Three fermions in a box at the unitary limit: universality in a lattice model

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Three fermions in a box at the unitary limit:

universality in a lattice model

Ludovic Pricoupenko, Yvan Castin

To cite this version:

Ludovic Pricoupenko, Yvan Castin. Three fermions in a box at the unitary limit: universality in a

lattice model. Journal of Physics A General Physics (1968-1972), Institute of Physics (IOP), 2007,

pp.12863. �hal-00145589v2�

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hal-00145589, version 2 - 26 Sep 2007

L. Pricoupenko¹ and Y. Castin²

1Laboratoire de Physique Théorique de la Matière Condensée, Université Pierre et Marie Curie, case courier 121, 4 place Jussieu, 75252 Paris Cedex 05, France.

2Laboratoire Kastler Brossel, Ecole normale sup´erieure, UPMC, CNRS, 24 rue Lhomond, 75231 Paris Cedex 05, France.

(Dated: September 26, 2007)

We consider three fermions with two spin components interacting on a lattice model with an infinite scattering length. Low lying eigenenergies in a cubic box with periodic boundary conditions, and for a zero total momentum, are calculated numerically for decreasing values of the lattice period. The results are compared to the predictions of the zero range Bethe-Peierls model in continuous space, where the interaction is replaced by contact conditions. The numerical computation, combined with analytical arguments, shows the absence of negative energy solution, and a rapid convergence of the lattice model towards the Bethe-Peierls model for a vanishing lattice period.

This establishes for this system the universality of the zero interaction range limit.

PACS numbers: 03. 75. Ss, 05. 30. Fk, 21. 45. +v

Recent experimental progress has allowed to prepare a two-component Fermi atomic gas in the BEC-BCS crossover regime and to study in the lab many of its physical properties, such as the equation of state of the gas and other thermodynamic properties, the fraction of condensed particles, the gap in the excitation spectrum corresponding to the breaking of a pair, the superfluid properties and the formation of a vortex lattice, the effect of a population imbalance in the two spin-components and the corresponding possible quantum phases, . . . [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16].

The key to this impressive sequence of experimental results is the use of Feshbach resonances [17]: an external magnetic field (B) permits to tune the two-bodys-wave scattering length a almost at will, to positive or negative values, so that one can e.g. adiabatically transform a weakly attractive Fermi gas into a Bose condensate of molecules. Interestingly, close to the resonance, the scattering length diverges asa∝ −1/(B−B0) so that the infinite scattering length regime (|a|=∞) can be achieved.

When the typical relative momentum k of the particles further satisfies kb≪1, k|re| ≪1, where b is the range andrethe effective range of the interaction potential, the s-wave scattering amplitude between two particles takes the maximal modulus value fk = −1/(ik): This is the so-called unitary regime, where the gas is strongly, and presumably maximally, interacting.

The unitary regime is achieved in present experiments for broad Feshbach resonances, that is for resonances where the effective rangereis of the order of the Van der Waals range of the interatomic forces [18, 19]. Examples ofs-wave broad resonances are given for⁶Li atoms by the one at B0 ≃830 G [1, 4, 5, 7] or also for ⁴⁰K atoms at B0≃200 G [3]. On a more theoretical point of view, the unitary regime has the striking property of being universal: E.g., the zero temperature equation of state involves only~, the atomic massm, the atomic density and a numerical constant independent of the atomic species; this was checked experimentally, this also appears in fixed

node Monte Carlo simulations [20, 21] and more recently in exact Quantum Monte Carlo calculations [22, 23, 24].

In Refs.[22, 24], exact Quantum Monte Carlo simulations at the unitary regime are performed using a Hub- bard model. From the condensed matter physics point of view, this modelling of the system is a clever way to avoid the fermionic sign problem. But it is more than a theoretical trick in the case of ultra-cold atoms since it can be achieved experimentally by trapping atoms at the nodes of an optical lattice in the tight-binding regime [25]. The Bethe-Peierls zero range model is another commonly used way of modelling the unitary regime: pairwise interac- tions are replaced by contact conditions imposed on the many-body wave function [26, 27, 28, 29, 30, 31, 32, 33].

