details

we calculate the approximated memory function

$$\begin{eqnarray*} M_{\mu \nu }(\omega ) & = & \frac{\left\langle F_{\mu };F_{\nu... ...{\mu };F_{\nu }\right\rangle _{\omega =0}^{0}}{-\chi (0)\omega } \end{eqnarray*}$$

first we extract the current density from the effective electronic Lagrangian

$$\begin{eqnarray*} j_{x} & = & \left[\frac{\sqrt{2}}{\pi }ev_{F}g^{-1}\partial _{... ...y}\right)g^{-1}\partial _{x}u\right]\delta \left(y-u(x,t)\right) \end{eqnarray*}$$

assuming $\rho \gg m$we can set

$$\begin{eqnarray*} \Pi _{c} & = & \partial _{x}\theta _{c}\\ \Pi _{u} & = & \rho \partial _{t}u \end{eqnarray*}$$

thus to order $\left(\partial _{x}u\right)^{2}$

$$\begin{eqnarray*} j_{x} & = & \left[\frac{\sqrt{2}}{\pi }ev_{F}\left(1-\left(\pa... ...y}\right)g^{-1}\partial _{x}u\right]\delta \left(y-u(x,t)\right) \end{eqnarray*}$$

the memory elemnts are

$$\begin{eqnarray*} F_{\mu }(t) & = & \left[J_{\mu },H\right]=\left[J_{\mu },H_{f}... ...right\rangle \right\vert _{i\omega \rightarrow \omega +i\delta } \end{eqnarray*}$$

where the optical component of the current is

$$\begin{eqnarray*} J_{\mu } & = & \frac{1}{L_{x}L_{y}}\int dxdyj_{\mu } \end{eqnarray*}$$

and the scattering potentials

$$\begin{eqnarray*} H_{f} & = & -\frac{\sqrt{2}}{\pi }\int dxdy\eta (x,y)\partial ... ...os \sqrt{2}\phi _{s}(x,t)+h.c\right]\delta \left(y-u(x,t)\right) \end{eqnarray*}$$

we enumerate the following operators:

$$\begin{eqnarray*} \mathcal{O}_{1} & = & \frac{1}{L_{x}L_{y}}\int dxdy\delta \lef... ...ht)\partial _{x}\phi _{c}\left(\partial _{x}u\right)^{2}\Pi _{u} \end{eqnarray*}$$

and define the operators:

$$\begin{eqnarray*} C_{i} & = & C_{if}+C_{ib}\\ C_{if} & = & \left[\mathcal{O}_{i},H_{f}\right]\\ C_{ib} & = & \left[\mathcal{O}_{i},H_{b}\right] \end{eqnarray*}$$

next we define a matrix of correlators

$$\begin{eqnarray*} m_{ij}(\tau ) & = & \left\langle T_{\tau }C_{i}(\tau )C_{j}(0)\right\rangle =m_{ijf}(\tau )+m_{ijb}(\tau ) \end{eqnarray*}$$

the four basic commutators:

$$\begin{eqnarray*} C_{11f}(\tau ) & = & 0\\ C_{11b}(\tau ) & = & \frac{\sqrt{2}}... ...x,\tau )+h.c\right]\partial _{y}\delta \left(y-u(x,\tau )\right) \end{eqnarray*}$$

the non-zero diagonal temrs:

$$\begin{eqnarray*} m_{11b} & = & -\frac{2}{\left(\pi \alpha \right)^{2}L_{y}}\lef... ...^{2}}\left\vert\frac{v_{c}\tau }{\alpha }\right\vert^{-2}m_{55b} \end{eqnarray*}$$

the rest non-zero elements:

$$\begin{eqnarray*} m_{13} & = & \left\langle \left(\partial _{x}u\right)^{2}\righ... ...\left\langle \left(\partial _{x}u\right)^{2}\right\rangle m_{77} \end{eqnarray*}$$

so the memory elements are

$$\begin{eqnarray*} \left\langle T_{\tau }F_{x}(\tau )F_{x}(0)\right\rangle & = & ... ...left\vert\frac{v_{c}\tau }{\alpha }\right\vert^{-2}\right]m_{44} \end{eqnarray*}$$

the string correlations involved:

$$\begin{eqnarray*} \left\langle D_{\xi }\left(u(x,\tau )-u(x,0)\right)\right\rang... ...{x}^{2}G(0,\tau )\pi D_{\xi }}{4\left(\pi \Delta G\right)^{3/2}} \end{eqnarray*}$$

the f. transform has the typical integral

$$\begin{eqnarray*} I(i\omega ;\beta ) & = & \int _{0}^{\infty }d\tau \left(\frac{... ...nfty }dz\left(\frac{z}{\epsilon }+\omega \right)^{-\beta }e^{iz} \end{eqnarray*}$$

