L-9

Eigenstates

Without disorder, the eigenstates are delocalized plane waves.

In the presence of disorder, three scenarios can arise: Delocalized eigenstates, where the wavefunction remains extended; Localized eigenstates, where the wavefunction is exponentially confined to a finite region; Multifractal eigenstates, occurring at the mobility edge, where the wavefunction exhibits a complex, scale-dependent structure.

To distinguish these regimes, it is useful to introduce the inverse participation ratio (IPR).

$IPR(q)=\sum _{n}|\psi _{n}|^{2q}\sim L^{-\tau _{q}}$

Delocalized eigenstates

In this case, $|\psi _{n}|^{2}\approx L^{-d}$. Hence, we expect

$IPR(q)=L^{d(1-q)}\quad \tau _{q}=d(1-q)$

Localized eigenstates

In this case, $|\psi _{n}|^{2}\approx 1/\xi _{\text{loc}}^{1/d}$ for $\xi _{\text{loc}}^{d}$ sites and almost zero elsewhere. Hence, we expect

$IPR(q)={\text{const}},\quad \tau _{q}=0$

Multifractal eigenstates

The exponent $\tau _{q}$ is called the multifractal exponent. It is a non-decreasing function of $q$ with some special points:

• $\tau _{0}=-d$, since the wavefunction is defined on all sites. In general, $\tau _{0}$ represents the fractal dimension of the object under consideration and is purely a geometric property.

• $\tau _{1}=0$, imposed by normalization.

To observe multifractal behavior, we expect:

$|\psi _{n}|^{2}\approx L^{-\alpha }\quad {\text{for}}\;L^{f(\alpha )}\;{\text{sites}}.$

The exponent $\alpha$ is positive, and $f(\alpha )$ is called the multifractal spectrum. Its maximum corresponds to the fractal dimension of the object, which in our case is $d$. The relation between the multifractal spectrum $f(\alpha )$ and the exponent $\tau _{q}$ is given by:

$IPR(q)=\sum _{n}|\psi _{n}|^{2q}\sim \int d\alpha \,L^{-\alpha q}L^{f(\alpha )}$

for large $L$. From this, we obtain:

$\tau (q)=\min _{\alpha }(\alpha q-f(\alpha )).$

This implies that for $\alpha ^{*}(q)$, which satisfies

$f'(\alpha ^{*}(q))=q,$

we have

$\tau (q)=\alpha ^{*}(q)q-f(\alpha ^{*}(q)).$

Delocalized wavefunctions have a simple spectrum: for $\alpha =d$, we find $f(\alpha =d)=d$ and $f(\alpha \neq d)=-\infty$. This means that $\alpha ^{*}(q)=d$ is independent of $q$.

Multifractal wavefunctions exhibit a smoother dependence, leading to a continuous spectrum with a maximum at $\alpha _{0}$, where $f(\alpha _{0})=d$. At $q=1$, we have $f'(\alpha _{1})=1$ and $f(\alpha _{1})=\alpha _{1}$.

Larkin model

In your homewoork you solved a toy model for the interface:

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

For simplicity, we assume Gaussian disorder ${\overline {F(r)}}=0$, ${\overline {F(r)F(r')}}=\sigma ^{2}\delta ^{d}(r-r')$.

You proved that:

• the roughness exponent of this model is $\zeta _{L}={\frac {4-d}{2}}$ below dimension 4
• The force per unit length acting on the center of the interface is $f=\sigma /{\sqrt {L^{d}}}$
• at long times the interface shape is

${\overline {h(q)h(-q)}}={\frac {\sigma ^{2}}{q^{d+2\zeta _{L}}}}$

In the real depinning model the disorder is however a non-linear function of h. The idea of Larkin is that this linearization is correct up, $r_{f}$ the length of correlation of the disorder along the h direction . This defines a Larkin length. Indeed from

${\overline {(h(r)-h(0))^{2}}}=\int _{d}^{d}q({\overline {h(q)h(-q)}}(1-\cos(qr)\sim \sigma ^{2}r^{2\zeta _{L}}$

You get

${\overline {(h(\ell _{L})-h(0))^{2}}}=r_{f}^{2}\quad \ell _{L}=\left({\frac {r_{f}}{\sigma }}\right)^{1/\zeta _{L}}$

Above this scale, roguhness change and pinning starts with a crtical force

$f_{c}={\frac {\sigma }{\ell _{L}^{d/(2\zeta _{L})}}}$

In $d=1$ we have $\ell _{L}=\left({\frac {r_{f}}{\sigma }}\right)^{2/3}$