One dimensional gas of bosons with integrable resonant interactions
OOne dimensional gas of bosons with integrable resonant interactions
V. Gurarie
Department of Physics, CB390, University of Colorado, Boulder CO 80309, USA (Dated: April 8, 2019)We develop an exact solution to the problem of one dimensional chiral bosons interacting via an s -wave Feshbach resonance. This problem is integrable, being the quantum analog of a classical two-wave model solved by the inverse scattering method thirty years ago. Its solution describes one ortwo branches of dressed chiral right moving molecules depending on the chemical potential (particledensity). We also briefly discuss the possibility of experimental realization of such a system. I. INTRODUCTION
The advances of the last decade in the techniques ofatomic physics allowed to realize a variety of exactly solv-able many-body models experimentally. In particular,the Tonks-Girardeau [1] gas, the dilute gas of repulsiveone-dimensional bosons has been realized [2]. More gen-erally, the Lieb-Liniger model [3] of interacting bosons inone dimensions can now be studied experimentally.In the field of cold atomic gases it is customary touse Feshbach resonances to control interactions betweenthe atoms. On some level, Feshbach resonances can bethought of as simply a tool to change the interactionstrength. But on a deeper level, they are a way to convertpairs of atoms into molecules whose binding energy canbe controlled. When confined to one dimensions, such asystem is then described by the Hamiltonian [4] H F = (cid:90) dx (cid:20) m a ∂ x ˆ a † ∂ x ˆ a + 12 m b ∂ x ˆ b † ∂ x ˆ b + (cid:15) ˆ b † ˆ b ++ g √ (cid:16) ˆ b ˆ a † + ˆ b † ˆ a (cid:17)(cid:21) , (1)Here ˆ a † , ˆ a are the creation and annihilation operatorsof atoms, ˆ b † , ˆ b are those of molecules, m a , m b are theirrespective masses, and (cid:15) and g are two parameters con-trolling the resonance. Throughout the paper we takeboth atoms and molecules to be bosons.The model described by Eq. (1) is unlikely to be inte-grable even classically [5] and cannot be solved exactly. Aparticularly straightforward argument against quantumintegrability involves calculating the amplitude of threeatom scattering with incoming momenta p , p , p intooutgoing momenta p (cid:48) , p (cid:48) , p (cid:48) , distinct from any permuta-tion of p , p , p , in the first nonvanishing Born approx-imation. This amplitude can be verified to be nonzero,while integrability would require it to be zero [6].The bosonization techniques of developed in the con-text of Luttinger liquid theory can be applied to under-stand Eq. (1) [7]. An approximate technique based on theideas of the asymptotic Bethe ansatz can also be used [8].Yet there exist an exactly solvable model closely re-lated to Eq. (1). It is the model of chiral atoms andmolecules interacting via a Feshbach resonance. Its Hamiltonian takes the following formˆ H = (cid:90) dx (cid:34) − iu ˆ a † d ˆ adx − iv ˆ b † d ˆ bdx + g √ (cid:16) ˆ b ˆ a † + ˆ b † ˆ a (cid:17)(cid:35) . (2)Eq. (2) describes the atoms which move in one directionwith the velocity u , independent of their wave vector,and molecules which also move in one direction with thevelocity v , also independent of their wave vector. Thecase of u = v is degenerate, as we will see below. In whatfollows, it will be assumed that u (cid:54) = v .The classical version of this model is called the 2-wavemodel in the literature. It is known to be integrable anddescribes various phenomena, for example, in non-linearoptics [9].In this paper we demonstrate that this problem is alsointegrable quantum mechanically. We do that by em-ploying the coordinate Bethe Ansatz and developing theexact solution of the