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‘Baryonic’ bound-state instability in trapped fermionic atoms A. S. Stepanenko and J. M. F. Gunn School of Physics and Astronomy, University of Birmingham, Edgbaston, Birmingham B15 2TT, United Kingdom. (February 1, 2008) 9 9 9 1 n Abstract a J 7 We consider a homogeneous gas of spin-S fermionic atoms, as might occur 2 near the center of an optical trap. In the case where all scattering lengths 1 v 7 are negative and of the same magnitude we demonstrate the instability of 1 3 the Fermi sea to the condensation of bound ‘baryonic’ composites containing 1 0 9 2S +1 atoms. The gap in the excitation spectrum is calculated. 9 / t a m I. INTRODUCTION - d n o Sympathetic cooling of trapped fermionic gases [1] is likely to lead to the creation of c : v degenerate Fermi gases in the near future. In magnetic traps, the fermion’s spin is locked to i X the field direction. However the advent [2,3] of optical traps holds the promise of degenerate r a Fermi gases with unquenched spins and hence a variety of nonideal Fermi gases. Already there has been some discussion of the possibility of pairing [4–6] in the presence of attractive interactions, which may be of an exotic type [7,8]. In part the exotic possibilities occur because alkali fermions exist which have larger total spins than spin 1/2: 22Na, 86Rb, 132Cs, 134Cs, 136Cs have S = 5/2, 5/2, 3/2, 7/2, 9/2 respectively [9]. In this paper we will point out that in addition to pairing there are other possible ground states, in particular when the atoms experience spin-independent attractive interactions. One-dimensional models of interacting fermions with spins greater than one half have been considered for a substantial time, both for repulsive [10] and attractive [11–13] inter- 1 actions. In the attractive case, the ground state has been found to contain bound states of fermionswithupto2S+1constituents (withequality intheabsenceofanapplied‘magnetic’ field which would distinguish the different spin states). One may rationalise these results by noting that the Pauli principle does not militate against binding extra fermions until a spin state must be doubly occupied, and hence extra nodes in the spatial wave function occur. In the light of these results we will examine whether the ground state of the three-dimensional homogeneous S > 1/2 weakly interacting attractive Fermi gas is characterised by a condensation of composites of 2S + 1 fermions. The possible condensation of alpha particles as four-particle composites in nuclear matter and at the surface of nuclei has a long history whose relation to the current work we will discuss at the end of this paper. II. BARYONIC GROUND STATE ENERGY In three dimensions we use a variational approach similar to that used in the original BCS paper [14]. We will assume that the composites (‘baryons’) have zero centre of mass momentum (as in BCS) and are also total spin singlets, as in the one-dimensional results. (The latter can be understood physically as providing the lowest kinetic energy associated with relative motion in the bound state.) The condensed state, with all baryons having centre of mass momentum zero, is of the form (here n = 2S +1): N B = ϕ(k , ,k )c† c† | i  1 ··· n k1,σ1 ··· kn,σn {Xki}   where all the spin states in the baryon, σ , (with i = 1, ,n) are different. ϕ(k , ,k ) i 1 n ··· ··· is the Fourier transform of the completely symmetric relative wavefunction ϕ(r , ,r ). 