D. Sun, A. Abanov, and V. Pokrovsky Department of Physics, Texas A&M University

Molecular production at broad Feshbach resonance in cold Fermi-gas D. Sun, A. Abanov, and V. Pokrovsky Department of Physics, Texas A&M University College Station, Wednesday, Dec 5, 007

OUTLINE Alkali Atom structure, hyperfine zeeman Feshbach Resonance, broad, narrow Hamiltonian to Spin Problem. Broad Resonance Approximation Molecular Production and reverse process Conclusion

Easily controllable by Laser, Magnetic field

Atoms in Magnetic Field Hyperfine Interaction for alkali atoms Orbiting Electron L=0,S=1/; 1 1 H hf =A IiS; F=I+S = I ± ; Δ Ehf = A( I + ) Zeeman Splitting in Magnetic Field H = AIiS + CS spin C = μ B B z NATv46p537;K40@JILA; lowest lying two channels; m_f=-7/------pseudospin UP m_f=-9/------pseudospin DOWN

Ultracold, Superfluid To be simple, the system is dilute gas, otherwise solidify (except Helium, dense & strongly interacting) 14 3 1/ vu f 1/ kf a ~ 10 / C ~ 0.57 0.57 ~ 10 n cm T μe = μe nk μk U can be tuned via Feshbach Resonace T C /T F : 10-4 - conventional superconductors; 10-3 for Helium-3; 10 - for High-T C superconductors; 0.1 fermionic gas superfluid! ---------The STRONG interactions induced by Feshbach resonance

Feshbach Resonance Field B Atoms entering in the triplet potential are coupled to a singlet bound molecular state. The bare molecule state of the closed channel tunes differently with magnetic field than the open-channel threshold. This can lead to a crossing of the two levels.

Feshbach Resonance. Adiabatic, reversible conversion S wave BCS regime Isentropic process Carr et al., PRL 04 k F a = 0.5 BCS forms if T low, say T/Tf~0.0

Feshbach Resonance. Adiabatic, reversible conversion na 3,k F a = Isentropic process Carr et al., PRL 04 k F a = 0.4

Feshbach Resonance. Adiabatic, reversible conversion na 3,k F a = Molecule formed BEC forms if T low Isentropic process na 3 = 0.001 Carr et al., PRL 04 k F a = 0.4

Broad v.s. Narrow Li-6; the two lowest hyperfine states; wide/narrow at 834.G/543G. 8 Γ 8 γ = = Γ 0 = 4 m( μ BabgH w / ) π ε π k r 0 BROAD: 1 F F 0 r 8π = m g 0

S-wave two channel Hamiltonian --->Spin problem g H= (() ε t + ε ) a a + ω b b + ( ba a + b a a ) + + + + + p pσ pσ p p p q p+ q, p, q p, p+ q, p, σ p V p, q Broad Resonance Δ B ε F ε 0 ( mg / ) g n ε F Γ n 4/3 (3 π ) Single Mode Approximation ωp 0 1/3 1; p Anderson Transformation 1 S = ( n + n 1); S = a a z p p p p p p [ S, S ] = S ;[ S, S ] =± S + z z ± ±

S-wave two channel Hamiltonian Approximation made g + H = () ε tsz+ ( bs+ + bs ) V Commutes with, S + b + b, related to the total number of atoms N z Commutes with, S, related to the total spin Heisenberg Eq. ~ b= igs ; S = iε ( t) S + igb S z ; ~ + ^ 1 ^ ^ ^ ^ Sz = ( np + n p 1) Sz = ( N F NS)/ p No way to solve it analytically. Conjecture: N F N ; Sz = N / S S which turns out to be self-consistent.