This model is very well adapted to analytical calculations in few-body problems [27, 32] but can also be useful to predict many-body properties like time-dependent scaling solution [34], the link between short range scaling properties and the energy of the trapped gas [31], and hidden symmetry properties [35] of the trapped gas.

However, there is to our knowledge no general rigorous result concerning the equivalence between the discrete (Hubbard model) and the continuous (Bethe-Peierls) models for the unitary gas. As a crucial example, one may wonder if there is any few- or many-body bound state in a discrete model at the infinite scattering length limit. This is a non-trivial question, since the infinite scattering length corresponds to an attractive on-site interaction in the discrete model.

In this paper, we address this question for two and three fermions in a cubic box with periodic boundary conditions, when the interaction range tends to zero for a fixed infinite value of the scattering length. Our results for the equivalence of the lattice model and the Bethe- Peierls approach are analytical for two fermions but still rely on a numerical step for three fermions. In this few body problem, the grid spacing can however be made very small in comparison to the grids currently used in Quantum Monte Carlo many-body calculations, thus al-

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2 lowing a more precise study of the zero lattice step limit

and a test of the linear scaling of thermodynamic quanti- ties with the grid spacing used in [22]. Our computations also exemplify the remarkable property that short range physics of the binary interaction does not play any signif- icant role in the unitary two-component Fermi gas, and the fact that the Bethe-Peierls model is well behaved for three equal mass fermions.

Our model is the lattice model used in the Quantum Monte Carlo simulations of [24]. It has already been de- scribed in details in Refs. [36, 37] so that we recall here only its main features. The positionsr_i of each particlei are discretized on a cubic lattice of periodb. The Hamil- tonian contains the kinetic term of each particle,p²/2m, such that the plane wave of wave vectorkhas an energy

ǫk= ~²k²

2m . (1)

Here the wave vector is restricted to the first Brillouin zone of the lattice :

k∈ D ≡[−π/b, π/b[³. (2) We enclose the system in a cubic box of size L with periodic boundary conditions, so that the components {kα}α∈{x,y,z} of k are integer multiples of 2π/L. In what follows we shall for convenience restrict our computations to the case where the ratio L/b= 2N+ 1 is an odd integer, so that kα= 2πnα/L withnα∈ {−N,−N+ 1, . . . , N}. The Hamiltonian also contains the interaction potential between opposite spin fermionsiandj, which is a discrete delta on the lattice:

V(ri,r_j) =g0

b³δr_i,rj. (3) In [37] the matrix elements of the two-body T-matrix hk|T(E+i0⁺)|k^′i for an infinite box size are shown to depend only on the energy E, not on the plane wave momenta, which would imply in a continuous space a pures-wave scattering. The bare coupling constantg0 is then adjusted in order to reproduce in the zero energy limit the desired value of thes-wave scattering length a between two opposite spin particles [33, 36, 37]:

1 g0 −1

g =− Z

D

d³k (2π)³

1 2ǫk

=− mK

4π~²b, (4) where

K=12 π

Z π/4 0

dθ ln(1 + 1/cos²θ) = 2.442749. . . (5) may be expressed in terms of the dilog function, and g= 4π~²a/mis the usual effectives-wave coupling constant. From the calculated energy dependence of theT- matrix, one may also extract the effective rangereof the interaction in the lattice model; re is found to be pro- portional to the lattice period,re≃0.337b[33], and the

limitb→0 is equivalent to the limit of both zero range and zero effective range for the interaction [38]. As men- tioned in the introduction, this is the desired situation to reach the unitary limit when|a|=∞.

We first solve the problem for two opposite spin fermions in the box, in the singlet spin state|si= (| ↑↓

i − | ↓↑i)/√

(2ǫk−E)hk,−k|φi+ g0

L^3/2hr,r|φi= 0, (6) where the last term does not depend on a common positionr of the two particles. A first type of eigenstates corresponds tohr,r|φi= 0: these eigenstates have a zero probability to have two particles at the same point, and are also eigenstates of the non-interacting case. An example of such a state with the correct exchange symmetry is given by the wavefunction:

φ(r1,r2)∝cos 2π

L(x1−x2)

−cos 2π

L(y1−y2)

. (7) We are interested here in the states of the second type, what we call ‘interacting’ states, such that hr,r|φi 6= 0.

