Density profile

The wave function of the perturbation, $\delta\psi_k$, was obtained in the momentum representation and is given by expression (7.5). The spatial dependence, $\delta\psi(x)$, is related to $\delta\psi_k$ by means of the Fourier transformation. We will find the density profile $n(x)= \vert\psi(x)\vert^2$. Within the same level of accuracy, as in the calculations above, $n(x)$ is given by

$$n(x) \approx \phi_0^2+\phi_0(\delta\psi(x)+\delta\psi^*(x))$$ (7.43)

In terms of Fourier components one has $\delta\psi(x) = \int e^{ikx}\delta\psi_k\frac{dk}{2\pi}$

$$n(x) = n_0 +\phi_0\int e^{ikx}(\delta\psi_k+(\delta\psi_{-k})^*)\frac{dk}{2\pi}$$ (7.44)

where we used property of the Fourier transformation $(\delta\psi^*)_k = (\delta\psi_{-k})^*$. Together with (7.5) and (7.40) we obtain a simple expression

$$n(x) = n_0\left(1 + \frac{4}{b_{1D}}\int\limits_{-\infty}^\infty \frac{e^{ikx}}{k^2+\frac{4m^2(c^2-V^2)}{\hbar^2}}\frac{dk}{2\pi} \right)$$ (7.45)

There are two cases to be considered separately:

1)

The impurity moves with velocity smaller than the speed of sound. We introduce the notation $\varkappa=2m\sqrt{c^2-V^2}/\hbar>0$ and note that the integral has form of the inverse Fourier transform of the Yukawa potential:

$$\int\limits_{-\infty}^\infty \frac{e^{ikx}}{k^2+\varkappa^2}\frac{dk}{2\pi} =\frac{\exp\{-\varkappa\vert x\vert\}}{2\varkappa}$$ (7.46)

Thus, the density perturbation has a form of a bump and decays exponentially fast:

$$n(x) = n_0 \left(1 + \frac{2e^{-\varkappa\,\vert x\vert}}{\varkappa b_{1D}} \right)$$ (7.47)

For a repulsive interaction with the impurity the scattering length is negative $b_{1D}<0$ and the density is suppressed by the presence of the impurity. Instead an attractive interaction $b_{1D}>0$ leads to an increase in the density.

2)

The impurity moves with velocity larger than the speed of sound. In this case we introduce $\varkappa$ in the following way $\varkappa=2m\sqrt{V^2-c^2}/\hbar>0$. There are poles appearing in the function in the integral. We use Landau casuality rule $k\to k+i 0$ in order to modify the integration contour.

In this case for $x>0$ the pole is absent and the integral vanishes. This means that there is no perturbation in front of the impurity (impurity moves to the right).

Instead for $x<0$ the pole is present and the integral is different from zero.

$$\int\limits_{-\infty}^\infty \frac{e^{ikx}}{k^2-\varkappa^2}\frac{dk}{2\pi} = \frac{\sin \varkappa x}{\varkappa}$$ (7.48)

so the density profile behind the perturbation is oscillating and corresponds to the wake generated by the moving impurity

$$n(x)= \left\{ \begin{array}{ll} n_0 \left(1 + \frac{4}{\varkappa\,b_{1D}}\sin\varkappa x\right)& x<0\\ n_0 & x>0 \\ \end{array}\right.$$ (7.49)

The condition of the applicability of the perturbation theory demands the perturbation $\vert n(x)-n_0\vert$ be small compared to the unperturbed solution $n_0$. This condition is satisfied if the velocity of the impurity $V$ is not to close to the speed of sound $c$.