Anisotropic exactly solvable models in the cold atomic systems（PPT讲稿）

Anisotropic exactly solvable models in the cold atomic systems Junpeng Cao Jiang, Guan, Wang lin

Anisotropic exactly solvable models in the cold atomic systems Jiang, Guan, Wang & Lin Junpeng Cao

Content Spin-1/2 bose gas IL. Spin-1 bose gas Spin-3/2 fermi gas

Content I. Spin-1/2 bose gas II. Spin-1 bose gas III. Spin-3/2 fermi gas

Anisotropic exactly solvable cold atomic model N amiltonian H a2+ Wi G x -x (pseudo-)spin Interaction symmetry ★00os0)g=c U(l)Lieb,eta,P130.605(1963 ★( fermion)go=c SU(2) Yang, PRL19.1312(1967) ★12(og SU(2)Li,EPL61.368(2003 12(boson) g1 81=c2810=0.U(1) ★1( boson) go=C,82=c SU(3)Zhou,JPA21.2391:2399(99 ■1(0m)80=-c,82=2c.SU(2)actE179300020 1( boson)800=c,g21=0.821=0 U() g 2.0 g ★1( fermion) SU(3) Sutherland. PRL2098(1968) ★32( mion)g=c,g2=c SU(4)Sutherland, PRL2098(1968) ■32(emin)8o=3c,g2=c Sp(4) Jiang, eta/, EPL87 10006(2009) 3/2( fermion)go0=0,g22=c1,g2=c2,U(1) g 0,g2.0=0,g2-1=0 Integer s(boson) 8o=-(s-1/2)c, g24.=c SO(2s +1)Jiang, et al, JPA44.345001(2011) Half-odd s(fermion)go=(S+3/2)C,82.4.=C. Sp(2s +1)Jiang et al, JPA44. 345001(2011)

Li, EPL 61. 368 (2003) Zhou, JPA 21. 2391; 2399 (1988) Anisotropic exactly solvable cold atomic model

I. Anisotropic spin-1/ 2 bose gas Anisotropic spin-exchanging interaction Motivation 1. Kondo problems: spin-1 fermions =∑。+∑(07+0+△076(x-x) Contact interaction: non-integrable Heisenberg long range interactions i e 1/r&1/r2: integrable Spin-1/2 bosons: non-integrable 2. Cold atoms: spin -12 bosons =-∑+∑ (CIi )6(x-x)+∑ i≠ 202+21+2++()0+(x=x)-

I. Anisotropic spin-1/2 bose gas Anisotropic spin-exchanging interaction Motivation 1. Kondo problems : spin-½ fermions Contact interaction: non-integrable; Heisenberg & long range interactions, i.e. 1/r & 1/r2 : integrable. 2. Cold atoms : spin-½ bosons Spin-1/2 bosons : non-integrable

Exact so|utⅰons Sab() k+ic ab kV2 b +po,o k-icI ali k-ic2 pl 1,0 ab tpab E=∑∑k-MF,K=∑∑k i=1j=1 i=lj=l N +Ic e kg-ko-ici j=1,2,…,N-M,讠=1,2 M2=(N1-N2)/2

Exact solutions

Densities distribution of quasi-momentum C1=1,c2=0.5,n=1andh=0. 0 Magnetization mz=(n1-12)/2 interaction 0.6 Spontaneous magnetization when h=o Phase transition from fully polarized 0.2 state to partially polarized state Critical points with strong repulsion hc+=m2n2-8m2n3/3c1, he=n2n2-8m2n3/3 -0.500.511.5 The critical points are different because the couplings cl and c2 are different

Densities distribution of quasi-momentum Magnetization Critical points with strong repulsion Spontaneous magnetization when h=0 Phase transition from fully polarized state to partially polarized state. The critical points are different because the couplings c1 and c2 are different. interaction

The pressures magnetization in the strong coupling limit h22 h n+ 12C1 (3n-57n2) 3n+5|+01/c2)+01/c2 12C2 h h mz= n+ 3n2 +O(1/c1)+O(1/c2) (b n=08 n=0.04 0 02 =n=0.8 n=0.6 n=0.04 10 0 510 10 0 h When the external field h is zero, the pressure takes its minimum. With the increasing h, the pressure increases. At the fully polarized state, the pressure arrives at its maximum

The pressures & magnetization in the strong coupling limit When the external field h is zero, the pressure takes its minimum. With the increasing h, the pressure increases. At the fully polarized state, the pressure arrives at its maximum

The ground state energy density 0.02 (上) 0.02 0.04 0.06 -0.08 n=1 -0.1 n=1 n=08 n=0.8 n=06 -0.12 =!=n=0.6 11n=0 -0.14 ……n=04 3 10 510 C1=100andc2=50 At the critical point, the second E=∑∑k-hM2 order derivative of the ground state i=1j=1 energy density is not continue, thus it is a second order phase transition

The ground state energy density At the critical point, the second order derivative of the ground state energy density is not continue, thus it is a second order phase transition

Entropy at finite temperature 0.1 0.6 008 05 006 04 03 004 0.2 002 0.1 -0.5 k0.5 15 C1=1,C2=0.5andn=1

Entropy at finite temperature

Dressed energy Ei(k)=TIn h()/ni(k) 0.4 (a) △E1 0.2 E2 0.8 △E2 C2=0.5 04 =1 0 2 0.5 0.5 k K/2 Strong repulsion E1(k)=k2-1-c17T32m12F1/2(u/T)+Oc12 Fermi-Dirac function (x)=T1(+1)0y/eyx+1dy

Dressed energy Strong repulsion Fermi-Dirac function