Treating the interacting term in Eq.(6) as a source term, one expresses|φiin terms of hr,r|φiand a sum over k.

Projecting the resulting expression onto|r,rileads to a closed equation (nowE6= 2ǫk):

1 g0

+ 1 L³

X

k∈D

1

2ǫk−E = 0. (8)

The resulting implicit equation forE, of the formu(E) = 0, whereu(E) is the left hand side of Eq.(8), is then read- ily solved numerically; to this end, one notes thatu(E) has poles in eachE= 2ǫk, and that it varies monotoni- cally from−∞to +∞ between two poles, so thatu(E) vanishes once and only once between two successive values of 2ǫk. In Fig.1, we show for|a|=∞the first low lying eigenenergies as functions of the lattice spacing; one observes a convergence to finite values in the b/L→0 limit, with a first correction scaling asb/L. A rewriting of the implicit equation forE that will reveal convenient in theb= 0 limit is:

πL

a = (2π~)² mL²

"

1

E + X

k∈D−0

1

E−2ǫk + 1 2ǫk

#

+C(b), (9) where the functionC(b) is defined by:

C(b) = (2π~)²L 2m

Z

D

d³k (2π)³

1 ǫk − 1

L³ X

k∈D−0

1 ǫk

! , (10)

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2.8 2.85 2.9 2.95 3

E [E 0]

0.9 0.95 1 1.05

E [E 0]

0 0.05 0.1

b/L

-0.2 -0.195 -0.19

E [E 0]

FIG. 1: First three eigenenergies for the interacting states of two fermions in a box of size L for an infinite scattering length in the lattice model, as functions of the lattice periodb.

The total momentum of the eigenstates is fixed to zero. The computed eigenenergies are given by the plotting symbols, in units ofE0 = (2π~)²/2mL²; the straight lines are linear fits performed on the data withb/L <2×10⁻².

and has a finite limit forb→0 which is given byC(0)≃ 8.91364.

We now briefly check that the b = 0 limit in Eq.(9) coincides with the prediction of the Bethe-Peierls model, which is a continuous space model where one replaces the interaction potential by the following contact conditions on the wavefunction [26, 27, 28, 29, 30, 31, 32, 33]: there exists a functionS(R) such that,

φ(r1,r₂) =S(R) 1

r−1 a

+O(r), (11) where r=|r₁−r₂| →0 is the distance between the two particles and the center of mass positionR= (r1+r2)/2 is fixed. At positionsr16=r2, the wavefunction solves the free Schr¨odinger equation. Using this model we arrive at an implicit equation for the energy of an interacting state exactly of the form obtained by taking theb= 0 limit in Eq.(9), except that the constantC(0) in the right hand side is replaced by [40]:

CBP = lim

σ→0

Z

d³ue^−u²^σ² u² − X

n∈Z³^∗

e⁻ⁿ²^σ² n²

!

. (12)

We expect the identityCBP=C(0) from the general result that the Bethe-Peierls model for the two-body problem reproduces the zero range limit of a true interaction potential [28, 41]. It is however instructive to check this property explicitly for the lattice model. One can show that:

CBP−C(0) = lim

σ→0

X

n∈Z³^∗

Z

I

d³u[hσ(n+u)−hσ(n)]

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where hσ(q) = [exp(−q²σ²)−1]/q² and the integration domain isI= [−1/2,1/2]³. The desired identityC(0) = CBPresults from the fact that one can exchange theσ= 0 limit and the summation over n in the above equation [42, 43].

In the lattice model, it is possible to show analytically that the spectrum of the two-body problem for an infinite scattering length is bounded from below in the b→0 limit. Sinceg0<0 for|a|=∞, there exists at least one non-positive energy solution, by a variational argument.

