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Repulsive Fermi gas

This Hamiltonain can be solved for an arbitrary number of particles spin up $N_{\uparrow}$ and spins down $N_\downarrow$. The corresponding integral equations are ( $\gamma = \frac{g_{1D}m}{\hbar^2n}>0$)[Yan67]:

$\displaystyle \left\{
{\begin{array}{ccc}
\sigma(k) &=&
-\int\limits_{-B}^B\fra...
...kappa)}{\gamma^2+4(k-\varkappa)^2}
\frac{d\varkappa}{2\pi}
\end{array}}
\right.$     (10.8)

The limit $B\to\infty$ correspond to $N_\downarrow = N_\uparrow$. In this limit one can simplify further the system of integral equations by introducing a Fourier transformation (see also discrete lattice model [Col74]):

$\displaystyle \sigma(x) = \int\limits_{-\infty}^\infty e^{-ikx}\sigma(k)\frac{dk}{2\pi}$     (10.9)
$\displaystyle \sigma(k) = \int\limits_{-\infty}^\infty e^{ikx}\sigma(x)\,dx$     (10.10)

By multiplying first equation from (A.8) by $e^{-ikx}/2\pi$ and integrating over $\k$ one obtains expression for the $\sigma(x)$10.1


$\displaystyle \sigma(x) = \frac{1}{2\mathop{\rm ch}\nolimits \frac{\gamma\vert x\vert}{2}} \int\limits_{-Q}^Q e^{-ikx}\rho(k)\frac{dk}{2\pi}$     (10.11)

Setting $x=0$ one immediately sees that number of spin-down particles is half of the total number of particles $\int\sigma(k)\,dk = \frac{1}{2}\int\rho(k)\,dk$.

Inserting (A.10) into second equation from (A.8), taking into account formula (A.11) and carrying out two integrations one obtains the integral equation involving only $\rho(x)$

$\displaystyle \rho(k) = \frac{1}{2\pi} + \int\limits_{-Q}^QK(k-\varkappa)\rho(\varkappa)\,\frac{d\varkappa}{2\pi},$     (10.12)

where the kernel is


$\displaystyle K(\xi) = %%\frac{4c}{c^2+4\xi^2}-\frac{\pi}{c\ch \frac{\pi x}{c}}
2\int\limits_0^\infty \frac{\cos\xi x}{1+e^{\gamma x}}\,dx$     (10.13)

Ones this equation is solved the density and energy are given by

$\displaystyle na_{1D} = \int\limits_{-Q}^Q\rho(k)\,dk,$     (10.14)
$\displaystyle e(\gamma) = \frac{1}{na_{1D}}\int\limits_{-Q}^Qk^2\rho(k)\,dk$     (10.15)

In the strongly interacting limit $\gamma\to\infty$ and the kernel can be simplified

$\displaystyle K(\xi) = 2\int\limits_0^\infty \frac{\cos\xi x}{1+e^{\gamma x}}\,...
...{2}{\gamma}\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n}
=\frac{2\ln 2}{\gamma}$     (10.16)

The energy per particle in units of $\left[\frac{\hbar^2}{2ma^2}\right]$ is given by

$\displaystyle E = \frac{\pi^2 n^2}{3} - \frac{2\ln(2)\pi^3 n}{3}$     (10.17)

which equals to the energy of gas of $N$ free fermions of the same spin.

It is possible to express the kernel in terms of $\beta$-function (see Gradstein-Ryzhik). Taking into account the series representation $\beta(z) =
\sum\limits_{k=0}^\infty\frac{(-1)^k}{z+k}$ one obtain following result from the (exact) sum (A.16)

$\displaystyle K(\xi) = -\frac{1}{\gamma}\left(\beta\left(\frac{i\xi}{\gamma}\right)
+\beta\left(-\frac{i\xi}{\gamma}\right)\right).$     (10.18)

The $\beta$-function is defined using the digamma function $\beta(z) =
\frac{1}{2}\left(\Psi(\frac{x+1}{2})-\Psi(\frac{x}{2})\right)$. The digamma function is defined as logarithmic derivative of the Gamma function $\Psi(z) =
\frac{\partial}{\partial z} \ln \Gamma(z)$.

The kernel can be expanded at small and large values of the argument:

$\displaystyle K(\xi) = \frac{2}{\gamma}\ln 2 -\frac{3}{2\gamma^3} \xi^2 + {\cal O}(\xi^4),$     (10.19)
$\displaystyle K(\xi) = \frac{\gamma}{2}\xi^{-2} +\frac{2\gamma^3}{4}\xi^{-4} + {\cal O}(\xi^{-6})$     (10.20)


next up previous contents
Next: Numerical solution Up: Bethe ansatz solutions Previous: Attractive Fermi gas   Contents
G.E. Astrakharchik 15-th of December 2004