problem defined by Eq. (2).A closely related integrable model, called the 3-wavemodel [9] (which differs from Eq. (2) by having 3 chi-ral fields two of which can fuse into the third one) wasstudied and demonstrated to be integrable quantum me-chanically in Ref. [10]. This model is also interesting forthe applications in quantum optics and atomic systems.Ref. [10] did not work out the finite density behavior forthe 3-wave model, which is something we do here for the2-wave model. Studying this is thus an interesting direc-tion of further work.One can argue that it is not entirely straightforwardto realize the Hamiltonian Eq. (2) using real atoms andmolecules. However, this Hamiltonian can be thoughtof as an approximation to a true problem of atoms andmolecules propagating in one dimension, given by Eq. (1),if we can restrict our attention to atoms and moleculeswhose momentum is close to a specially chosen momen-tum p F (for atoms) and 2 p F (for molecules). Indeed, inthat case their spectrum is linear, as in (cid:15) = ( p F + δp ) m ≈ p F m + δp p F m . (3)We can now interpret p F /m as the velocity v the atoms inEq. (2). One can immediately see a particular difficultywith this interpretation: the mass of the molecules hasto be twice that of an atom, and so are their “Fermi”momenta. Thus the velocity of an atom and a molecule a r X i v : . [ c ond - m a t . qu a n t - g a s ] J a n must be equal. To justify the assumption that u (cid:54) = v ,we may have to place the atoms and the molecules onan optical lattice where the molecular and atomic matrixelements are distinct from each other. Then the effectivemasses of atoms and molecules no longer have to be thesame.The interaction in the Hamiltonian Eq. (2), controlledby the cubic Feshbach term, occurs at a point in space.In reality, however, the interactions between the atomswhich lead to this term have a finite range which wedenote r . Remembering that r is not zero is importantin what follows. II. THE COORDINATE BETHE ANSATZ
We begin by constructing a few body eigenstates ofthe Hamiltonian Eq. (2) and then proceed to generalizethem to many body states made possible by the inte-grability of this problem. Then we impose the periodicboundary conditions to find the excitation spectrum ofthe system. In all steps we closely follow the standardtechniques originally developed for the solution of theLieb-Liniger model [3] as explained in Ref. [11].
A. Single atom state
First of all, we observe that a single atom represents anexact eigenstate of the Hamiltonian Eq. (2). This statecan be written as (cid:90) dx e ikx ˆ a † ( x ) | (cid:105) (4)where | (cid:105) is the vacuum, and its energy is given by E = uk . Indeed, all terms in the Hamiltonian besides the veryfirst one annihilate this state. B. Two atom states
Now we seek the two-atom states in the following gen-eral form (cid:90) dx dx ψ ( x , x ) ˆ a † ( x )ˆ a † ( x ) | (cid:105) + (cid:90) dy φ ( y ) ˆ b † ( y ) | (cid:105) (5)Acting on it by the Hamiltonian Eq. (2) we find the fol-lowing first-quantized Schr¨odinger equation − iu (cid:88) i =1 dψ ( x , x ) dx i + g √ δ ( x − x ) φ ( x ) = Eψ ( x ) , − iv dφ ( y ) dy + √ g ψ ( y, y ) = Eφ ( y ) . (6)This equation has two classes of solutions.On the one hand, two atoms separated by some dis-tance will forever move at equal velocity u regardless of their momenta. Thus they will never interact. This isreflected in the existence of an exact eingenstate of theHamiltonian given by ψ ( x , x ) = f ( | x − x | ), φ = 0,or (cid:90) dx dx f ( | x − x | ) e ik x + ik x a † ( x ) a † ( x ) | (cid:105) . (7)Here f ( x ) is an arbitrary function of its argument suchthat f (0) = 0. The energy of this state is simply E = u ( k + k ). We call this