Dynamic form factor

The dynamic form factor is related to the time-dependent density-density correlation function by the means of the Fourier transformation:

$$S(k,\omega) = \frac{\rho_0}{\hbar}\int\!\!\!\!\int e^{i(\omega t-... ...ft[\frac{\langle\hat\rho(x,t)\hat\rho(0,0)\rangle}{\rho_0^2}-1\right]\, dx\, dt$$ (1.181)

The correlation function was calculated with non-logarithmic accuracy and is given by formula (1.180). The evaluation of direct Fourier transformation (1.181) would give us the expression for the dynamic form factor. It turns out that it is easier to go other way around, i.e. guess the form of the $m$-th component of $S(k,\omega)$

$$S(k,\omega) = A (\omega^2-c^2(k-2mk_F)^2)^{\frac{m^2\eta}{2}-1},$$ (1.182)

make the inverse Fourier transformation to go back from $(k,\omega)$ to $(x,t)$ and comparing obtain result to (1.180) fix the value of the constant $A$. Here we use notation $k_F = \pi\rho_0$. We start by doing the integration over momentum. $S(x,\omega) = \int\limits_{-\infty}^\infty e^{ikx} S(k,\omega)\frac{dk}{2\pi}$. We introduce notation $\triangle k = k+2mk_F$ the integral is limited to the region $(-\omega/c,\omega/c)$:

$$S(x,\omega) = A c^{m^2\eta-2} e^{i2mk_Fx} \int\limits_{-\omega / ... ...ght)^2-(\triangle k)^2\right]^{\frac{m^2\eta}{2}-1} \frac{d(\triangle k)}{2\pi}$$ (1.183)

As a reference we use formula 3.771(464/465) from Gradstein-Ryzhik book [GR80]:

$$\int_{0}^{u}{(u^{2}-x^{2})^{\nu-{1\over2}}} \cos (ax) dx = {\sqrt... ...1\over2}\right)J_{\nu}(au),\quad \left[a>0, u>0,\mbox{Re} \nu>-{1\over2}\right]$$

The substitution $u=\omega/c$, $a=x$, $\nu = (m^2\eta-1)/2$ gives an expression in terms of $(x,\omega)$

$$S(x,\omega) = \frac{A e^{i2mk_Fx}}{2\sqrt\pi c} \left(\frac{2\ome... ...frac{m^2\eta}{2}\right) J_{\frac{m^2\eta-1}{2}}\left(\frac{\omega x}{c}\right),$$ (1.184)

where $J_n(x)$ is the Bessel function of the first kind. The integration over the frequencies $S(x,t) = \int\limits_{0}^\infty e^{-i\omega t} S(x,\omega) \frac{d\omega}{2\pi}$ can be done easily done by using the formula (6.699.5) from [GR80]

$$\int_{0}^{\infty}x^{\nu}\cos (ax) J_{\nu}(bx)\,dx =2^\nu \frac{b^... ...frac{1}{2}}\quad \left[0< a<b,\quad \vert\mbox{Re}\,\nu\vert< {1\over 2}\right]$$

and gives

$$S(x,t) = A\frac{(2c)^{m^2\eta-1}}{2\pi^2} \Gamma^2\left(\frac{m^2\eta}{2}\right)\frac{\cos(2mk_F)}{(x^2-c^2t^2)^{\frac{m^2\eta}{2}}}$$ (1.185)

Comparing this result with (1.180) we fix so far unknown coefficient of the proportionality to the value $A = \frac{8\pi^2c\rho_0}{\hbar\Gamma^2\left(m^2\eta/2\right)} \left(\frac{\hbar}{8Cmc^2}\right)^{m^2\eta}$ and, finally, obtain

$$S(k, \omega) = \!\!\sum\limits_{m=1}^\infty\! \frac{8\pi^2\rho_0 ... ...\!+\!(\omega^2\!\!-\!\!c^2(k\!\!+\!\!2mk_F)^2)^{\frac{m^2\eta}{2}-1}}{2}\right]$$ (1.186)