One then notes that the right hand side in Eq.(9) is a strictly decreasing function ofEover ]− ∞,0[, that tends to−∞in E = 0⁻, so that at most one negative energy solution may exist. Furthermore one can show that the b→0 limit of the right hand side tends to +∞ when E→ −∞ [44], whence this negative energy solution is finite [45].

We now turn to the problem of three interacting fermions in the box. Schr¨odinger’s equation is obtained without loss of generality by considering the particular spin component (1:↑; 2:↑; 3:↓), so that the interaction takes place only among the pairs (1,3) and (2,3), and in the lattice model one obtains:

" ₃ X

i=1

p²_i 2m +g0

b³(δr1,r3+δr2,r3)−E

#

ψ(r1,r₂,r₃) = 0.

(14) We restrict to a zero total momentum modulo 2π/balong each direction [39]; using the fermionic antisymmetry condition for the transposition of particles 1 and 2, we express the part of Eq.(14) involving the interaction in terms of a function of the position of a single particle:

ψ(r1,r₂,r₁) = f(r2−r₁) (15) ψ(r₁,r₂,r₂) = −f(r₁−r₂). (16) We then project Eq.(14) on the plane waves in the box, which leads to:

hk1,k2,k3|ψi= g0δ_k^mod₁_+k₂_+k₃_,0 E−ǫk1−ǫk2−ǫk3

(fk2−fk1) (17) where δ^mod is a discrete delta modulo 2π/b along each direction, and where the Fourier transform off(r) is defined as:

fk=hk|fi= b³ L^3/2

X

r∈[0,L[³

exp (−ik·r)f(r). (18) Replacing f(r) in the right-hand side of this equation by its expression in terms ofhk₁,k₂,k₃|ψideduced from Eq.(15), we obtain a closed equation forfk:

L³

g0fk=fk

X

q∈D

ak,q−X

q∈D

ak,qfq (19) where we have introduced the matrix:

ak,q= 1

E−ǫk−ǫq−ǫ[k+q]FBZ

, (20)

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4 and for an arbitrary wavevector u, [u]FBZ denotes the

vector in the first Brillouin zone that differs fromu by integer multiples of 2π/balong each direction. The eigenvalues E of the three-body problem are such that the linear system (19) admits a non-identically vanishing so- lutionfk, that is the determinant of this linear system is zero. Note that from Eq.(19), one hasf(0)∝P

q∈Dfk= 0, a consequence of Pauli exclusion principle.

For|a|=∞, we have computed numerically the first eigenenergies of the system, by calculating the determinant as a function ofE. In Fig.2 we give these eigenenergies as functions of the ratiob/L. A rapid convergence in the zero-b limit is observed, with a linear dependence inb/L.

This rapid convergence illustrates the fact that equal mass fermions easily exhibit universal properties, as re- vealed by experiments; here b plays the role of the finite Van der Waals range of the true potential [given by (mC6/~²)^1/4, whereC6is the Van der Waals coefficient], and L is of the order of the mean interparticle distance in a real gas. As an example, for⁶Li atomsb∼3 nm and in experiments for the broad Feshbach resonance in the s-wave channel at∼830 G the atomic density is of the order of 10¹³ cm⁻³, so that the ratiob/Lis of the order of 10⁻²which is well within the zero-blimit.

The absence of negative three-body eigenenergies in the unitary limit can be obtained numerically very effi- ciently through a formal analogy between Eq.(19) and a set of rate equations on fictitious occupation numbers of the single particle modes in the box. AssumingE≤0, we note Πk the fictitious occupation number in the mode k and Γk→q=g0aq,k/L³the transition rate from the mode kto the modeq. From Eq.(20), one obtains the property Γk→q= Γq→k, and the rate equation can be written as:

dΠk

dt =−



 X

q6=k

Γk→q



Πk+X

q6=k

Γq→kΠq. (21) The symmetric matrixM(E), which defines the first order linear system in Eq.(21),dΠ/dt=M(E)Π, has the following properties: 1) its eigenvalues are non-positive, since it is a set of rate equations; 2) its eigenvalues are decreasing function of the energy E, which can be de- duced from the fact that dM(E)/dE is a matrix of rate equations and obeys property 1), and from the Hellman- Feynman theorem; and 3) eigenmodes of Eq.(21) with an eigenvalue equal to −1 correspond to solutionsfk of Eq.(19) with Πk=fkexp(−t). Therefore, in order to check that there is no non-zero solution of Eq.(19) for E < 0, it is sufficient to check that all eigenvalues of M(E= 0) are strictly larger than−1.

