Binary Mixture of Quasi-One-Dimensional Dipolar Bose–Einstein Condensates with Tilted Dipoles

Abstract

We consider a $^{168}$Er–$^{164}$Dy dipolar mixture, trapped by a cigar-shaped harmonic potential. We derive the quasi-1D interspecies effective potential exhibiting the tilting angles and show that it is a quite natural generalization of a single dipolar gas. By solving the coupled Gross–Pitaevskii equations, we observe a transition from miscible to immiscible mixture as the orientations of the magnetic moments are varied. The atom numbers are also shown to lead to noticeable effects on the characteristics of this transition.

Appendix

In this appendix, we provide the details for the computation of quasi-1D effective potential (2.11) which appears in Eqs. (2.8–2.9). It is given by the integral expression:

$$\begin{aligned} U^{(1D)}_{dd_{12}}(x_1-x_2)&=\frac{1}{\pi ^{2}l_{1}^2 l_{2}^2}\int d^2\rho _1 d^2\rho _2 \mathrm{d}x_2\nonumber \\&\quad U_{dd_{12}}^{(3D)}(\mathbf r _1-\mathbf r _2)\exp (-\rho _1^2 /l_1^2)\exp (-\rho _2^2 /l_2^2), \end{aligned}$$

(1)

where, up to a factor $\mu _0\mu _1\mu _2/4\pi$, the 3D potential is given by [10]

$$\begin{aligned} U_{dd_{12}}^{(3D)}(\mathbf r )=\frac{(\mathbf e _{1}.\mathbf e _{2})r^2 -3(\mathbf e _{1}.\mathbf r )(\mathbf e _{2}.\mathbf r )}{r^{5}}. \end{aligned}$$

(2)

$\mathbf e _{1}$ and $\mathbf e _{2}$ are the unit vectors along the dipole directions with $\frac{\mathbf{e }_{i}.\mathbf{r }}{r}=\cos \theta _{i}$ and $\mathbf e _{1}.\mathbf e _{2}=\cos (\theta _{1}-\theta _{2})$ which leads to (2.4).

In order to compute the integrals appearing in (1), we introduce the relative (y, z) and the center of mass (CM) Cartesian coordinates (Y, Z) in the $y-z$ plane:

$$\begin{aligned} y_{1,2}=&Y\pm \frac{m_{2,1}}{m_1+m_2}y\\ z_{1,2}=&Z\pm \frac{m_{2,1}}{m_1+m_2}z \end{aligned}$$

(3)

It is now straightforward to notice that the CM coordinates can be integrated out to yield an expression depending solely on the relative coordinates. Noting $x=x_1-x_2$, we get

$$\begin{aligned} U^{(1D)}_{dd_{12}}(x)={\displaystyle 1\over \displaystyle \pi (l_1^2+l_2^2)}\int \mathrm{d}y\mathrm{d}z\, U_{dd_{12}}^{(3D)}(x, y, z) \exp \left( -{\displaystyle y^2+z^2\over \displaystyle l_1^2+l_2^2}\right) . \end{aligned}$$

(4)

Now, assuming that the dipoles are in the $x-z$ plane, $\mathbf e _i=(\cos \alpha _i,0,\sin \alpha _i)$, we may change back to polar coordinates ($y=\rho \cos \phi$, $z=\rho \sin \phi$). One obtains the simple relations

$$\begin{aligned}&\cos (\theta _1-\theta _2)=\cos (\alpha _1-\alpha _2),\\&\cos \theta _i={\displaystyle x\cos \alpha _i+\rho \sin \phi \sin \alpha _i\over \displaystyle (x^2+\rho ^2)^{1/2}}, \end{aligned}$$

(5)

which allow us to simplify (4)

$$\begin{aligned} U^{(1D)}_{dd_{12}}(x)&={\displaystyle \cos (\alpha _1-\alpha _2)+3\cos (\alpha _1+\alpha _2)\over \displaystyle 2(l_1^2+l_2^2)}\nonumber \\&\quad \int _0^{\infty }\mathrm{d}\rho \,\rho {\displaystyle \rho ^2-2x^2 \over \displaystyle (x^2+\rho ^2)^{5/2}}\exp \left( -{\displaystyle \rho ^2\over \displaystyle l_1^2+l_2^2}\right) , \end{aligned}$$

(6)

and yield result (2.11).