a two-atom state.On the other hand, two atoms located at the samepoint will forever be located at the same point sincethey move at an equal velocity. They hybridize with amolecule to form a dressed molecule state. It is given by ψ ( x , x ) = Aδ ( x − x ) e ikx , φ ( y ) = M e iky , equivalentto the following form (cid:90) dx e ikx (cid:104) A ˆ a † ( x ) + M ˆ b † ( x ) (cid:105) | (cid:105) , (8)where A and M are the atom-mlecule amplitudes. Sub-stituting these into Eq. (6) gives ukA + g √ M = EA,vkM + √ gδ (0) A = EM. (9)Notice the appearance of δ (0). Treated naively, this couldbe interpreted as infinity. However, for the regularizedmodel where the interactions happen at some length scale r , this term should be interpreted as δ (0) = 1 /r , whichis what we do in what follows. We now find from Eq. (9) E ± = ( v + u ) k ± (cid:113) ( v − u ) k + 4 g r , (10)as well as M ± A ± = k ( v − u ) ± (cid:113) ( v − u ) k + 4 g r √ g . (11)Thus the dressed molecules come in two different species,labelled by the superscripts + and − . These two solutionare easiest to parametrize if one introduces new variables λ ( k ) which we call repidities (for reasons which will be-come clear later). We define those as in λ ≡ ( v − u ) ( v − u ) k ± (cid:113) ( v − u ) k + g r g . (12)Assuming, without the loss of generality, that v > u , λ is positive if the plus sign is chosen in front of the squareroot and it is negative if the minus sign is chosen, thusin terms of λ , the two species of dressed molecules arestraightforward to distinguish (and so, the superscript ± is not necessary for λ ). In terms of these, one can write M ± A ± = 2 √ gv − u λ, (13)as well as E ± = (cid:15) ( λ ) , (cid:15) ( λ ) = 2 g vλ ( v − u ) − u r λ . (14)Here positive λ correspond to the choice of + in the su-perscript and negative λ to the choice of − .For completeness, we also note that k can be expressedin terms of λ via k = 2 g λ ( v − u ) − r λ . (15)As seen from this relation, the energy of the dressedmolecule is no longer a linear function of k , thus the dressed molecules have a nontrivial dispersion. As a re-sult, their velocity ∂(cid:15) /∂k depends on their wave vector k , and unlike atoms, the molecules can catch up witheach other and undergo scattering, as we will see in thenext subsection. C. Many-atomic states
As happens in all integrable systems, more generalstates can be reduced to a combination of two bodystates. The most general eigenstate of the HamiltonianEq. (2) state can be written in the following form (cid:90) N (cid:89) n =1 (cid:104) dx n e ik n x n (cid:16) A n ˆ a † ( x n ) + M n ˆ b † ( x n ) (cid:17)(cid:105) M (cid:89) n =1 (cid:2) dy n e ip n y n ˆ a † ( y n ) (cid:3) f ( | y α − y β | ) | (cid:105) × (cid:89) n
0. This state represents N dressed molecules and M free atoms scattering off eachother. A n and M n are the amplitudes for the dressedmolecules corresponding to the rapidities λ n , which inturn depend on the momenta k n (and on the species ofthe molecule, or on whether λ n is positive or negative). f ( | y α − y β | ) is an arbitrary function of all possible differ-ences of the atomic coordinates, such that it vanishes ifany two atomic coordinates coincide. S nm is an S-matrixrepresenting the scattering of two dressed molecules, n and m , off each other. Finally S amn is an S-matrix forscattering between an atom and a dressed molecule, withthe only index n used to emphasize that it depends onlyon the rapidity of the molecule and not of the atom. In-deed, these S matrices take the following form S nm = i ( λ n − λ m ) + 1 i ( λ n − λ m ) − , S amn = 2 iλ n − iλ n + 1 . (17)This form of the scattering matrices justifies the term“rapidity” for the parameter λ . It is now a matter ofa straightforward algebra to check that Eq. (16) is aneigenstate of the Hamiltonian Eq. (2) with the energy E = N (cid:88) n =1 (cid:15) ( λ n ) + u M (cid:88) n =1 p n . (18) D. Bethe equations