We have computed the lowest eigenvalue m0 of the matrixM(E = 0) as a function of the ratio b/L. A fit of m0 as a function of b/L suggests limb→0m0 ≃ −1.

To better see what happens in the zero b/L limit, we note that havingm0>−1 is equivalent to having (m0+ 1)/g0<0, or more simply (m0+ 1)/(b/L)>0. We have thus plotted in Fig.3 the ratio (m0+ 1)/(b/L), which is

0 0.05 0.1

b/L

0.75 0.8 1.44 1.46 2.48 2.52

E [ E0 ]

2.89 2.895 3.15 3.2

1.72 1.73 2.64 2.66

FIG. 2: First eigenenergies of three fermions in a box of size L for an infinite scattering length in the lattice model, for a zero total momentum. The computed eigenenergies (dia- monds) are given in units ofE0= (2π~)²/2mL² for different values of the lattice periodb. For functionsf(r) invariant by reflection alongx, y, zand by arbitrary permutation ofx, y, z we have computed the eigenenergies down to smaller values ofb/L. The straight lines are a linear fit performed on the data over the rangeb/L≤1/15, except for the energy branch E≃2.89E0 which becomes more slowly linear than the other branches. The eigenenergies predicted by the Bethe-Peierls model are given by stars inb= 0.

seen to tend to a positive value forb→0,≃1.085, with a negative slope; this excludes the existence of negative eigenenergies for the three fermions at infinite scattering length even in the smallblimit [46].

In a last step, we compare the results of the lattice model to the predictions of the Bethe-Peierls approach for three fermions in a continuous space, which was shown to be a successful model in free space [29, 30] and in a har- monic trap at the unitary limit [32]. For this purpose, we introduce the functionFkwhich is the Fourier transform of the regular part of the wave function as|r₁−r₃| →0:

F(R) = lim

r→0

hrψ R+r

2,0,R−r 2

i, (22) where we have used the translational invariance. By re- producing a calculation procedure analogous to what we have done for the lattice model, we obtain the following

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0 0.05 0.1

b/L

1.04 1.05 1.06 1.07 1.08

(m

0

+1)/(b/L)

FIG. 3: Quantity (m0+ 1)/(b/L) as a function of the lattice periodb. Herem0is the lowest eigenvalue of the matrixM(E) defining the linear system Eq.(21), for E = 0 and for an infinite scattering length. The fact thatm0+1>0 shows that there is no negative eigenenergy for the three fermions, see text. The symbols are obtained from a numerical calculation of m0. The solid line is a linear fit over the range b/L ≤ 1/29, not including the point withb/L= 1/81: for this point, the matrixM has more than half a million lines so thatm0

was obtained by a computer memory-saving iterative method rather than by a direct diagonalisation.

infinite dimension linear system:

L³

g Fk = Fk



Ak,0+X

q6=0

Ak,q+ 1 2ǫq

+mL²CBP

(2π~)²





−X

q

Ak,qFq , (23)

where the wavevectorsk andq now run over the whole space (2π/L)Z³, and:

Ak,q= 1

E−ǫk−ǫq−ǫk+q . (24)

The similarity between the structure of (19) and (23) is apparent. Numerically, at|a|=∞, we have verified the convergence between the two models asb→0 in Eq.(19), see Fig.2. Analytically, one can even formally check the equivalence between the two sets of equations (19) and (23): First, we eliminate the integral of 1/ǫk between (4) and (10), to express 1/g0 in terms of 1/gand C(b).

