Collision clock shift of two Fermi atoms in harmonic potentials

Chen Ze-Rui; Liu Guang-Cun; Yu Zhen-Hua

doi:10.7498/aps.70.20210243

Abstract

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1. 引　言
2. 理论模型
3. 运动方程
4. 光钟频移
5. 有限温平均
6. 结　论
附录A　运动方程(13)的零阶解和一阶解
附录B　(29)式的推导
附录C　高阶分波对钟频移的贡献

References

Abstract

Atomic clocks provide the most accurate definition for time. The precision of atomic clock has been improved by many orders of magnitude since the first atomic clock was built. However, the interatomic interaction usually suppress the precision of atomic clock. As a result, it is especially meaningful to study the interaction effect in atomic clock, which is considered to be helpful in improving the precision and accuracy of atomic clock. In order to characterize the collision effect induced clock shift, we theoretically study the collision clock shift in the Rabi spectrum, caused by the short-range interaction between two Fermi atoms in harmonic potential. Given that the short-range interatomic interaction is generally weak, and that the parameters of external lattice laser field are in the Lamb-Dicke regime, we make an approximation that the spatial wave-function of the Fermi atoms does not change, and then derive the motion equation for the internal wave-function under the external Rabi driving field. We solve the equation of motion by the perturbative method, and obtain the solution to first order, and thus derive the expression of the collision clock shift of the Rabi spectrum in terms of the interatomic interaction and the external Rabi driving laser field parameters for specific spatial wave-functions of atoms. Finally, we use the exact expression of the Green’s function in harmonic potential to obtain the averaged clock shift of collision at finite temperatures. Our results relate the atomic interaction with atomic clock shift, and provide a unified description of all partial waves of atomic interaction induced clock shift. Therefore, it becomes much more convenient to study the contributions of different partial waves to atomic clock shift. On the other hand, our results indicate that through precisely measuring the clock shift, the information about the interatomic interactions can also be obtained. In addition, our results for two interacting atoms can inspire the future study of real many-body interacting system which will be the next research topic.

Keywords:

PACS:

06.30.Ft	(Time and frequency)
34.20.Cf	(Interatomic potentials and forces)
34.50.-s	(Scattering of atoms and molecules)
67.85.-d	(Ultracold gases, trapped gases)

Authors and contacts

Guangdong Provincial Key Laboratory of Quantum Metrology and Sensing, School of Physics and Astronomy, Sun Yat-Sen University, Zhuhai 519082, China

Corresponding author: Yu Zhen-Hua, yuzhh5@mail.sysu.edu.cn

Funds: Project supported by the Key Area Research and Development Program of Guangdong Province, China (Grant No. 2019B030330001) and the National Natural Science Foundation of China (Grant Nos. 11474179, 11722438, 91736103, 12074440)

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1. 引　言

作为定义时间的科学标准, 原子钟从上个世纪开始就是一个众多学者致力研究的活跃领域^[1,2]. 而近期迅速发展的光频原子钟预示着新一代的时间标准. 原子钟性能的每一步提升都难能可贵, 人们通过大量的理论研究和实验探索, 逐渐实现了将抑制原子钟精度的各种效应所引发的时钟频移限制在相对更低的量级下, 诸如多普勒效应、光晶格频移、光子反冲等. 长期以来, 原子间短程相互作用导致的碰撞频移在各个阶段往往是抑制原子钟性能的重要因素. 所以, 对原子光钟碰撞频移的研究, 不仅对于更高精度原子钟的研制具有重要意义, 而且能够加深我们对原子相互作用的认识^[3-9].

从原子相干性对Ramsey条纹的影响^[10], 到费米原子气体中冷碰撞频移免疫论断的推翻^[11,12], 及玻色原子气体中碰撞频移令人困惑的 $2-$ 因子的问题^[13-17], 再到实验上测量Yb晶格光钟的冷碰撞频移获得p波相互作用的贡献^[18,19], 人们关于原子光钟碰撞频移的认知越来越丰富. 而最近实现的三维光晶格中的费米原子气体可以达到晶格格点上的单位填充^[20,21], 使得完全避免由原子间短程相互作用引起的碰撞频移成为可能.

另一方面, 对原子光钟碰撞频移的精密测量可以提供原子间相互作用的宝贵信息, 本研究也明显地展示了这一点. 本文理论研究了谐振子势阱中双费米原子光钟由于原子间的短程相互作用而在拉比频谱中引起的碰撞频移, 从而得到了拉比频谱的碰撞频移依赖于拉比探测光参数与在原子特定外态中相互作用的表达式. 并且, 利用谐振子势阱中格林函数的解析表达式, 得到了有限温下碰撞频移与原子间相互作用的关系. 这些结果都表明精密测量原子光钟的频移可以成为探测原子间相互作用势能性质的重要实验手段.

2. 理论模型

本文研究的物理过程是谐振子势阱中费米原子光钟在拉比脉冲作用下的光钟跃迁. 为揭示原子间短程相互作用对光钟跃迁的影响, 考虑在谐振子势阱中的双费米原子系统. 对于该问题, 系统的哈密顿量如下:

$H = H_0+H_{\rm{int}}+H_{\rm{drive}},$

(1)

$\begin{split} H_0 =\;& \sum\limits_{j = 1,2}\bigg[\frac{{\boldsymbol{P}}^2_j}{2m}+V_{\rm{ho}}({\boldsymbol{r}}_j)\\ &+E_g|g_j\rangle\langle g_j| +E_e|e_j\rangle\langle e_j| \bigg], \end{split}$

(2)

$\begin{split} H_{\rm{int}} =\;& U_{s}({\boldsymbol{r}})|s\rangle\langle s|+U_{d}({\boldsymbol{r}})|d\rangle\langle d|\\ &+U_{z}({\boldsymbol{r}})|z\rangle\langle z|+U_{u}({\boldsymbol{r}})|u\rangle\langle u|, \end{split}$

(3)

$H_{\rm{drive}} = \sum\limits_{j = 1,2}\left[\frac{\varOmega({\boldsymbol{r}}_j)}{2}{\rm e}^{-{\rm{i}}\omega_{\rm{L}} t}|e_j\rangle\langle g_j|+{\rm{h.c.}}\right],$

(4)

$V_{\rm{ho}}({\boldsymbol{r}}) = \frac{1}{2}m\omega^2_xx^2+\frac{1}{2}m\omega^2_y y^2+\frac{1}{2}m\omega^2_z z^2,$

(5)