Construction of the exact eigenstates is but the firststep towards exact solution of an integrable problem us- ing the coordinate Bethe ansatz. The next step is theimposition of the appropriate boundary conditions, thedetermination of the ground state energy and of the en-ergy of the excitations above the ground states. As usual,we impose the periodic boundary conditions to arrive atthe Bethe equations (here L is the system size) e ik j L (cid:2) S amj (cid:3) M (cid:89) l (cid:54) = j S jl = 1 , e ip j L (cid:89) j S amj = 1 . (19) S amj depends on λ j only and is p j independent. Thus itis always possible to choose p j in such a way that the sec-ond equation in (19) is satisfied. Then the first equationreduces to k j L + (cid:88) l θ jl − M θ amj = 2 πn j , (20)where iθ jl = ln S jl , iθ amj = ln S amj . Following Ref. [11]it is straightforward to prove that the solution to theseequations are unique and real, and all n j are distinct.This last claim is the consequence of the 1D “Pauli”principle (at work here, as well as in the standard Lieb-Liniger model), which says that no two λ can be thesame, or the wave function Eq. (16) vanishes if λ j = λ k for j (cid:54) = k as can be checked directly.As a next step, we take n j to be a continuous variable n ( j ), with λ j and k j becoming functions of n . This gives Lk ( n ) + (cid:88) l θ ( λ ( n ) − λ ( l )) − M θ am ( λ ( n )) = 2 πn. (21)Here θ ( x ) = ln [( ix + 1) / ( ix − /i and θ am ( x ) =ln [(2 ix − / (2 ix + 1)] /i Finally, we differentiate with respect to λ ( n ), introducethe function ρ = 1 L dndλ (22)playing the role of the density of λ , and replace summa-tion by integration to arrive at ρ ( λ ) − π (cid:90) dµ ρ ( µ )( λ − µ ) + 1 = 12 π dkdλ + M πL
41 + 4 λ . (23) E. Yang-Yang equation
There are two ways to make further progress in thedetermination of the ground state energy and the exci-tation spectrum of the system. One follows explicit con-structions of the excitations, by exciting a state with aparticular rapidity λ , while shifting the rest of the rapidi-ties to accommodate the Bethe equations Eq. (20). Thesecond is by studying the excitations at finite tempera-ture and then taking the limit T →
0. Both methods aredescribed in Ref. [11]. It is technically easier to use thesecond approach. Although this method is well known,we go over it briefly in the particular case of interest here.First we note that in a general state n j takes valuesin some subset of all possible integer numbers. We in-troduce ρ p as the density of λ among the values of n which are taken (“occupied”), and ρ h as the density of λ where these values are unoccupied, with ρ t = ρ p + ρ h (seeRef. [11] for the discussion on how this is done). Thenwe find ρ t ( λ ) − π (cid:90) dµ ρ p ( µ )( λ − µ ) + 1 = 12 π dkdλ + M πL
41 + 4 λ . (24)Next we construct the energy, the entropy, and the par-ticle number of such configuration, given by E = L (cid:90) dλ ρ p ( λ ) (cid:15) ( λ ) ,S = L (cid:90) dλ ( ρ t ln ρ t − ρ p ln ρ p − ρ h ln ρ h ) ,N = L (cid:90) dλ ρ ( λ ) . (25)Then we minimize the thermodynamic potential Ω = E − T S − hN ( T is the temperature, and h is chemicalpotential) with respect to ρ p , while remembering that thevariation of δρ p is related to δρ t by δρ t ( λ ) = 12 π (cid:90) dµ λ − µ ) + 1 δρ p . (26)Following standard methods [11], we introduce ρ h ρ p = e (cid:15) ( λ ) T . (27) (cid:15) ( λ ) plays the role of the excitation spectrum of the sys-tem. It satisfies, as a result of the minimization of Ω, (cid:15) ( λ ) + T