Second, we replace 1/g0 by the resulting expression in Eq.(19). Third, we take the limitb→0: we exactly re- cover the system (23). Hence, if the eigenenergyE and the corresponding function f in the lattice model have a well defined limit forb = 0, this shows that the limit is given by the Bethe-Peierls model. Of course, the real mathematical difficulty is to show the existence of the limit, in particular for all eigenenergies. This property is not granted: For example, the present lattice model generalized to the case of a↓ particle of a massm3 different from the massm of the two↑ particles leads, for a large enough mass ratiom/m3, to a three-body energy spectrum not bounded from below in the b = 0 limit, even though the Pauli exclusion principle prevents from having the three particles on the same lattice site [47].

In conclusion, we have computed numerically the low lying eigenenergies of three spin-1/2 fermions in a box, interacting with an infinite scattering length in a lattice model, for a zero total momentum and for decreasing values of the lattice period. Our results show numerically the equivalence between this model and the Bethe-Peierls approach in the limit of zero lattice period. This is re- lated to the fact that the eigenenergies E are bounded from below in the zero lattice period limitb →0, more preciselyE >0. Such a convergence of the eigenstates of fermions in a lattice model towards universal states when b→0 is a key property used in Monte Carlo simulations at theN-body level [22, 23, 24].

We thank F. Werner for interesting discussions on the subject. Laboratoire de Physique Théorique de la Matière Condensée is the Unité Mixte de Recherche 7600 of Centre National de la Recherche Scientifique (CNRS).

The cold atom group at LKB is a member of IFRAF.

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[32] F. Werner, Y. Castin, Phys. Rev. Lett.97, 150401 (2006).

[33] Y. Castin, in Proceedings of the Enrico Fermi Varenna School on Fermi gases, June 2006, edited by M. Inguscio, W. Ketterle, C. Salomon, SIF (2007).

[34] Y. Castin, Comptes Rendus Physique5, 407 (2004).

[35] F. Werner and Y. Castin, Phys. Rev. A 74, 053604 (2006).

[36] C. Mora, Y. Castin, Phys. Rev.A67, 053615 (2003).

[37] Y. Castin, in Proceedings of the school“Quantum Gases in Low Dimensions”, J. Phys. IV (France)116, 89 (2004).

[38] In contrast, in a two-channel model for a Feshbach resonance, one finds that the effective range has a finite (and negative) limit in the zero potential range limit (the so- called narrow Feshbach resonance limit) [33].

[39] One thus cannot conclude that the corresponding minimal eigenenergy is the absolute ground state energy.

[40] Usually one expresses the Green’s function of the Lapla- cian in a cubic box in terms of plane waves. This leads to CBP= limx→0v(x), withv(x) =R

d³uexp(iu·x)/u²− P

n∈Z³^∗exp(in·x)/n². This definition ofv(x) should be understood within the frame of the theory of distribu- tions. We define thex= 0 limit ofv(x) as the limit for σ→0 ofR

d³xv(x)φ(x/σ)/σ³, whereφis a C^∞ rapidly decreasing function with R

d³xφ(x) = 1. In Eq.(12) we have taken for simplicity φ to be a Gaussian, but we have shown thatCBPis independent of this choice.

[41] M. Olshanii, L. Pricoupenko, Phys. Rev. Lett.88, 010402 (2002).

[42] One uses the rewriting hσ(n+u)−hσ(n) = T1+T2, withT1= [ ˆφ[σ(n+u)]−φ(σn)]/(nˆ +u)²,T2= [ ˆφ(σn)− 1][1/(n+u)²−1/n²] and ˆφ(x) = exp(−x²). Using a large n expansion, one finds that the integral of T2 over the symmetric integration domainI isO(1/n⁴), so that the theorem of dominated convergence applies. ForT1, one uses the Taylor-Lagrange formula up to second order for the numerator: For a givenu, there exists a vectorxuon the line connectingσnandσ(n+u) such that ˆφ[σ(n+ u)]−φ(σn) =ˆ P

iσui∂iφ(σn) +ˆ ¹₂P

i,jσ²uiuj∂i∂jφ(xˆ ^u).