According to Mathematica

$$\begin{eqnarray*} \int _{0}^{\infty }d\tau \frac{e^{i\omega \tau }-1}{\left(\fra... ...\omega \right)^{\beta -1}+o\left(\epsilon ^{2}\omega ^{2}\right) \end{eqnarray*}$$

The integral can be estimated as

$$\begin{eqnarray*} \int _{0}^{\infty }d\tau \frac{e^{i\omega \tau }}{\left(\frac{... ...psilon }^{\infty }d\tau \frac{e^{i\omega \tau }}{\tau ^{\beta }} \end{eqnarray*}$$

shifting the path to the imaginary axis

$$\begin{eqnarray*} \int _{\epsilon }^{\infty }d\tau \frac{e^{i\omega \tau }}{\tau... ...ac{\omega \epsilon }{2-\beta }\left(i^{2-\beta }-1\right)\right] \end{eqnarray*}$$

thus

$$\begin{eqnarray*} \int _{0}^{\infty }d\tau \frac{e^{i\omega \tau }-1}{\left(\fra... ...1-\beta ;\omega \epsilon \right)\left(i\omega \right)^{\beta -1} \end{eqnarray*}$$

the answer we expect (hopefully as a limit from finite temperature and taking it to zero)

$$\begin{eqnarray*} \int _{0}^{\infty }d\tau \frac{e^{i\omega \tau }-1}{\left(\fra... ...1-\beta ;\omega \epsilon \right)\left(i\omega \right)^{\beta -1} \end{eqnarray*}$$

generic form

$$\begin{eqnarray*} \left\langle T_{\tau }F_{x}(\tau )F_{x}(0)\right\rangle & = & ... ...\rangle \right)\left\vert\frac{\tau }{\epsilon }\right\vert^{-K} \end{eqnarray*}$$

$$\begin{eqnarray*} \epsilon & = & \frac{\alpha }{v_{s}}\\ K & = & K_{c}+K_{s}\\ ... ...{v_{s}}{v_{c}}\right)^{2}\\ z & = & \epsilon (\omega +i\delta ) \end{eqnarray*}$$

stationary string

$$\begin{eqnarray*} \left\langle T_{\tau }F_{x}(\tau )F_{x}(0)\right\rangle & = & -AD_{\xi }\left\vert\frac{\tau }{\epsilon }\right\vert^{-K} \end{eqnarray*}$$

$$\begin{eqnarray*} M_{xx}(z) & = & A\epsilon ^{2}D_{\xi }e^{-\pi i(K-1)}\Gamma \left(1-K\right)z^{K-2} \end{eqnarray*}$$

rigid string

$$\begin{eqnarray*} \left\langle D_{\xi }\left(u(x,\tau )-u(x,0)\right)\right\rang... ...c{D_{\xi }}{\sqrt{1-e^{-\omega _{0}\left\vert\tau \right\vert}}} \end{eqnarray*}$$

$$\begin{eqnarray*} \left\langle T_{\tau }F_{x}(\tau )F_{x}(0)\right\rangle & = & ... ...-K}\frac{1}{\sqrt{1-e^{-\omega _{0}\left\vert\tau \right\vert}}} \end{eqnarray*}$$

$$\begin{eqnarray*} A & = & \frac{1}{L_{y}}\left(\frac{2ev_{F}}{\pi ^{2}\alpha }\r... ...ht)^{K_{c}}\left(\frac{m\omega _{0}}{2\pi }\right)^{\frac{3}{2}} \end{eqnarray*}$$

$$\begin{eqnarray*} \int _{0}^{\infty }d\tau \frac{e^{i\omega \tau }-1}{\left(\fra... ...omega \tau }-1}{\left(\frac{\tau }{\epsilon }+1\right)^{\beta }} \end{eqnarray*}$$

