Finite-Size Effects of Surface Tension in Two Segregated BECs Confined by Two Hard Walls

Abstract

The finite-size effects of the surface tension in two segregated Bose–Einstein condensates limited by two hard walls are studied respectively in canonical ensemble and grand canonical ensemble by means of the Gross–Pitaevskii theory in the modified double-parabola approximation. The analytical formulae of surface tensions and their finite-size effects are found together with a new type of long-range forces acting on two walls.

Appendices

Appendix 1: The Analytics Expression for $A_j, B_j$

The constants in Eqs. (15) and (16) are obtained by substitution of them in Robin boundary conditions and continuity conditions for $\phi _j$ yielding

$$\begin{aligned}&A_1=\frac{e^{\sqrt{2}z_0}[\sqrt{\eta }-\sqrt{\eta }_- e^{\sqrt{2}h_-}+e^{2\sqrt{\eta }h_+}(\sqrt{\eta }-\sqrt{\eta }_+ e^{\sqrt{2}h_-})]}{X}, \end{aligned}$$

(30)

$$\begin{aligned}&A_2=\frac{\sqrt{2}(e^{\sqrt{2}h}-e^{\sqrt{2}z_0})^2e^{\sqrt{\eta }(2h+z_0)}}{X},\end{aligned}$$

(31)

$$\begin{aligned}&B_1=\frac{\sqrt{2}(e^{\frac{\sqrt{2}h_+}{\xi }}-1)^2e^{\frac{\sqrt{\eta }z_0}{\xi }}}{Y},\end{aligned}$$

(32)

$$\begin{aligned}&B_2=\frac{e^{\frac{\sqrt{2}h}{\xi }}(\sqrt{\eta }_- e^{\frac{2\sqrt{\eta }h}{\xi }}+\sqrt{\eta }_+e^{\frac{2\sqrt{\eta }z_0}{\xi }})}{Y}-\frac{e^{\frac{\sqrt{2}h}{\xi }}\sqrt{\eta }(e^{\frac{2\sqrt{\eta }h+\sqrt{2}h_+}{\xi }}+e^{\frac{\sqrt{2}h_++2\sqrt{\eta }z_0}{\xi }})}{Y}\qquad \quad \end{aligned}$$

(33)

with

$$\begin{aligned}&h_{\pm }=h\pm z_0, \sqrt{\eta }_{\pm }=\sqrt{\eta }\pm \sqrt{2},\\&X=\sqrt{\eta }_- e^{2\sqrt{2}h}-\sqrt{\eta }_+e^{2\sqrt{2}z_0}+2 e^{(2\sqrt{\eta }+\sqrt{2})h_+}(\sqrt{\eta }\text {sinh}[\sqrt{2}h_-]+\sqrt{2}\text {cosh}[\sqrt{2}h_-]),\\&Y=e^{\frac{2\sqrt{\eta }z_0}{\xi }}[\sqrt{\eta }_+ -\sqrt{\eta }_- e^{\frac{2\sqrt{2}h_+}{\xi }}]- e^{\frac{2\sqrt{\eta }h}{\xi }}[-\sqrt{\eta }_-+\sqrt{\eta }_+ e^{\frac{2\sqrt{2}h_+}{\xi }}]. \end{aligned}$$

Appendix 2: The Integrals

Using the wave functions for ground state (15), (16) with the constants in “Appendix 1” we arrive

$$\begin{aligned} I_1= & {} \frac{1}{2}(2A_1^2e^{2\sqrt{2}h}(\sqrt{2}\text {sinh}(2\sqrt{2}h_-)+4h_-)+2\sqrt{2}A_1e^{\sqrt{2}(3h-2z_0)}\nonumber \\&-\,2A_1e^{\sqrt{2}h}(\sqrt{2}-4h_-)+\sqrt{2}(e^{2\sqrt{2}h_-}-1)) \nonumber \\&+\,A_2^2\sqrt{\eta }e^{-2\sqrt{\eta }h}(2\sqrt{\eta }h_++\text {sinh}(2\sqrt{\eta }h_+)), \end{aligned}$$