The term involving the first order derivatives of ˆφvan- ishes after integration over u. Since the second order derivatives of ˆφ(x) are rapidly decreasing functions, they are in particular≤A/x² at largexfor some numberA, so that the integral ofT1overIis bounded byA/n⁴and the theorem of dominated converge applies again.

[43] The value ofCBP disagrees with the one (7.44 ≃ π× 2.37. . .) given in Eq.(53) of K. Huang, C.N. Yang, Phys.

Rev.105, 767 (1957).

[44] One uses the fact that forn∈N^∗,ǫ/[n²(n²+ǫ)] is positive whenǫ >0, and tends to 1/n²forǫ→+∞. The fact thatP

n∈Z³^∗1/n² = +∞gives the result.

[45] In the limit b→0, there exists a negative energy solution E < 0 for all a. Its energy can be calculated ac- curately directly from the Bethe-Peierls model from a more convenient representation of the function v(x) in [40], using Poisson’s summation formula applied to the function u→eⁱ^u·x/(u²+λ²) where λ >0 is arbitrary.

One obtains CBP =λ⁻²+ 2π²λ−P

n∈Z³^∗[λ²n⁻²(λ²+ n²)⁻¹ +πexp(−2πλn)/n], an expression whose value does not depend onλ. Specializing to the unitary limit, and taking λ = α/(2π), where α = 1.945766. . . solves α=P

n∈Z³^∗exp(−αn)/n, one finds a minimal eigenen- ergyE=−α²~²/mL².

[46] One may fear at this stage that an eigenvalue mx of M(E = 0), although not being the lowest one for the values ofb/Lconsidered in the figure, may be such that (mx+ 1)/(b/L) varies rapidly withb/L,e.g.with a large and positive slope, so as to converge for b/L →0 to a lower value than 1.08. To test this possibility, we have considered the lowest twenty eigenvalues of M(E = 0) in each symmetry sector with respect to reflections along x, y, z. All these eigenvalues mi are found to lead to (mi+ 1)/(b/L) having a negative slope as functions of b/Land converging for b/L→0 to values≃2.13, 2.27, 2.51,. . ., larger than 1.08.

[47] For an arbitrary mass ratio, the coupling constantg0 for an infinite scattering length isg0 =−2π~²b/(µK) where 1/µ = 1/m+ 1/m3 is the inverse of the reduced mass.

One may take as a simple variational ansatz the ground state of the three-body problem form=∞, of the form

|ψ∞i= [|r1i|r2i − |r2i|r1i]|χi, with r1−r2 =bex and ex the unit vector along x; |χi has a simple expression in momentum space and one finds χ(r1) = χ(r2). For a finite value of m the expectation value of H in |ψ∞i gives an upper boundEvon the ground state three-body energy,

Ev= ~²π² 2m3b²

A+Bm3

m

whereAis the smallest root ofF(A) = 1+R

[−1,1]³d³q[1+

(8)

cos(πq·ex)]/[2πK(A−q²)] and B = 2 + 1/F^′(A). One findsA≃ −0.042088 andB ≃1.75762. ThenEv→ −∞

when b → 0 for a mass ratio m/m3 above the critical value ≃ 41.8. Actually the exact critical mass ratio is expected to be below 13.6069. . . since the Efi- mov phenomenon takes place for m/m3 > 13.6069. . . [48].Note: in the bosonic case, for the lattice model at

|a| = ∞ with NB bosons of mass m in the same spin state, one may take as a variational ansatz the state vector where all the NB bosons are on the same lat-

tice site; one then finds an upper bound on the ground state energyEv^B =g0NB[NB−(1 +πK/4)]/(2b³), with 1+πK/4≃2.9185, so that the ground state energy tends to−∞forb→0 ifNB≥3.

[48] From [29] we find that the minimal mass ratio m/m3

leading to the Efimov phenomenon solves−^π₂ sin²(2θ) + cot 2θ+ 2θ= 0, excluding the trivial root θ=π/4, with θ= arctan[(1 + 2m/m3)^1/2].