式中, 下标j代表第j个原子, m为原子质量, $|g\rangle$ 和 $|e\rangle$ 为原子内态的两个钟态, $E_g$ 和 $E_e$ 表示对应能级. 原子间的相互作用 $H_{\rm{int}}$ 被分解到两个原子内态的单重态和三重态子空间中:

(6)

$|d\rangle = |g_1\rangle|g_2\rangle,$

(7)

(8)

$|u\rangle = |e_1\rangle|e_2\rangle.$

(9)

假设各子空间中的势场 $U_{\alpha}({\boldsymbol{r}})$ ( $\alpha = d,$ $z, u$ )为中心势场. 本文考虑的谐振子势阱为一般情况, 即3个方向的束缚频率 $\omega_x,\; \omega_y,\; \omega_z$ 各不相同. 驱动哈密顿量 $H_{\rm{drive}}$ 来源于拉比探测光与原子电偶极矩的耦合. 由于选择激光频率 $\omega_{\rm{L}}$ 接近原子的光钟跃迁频率 $\omega_0\equiv E_e-E_g$ , 即失谐量 $\delta\equiv \omega_{\rm{L}}-$ $\omega_0$ 的模量远小于 $\omega_0$ , 我们已采用了旋转波近似. 原则上, 拉比频率 $\varOmega({\boldsymbol{r}})$ 依赖于原子位置 ${\boldsymbol{r}}$ .

3. 运动方程

一般情况下, 原子间的相互作用能比谐振子势阱的束缚频率小很多. 假设先不考虑拉比探测光的驱动, 那么对于两个费米原子而言, 如果内态处在三重态, 则外态波函数应为 $[\phi_a( {\boldsymbol{r}}_1)\phi_b({\boldsymbol{r}}_2)- \phi_a({\boldsymbol{r}}_2)\times$ $\phi_b({\boldsymbol{r}}_1)]/\sqrt2$ ; 相应地, 如果内态处在单重态, 那么外态波函数应为 $[\phi_a({\boldsymbol{r}}_1)\phi_b({\boldsymbol{r}}_2)+ \phi_a({\boldsymbol{r}}_2)\phi_b({\boldsymbol{r}}_1)] /\sqrt2$ . 这里的 $\phi_a({\boldsymbol{r}})$ 是谐振子势阱中单原子的本征波函数, 具有本征能量 $\epsilon_a$ . 而当加上拉比探测光时, 为防止在拉比探测光作用下原子光钟跃迁的同时发生反冲, 谐振子势阱的束缚频率往往被调得很大, 使系统处在Lamb-Dicke区域, 即原子的外态不会发生改变.

下面考虑系统的初态被制备为 $[\phi_a({\boldsymbol{r}}_1)\phi_b({\boldsymbol{r}}_2)-$ $\phi_a({\boldsymbol{r}}_2)\phi_b({\boldsymbol{r}}_1)] |d\rangle/\sqrt2$ , 研究系统在拉比探测光作用下的随时演化. 之前的研究表明, 如果拉比探测光的场强存在空间非均匀性, 此非均匀性会影响光钟的频移^[12]. 这里假设探测光场强的空间非均匀性可以忽略不记, 即单个原子感受到的拉比频率大小不依赖该原子空间位置. 因此可以把(4)式中的 $\varOmega({\boldsymbol{r}})$ 替换成 $\varOmega$ (在将相位差吸收到内态后), 且 $\varOmega$ 为实数. 如此, 系统内态 $|\chi(t)\rangle$ 会局限在三重态子空间里演化^[12]. 由 $|\chi(t)\rangle$ 满足的含时薛定谔方程为

${\rm i}\partial_t |\chi(t)\rangle =\tilde H |\chi(t)\rangle,$

(10)

$\begin{split} \tilde H =&\; \dfrac12\int {\rm d}{\boldsymbol{r}}_1\int {\rm d}{\boldsymbol{r}}_2 \\ &\times [\phi_a({\boldsymbol{r}}_1)\phi_b({\boldsymbol{r}}_2)-\phi_a({\boldsymbol{r}}_2)\phi_b({\boldsymbol{r}}_1)]^* \\ & \times H[\phi_a({\boldsymbol{r}}_1)\phi_b({\boldsymbol{r}}_2)-\phi_a({\boldsymbol{r}}_2)\phi_b({\boldsymbol{r}}_1)], \end{split}$

(11)

将系统的内态表示成

$\begin{split} |\chi(t)\rangle =&\; {\rm e}^{-{\rm{i}}(E_e+E_g+\epsilon_a+\epsilon_b)t}\left[{\rm e}^{{\rm{i}}\omega_{\rm{L}} t}d(t)|d\rangle\right.\\ &+\left.z(t)|z\rangle+{\rm e}^{-{\rm{i}}\omega_{\rm{L}} t}u(t)|u\rangle\right], \end{split}$

(12)

随即得到运动方程

$\begin{split} {\rm{i}}\frac{\rm d}{{\rm d}t} \left( \begin{array}{c} d\\z\\u \end{array} \right) =&\; \left( \begin{array}{ccc} \delta & \varOmega/\sqrt2 & 0\\ \varOmega/\sqrt2 & 0 & \varOmega/\sqrt2\\ 0 & \varOmega/\sqrt2 & -\delta \end{array} \right)\\ &\times\left( \begin{array}{c} d\\z\\u \end{array} \right)+ \left( \begin{array}{c} U_{d,ab}d\\[3mm] U_{z,ab}z\\ U_{u,ab}u \end{array} \right), \end{split}$

(13)

其中原子相互作用能 $U_{\alpha, ab}$ ( $\alpha = d, z, u$ )为

$\begin{split} U_{\alpha,ab} =&\; \frac12\int {\rm{d}}{\boldsymbol{r}}_1\int {\rm{d}}{\boldsymbol{r}}_2 U_{\alpha}(|{\boldsymbol{r}}_1-{\boldsymbol{r}}_2|)\\ &\times|\phi_a({\boldsymbol{r}}_1)\phi_b({\boldsymbol{r}}_2)-\phi_a({\boldsymbol{r}}_2)\phi_b({\boldsymbol{r}}_1)|^2. \end{split}$

(14)

显然, 相互作用能的大小依赖于两个原子空间波函数的交叠情况. 由于原子间的相互作用能很小, 连同初始值 $d(t = 0) = 1$ , $z(t = 0) = u(t =$ $0) = 0$ 解析求解(13)式到 $U_{\alpha, ab}$ 的一阶(详细结果见附录A).