Equation validated Heisenberg Eq. b+ iε ( t) b+δ b= 0;where Δ= g ns If ε( t) is linear in time, this equation becomes parabolic cylinder equation. Comparing the asymptotic solution with the Landau Zener probablity, we conclude that Δ= g n is the LZ gap. s p p model neglecting dispersion is valid: Δ ετ ( LZ) ; say, Δ= ; m m 3 ( p/ ) 3 together with n =, we have n s = nγ. 3π Γ 1 n n; N N! s S s

Estimate of Δ from experiment curve 40 K NATUREv46p537; @JILA; lowest lying two channels; Broad; m_f=-7/,m_f=-9/;b_i=0.78g, B_0=0.1G,B_f >01.54G; Δ:10 K ~10 7 6 NATv44p47; 40 K @JILA; first excited levels; m_f=-5/,m_f=- 9/;B_i=7.81G, B_0=4.1G,B_f >15G 6 5 Δ:10 K ~10 K PRLv91p00401; 6 Li @ENS,Paris; lowest lying two states; B_i=1060G,B_0=830G,B_f=650-850G;w~100G; Δ:10 K ~10 6 5 K K 1G 10 since Δ< range of sweeping, asymptotic solution of the differential equation is made practical 4 K

General Solution to the equation Heisenberg Eq. b+ iε ( t) b+δ b= 0;where Δ= g ns m 0 0 N () t = u() t N ( t ) + g v() t F ( t ) m = m 0 + 0 g F () t u() t N ( t ) g v() t F ( t ) ~ 1 ~ * * m 0 0 Ct () = ig uun ( t) + igvvf ( t) where u(t,t ) and vtt (, 0 0 with proper initial conditions, + N () t = b b () t is the number of molucules, m F ( ) ) are the two standard solutions of the same equation + t = S S ( t) is the square of BCS condensate amplitude, + C(t)= b S ( t) is the coherence factor and C(t ) = 0. 0

Discussion of the solution Conservation law: s m() () N N t + F t = const If N ( t ) = 0, then const = F ( t ), and N ( t) F ( t ) / N m 0 0 m 0 s Only fermions and no molecules N ( t) = g v( t) F ( t ), strong inital state dependence, m large F(t ) to get large conversion rate, λ BCS = F 0 v g 0 / ε, might explain different exps.

B = 8G B = 15G B = 4G 1 0 Δ:10 K ~10 6 5 K

Δ:10 K ~10 7 6 K

Discussion of the solution Conservation law: NsNm() t + F () t = const If Nm( t0) = 0, then const = F ( t0), and Nm( t) F ( t0) / Ns Only fermions and no molecules N ( t) = g v( t) F ( t ), strong inital state dependence, m large F(t ) to get large conversion rate, λ BCS = F 0 v g 0 / ε, might explain different exps. since Δ< range of sweeping, we could look for asymptotic solution N ( t = ) = F N [1 exp( πδ / ε )] m 1 0 S LZ probability multiplied by an effective number. Does not depend on atom density, contrary to other theories. nh αnhwabg / B 3 S Wabg B C. E. Wieman, 005 π ( ) / m

Discussion of the solution Conservation law: s m( ) ( ) No fermions and only molecules b b S ( ) = N /; ( + ) = N /exp( Δ / ); π ε N N = [1 exp( πδ / ε)]; N π ε S NN t + F t = const Physics: [1 exp( Δ / )] cooper pairs equally distributed between N S available states with probability w= [1 exp( πδ / ε )]. NS S = N w S 1/3 Compact size rpair = n quasimolecule p s Runaway pairs with opposite momenta but parallel spin. N

Conclusion we have considered molecule formation and dissociation in a cooled Fermi-gas when the magnetic field is swept across the broad Feshbach resonance. We derived the equation in broad resonance approximation and discussed the asymptotic solutions. Experiment verifiable predictions: Molecular production strongly depends on the initial value of magnetic field. Coefficient in the LZ exponent for the molecular and the BCS condensate productions is independent of initial density of atoms (molecules).

Thanks! The End!

Landau Zener Problem Δ= g n s

//not slide//question for me: To get large condensation, start with small detuning. Which means larger a, more condensation. Read everywhere BCS limit requires k a <<1.. What does this mean?

Fermions Many-body aspects (BEC-BCS crossover) BEC regime unitary limit BCS regime

E B =-h /ma a 0 a = 00 scattering length [nm] 100 0-100 -00 0,0 0,5 1,0 1,5,0 Magnetic field [kg] Lithium 6 Feshbach resonance In In F=1/ manifold a 0