potential $V(x(\tau ),\tau )$, evolving from time zero to time ${\tilde {t}}$. From the Feymann path integral formulation, $Z[x,{\tilde {t}}]$ is the propagator of the quantum particle.

In absence of disorder, one can find the propagator of the free particle, that, in the original variables, writes:

$Z_{\text{free}}(x,t)={\frac {e^{-x^{2}/(2Tt)}}{\sqrt {2\pi Tt}}}$

Feynman-Kac formula

Let's derive the Feyman Kac formula for $Z(x,t)$ in the general case:

• First, focus on free paths and introduce the following probability

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

• Second, the moments generating function

$Z_{p}(x,t)=\int _{-\infty }^{\infty }dAe^{-pA}P[A,x,t]=\int _{x(0)=0}^{x(t)=x}{\cal {D}}x(\tau )e^{-{\frac {1}{T}}\int _{0}^{t}d\tau {\frac {1}{2}}(\partial _{\tau }x)^{2}-p\int _{0}^{t}d\tau V(x(\tau ),\tau )}$

• Third, the backward approach. Consider free paths evolving up to $t+dt$ and reaching $x$ :

$Z_{p}(x,t+dt)=\left\langle e^{-p\int _{0}^{t+dt}d\tau V(x(\tau ),\tau )}\right\rangle =\left\langle e^{-p\int _{0}^{t}d\tau V(x(\tau ),\tau )}\right\rangle e^{-pV(x,t)dt}=[1-pV(x,t)dt+\dots ]\left\langle Z_{p}(x-\Delta x,t)\right\rangle _{\Delta x}$

Here $\langle \ldots \rangle$ is the average over all free paths, while $\langle \ldots \rangle _{\Delta x}$ is the average over the last jump, namely $\langle \Delta x\rangle =0$ and $\langle \Delta x^{2}\rangle =Tdt$.

• At the lowest order we have

$Z_{p}(x,t+dt)=Z_{p}(x,t)+dt\left[{\frac {T}{2}}\partial _{x}^{2}Z_{p}-pV(x,t)Z_{p}\right]+O(dt^{2})$

Replacing $p=1/T$ we obtain the partition function is the solution of the Schrodinger-like equation:

$\partial _{t}Z(x,t)=-{\hat {H}}Z=-\left[-{\frac {T}{2}}{\frac {d^{2}}{dx^{2}}}+{\frac {V(x,\tau )}{T}}\right]Z(x,t)$

$Z[x,t=0]=\delta (x)$

Remarks

Remark 1:

This equation is a diffusive equation with multiplicative noise. Edwards Wilkinson is instead a diffusive equation with additive noise. The Cole Hopf transformation allows to map the diffusive equation with multiplicative noise in a non-linear equation with additive noise. We will apply this tranformation and have a surprise.

Remark 2: This hamiltonian is time dependent because of the multiplicative noise $V(x,\tau )/T$. For a time independent hamiltonian, we can use the spectrum of the operator. In general we will have to parts:

$Z[x,t]=\left(e^{-{\hat {H}}t}\right)_{0\to x}=\sum _{n}\psi _{n}(0)\psi _{n}^{*}(x)e^{-E_{n}t}+\int _{0}^{\infty }dE\,\rho (E)\psi _{E}(0)\psi _{E}^{*}(x)e^{-Et}.$