4. 光钟频移

光钟拉比频谱的工作程序是将原子制备到 $|g\rangle$ 初态, 然后拉比探测光施加一个时长为 $t_\pi$ 的 $\pi$ 脉冲, 即 $\varOmega t_\pi = \pi$ , 接着再测量激发到 $|e\rangle$ 态的原子概率 $P_e$ . 对于本文研究的双费米原子的问题

$P_e = \left(|z|^2+2|u|^2\right)/2.$

(15)

通过解析微扰求解(13)式, 同样可以得到 $P_e$ 的零阶解 $P_e^{(0)}$ 和一阶解 $P_e^{(1)}$ , 形式为

$P_e^{(0)}(\delta, t) = \frac{\varOmega^2}{\varDelta^2}\sin^2(\varDelta t/2),$

(16)

$\begin{split} P_e^{(1)}(\delta, t) =&\; (U_{d,ab}-U_{z,ab})\mathcal C(\delta,t)\\ &+(U_{u,ab}-U_{z,ab})\mathcal D(\delta,t), \end{split}$

(17)

$\mathcal C(\delta,t) = \dfrac1{4 \varDelta ^6}\left(\delta \varOmega ^2 \sin \left(\dfrac{t \varDelta }{2}\right) \Big\{t \varDelta \left(4 \delta ^2+\varOmega ^2\right) \cos \left(\dfrac{t \varDelta }{2}\right)+\sin \left(\dfrac{t \varDelta }{2}\right) \big[-8 \delta ^2+\varOmega ^2 \cos (t \varDelta)-3 \varOmega ^2\big]\Big\}\right),$

(18)

$\mathcal D(\delta,t) = -\dfrac1{16 \varDelta ^6}\left\{\delta \varOmega ^4 \big[6 t \varDelta \sin (t \varDelta)+8 \cos (t \varDelta)+\cos (2 t \varDelta)-9\big]\right\},$

(19)

其中 $\varDelta = \sqrt{\delta ^2+\varOmega ^2}$ .

拉比频谱确定频移的方法是找到两个拉比探测光的失谐量 $\delta_{\rm r}( < 0)$ 和 $\delta_{\rm b}( > 0)$ , 使得在 $\pi$ 脉冲后

$P_e(\delta_{\rm r},t_\pi) = P_e(\delta_{\rm b},t_\pi),$

(20)

频移 $\delta_{\rm s}$ 就被定义为

$\delta_{\rm s} = (\delta_{\rm r}+\delta_{\rm b})/2.$

(21)

对于单原子而言, $P_e(\delta, t) = P_e^{(0)}(\delta, t)$ , $P_e(\delta, t)$ 是 $\delta$ 的偶函数, 所以必定有 $\delta_{\rm s} = 0$ . 但原子间的相互作用导致了修正项 $P_e^{(1)}(\delta, t)$ , 而且 $P_e^{(1)}(\delta, t)$ 显然不是 $\delta$ 的偶函数, 相应的 $\delta_{\rm s}$ 不再为 $0$ .

要从(16)式和(19)式推导出 $\delta_{\rm s}$ 的解析表达式, 我们在蓝失谐侧选取一个较小的参考失谐量 $\delta_{\rm R}( > 0)$ . 根据(20)式的定义, 有

$P_e(-\delta_{\rm R}+2\delta_{\rm s}, t_\pi) = P_e(\delta_{\rm R},t_\pi).$

(22)

仍然利用原子间的相互作用能比较小的特点, 这样频移 $\delta_{\rm s}$ 也会比较小. 泰勒展开(22)式左边到 $\delta_{\rm s}$ 的一阶, 得到

$\delta_{\rm s} = - \left.\frac{P^{(1)}_{e}(\delta,t_\pi)-P^{(1)}_{e}(-\delta,t_\pi)} {2\big[\partial P^{(0)}_{e}(\delta_,t_\pi)/\partial\delta\big]}\right|_{\delta = \delta_{\rm R}},$

(23)

将(16)式和(19)式代入(23)式, 计算得到

$\begin{split} \delta_{\rm s} = &\;(U_{d,ab}-U_{z,ab})A(\delta_{\rm R}/\varOmega)\\ &+(U_{u,ab}-U_{d,ab})B(\delta_{\rm R}/\varOmega), \end{split}$

(24)

其中

$A(\nu) = \frac{\sqrt{\nu ^2 + 1} \sin \left(\dfrac{1}{2} \pi \sqrt{\nu ^2 + 1}\right) \big[-\cos \left(\pi \sqrt{\nu ^2 + 1}\right) + 8 \nu ^2+3\big]-\pi (\nu ^2 + 1) (4 \nu ^2 + 1) \cos \left(\dfrac{1}{2} \pi \sqrt{\nu ^2+1}\right)}{4 \left(\nu ^2+1\right) \Big[\pi \left(\nu ^2+1\right) \cos \left(\dfrac{1}{2} \pi \sqrt{\nu ^2+1}\right)-2 \sqrt{\nu ^2+1} \sin \left(\dfrac{1}{2} \pi \sqrt{\nu ^2+1}\right)\Big]},$

(25)

$B(\nu) =\frac{6 \pi \left(\nu ^2+1\right) \cos \left(\dfrac{1}{2} \pi \sqrt{\nu ^2+1}\right)-\sqrt{\nu ^2+1} \Big[9 \sin \left(\dfrac{1}{2} \pi \sqrt{\nu ^2+1}\right)+\sin \left(\dfrac{3}{2} \pi \sqrt{\nu ^2+1}\right)\Big]}{8 \left(\nu ^2+1\right) \Big[\pi \left(\nu ^2+1\right) \cos \left(\dfrac{1}{2} \pi \sqrt{\nu ^2+1}\right)-2 \sqrt{\nu ^2+1} \sin \left(\dfrac{1}{2} \pi \sqrt{\nu ^2+1}\right)\Big]}.$

(26)

(24)式反映了 $\pi$ 脉冲激光参数 $\nu = \delta_{\rm R}/\varOmega$ 和原子间相互作用能 $U_{\alpha, ab}$ ( $\alpha = d, z, u$ )对频移 $\delta_{\rm s}$ 的贡献. $\delta_{\rm s}$ 只依赖于 $U_{\alpha, ab}$ ( $\alpha = d, z, u$ )的差值, 因为如果 $U_{\alpha, ab}$ ( $\alpha = d, z, u$ )都相同的话, (13)式中相互作用能只是把三重态的能量整体频移了一下, 不可能影响光钟频移. 图1显示在实验关心的区间 $\delta_{\rm R}/\varOmega\sim1/2$ , A和B的模量都是1的量级. 从另一个角度看, (24)式表明通过测量 $\delta_{\rm s}$ 在两个不同 $\delta_{\rm R}$ 时的值, 就可以反推出 $U_{\alpha, ab}$ ( $\alpha = d, z, u$ )之间的差值. 如要获得单独 $U_{\alpha, ab}$ ( $\alpha = d, z, u$ )的值, 实验则需选取特定的量子态使得至少一个 $U_{\alpha, ab} ( \alpha =$ $d, z, u )$ 的值为 $0$ .