for $\omega \ll \omega _{0}$

$$\begin{eqnarray*} M_{xx}(z) & = & A\epsilon ^{2}D_{\xi }e^{-\pi i(K-1)}\Gamma \left(1-K\right)z^{k-2} \end{eqnarray*}$$

for $\omega \gg \omega _{0}$

$$\begin{eqnarray*} M_{xx}(z) & = & \frac{A\epsilon ^{2}D_{\xi }}{\sqrt{\epsilon \... ...{0}}}e^{-\pi i(K-1/2)}\Gamma \left(\frac{1}{2}-K\right)z^{K-3/2} \end{eqnarray*}$$

gaussian string

$$\begin{eqnarray*} \left\langle D_{\xi }\left(u(x,\tau )-u(x,0)\right)\right\rang... ...rac{v_{u}\tau }{\alpha }\right\vert^{-2-k^{2}\lambda ^{2}/2\pi } \end{eqnarray*}$$

$$\begin{eqnarray*} \left\langle T_{\tau }F_{x}(\tau )F_{x}(0)\right\rangle & = & ... ...au }{\epsilon }\right\vert^{-2-K-k^{2}\lambda ^{2}/2\pi }\right] \end{eqnarray*}$$

$$\begin{eqnarray*} A & = & \frac{1}{L_{y}}\left(\frac{2ev_{F}}{\pi ^{2}\alpha }\r... ...o }\right)^{2}\left(\frac{v_{s}}{v_{c}}\right)^{K_{c}}\eta ^{-2} \end{eqnarray*}$$

$$\begin{eqnarray*} M_{xx}(z) & = & \epsilon ^{2}\int \frac{dk}{2\pi }D_{\xi }(k)\... ...)\left(K+\frac{k^{2}\lambda ^{2}}{2\pi }+2\right)}\right)\right] \end{eqnarray*}$$

if we insert $D(k)=D_{0}exp(2l-K_{c}l-\lambda ^{2}k^{2}l/2\pi )$ we have

$$\begin{eqnarray*} M_{xx}(z) & = & \epsilon ^{2}D_{\xi 0}e^{\left(2-K_{c}\right)l... ...(1+\frac{rz^{2}}{\left(K+3\right)\left(K+2\right)}\right)\right] \end{eqnarray*}$$

floppy string

$$\begin{eqnarray*} \pi \Delta G & = & \frac{\Gamma \left(\frac{1}{n}\right)\lambd... ...left\vert s\frac{v_{s}\tau }{\alpha }\right\vert^{1-\frac{3}{n}} \end{eqnarray*}$$

$$\begin{eqnarray*} \left\langle D_{\xi }\left(u(x,\tau )-u(x,0)\right)\right\rang... ...frac{v_{s}\tau }{\alpha }\right\vert^{-\frac{1}{2}-\frac{3}{2n}} \end{eqnarray*}$$

$$\begin{eqnarray*} \left\langle T_{\tau }F_{x}(\tau )F_{x}(0)\right\rangle & = & ... ...\frac{\tau }{\epsilon }\right\vert^{-\frac{1}{2}-\frac{3}{2n}-K} \end{eqnarray*}$$

$$\begin{eqnarray*} A_{n} & = & \frac{1}{\lambda L_{y}}\left(\frac{2ev_{F}}{\pi ^{... ...{2}}\left(\frac{\lambda }{s^{1/n}\alpha }\right)^{\frac{n+3}{2}} \end{eqnarray*}$$

$$\begin{eqnarray*} M_{xx}(z) & = & \epsilon ^{2}D_{\xi }e^{-\pi i(K-1/2n-1/2)}z^{... ...{2}\right)\left(K+\frac{3}{2n}+\frac{1}{2}\right)}\right)\right] \end{eqnarray*}$$

floppy string $n=2$

$$\begin{eqnarray*} M_{xx}(z) & = & \epsilon ^{2}D_{\xi }e^{-\pi i(K-3/4)}\Gamma \... ...ft(K+\frac{9}{4}\right)\left(K+\frac{5}{4}\right)}\right)\right] \end{eqnarray*}$$

$$\begin{eqnarray*} A_{2} & = & \frac{1}{\lambda L_{y}}\left(\frac{2ev_{F}}{\pi ^{... ... }{\alpha }\right)^{\frac{3}{2}}\frac{\lambda }{\sqrt{s}\alpha } \end{eqnarray*}$$

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