π (cid:90) dµ λ − µ ) + 1 ln (cid:16) e − (cid:15) ( µ ) T (cid:17) = (cid:15) ( λ ) − h. (28)Finally, we take the limit of zero temperature, T → (cid:15) ( λ ) − π (cid:90) (cid:15) ( µ ) < dµ (cid:15) ( µ )( λ − µ ) + 1 = (cid:15) ( λ ) − h. (29)All of these steps are standard, with the exception ofEq. (24), leading to the equation Eq. (29) which is againstandard with the exception of its nonstandard right-hand side. Here h is the chemical potential, and theintegral is taken over only the region of µ where (cid:15) ( µ ) < (cid:15) ( λ ) produces the excitationspectrum of the system, which is the quantity we wouldlike to compute. Notice that the coupling constant g isnot explicitly present, except through the definition of (cid:15) ( λ ) in Eq. (14).If (cid:15) ( λ ) >
0, then the excitation at this λ is a particle.If, on the other hand, (cid:15) ( λ ) <
0, then the excitation is ahole whose energy is − (cid:15) ( λ ). III. THE EXCITATION SPECTRUMA. Dimensionless parameters
The excitation spectrum can be found by solving theequation Eq. (29). This can only be done numerically.To do this in a meaningful way, let us first study thescale of the parameters involved in Eq. (29).Since the interactions occur at a finite range r , we willrestrict the possible values of momenta k to the range k ∈ (cid:20) − πr , πr (cid:21) (30)(as if the model Eq. (2) is defined on a lattice of latticespacing r ). We would also like to make sure that the in-teractions are sufficiently weak so that particles movingwith momenta close to π/r would be close to noninter-acting. This can be achieved if the 4 g /r is much smallerthan ( v − u ) k where k ∼ π/r in Eq. (10). This gives g r ( u − v ) (cid:28) . (31)From now on, we adopt this assumption.Second, it is convenient to rescale the rapidity λ tosimplify the expression for the energy spectrum (cid:15) ( λ ).We introduce a parameter c = 2 g √ r | u − v | (cid:114) vu (cid:28) λ = λc. (33)We also introduce the dimensionless rescaled energy spec-trum ˜ (cid:15) = | v − u |√ r g √ uv (cid:15), ˜ (cid:15) = | v − u |√ r g √ uv (cid:15) . (34)The equation Eq. (29) gets simplified to˜ (cid:15) (˜ λ ) − π (cid:90) ˜ (cid:15) (˜ µ ) < d ˜ µ c ˜ (cid:15) (˜ µ )(˜ λ − ˜ µ ) + c = ˜ λ − λ − ˜ h. (35)This equation can only be solved numerically, even in thephysical limit of c (cid:28) λ . Eq. (12) together with Eq. (30) gives λ ∈ (cid:20) − ( v − u ) π g r , − π (cid:21) (cid:91) (cid:20) π , ( v − u ) π g r (cid:21) (36) In turn, this gives for ˜ λ ˜ λ ∈ (cid:20) − πvuc , − c π (cid:21) (cid:91) (cid:20) c π , πvuc (cid:21) . (37)The integration range over ˜ µ in Eq. (35) is over these twocombined intervals.Now we are in the position to solve Eq. (35) numeri-cally. The standard method is by interacting the relation˜ (cid:15) n +1 (˜ λ ) = 12 π (cid:90) ˜ (cid:15) n (˜ µ ) < d ˜ µ c ˜ (cid:15) n (˜ µ )(˜ λ − ˜ µ ) + c +˜ λ − λ − ˜ h. (38)This leads to ˜ (cid:15) n (˜ λ ) quickly diverging to negative infin-ity as n increases. And indeed, the proof given in Ref.[11] regarding the convergence of this procedure is notapplicable to Eq. (35).Instead, we use a different technique. We define a func-tional Q = 18 (cid:90) d ˜ λ (cid:16) ˜ (cid:15) (˜ λ ) − (cid:12)(cid:12)(cid:12) ˜ (cid:15) (˜ λ ) (cid:12)(cid:12)(cid:12)(cid:17) − π (cid:90) d ˜ λd ˜ µ c (cid:16) ˜ (cid:15) (˜ λ ) − (cid:12)(cid:12)(cid:12) ˜ (cid:15) (˜ λ ) (cid:12)(cid:12)(cid:12)(cid:17) (˜ (cid:15) (˜ µ ) − | ˜ (cid:15) (˜ µ ) | ) c + (˜ λ − ˜ µ ) − (cid:90) d ˜ λ (cid:18) ˜ λ − λ − ˜ h (cid:19) (cid:16) ˜ (cid:15) (˜ λ ) − (cid:12)(cid:12)(cid:12) ˜ (cid:15) (˜ λ ) (cid:12)(cid:12)(cid:12)(cid:17) (39)such that δQδ ˜ (cid:15) (˜ λ ) = 0 (40)is equivalent to Eq. (35), up to a multiplication by1 − sign ˜ (cid:15) (˜ λ ). Then we introduce an extra fictitious pa-rameter τ , and construct the solution to the equation ∂ ˜ (cid:15) (˜ λ, τ ) ∂τ = − δQδ ˜ (cid:15) (˜ λ, τ ) (41)in the limit where τ → ∞ .This procedure allows us to compute ˜ (cid:15) (˜ λ ) for all such˜ λ that ˜ (cid:15) (˜ λ ) <