$图 1 光钟频移系数(a) $A\;(\delta_{\rm R}/\varOmega)$与(b) $ B\;(\delta_{\rm R}/\varOmega) $\r\nFig. 1. Coefficient (a) $A\;(\delta_{\rm R}/\varOmega)$ and (b) $ B\;(\delta_{\rm R}/\varOmega) $ for clock shift.$

图 1 光钟频移系数(a)

$A\;(\delta_{\rm R}/\varOmega)$

与(b)

$B\;(\delta_{\rm R}/\varOmega)$

Fig. 1. Coefficient (a)

$A\;(\delta_{\rm R}/\varOmega)$

and (b)

$B\;(\delta_{\rm R}/\varOmega)$

for clock shift.

下载: 全尺寸图片幻灯片

在此值得强调的是, 以上求解钟频移的方法要求拉比谱的线形是光滑的. 这一前提对于本文考虑的弱相互作用情形是满足的, 但当相互作用足够强时, 拉比谱会出现边带峰^[18], 这种情况下以上求解钟频移的方法就不再适用了.

5. 有限温平均

有限温度 $T\equiv1/\beta$ 不为 $0$ 的情况下, 两个费米原子会服从正则系综占据谐振子势的各个单粒子本征态. 所以要取得 $\delta_{\rm s}$ 在有限温度下的平均值 $\bar\delta_{\rm s}$ , 需要对 $U_{\alpha, ab}$ 进行正则系综平均以取得其平均值 $\bar U_{\alpha}$ . 通过转换到两个费米原子的质心坐标 ${\boldsymbol{R}} =$ $({\boldsymbol{r}}_1+{\boldsymbol{r}}_2)/2$ 和相对坐标 ${\boldsymbol{r}} = {\boldsymbol{r}}_1-{\boldsymbol{r}}_2$ , 两个费米原子外态波函数可以写成

$\psi({\boldsymbol{r}}_1,{\boldsymbol{r}}_2) = \varPhi_\rho({\boldsymbol{R}})\varPsi_\sigma({\boldsymbol{r}}),$

(27)

其中, $\varPhi_\rho({\boldsymbol{R}})$ 是质量为 $M = 2 m$ 的谐振子本征态, $\varPsi_\sigma({\boldsymbol{r}})$ 是质量为 $\mu = m/2$ 的谐振子本征态; 谐振子3个方向的振动频率仍为 $\omega_x,\; \omega_y, \;\omega_z$ . 如此, (24)式中的 $U_{\alpha, ab}$ 应被替换成

$U_{\alpha,\sigma} = \int U_{\alpha}({\boldsymbol{r}})|\varPsi_\sigma({\boldsymbol{r}})|^2 \,{\rm d}{\boldsymbol{r}}.$

(28)

注意到由于费米统计, 内态为三重态时, 外态必须是奇宇称态, 即 $\varPsi_\sigma({\boldsymbol{r}}) = -\varPsi_\sigma(-{\boldsymbol{r}})$ .

为了用低能散射参数描述 $U_{\alpha}({\boldsymbol{r}})$ , 将 $U_{\alpha, \sigma}$ 表示成(详细推导见附录B)

$\begin{split} U_{\alpha,\sigma} =&\; \sum\limits_{\ell = 0}^{\infty}\frac{a_{\alpha,\ell}}{(2\ell+1)!} \lim\limits_{\lambda\to0}(-)^\ell \frac{\partial^\ell}{\partial\lambda^\ell}\frac{1}{2^3(\pi\lambda)^{3/2}}\\ &\times\int {\rm{d}}{\boldsymbol{r}} |\varPsi_\sigma({\boldsymbol{r}})|^2 {\rm{e}}^{-{\boldsymbol{r}}^2/4\lambda}, \end{split}$

(29)

其中不同的 $\ell$ 对应不同的展开阶数. 这里的参数

$a_{\alpha,\ell} = 4\pi\int_0^\infty {\rm{d}}{\boldsymbol{r}} {\boldsymbol{r}}^{2(\ell+1)} U_{\alpha}({\boldsymbol{r}})$

(30)

包含了相互作用势 $U_{\alpha}({\boldsymbol{r}})$ 的信息. 而非零温平均值 $\bar U_\alpha$ 则变成

$\begin{split} \bar U_{\alpha} =&\;\sum\limits_{\ell = 0}^{\infty}\frac{a_{\alpha,\ell}}{(2\ell+1)!} \lim\limits_{\lambda\to0}(-)^\ell \frac{\partial^\ell}{\partial\lambda^\ell}\frac{1}{2^3(\pi\lambda)^{3/2}}\\ &\times\int {\rm{d}}{\boldsymbol{r}} \sum\limits_\sigma^{\prime} P_\sigma |\varPsi_\sigma({\boldsymbol{r}})|^2 {\rm{e}}^{-{\boldsymbol{r}}^2/4\lambda}, \end{split}$

(31)

式中 $P_\sigma$ 是正则系综的概率分布, 求和号上打 ' 是强调在费米量子统计下只计奇宇称本征态. 可见有限温平均关键是要求得 $\displaystyle\sum\limits_\sigma^{\prime} P_\sigma |\varPsi_\sigma({\boldsymbol{r}})|^2$ .