(34)

$$\begin{aligned} I_2= & {} \frac{B_1^2}{\xi ^2}\sqrt{\eta }e^{\frac{2\sqrt{\eta }h}{\xi }} \Bigg [2\sqrt{\eta }h_--\xi \text {sinh}\frac{-2\sqrt{\eta }h_-}{\xi }\Bigg ] \nonumber \\&+\,\frac{e^{\frac{-2\sqrt{2}(3h+z_0)}{\xi }}}{2\xi ^2}\Bigg [B_2^2\left( -\sqrt{2}\xi e^{\frac{2\sqrt{2}h}{\xi }}+8h_+e^{\frac{2\sqrt{2}(2h+z_0)}{\xi }}+\sqrt{2}\xi e^{\frac{2\sqrt{2}(3h+2z_0)}{\xi }}\right) \nonumber \\&-\,2\sqrt{2}B_2\xi e^{\frac{3\sqrt{2}h}{\xi }}+2B_2e^{\frac{\sqrt{2}(5h+2z_0)}{\xi }}(4h_++\sqrt{2}\xi )-\sqrt{2}\xi \left( e^{\frac{4\sqrt{2}h}{\xi }}-e^{\frac{2\sqrt{2}(3h+z_0)}{\xi }}\right) \Bigg ],\nonumber \\ \end{aligned}$$

(35)

and analytical form of the normalization constants

$$\begin{aligned} \mathcal {N}_1= & {} \frac{1}{4}(2A_1^2e^{2\sqrt{2}h}(\sqrt{2}\text {sinh} (2\sqrt{2}h_-) -4h_-)+ 2A_1(e^{\sqrt{2}h}(3\sqrt{2}-4h_-) \nonumber \\&-\,2\sqrt{2}e^{\sqrt{2}(2h-z_0)}+\sqrt{2}e^{\sqrt{2}(3h-2z_0)}-2\sqrt{2}e^{\sqrt{2}z_0})\nonumber \\&+\,\sqrt{2}e^{\sqrt{2}h_-}(e^{\sqrt{2}h_-}-4)+4h_-+3\sqrt{2})\nonumber \\&+\,\frac{A_2^2e^{-2\sqrt{\eta }h}(\text {sinh}(2\sqrt{\eta }h_+)-2\sqrt{\eta }h_+)}{\sqrt{\eta }},\end{aligned}$$

(36)

$$\begin{aligned} \mathcal {N}_2= & {} \frac{1}{4}e^{-\frac{2\sqrt{2}(3h+z_0)}{\xi }} \left( B_2^2\left( -\left( \sqrt{2}\xi e^{\frac{2\sqrt{2}h}{\xi }} +8h_+e^{\frac{2\sqrt{2}(2h+z_0)}{\xi }} -\sqrt{2}\xi e^{\frac{2\sqrt{2}(3h+2z_0)}{\xi }}\right) \right) \right. \nonumber \\&\left. +\,4\sqrt{2}\xi e^{\frac{\sqrt{2}(5h+z_0)}{\xi }}-2\sqrt{2}B_2\xi \left( e^{\frac{3\sqrt{2}h}{\xi }}-2e^{\frac{3\sqrt{2}(2h+z_0)}{\xi }}-2e^{\frac{\sqrt{2}(4h+z_0)}{\xi }}\right) \right. \nonumber \\&\left. -\,2B_2e^{\frac{\sqrt{2}(5h+z_0)}{\xi }}(4h_+ +3\sqrt{2}\xi )-\sqrt{2}\xi e^{\frac{4\sqrt{2}h}{\xi }}+e^{\frac{2\sqrt{2}(3h+z_0)}{\xi }}(4h_+ -3\sqrt{2}\xi )\right) \nonumber \\&-\,\frac{B_1^2}{\sqrt{\eta }}e^{\frac{2\sqrt{2}h}{\xi }}\xi \text {sinh}\Bigg (\frac{-2\sqrt{\eta }h_-}{\xi }\Bigg )-2h_-. \end{aligned}$$

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