Popov's coefficient

The introduced above approach allows us to find the asymptotic behavior of the one-body density matrix and estimate the coefficient of its decay. Within the first order of accuracy we split the average as

$$g_1(x) =\langle \sqrt{\hat\rho(x)\hat\rho(0)} e^{i(\hat\varphi(x)... ...t\rho(x)\hat\rho(0)}\rangle \langle e^{i(\hat\varphi(x)-\hat\varphi(0))}\rangle$$ (1.187)

We first calculate the contribution coming from the phase fluctuations

$$g_{1}^{phase}(x)=\langle e^{i(\hat{\varphi}(x)-\hat{\varphi}(0))}... ...\hat{b}_{k}(e^{ikx}-1)-\hat{b}_{k}^{\dagger }(e^{-ikx}-1)\right\} \right\rangle$$ (1.188)

The average of the exponent can be further developed by using the relation for the gaussian average $\left\langle \exp A\right\rangle = \exp \left\langle \frac{1}{2}A^{2}\right\rangle$. At zero temperature excitations are absent and the only nonzero average is $\left\langle \hat{b}%% _{k}\hat{b}_{k}^{\dagger }\right\rangle =1$. Thus we obtain

$$g_{1}^{phase}(x)=\exp \left\{ -\frac{1}{2}\sum_{k}\frac{\pi}{\eta... ...infty}^{\infty} \frac{\pi(1-\cos kx)}{\eta\vert k\vert}\frac{dk}{2\pi }\right\}$$ (1.189)

At this point we substitute the phononic excitation spectrum with the proper Bogoliubov dispersion. This can be done by changing $\eta\rightarrow \eta/\sqrt{ 1+(\hbar k/2Mc)^{2}}$.

$$g_{1}^{phase}(x)=\exp \left\{ -\frac{1}{\eta }\int\limits_{0}^{\infty }\frac{%% \sqrt{1+(\hbar k/2Mc)^{2}}(1-\cos kx)dk}{k}\right\}$$ (1.190)

Formally this integral is diverging. However, we will take use of the properties of the $\delta$-function

$$\int\limits_{-\infty }^{\infty }\cos \frac{kx}{2\pi }=\delta (x)$$ (1.191)

As we are interested in long range asymptotical behavior we subtract (1.191) from the exponent of (1.190) and consider a well-convergent expression

$$g_{1}^{phase}(x)=\exp \left\{ -\frac{1}{\eta }\int\limits_{0}^{\infty }\left( \frac{\sqrt{1+(\hbar k/2Mc)^{2}}}{k}-1\right) (1-\cos kx)\right\} dk$$ (1.192)

Partial integration together with the notation $\varepsilon =\hbar/(2xMc)\ll 1$ and $z=kx$ gives

$$g_{1}^{phase}(x)=\exp \left\{ -\frac{1}{\eta }\int\limits_{0}^{\infty} \frac{z-\sin z}{z^{2}\sqrt{1+\varepsilon ^{2}z^{2}}}\right\}\,dz$$ (1.193)

This equation can be calculated with non-logarithmic accuracy at $x\gg \xi$ by splitting the integral in three parts $1\ll N \ll 1/\varepsilon$

$$\int\limits_{0}^{\infty }\frac{z-\sin z}{z^{2}\sqrt{1+\varepsilon... ...y }\frac{1}{z\sqrt{1+\varepsilon ^{2}z^{2}}}dz=\gamma -1-\ln \frac{\hbar}{4xMc}$$ (1.194)

The calculation gives the result

$$g_{1}^{phase}(x) =\left( \frac{e^{1-\gamma }\eta }{8\pi \rho _{0}x}\right) ^{\frac{1}{\eta }}$$ (1.195)

In order to take into account the density fluctuations we develop (1.187) using Taylor expansion $\sqrt{1+x} = 1+x/2-x^2/8+{\cal O}(x^3)$, so $g_1^{\rho}(x) =\langle \sqrt{\hat\rho(x)\hat\rho(0)}\rangle \approx \rho_0+\fra... ...}\langle [\hat\rho^{\prime}(x)-\hat\rho^{\prime}(0)]\hat\rho^{\prime}(0)\rangle$. Writing the density operator in terms of creation and annihilation operators (eq. 1.156) we obtain

$$g_1^{\rho}(x) =\rho_0+ \frac{1}{4} \sum_k \frac{\rho_0\hbar\vert ... ...int\limits_0^\infty \frac{\rho_0\hbar k}{Mc(k)} (\cos kx\!-\!1) \frac{dk}{2\pi}$$