为方便计算, 需要先处理 $\displaystyle\sum\limits_\sigma^{\prime} P_\sigma \varPsi^ {*}_\sigma({\boldsymbol{r}}') \varPsi_{\sigma}({\boldsymbol{r}})$ . 对于简谐振子而言, 虚时间传播子为

$\begin{split} G({\boldsymbol{r}},{\boldsymbol{r}}'|\beta) =&\; \langle{\boldsymbol{r}}|{\rm{e}}^{-\beta H_{{\rm {ho}}}}|{\boldsymbol{r}}'\rangle \\ =&\sum\limits_{n}{\rm{e}}^{-\beta H_{{\rm {ho}}}} \varPsi^ {*}_n({\boldsymbol{r}}')\varPsi_n({\boldsymbol{r}}), \end{split}$

(32)

这里的 $H_{\rm{ho}}$ 表示简谐振子的哈密顿量. 简谐振子的虚时间传播子的解析形式为

$\begin{split} & G({\boldsymbol{r}},{\boldsymbol{r}}'|\beta) = \prod\limits_{i = x,y,z}\left[ \frac{\mu\omega_i}{2\pi \sinh(\beta\omega_{i})}\right]^{1/2}\\ &~~~~ \times {\rm{e}}^{-\textstyle\frac{\mu\omega_{i}}{2\pi \sinh(\beta\omega_{i})}[({\boldsymbol{r}}^2_i + {{\boldsymbol{r'}}_{i}}^2)\cosh(\beta\omega_{i})-2{\boldsymbol{r}}^{2}_{i}{\boldsymbol{r'}}^2_{i}]}, \end{split}$

(33)

其中, $\mu = m/2$ 是约化质量. 根据费米子统计, 我们只想提取 $G({\boldsymbol{r}}, {\boldsymbol{r'}}|\beta)$ 中奇宇称部分的贡献. 于是用一个小技巧得到

$\begin{split} & \sum\limits_n^\prime P_n\phi^ {*}_n({\boldsymbol{r}}')\phi_{n}({\boldsymbol{r}}) = [ G({\boldsymbol{r}},{\boldsymbol{r}}'|\beta) -G(-{\boldsymbol{r}},{\boldsymbol{r}}'|\beta)]/Z, \end{split}$

(34)

其中, 归一化系数Z为

$\begin{split} Z =\;& \int {\rm{d}}{\boldsymbol{r}}[G({\boldsymbol{r}},{\boldsymbol{r}}|\beta) - G(-{\boldsymbol{r}},{\boldsymbol{r}}|\beta)]\\ =&\; \prod\limits_{i = x,y,z}({\rm{e}}^{\beta\omega_i/2}-{\rm{e}}^{-\beta\omega_i/2})^{-1}\\ &-\prod\limits_{i = x,y,z}({\rm{e}}^{\beta\omega_i/2}+{\rm{e}}^{-\beta\omega_i/2})^{-1} . \end{split}$

(35)

结合(31)式—(35)式, 可以得到

$\bar\delta_{\rm s} = (\bar U_{d}-\bar U_{z})A(\delta_{\rm R}/\varOmega)+(\bar U_{u}-\bar U_{d})B(\delta_{\rm R}/\varOmega),$

(36)

$\bar U_{\alpha} = \sum \limits_{\ell = 0}^\infty \frac{a_{\alpha,\ell}}{(2\ell+1)!}C_\ell,$

(37)

$\begin{split} &\; C_\ell =\lim\limits_{\lambda \to 0}(-)^\ell \frac{\partial^\ell}{\partial\lambda ^\ell} \frac{1}{2^3(\pi\lambda)^{3/2}}\frac{1}{Z}\\ & ~~\times \Bigg\{\prod\limits_{i = x,y,z}\frac{1}{\sqrt{2\sinh(\beta\omega_i)\Big[\tanh(\beta\omega_i/2)+ \dfrac14\mu\omega_{i}\lambda\Big]}}\\ &~~ -\prod\limits_{i = x,y,z}\frac{1}{\sqrt{2\sinh(\beta\omega_i)\Big[\coth(\beta\omega_i/2)+ \dfrac14\mu\omega_{i}\lambda\Big]}}\Bigg\}. \end{split}$

(38)

直接计算可知 $C_0 = 0$ , 这是由于当内态处在三重态子空间中时, 相对坐标的空间波函数反对称, 感受不到零程相互作用. 图2给出了 $C_1$ 和 $C_2$ 随温度的变化. 作图时选取 $\omega_y = (1+\epsilon)\omega_x$ , $\omega_z = (1- \epsilon)\omega_x$ , 并且 $\epsilon = 1/10$ . 可以看出, 随着温度升高 $C_\ell$ 变大. 这是由于温度升高, 双费米原子可以占据更高能量的外态. 同时发现, 当 $1/(\beta\omega_x)$ 较大时, $C_\ell$ 随温度变化在双对数图中呈现线性, 如图3所示, 这表明当温度较高时, $C_\ell$ 与温度呈幂次关系, 幂次分别为5.54203 $(\ell = 1)$ 和6.5296 $(\ell = 2)$ . 虽然在 $1/\beta\omega_x \sim 8$ 处, $C_2/(\mu\omega_x)^{7/2}$ 比 $C_1/(\mu\omega_x)^{5/2}$ 大两个数量级. 但考虑到 $a_{\alpha, 2}/a_{\alpha, 1}\sim r_{\rm{vdW}}^2$ , 这里 $r_{\rm{vdW}}$ 是原子的范德瓦耳斯半径. 一般情况下 $r^2_{\rm{vdW}}/\left[1/(\mu\omega_x)\right]\sim 10^{-4}$ . 根据(37)式, 上述温度范围中, 总体而言 $\ell = 2$ 的贡献仍小于 $\ell = 1$ . 当然在更高温度下 $\ell$ 取值大的项会贡献越来越大(参看附录C).

$图 2 系数$ C_\ell $随温度的变化　(a) C1; (b) C2\r\nFig. 2. Coefficient $ C_\ell $ versus temperature: (a) C1; (b) C2.$

图 2 系数

$C_\ell$

随温度的变化　(a) C₁; (b) C₂

Fig. 2. Coefficient

$C_\ell$

versus temperature: (a) C₁; (b) C₂.

