Incommensurate Filling of Ultracold Spin-1 Atoms in Optical Superlattice with a Weak Magnetic Field

Abstract

The ground states of ultracold spin-1 atoms trapped in an optical superlattice in a weak magnetic field with incommensurate filling of three atoms in one double-well are obtained. It is shown that the ground-state diagrams of the reduced double-well model are remarkably different for the antiferromagnetic and ferromagnetic atoms. These novel spin-states can be controlled easily and exactly by modulating the tunneling parameter and the quadratic Zeeman energy, which may be a tool for the study of spin-entanglement.

Appendix

To find the intermediate states and corresponding transition matrix elements, it is necessary to know what they will be when the operators $\hat{a}_{\pm1,0}^{\dagger}$ and $\hat{a}_{\pm1,0}$ work on the spin states |S,S _z ;n〉, which have been done in Ref. [25]. But they obtained the formulas only for S≤2. We have obtained those for S=3, which are needed in this paper, as follows,

$$ \hat{a}_{\pm1}^{\dagger}\vert 3,\pm3;n \rangle=\frac{2}{3} \sqrt{n+6}\vert 4,\pm4;n+1 \rangle, $$

$$ \hat{a}_{0}^{\dagger}\vert 3,\pm3;n \rangle=\frac {1}{3} \sqrt{n+6}\vert 4,\pm3;n+1 \rangle, $$

$$ \hat{a}_{\mp1}^{\dagger}\vert 3,\pm3;n \rangle=\frac{1}{3} \sqrt{\frac{n+6}{7}}\vert 4,\pm2;n+1 \rangle-\sqrt{\frac {3 ( n-1 ) }{7}}\vert 2,\pm2;n+1 \rangle, $$

$$ \hat{a}_{\pm1}^{\dagger}\vert 3,\pm2;n \rangle=\sqrt{ \frac {n+6}{3}}\vert 4,\pm3;n+1 \rangle, $$

$$ \hat{a}_{0}^{\dagger}\vert 3,\pm2;n \rangle=2\sqrt{ \frac {n+6}{21}}\vert 4,\pm2;n+1 \rangle+\sqrt{\frac{n-1}{7}}\vert 2,\pm 2;n+1 \rangle, $$

$$ \hat{a}_{\mp1}^{\dagger}\vert 3,\pm2;n \rangle=\sqrt{ \frac {n+6}{21}}\vert 4,\pm1;n+1 \rangle-\sqrt{\frac{2 ( n-1 ) }{7}}\vert 2, \pm1;n+1 \rangle, $$

$$ \hat{a}_{\pm1}^{\dagger}\vert 3,\pm1;n \rangle=\sqrt{ \frac {5 ( n+6 ) }{21}}\vert 4,\pm2;n+1 \rangle-\sqrt {\frac{n-1}{35}}\vert 2, \pm2;n+1 \rangle, $$

$$ \hat{a}_{0}^{\dagger}\vert 3,\pm1;n \rangle=\sqrt{ \frac {5 ( n+6 ) }{21}}\vert 4,\pm1;n+1 \rangle+2\sqrt{\frac {2 ( n-1 ) }{35}}\vert 2, \pm1;n+1 \rangle, $$

$$ \hat{a}_{\mp1}^{\dagger}\vert 3,\pm1;n \rangle=\sqrt{ \frac {2 ( n+6 ) }{21}}\vert 4,0;n+1 \rangle-2 \sqrt{\frac{6 ( n-1 ) }{35}}\vert 2,0;n+1 \rangle, $$

$$ \hat{a}_{\pm1}^{\dagger}\vert 3,0;n \rangle=\frac {1}{3}\sqrt{ \frac{10 ( n+6 ) }{7}}\vert 4,\pm1;n+1 \rangle -\sqrt{\frac{3 ( n-1 ) }{35}}\vert 2, \pm1;n+1 \rangle, $$

$$ \hat{a}_{0}^{\dagger}\vert 3,0;n \rangle=\frac{4}{3}\sqrt {\frac{n+6}{7}}\vert 4,0;n+1 \rangle+3\sqrt{\frac{n-1}{35}}\vert 2,0;n+1 \rangle, $$

$$ \hat{a}_{\pm1}\vert 3,\pm3;n \rangle=-\frac{1}{3}\sqrt { \frac{n-3}{7}}\vert 4,\pm2;n-1 \rangle+\sqrt{\frac{3 ( n+4 ) }{7}}\vert 2, \pm2;n-1 \rangle, $$

$$ \hat{a}_{0}\vert 3,\pm3;n \rangle=\frac{1}{3}\sqrt{n-3} \vert 4,\pm3;n-1 \rangle, $$

$$ \hat{a}_{\mp1}\vert 3,\pm3;n \rangle=-\frac{2}{3}\sqrt{n-3} \vert 4,\pm4;n-1 \rangle, $$

$$ \hat{a}_{\pm1}\vert 3,\pm2;n \rangle=-\sqrt{\frac{n-3}{21}} \vert 4,\pm1;n-1 \rangle+\sqrt{\frac{2 ( n+4 ) }{7}}\vert 2,\pm1;n-1 \rangle, $$

$$ \hat{a}_{0}\vert 3,\pm2;n \rangle=2\sqrt{\frac{n-3}{21}} \vert 4,\pm2;n-1 \rangle+\sqrt{\frac{n+4}{7}}\vert 2,\pm 2;n-1 \rangle, $$

$$ \hat{a}_{\mp1}\vert 3,\pm2;n \rangle=-\sqrt{\frac{n-3}{3}} \vert 4,\pm3;n-1 \rangle, $$

$$ \hat{a}_{\pm1}\vert 3,\pm1;n \rangle=-\sqrt{\frac{2 ( n-3 ) }{21}}\vert 4,0;n-1 \rangle+\sqrt{\frac{6 ( n+4 ) }{35}}\vert 2,0;n-1 \rangle, $$

$$ \hat{a}_{0}\vert 3,\pm1;n \rangle=\sqrt{\frac{5 ( n-3 ) }{21}}\vert 4, \pm1;n-1 \rangle+2\sqrt{\frac{2 ( n+4 )}{35}}\vert 2,\pm1;n-1 \rangle, $$

$$ \hat{a}_{\mp1}\vert 3,\pm1;n \rangle=-\sqrt{\frac{5 ( n-3 ) }{21}}\vert 4,\pm2;n-1 \rangle+\sqrt{\frac {n+4}{35}}\vert 2,\pm2;n-1 \rangle, $$

$$ \hat{a}_{\pm1}\vert 3,0;n \rangle=-\frac{1}{3}\sqrt{ \frac{10 ( n-3 ) }{7}}\vert 4,\mp1;n-1 \rangle+ \sqrt{\frac{3 ( n+4 ) }{35}}\vert 2, \mp1;n-1 \rangle, $$

$$ \hat{a}_{0}\vert 3,0;n \rangle=\frac{4}{3}\sqrt{ \frac {n-3}{7}}\vert 4,0;n-1 \rangle+3\sqrt{\frac{n+4}{35}} \vert 2,0;n-1 \rangle. $$