1 n ··· Unlike the BCS case, there are several grand canonical states, BCS , corresponding to B . | i | i We will construct the simplest one. This family of variational states allows us to use 1/n as a small parameter. (Indeed if the coupling between the atoms were strong, there would be similarities with work [15] on the structure of baryons in QCD using a 1/N expansion.) 2 Consider the following (non-normalized) ground state: ψ˜ = exp ϕ(k ,...,k )c†(k )...c†(k ) 0 | bi  1 n 1 1 n n | i k1X,...,kn   Below we show that the corresponding normalized state has the form ψ = e(n−2)Nb/2 un−1(k) exp ϕ (k ,...,k )c†(k )...c†(k ) 0 | bi ·  α 1 n 1 1 n n | i Yk k1X,...,kn   n = e(n−2)Nb/2 u(k )+v(k ,...,k )c†(k )...c†(k ) 0 (1)  i 1 n 1 1 n n | i k1+..Y.+kn=0iYi6==α1    wherec†(k),c (k),α = 1,...,n = 2S+1thecreationandannihilationoperatorsofafermion α α with α-th projection of the spin and momentum k, N N/n is the number of baryons b ≡ (with N being the number of atoms) and v(k ,...,k )u(k ) 1 n α ϕ (k ,...,k ) δ (2) α 1 n ≡ u(k ) ... u(k ) ∝ k1+...+kn,0 1 n · · The Kronecker symbol on the r.h.s. of (2) means that we consider baryonic states with total momentum equal to zero. The expression k + ...+k = 0 as a subscript to the product 1 n in (1) implies that the product is taken over all sets k ,...,k with the total momentum 1 n { } equal to zero. We will apply a ‘normalization’ condition similar to that used in BCS states: u(k) 2+ v(k) 2 = 1 , v(k) 2 v(k,k ,...,k ) 2 (3) 2 n | | | | | | ≡ | | k2X,...,kn This, unlike the BCS one, does not imply that the total state is normalised. The vector conjugate to ψ is b | i ψ = e(n−2)Nb/2 u¯n−1(k) 0 exp ϕ¯ (k ,...,k )c (k )...c (k ) (4) b α 1 n n n 1 1 h | ·h |   Yk k1X,...,kn   To calculate the normalization ψ ψ we make use of the following identity b b h | i exp ϕ (k ,...,k )c†(k )...c†(k )  α 1 n 1 1 n n  k1X,...,kn  n−1  = Dξ¯Dξ exp ξ¯(k)ξ (k)+c†(k)ξ (k) " − i i i i Z Xi=1 Xk n o ϕ (k ,...,k )c†(k )ξ¯(k )...ξ¯ (k ) (5) − α 1 n n n 1 1 n−1 n−1 # k1X,...,kn 3 ¯ where ξ (k),ξ (k) are Grassmann variables and the measure of integration is denoted by i i n−1 ¯ ¯ DξDξ dξ (k)dξ (k) i i ≡ i=1 k Y Y As c† enters linearly in the exponentials in (5) then the operator averaging in ψ ψ can b b h | i be easily fulfilled and we find n−1 ¯ ¯ ψ ψ = C DξDξ exp ξ (k)ξ (k) b b i i h | i "− Z i=1 k XX ¯ ¯ ϕ¯ (k ,...,k )ϕ (p ,...,p )ξ (p )ξ (k )...ξ (p )ξ (k )δ − α 1 n α 1 n 1 1 1 1 n−1 n−1 n−1 n−1 kn,pn# k1X,...,kn p1,...,pn where C denotes the factor C = e(n−2)Nb u(k) n−1 | | k Y For the second term in the exponential we make use of the following identity: ¯ ¯ exp ϕ¯ (k ,...,k )ϕ (p ,...,p )ξ (p )ξ (k )...ξ (p )ξ (k )δ "− n 1 n n 1 n 1 1 1 1 n−1 n−1 n−1 n−1 kn,pn# k1X,...,kn p1,...,pn n−1 n−1 ¯ = Dσ¯Dσ exp σ¯ (k,p)σ (k,p) σ¯ (k,p)ξ (k)ξ (p) i i i i i "− − Z i=1 k,p i=1 k,p XX XX n−1 + ϕ¯ (k ,...,k )ϕ (p ,...,p ) σ (p ,k )δ (6) n 1 n n 1 n i i i kn,pn # k1X,...,kn iY=1 p1,...,pn where n−1 Dσ¯Dσ dσ¯ (k,p)dσ (k,p) i i ≡ i=1 k,p Y Y The meaning of σ-fields is not clear at this point, but we will discuss this presently. Integrating over the Grassmann fields we obtain n−1 n−1 ψ ψ = C Dσ¯Dσ exp σ¯ (k,p)σ (k,p)+ trlog[1+σ¯ ]+K b b i i i h | i "− # Z i=1 k,p i=1 XX X where K denotes the following expression n−1 K = ϕ¯ (k ,...,k )ϕ (p ,...,p ) σ (p ,k )δ n 1 n n 1 n i i i kn,pn k1X,...,kn iY=1 p1,...,pn 4 We calculate the integral over σ¯,σ using the saddle point method, where n is the large parameter. We assume that the symmetry of the exponential over σ -fields is not broken i and put σ = σ,σ¯ = σ¯, so the exponent becomes i i (n 1)σ¯σ +(n 1)trlog[1+σ¯]+K − − − The saddle point equations are σ(k,p) = (1+σ¯)−1(k,p) (7) n−1 σ¯(k,p) = ϕ¯ (k,k ,...,k )ϕ (p,p ,...,p ) σ(p ,k )δ (8) n 2 n n 2 n i i kn,pn k2X,...,kn iY=2 p2,...,pn In the limit n we expect a mean field approximation to be valid for the description of → ∞ the composites: n v(k ,...,k ) = B δ v(k ) (9) 1 n · k1+...