下载: 全尺寸图片幻灯片

$图 3 当温度较高时, 系数$ C_\ell $随温度的变化, 这里横轴和纵轴都取对数　(a)线性拟合表达式为$y = 5.54203 x- 3.83918$; (b)线性拟合表达式为$ y = 6.5296 x-0.119146 $\r\nFig. 3. Coefficient $ C_\ell $ versus temperature for higher temperature. The horizontal coordinate and the vertical coordinate are logarithmic here: (a) Linear fitting function is $ y = 5.54203 x-3.83918 $; (b) linear fitting function is $ y = 6.5296 x-0.119146 $.$

图 3 当温度较高时, 系数

$C_\ell$

随温度的变化, 这里横轴和纵轴都取对数　(a)线性拟合表达式为

$y = 5.54203 x- 3.83918$

; (b)线性拟合表达式为

$y = 6.5296 x-0.119146$

Fig. 3. Coefficient

$C_\ell$

versus temperature for higher temperature. The horizontal coordinate and the vertical coordinate are logarithmic here: (a) Linear fitting function is

$y = 5.54203 x-3.83918$

; (b) linear fitting function is

$y = 6.5296 x-0.119146$

下载: 全尺寸图片幻灯片

6. 结　论

本文研究了双费米原子拉比激发过程中短程相互作用引起的碰撞频移. 计算结果给出了碰撞频移依赖于原子相互作用的关系. 相较以前碰撞频移的理论研究只考虑了s波及p波, 本文提供了各阶贡献的统一表达式. 本文的结果意味着通过精密测量碰撞频移可以获得原子间相互作用的重要信息. 如果要将本文双费米原子的结果拓展到多费米原子的情况, 可以预见的是, 在弱相互作用极限下, 多原子体系的相互作用效应仍应来源于原子两两之间的碰撞, 即相互作用能是所有粒子对的总和, 形式为 $\displaystyle\sum\nolimits_{ij}g_{ij}f_if_j$ . 这里 $f_i$ 为费米原子在第i个谐振子本征态上的占据数, $g_{ij}$ 是占据i和j本征态原子发生碰撞而导致的耦合系数. 对多费米原子情况的研究将是我们下一步的课题.

附录A　运动方程(13)的零阶解和一阶解

通过微扰的方法可解得运动方程(13)的零阶解: $\;$

$d^{(0)}(t) =\frac{-2 {\rm{i}} \delta \sqrt{\delta ^2+\varOmega ^2} \sin \left(t \sqrt{\delta ^2+\varOmega ^2}\right)+\left(2 \delta ^2+\varOmega ^2\right) \cos \left(t \sqrt{\delta ^2+\varOmega ^2}\right)+\varOmega ^2}{2 \left(\delta ^2+\varOmega ^2\right)}, \tag{A1}$

$z^{(0)}(t) = \frac{\varOmega \big[-\delta -{\rm{i}} \sqrt{\delta ^2+\varOmega ^2} \sin \left(t \sqrt{\delta ^2+\varOmega ^2}\right)+\delta \cos \left(t \sqrt{\delta ^2+\varOmega ^2}\right)\big]}{\sqrt{2} \left(\delta ^2+\varOmega ^2\right)}, \tag{A2}$

$u^{(0)}(t) = \frac{\varOmega ^2 \big[\cos \left(t \sqrt{\delta ^2+\varOmega ^2}\right)-1\big]}{2 \left(\delta ^2+\varOmega ^2\right)} , \tag{A3}$

一阶解的形式为

$\begin{split} {d^{(1)}}(t) =& - \frac{1}{{8{\varDelta ^5}}}{\rm{i}}\Big( 2{\varOmega ^2}\varDelta \big[t{\varOmega ^2}({U_{d,ab}} + {U_{u,ab}}) + 2\delta {U_{z,ab}}(\delta t + 2{\rm{i}}) - 4{\rm {i}}{\delta} {U_{d,ab}} \big] + \varDelta \cos (t\varDelta)\big\{ 8{\delta ^4}t{U_{d,ab}} + 4\delta {\varOmega ^2}\big[ \delta t(2{U_{d,ab}} + {U_{z,ab}}) \\ & + 2{\rm i}({U_{d,ab}} - {U_{z,ab}})\big] + t{\varOmega ^4}({U_{d,ab}} + 2{U_{z,ab}} + {U_{u,ab}}) \big\} + \sin (t\varDelta)\big\{ - 4{\rm{i}}\delta t{\varDelta ^2}\big[ 2{\delta ^2}{U_{d,ab}} \\ &+ {\varOmega ^2}({U_{d,ab}} + {U_{z,ab}}) \big] + 8{\delta ^2}{\varOmega ^2}({U_{d,ab}} - {U_{z,ab}}) + {\varOmega ^4}(5{U_{d,ab}} - 2{U_{z,ab}} - 3{U_{u,ab}}) \big\} \Big),\\[-12pt] \end{split} \tag{A4}$

$\begin{split} {z^{(1)}}(t) =& - \frac{1}{{4\sqrt 2 {\varDelta ^5}}}\varOmega \Big( - 2\varDelta \left\{ {2{\delta ^2}\left[ {{U_{d,ab}} + {U_{z,ab}}( - 1 + {\rm{i}}\delta t)} \right] + {\varOmega ^2}\left[ {{\rm{i}}\delta t({U_{d,ab}} + {U_{u,ab}}) - {U_{d,ab}} + {U_{u,ab}}} \right]} \right\} \\ &+ \varDelta \cos (t\varDelta)\left\{ {4{\delta ^2}({\rm{i}}\delta t{U_{d,ab}} + {U_{d,ab}} - {U_{z,ab}}) + {\varOmega ^2}\left[ {{\rm{i}}\delta t(3{U_{d,ab}} + 2{U_{z,ab}} - {U_{u,ab}}) - 2{U_{d,ab}} + 2{U_{u,ab}}} \right]} \right\} \qquad ~~\\ &+ 4{\delta ^3}\left[ {\delta t{U_{d,ab}} + {\rm{i}}({U_{z,ab}} - {U_{d,ab}})} \right] + \delta {\varOmega ^2}\left[ {\delta t(5{U_{d,ab}} + 2{U_{z,ab}} + {U_{u,ab}}) - {\rm{i}}({U_{d,ab}} + 2{U_{z,ab}} - 3{U_{u,ab}})} \right]\\ & + t{\varOmega ^4}({U_{d,ab}} + 2{U_{z,ab}} + {U_{u,ab}})\sin (t\varDelta)\Big), \end{split} \tag{A5}$

$\begin{split} {u^{(1)}}(t) =& \frac{1}{{8{\varDelta ^5}}}{\varOmega ^2}\Big( 4\delta \varDelta ({\rm{i}}\delta t{U_{z,ab}} + {U_{d,ab}} - {U_{u,ab}}) + 2{\rm{i}}t{\varOmega ^2}\varDelta ({U_{d,ab}} + {U_{u,ab}})- {\rm{i}}\varDelta \cos (t\varDelta)\big\{ t{\varOmega ^2}({U_{d,ab}} + 2{U_{z,ab}} + {U_{u,ab}}) \\ & + 2\delta \big[ \delta t({U_{d,ab}} + {U_{u,ab}}) - 2{\rm{i}}({U_{d,ab}} - {U_{u,ab}}) \big] \big\} + \big\{ 2{\delta ^2}\big[ {\delta t({U_{u,ab}} - {U_{d,ab}}) + {\rm{i}}({U_{d,ab}} - 2{U_{z,ab}} + {U_{u,ab}})} \big] \\ &+ {\varOmega ^2}\left[ {2\delta t({U_{u,ab}} - {U_{d,ab}}) - {\rm{i}}({U_{d,ab}} - 2{U_{z,ab}} + {U_{u,ab}}) \big]} \right\}{\sin (t\varDelta)} \Big). \end{split} \tag{A6}$