We substitute the speed of sound for the Bogoliubov dispersion relation $c(k) = c\sqrt{1+(\hbar k/2Mc)^2}$ and express the integral in dimensionless units $\varepsilon = \hbar/(2Mcx)$, $z=kx$

$$\langle \sqrt{\hat\rho(x)\hat\rho(0)}\rangle =\rho_0+ \frac{\rho_... ...c x^2} \int\limits_0^\infty \frac{(\cos z -1)z\,dz}{\sqrt{1+\varepsilon^2 z^2}}$$ (1.196)

The integral can be evaluated and expanded for small $\varepsilon$

$$\int\limits_0^\infty\frac{(\cos z -1)z\,dz}{\sqrt{1+\varepsilon^2... ... \approx \frac{1}{\varepsilon^2}+ (-1-3\varepsilon^2 +{\cal O}(\varepsilon^4)),$$ (1.197)

where $I_1(z)$ is modified Bessel function of first kind and $L_{-1}(z)$ is modified Struve function.

In terms of the parameter $\eta$ we have

$$\langle \sqrt{\hat\rho(x)\hat\rho(0)}\rangle =\rho_0\left(1+\frac{1}{\eta}-\frac{\eta}{16\pi\rho_0^2x^2}\right)$$ (1.198)

Combining together (1.195) and (1.198) we obtain finally the expression for the coefficient of the long-range asymptotics

$$g_1(x) = \rho_0\left(\frac{e^{1-\gamma}\eta}{8\pi}\right)^{\frac{... ...a}-\frac{\eta}{16\pi\rho_0^2x^2}\right) \left(\rho_0 x\right)^{-\frac{1}{\eta}}$$ (1.199)

In order to get an expression for the one-body density matrix at a finite temperature, one should account for thermal quasi-particle excitations. The long-range excitations are phonons and obey Bose-Einstein statistics $\langle \hat b^\dagger_k \hat b\rangle = (\exp(\hbar k c/k_B T)-1)^{-1}$. We are interested at the long-range behavior of the one-body density matrix, which corresponds to the limit $k\to 0$. In this conditions one can do a Taylor expansion and get $\langle \hat b^\dagger_k \hat b\rangle = k_B T/\hbar \vert k\vert c$. The calculation of the average (1.188) leads to appearance of an additional term, which depends on the temperature (compare with (1.189)):

$$g_{1}^{phase}(x)=\exp \left\{-\int\limits_{-\infty}^{\infty} \fra... ...os kx)(1+2k_B T/\hbar\vert k\vert c)}{\eta\vert k\vert}\frac{dk}{2\pi }\right\}$$ (1.200)

As we will show, the additional thermal suppression becomes dominating and will change the asymptotic behaviour of $g_1(x)$ significantly. The effect of the thermal phase fluctuations can be separated:

$$g_{1}^{phase}(x)= g_{1,T=0}^{phase}(x) \exp \left\{-\int\limits_{... ...(1-\cos kx)2k_B T/\hbar\vert k\vert c}{\eta\vert k\vert}\frac{dk}{2\pi}\right\}$$ (1.201)

where the zero temperature part $g_{1,T=0}^{phase}(x)$ is readily given by the formula (1.195). The integral in (1.201) is well behaved and can be easily calculated. It turns out to be proportional to $-\vert x\vert$ leading to exponential decay of the thermal fluctuation part at large distances:

$$g_{1}^{phase}(x) =\left( \frac{e^{1-\gamma }\eta }{8\pi\rho_0x}\right)^{\frac{1}{\eta}} \exp\left(-\frac{\vert x\vert}{\xi_T}\right)$$ (1.202)

The characteristic thermal decay length $\xi_T$ is inversely proportional to the temperature:

$$\xi_T = \frac{2\hbar^2\rho_0}{mk_BT}=(4\pi\rho_0\lambda^2)^{-1},$$ (1.203)

where the de Broglie thermal length is defined in the usual way $\lambda = \hbar/\sqrt{2\pi m k_B T}$.