+kn,0 i i=1 Y where B is chosen to satisfy the normalization condition (3). With this assumed form for v, it can be shown that the following anzatz can serve as a quite general solution of the saddle point equations: σ(k,p) = δ s (k)+c s¯ (k)s (p) , σ¯(k,p) = δ s (k)+c s¯ (k)s (p) . (10) k,p 1 2 2 2 k,p 3 4 4 4 with functions s ,i = 1,2,3,4 to be determined. If we further restrict ourselves by consid- i ering weak coupling case we obtain that the solution is (see Appendix for details) 2 v(k) σ(k,p) = δ u(k) 2 , σ¯(k,p) = δ (11) k,p k,p | | (cid:12)u(k)(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) Hence, (cid:12) (cid:12) ψ ψ = Cexp (n 1) v(k) 2 (n 1) log u(k) 2+ v(k,k ,...,k ) 2 b b 2 n h | i "− − | | − − | | | | # Xk Xk Xk k2X,...,kn Because of the normalization condition (3) and expression for C we finally obtain ψ ψ = 1+O(1/n) (12) b b h | i 5 Let us consider the following average: ψ c (k)c†(k′) ψ h b| α α | bi Repeating all the steps we have made before we can obtain the following result: ψ c (k)c†(k′) ψ = C Dσ¯Dσ (1+σ¯ )−1(k,k′) h b| α α | bi α Z n−1 n−1 exp σ¯ (k,p)σ (k,p)+ trlog[1+σ¯ ]+K i i i × "− # i=1 k,p i=1 XX X Calculating again the integral over σ-fields using the saddle point method and noting that the contribution of the first multiplier in the integrand to the saddle point equations can be neglected we obtain from (7): ψ c (k)c†(k′) ψ = σ(k,k′) h b| α α | bi So the σ-field may be interpreted as momentum distribution of the holes. In an analogous way we can obtain the following averages: ψ c†(k)c (k) ψ = v(k) 2+O(1/n) (13) h b| α α | bi | | ψ c†(k)c†( k +q)c ( k′ +q)c (k′) ψ h b| α β − β − α | bi q X = δ v(k) 2 v(q) 2+v¯(k)u(k)u¯(k′)v(k′)f(kˆ,kˆ′)+O(1/n) (14) k,k′ | | | | q X where kˆ k and the function f has the form ≡ |k| 3x x3 θ f(kˆ,kˆ′) f(x) = 1 + , x sin (15) ≡ − 2 2 ≡ 2 with θ being the angle between vectors k,k′. We are now in a position to consider a general Hamiltonian of the form: 1 Hˆ = ǫ(k)c†(k)c (k)+ V(k,k′)c†(k)c†( k +q)c ( k′ +q)c (k′) (16) α α 2Ω α β − β − α Xk,α kX,k′,q α,β where 6 k2 ǫ(k) = µ (17) 2m − and µ is the chemical potential. Then, using the results above for expectation values, we find the following expression for the ground state energy V(0)N2 n2 E = ψ Hˆ ψ = n ǫ(k) v(k) 2+ + V(k,k′)v¯(k)u(k)u¯(k′)v(k′)f(kˆ,kˆ′) b b b h | | i | | 2Ω 2Ω k k,k′ X X (18) with the condition ψ Nˆ ψ = n v(k) 2 = N (19) b b h | | i | | k X Note that if we consider an interaction potential depending only on the modulus of momentum: V(k,k′) = V( k , k′ ), and look for a solution for u,v also depending only on | | | | the modulus of momentum (that is in absence of a spontaneous breaking of the rotational symmetry) then the function f in (18) can be replaced by its average value: 1 3 dx f(x) = 8 Z0 and we have V(0)N2 n2 E = ψ Hˆ ψ = n ǫ(k) v(k) 2 + + V˜(k,k′)v¯(k)u(k)u¯(k′)v(k′) (20) b b b h | | i | | 2Ω 2Ω k k,k′ X X where V˜ = 3V. 8 The normalization condition, (3), allows the introduction of the following parameteriza- tion for u,v, by analogy with the usual procedure for the BCS case: v(k) = cosθ(k) , u(k) = sinθ(k) (21) and we obtain for the ground state energy V(0)N2 n2 E = n ǫ(k)cos2θ(k)+ + V(k,k′)cosθ(k)sinθ(k)cosθ(k′)sinθ(k′) (22) b 2Ω 2Ω k k,k′ X X We now minimize E with respect to θ(k) to find b 7 1 n tan[2θ(k)] = dk′ V˜(k,k′)sin[2θ(k′)] (23) 2ǫ(k)(2π)3 Z Introducing the following notation ∆(k) = n dk′ V˜(k,k′)sin[2θ(k′)] , E(k) = ǫ2(k)+∆2(k) 1/2 (24) −2(2π)3 Z h i such that ∆(k) ∆(k) ǫ(k) tan[2θ(k)] = , sin[2θ(k)] = , cos[2θ(k)] = (25) − ǫ(k) E(k) −E(k) we obtain the following equation for the gap ∆(k): n ∆(k′) ∆(k) = dk′ V˜(k,k′) (26) −2(2π)3 E(k′) Z and the chemical potential can be defined from n dk′ cos2θ(k) = ρ (27) (2π)3 Z Following Anderson and Morel [16], Eq.(24) can be rewritten in terms of an effective potential, U (k,k′): ξ 1 ∆(k′) ∆(k) = dk′ U (k,k′) , (28) −(2π)3 Z|ǫ(k′)|<ξ ξ 2E(k′) where ξ is some cut-off such that ξ µ kF2 and U (k,k′) satisfies the equation: ≪ ≡ 2m ξ 1 1 U (k,k′) = nV˜(k,k′) dq nV˜(k,q)U (q,k′) . (29) ξ − (2π)3 ξ 2 ǫ(q) Z|ǫ(q)|>ξ | | We take V(k,k′) to have the following separable (energy dependent) form: 4πa V(k,k′) = VΘ(µ ǫ(k) )Θ(µ ǫ(k′) ) , V = (30) −| | −| | m which, as we shall see, is consistent with low-energy approximation to the T-matrix. Here a is the scattering length (we put h¯ = 1), Θ(x) the step function. We assume that the interaction is weak (which is a good approximation experimentally) which means k a 1 (31) F ≪ 8 To check that the form of interaction (30) is a consistent low-energy approximation consider the equation for the T-matrix T(k,k′,z) of zero energy, z = 0 : 1 m T(k,k′,0) = V(k,k′) dq V(k,q)T(q,k′,0) (32) − (2π)3 q2 Z|ǫ(q)|>ξ Making the anzatz T(k,k′,0) = T Θ(µ ǫ(k) )Θ(µ ǫ(k′) ) 0 −| | −| | we obtain the following relation for T ,V: 0 1 1 mk = +√2 , ρ = F ρ T ρ V 0 2π2 0 0 0 As 1 π = 1 ρ V 2ak ≫ 0 F then T V (33) 0 ≈ Let us define U (k,k′). Assuming the form ξ U (k,k′) = U Θ(µ ǫ(k) )Θ(µ ǫ(k′) ) ξ ξ −| | −| | we get the following equation for U : ξ nV˜U q2 U = nV˜ ξ dq ξ − 4π2 ǫ(q) ξ<|ǫ(Zq)|<µ | | After some algebra we arrive at the relation 1 1 ξ = +log(√2+1) log (34) ρ U ρ nV˜ − 4µ 0 ξ 0 If we assume again that we can drop the second term in the right hand side (because the coupling is weak) and noting that V˜,U < 0 we finally obtain ξ 1 1 ξ = +log (35) ρ U ρ n V˜ 4µ 0| ξ| 0 | | Note that the energy can be expressed in terms of (24): 9 nΩ ǫ2(k) ∆2(k) E = dk ǫ(k) (36) b 2(2π)3 " − E(k) − 2E(k)# Z From (28) then we obtain 1 1 ∆ = 2ξexp = 8µexp (37) "−ρ0|Uξ|# "−ρ0n|V˜|# The energy of the normal state per unit volume can be written as E n n = dk ǫ(k) (38) Ω (2π)3 Z|k|<kF Hence, from (36,37) we find that the difference of the baryonic and the normal ground state energies is: ∆E E E 1 N mk b b n F − = 48ρµexp , ρ , ρ = (39) Ω ≡ Ω − "−ρ n V˜ # ≡ Ω 0 2π2 0 | | with N the number of atoms in the system. We conclude this section with a note about weakening the assumption of equality of scattering lengths in all channels. If the scattering lengths are different then the interaction will take the form [17]: [f] V = V (S S )n n 1 2 · n=0 X where [f]means theinteger partof f. In thecase where thescattering lengths inallchannels are equal, a = a, F, we have V = 4πa,V = 0,n = 0. If the scattering lengths are slightly F ∀ 0 m n 6 different then the terms in the interaction with V ,n = 0 are small and can be neglected n 6 but V is equal to some weighted average over the scattering lengths and the treatment in 0 this section will still be approximately valid. III. BCS GROUND STATE ENERGY We will now show that the conventional BCS ground state has an energy that is higher than the baryonic one, at least if we restrict ourselves to s-wave pairing. The BCS ground state has the form [18]: 10