附录B　(29)式的推导

正文中(29)式的推导过程如下:

$U_{\alpha,\sigma} = \int {\rm{d}}{\boldsymbol{r}}\; U_{\alpha\beta}(r)|\varPsi_\sigma({\boldsymbol{r}})|^2, \tag{B1}$

利用傅里叶变换得到

$\begin{split} U_{\alpha,\sigma} = \int \frac{{\rm{d}}{\boldsymbol{k}}}{(2\pi)^3}\int \frac{{\rm{d}}{\boldsymbol{k}}'}{(2\pi)^3} \varPsi_\sigma^*({\boldsymbol{k}})\varPsi_\sigma({\boldsymbol{k}}') \int {\rm{d}}{\boldsymbol{r}}\; U_{\alpha}(r){\rm{e}}^{{\rm{i}}\Delta{\boldsymbol{k}}\cdot{\boldsymbol{r}}}, \end{split} \tag{B2}$

其中 $\Delta{\boldsymbol{k}} = {\boldsymbol{k}}'-{\boldsymbol{k}}$ . 因为 $U_{\alpha\beta}( r)$ 是中心势场,

$\begin{split} U_{\alpha}(\Delta{\boldsymbol{k}}) =& \;\int {\rm{d}}{\boldsymbol{r}} U_{\alpha}(r){\rm{e}}^{{\rm{i}}\Delta{\boldsymbol{k}}\cdot{\boldsymbol{r}}} = \;\frac{4\pi}{\Delta k}\int_0^\infty {\rm{d}}r r U_{\alpha}(r)\sin(\Delta k r) = \;4\pi\sum\limits_{\ell = 0}^{\infty}\frac{\Delta k^{2\ell}}{(2\ell+1)!}\int_0^\infty {\rm{d}}r r^{2(\ell+1)} U_{\alpha}(r) = \;\sum\limits_{\ell = 0}^{\infty}\frac{\Delta k^{2\ell}}{(2\ell+1)!} a_{\alpha,\ell}, \end{split} \tag{B3}$

参数

$a_{\alpha,\ell} = 4\pi\int_0^\infty {\rm{d}}r r^{2(\ell+1)} U_{\alpha}(r) \tag{B4}$

包含了相互作用势 $U_{\alpha}(r)$ 的信息.

如此

$\begin{split} U_{\alpha,\sigma} =\;& \sum\limits_{\ell = 0}^{\infty}\frac{a_{\alpha,\ell}}{(2\ell+1)!} \int \frac{{\rm{d}}{\boldsymbol{k}}}{(2\pi)^3}\int \frac{{\rm{d}}{\boldsymbol{k}}'}{(2\pi)^3} \varPsi_\sigma^*({\boldsymbol{k}})\varPsi_\sigma({\boldsymbol{k}}')\Delta {\boldsymbol{k}}^{2\ell} = \sum\limits_{\ell = 0}^{\infty}\frac{a_{\alpha,\ell}}{(2\ell+1)!} \lim\limits_{\lambda\to0}(-)^\ell \frac{\partial^\ell}{\partial\lambda^\ell}\int \frac{{\rm{d}}{\boldsymbol{k}}}{(2\pi)^3} \int \frac{{\rm{d}}{\boldsymbol{k}}'}{(2\pi)^3} \varPsi_\sigma^*({\boldsymbol{k}})\varPsi_\sigma({\boldsymbol{k}}'){\rm{e}}^{-\lambda\Delta {\boldsymbol{k}}^2}\\ =\;&\sum\limits_{\ell = 0}^{\infty}\frac{a_{\alpha,\ell}}{(2\ell+1)!} \lim\limits_{\lambda\to0}(-)^\ell \frac{\partial^\ell}{\partial\lambda^\ell} \int {\rm{d}}{\boldsymbol{r}} \int {\rm{d}}{\boldsymbol{r}}'\varPsi_\sigma^*({\boldsymbol{r}})\varPsi_\sigma({\boldsymbol{r}}')\int \frac{{\rm{d}}{\boldsymbol{k}}}{(2\pi)^3}\int \frac{{\rm{d}}{\boldsymbol{k}}'}{(2\pi)^3} {\rm{e}}^{{\rm{i}}{\boldsymbol{k}}\cdot{\boldsymbol{r}}} {\rm{e}}^{-{\rm{i}}({\boldsymbol{k}}+{\boldsymbol{k}}')\cdot{\boldsymbol{r}}'}{\rm{e}}^{-\lambda{\boldsymbol{k}}'^{2}}\\ =\;& \sum\limits_{\ell = 0}^{\infty}\frac{a_{\alpha,\ell}}{(2\ell+1)!} \lim\limits_{\lambda\to0}(-)^\ell \frac{\partial^\ell}{\partial\lambda^\ell}\frac{1}{2^3(\pi\lambda)^{3/2}} \int {\rm{d}}{\boldsymbol{r}} |\varPsi_\sigma({\boldsymbol{r}})|^2 {\rm{e}}^{-{\boldsymbol{r}}^2/4\lambda}. \\[-12pt]\end{split} \tag{B5}$

附录C　高阶分波对钟频移的贡献

为了定量化高阶分波对钟频移的贡献随温度的变化, 由(36)式计算 ${\bar\delta}_{\rm s}$ 截断求和到第 $\ell$ 阶的值, 即 ${\bar\delta}_{\rm s}(\ell)$ . 为了完成定量计算, 采用方势阱模型势 $U_{\alpha}({\boldsymbol{r}}) =$ $U_{\alpha, 0}\theta(r-r_0)$ . 这里势阱力程 ${\boldsymbol{r}}_0$ 应为范德瓦耳斯半径的量级, 取成 ${{r}}_0 = 1$ nm. 而势阱深度的取值, 参考实验论文^[18], 分别为 $U_{d, 0} = 0$ , $U_{t, 0} = -4.4\pi$ Hz以及 $U_{u, 0} = -0.44\pi$ Hz. 当 $\delta_{\rm R}/\varOmega = 1/2$ 时, 如图C1所示, 当温度很低时, 同时考虑s波至d波(s, p, d)时的钟频移 ${\bar{\delta}}_{\rm s}\;(\ell = 2)$ 同只考虑s波至p波时的钟频移 ${\bar{\delta}}_{\rm s}\;(\ell = 1)$ 的差值极小, 即d波分量引起的钟频移接近零. 当温度升高时, 以上两者差值逐渐增大, 表明d波贡献的钟频移量也在逐渐增大. 由本文的解析表达式可推知, 更高分波的贡献也有相同的趋势.

$图 C1 同时考虑s波至d波时的钟频移$ {\bar\delta}_{\rm s}\;(\ell = 2) $与只考虑s波和p波时的钟频移$ {\bar\delta}_{\rm s}\;(\ell = 1) $的相对误差\r\nFig. C1. Relative error of clock shifts between considering s, p, d partial waves ($ {\bar\delta}_{\rm s}\;(\ell = 2) $) and only considering s, p partial waves ($ {\bar\delta}_{\rm s}\; (\ell = 1) $).$

图 C1 同时考虑s波至d波时的钟频移

${\bar\delta}_{\rm s}\;(\ell = 2)$

与只考虑s波和p波时的钟频移

${\bar\delta}_{\rm s}\;(\ell = 1)$

的相对误差

Fig. C1. Relative error of clock shifts between considering s, p, d partial waves (

${\bar\delta}_{\rm s}\;(\ell = 2)$

) and only considering s, p partial waves (

${\bar\delta}_{\rm s}\; (\ell = 1)$

下载: 全尺寸图片幻灯片

References

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[2]	Ludlow A D, Boyd M M, Ye J, Peik E, Schmidt P O 2015 Rev. Mod. Phys. 87 637 Google Scholar
[3]	Blatt S, Thomsen J W, Campbell G K, et al. 2009 Phys. Rev. A 80 052703 Google Scholar
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[8]	Burdick N Q, Tang Y, Lev B L 2016 Phys. Rev. X 6 031022
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图 1 光钟频移系数(a) $A\;(\delta_{\rm R}/\varOmega)$ 与(b) $B\;(\delta_{\rm R}/\varOmega)$

Figure 1. Coefficient (a) $A\;(\delta_{\rm R}/\varOmega)$ and (b) $B\;(\delta_{\rm R}/\varOmega)$ for clock shift.

DownLoad: Full-Size Img PowerPoint

图 2 系数 $C_\ell$ 随温度的变化　(a) C₁; (b) C₂

Figure 2. Coefficient $C_\ell$ versus temperature: (a) C₁; (b) C₂.

DownLoad: Full-Size Img PowerPoint

图 3 当温度较高时, 系数 $C_\ell$ 随温度的变化, 这里横轴和纵轴都取对数　(a)线性拟合表达式为 $y = 5.54203 x- 3.83918$ ; (b)线性拟合表达式为 $y = 6.5296 x-0.119146$

Figure 3. Coefficient $C_\ell$ versus temperature for higher temperature. The horizontal coordinate and the vertical coordinate are logarithmic here: (a) Linear fitting function is $y = 5.54203 x-3.83918$ ; (b) linear fitting function is $y = 6.5296 x-0.119146$ .

DownLoad: Full-Size Img PowerPoint

图 C1 同时考虑s波至d波时的钟频移 ${\bar\delta}_{\rm s}\;(\ell = 2)$ 与只考虑s波和p波时的钟频移 ${\bar\delta}_{\rm s}\;(\ell = 1)$ 的相对误差

Figure C1. Relative error of clock shifts between considering s, p, d partial waves ( ${\bar\delta}_{\rm s}\;(\ell = 2)$ ) and only considering s, p partial waves ( ${\bar\delta}_{\rm s}\; (\ell = 1)$ ).

DownLoad: Full-Size Img PowerPoint

[1]	阮军, 王叶兵, 常宏, 刘涛, 董瑞芳, 张首刚 2015 物理学报 64 160308 Google Scholar Ruan G, Wang Y B, Chang H, Jiang H F, Liu T, Dong R F, Zhang S G 2015 Acta Phys. Sin. 64 160308 Google Scholar
[2]	Ludlow A D, Boyd M M, Ye J, Peik E, Schmidt P O 2015 Rev. Mod. Phys. 87 637 Google Scholar
[3]	Blatt S, Thomsen J W, Campbell G K, et al. 2009 Phys. Rev. A 80 052703 Google Scholar
[4]	Zhu B, Gadway B, Foss-Feig M, Schachenmayer J, Wall M L, Hazzard K R A, Yan B, Moses S A, Covey J P, Jin D S, Ye J, Holland M, Rey A M 2014 Phys. Rev. Lett. 112 070404
[5]	Moses S A, Covey J P, Miecnikowski M T, Yan B, Gadway B, Ye J, Jin D S 2015 Science 350 659 Google Scholar
[6]	Labuhn H, Barredo D, Ravets S, de Léséleuc S, Macrì T, Lahaye T, Browaeys A 2016 Nature 534 667 Google Scholar
[7]	Aikawa K, Baier S, Frisch A, Mark M, Ravensbergen C, Ferlaino F 2014 Science 345 1484 Google Scholar
[8]	Burdick N Q, Tang Y, Lev B L 2016 Phys. Rev. X 6 031022
[9]	Kadau H, Schmitt M, Wenzel M, Wink C, Maier T, Ferrier-Barbut I, Pfau T 2016 Nature 530 194 Google Scholar
[10]	Heller E J, Falconer I S, Dewar R L 2002 Proceedings of the XVIII International Conference on Atomic Physics: the Expanding Frontier of Atomic Physics Cambridge, Massachusett, USA, July 28–August 2, 2002 p363
[11]	Zwierlein M W, Hadzibabic Z, Gupta S, Ketterle W 2003 Phys. Rev. Lett. 91 250404 Google Scholar
[12]	Campbell G K, Boyd M M, Thomsen J W, et al. 2009 Science 324 360 Google Scholar
[13]	Harber D M, Lewandowski H J, Mcguirk J M, Cornell E A 2002 Phys. Rev. A 66 053616
[14]	Fuchs J N, Gangardt D M, Laloe F 2002 Phys. Rev. Lett. 88 230404 Google Scholar
[15]	Kadio D, Band Y B 2006 Phys. Rev. A 74 053609
[16]	Kurt G 2009 Phys. Rev. Lett. 103 113202 Google Scholar
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Vol. 70, Issue 18 - 2021-09-20

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