0. One drawback of this procedure isthat once ˜ (cid:15) (˜ λ, τ ) = 0 for some ˜ λ and some τ , it willremain zero for larger τ . As a result, ˜ (cid:15) (˜ λ ) can become“trapped” at zero whereas it might actually be negative.We fix this problem by supplementing it with iterations(38). Once the initial ˜ (cid:15) (˜ λ ) used for iterations is close tothe solution of Eq. (35), subsequent iterations will notdiverge. Indeed, suppose˜ (cid:15) (˜ λ ) = ˜ (cid:15) s (˜ λ ) + δ ˜ (cid:15) (˜ λ ) , (42)where ˜ (cid:15) s is the solution of Eq. (35), and where δ ˜ (cid:15) (cid:28) ˜ (cid:15) s .Then we find δ ˜ (cid:15) n +1 (˜ λ ) = 12 π (cid:90) ˜ (cid:15) s (˜ µ ) < d ˜ µ cc + (˜ µ − ˜ λ ) δ ˜ (cid:15) n (˜ µ ) . (43) It is now fairly straightforward to prove that (cid:90) ∞−∞ d ˜ λ (cid:16) δ ˜ (cid:15) n +1 (˜ λ ) (cid:17) < (cid:90) ∞−∞ d ˜ λ (cid:16) δ ˜ (cid:15) n (˜ λ ) (cid:17) , (44)which proves that the iteration procedure is not diver-gent.Once we do that, we construct the rest of this functionby using Eq. (35) as a definition of ˜ (cid:15) (˜ λ ), or˜ (cid:15) (˜ λ ) = 12 π (cid:90) ˜ (cid:15) (˜ λ ) < d ˜ µ c ˜ (cid:15) (˜ µ )(˜ λ − ˜ µ ) + c + ˜ λ − λ − ˜ h. (45) B. Numerical solution
We now use this procedure to construct solutions toEq. (35). We take representative parameter values c = 0 . , vu = 2 . (46)We then take initial value˜ (cid:15) (˜ λ, τ ) (cid:12)(cid:12)(cid:12) τ =0 = − . (47)Then we run the Eq. (41) in steps of dτ = 0 .
01 up to τ = 100. The integrals are computed by discretizing therange of ˜ λ into 3200 intervals. After that, we use ˜ (cid:15) (˜ λ, − . The accuracy is defined as (cid:82) d ˜ λ (cid:16) ˜ (cid:15) n +1 (˜ λ ) − ˜ (cid:15) n (˜ λ ) (cid:17) (cid:82) d ˜ λ (cid:16) ˜ (cid:15) n (˜ λ ) (cid:17) . (48) (cid:45) (cid:45)
50 50 100
Λ(cid:142)(cid:45) (cid:45) (cid:45) (cid:45)
Ε(cid:142)
FIG. 1: ˜ (cid:15) (˜ λ ) for c = 0 . v/u = 2, h = 0. (cid:45) Λ(cid:142)(cid:45) (cid:45) (cid:45) (cid:45) (cid:45)
Ε(cid:142)
FIG. 2: Same figure as in Fig. 1 but the region close to ˜ λ = 0enlarged. First we illustrate the solution for h = 0. Fig. 1 shows˜ (cid:15) (˜ λ ) for the entire range of ˜ λ . Fig. 2 shows the region ofsmall ˜ λ enlarged.One sees that at ˜ λ <
0, ˜ (cid:15) <
0. So for the branch of exci-tation spectrum at negative rapidities, all the excitationsare holes, and they are all gapped. This is illustrated onFig. 3. For ˜ λ >
0, ˜ (cid:15) changes sign at some value of ˜ λ . Sohere we have gapless right moving excitations with thelinear spectrum. This is illustrated on Fig. 4.By decreasing the chemical potential h , it is possibleto make the ˜ λ < λ > (cid:45) (cid:45) (cid:45) Λ(cid:142)(cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) Ε(cid:142)
FIG. 3: Same figure as in Fig. 1 but the region of ˜ λ < λ = 0 enlarged. Λ(cid:142)(cid:45) (cid:45) (cid:45) Ε(cid:142)
FIG. 4: Same figure as in Fig. 1 but the region of ˜ λ > (cid:15) = 0 enlarged. as a function of the chemical potential (or of density),with the massless mode acquiring a gap. This phasetransition looks similar to the transition observed in theproblem defined by Eq. (1) in Ref. [7]. To elucidate thenature of this transition, it seems useful to study Eq. (2)using the bozonization techniques which we leave as asubject for future work. (cid:45) (cid:45) (cid:45)
Λ(cid:142)(cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) Ε(cid:142)
FIG. 5: Same figure as in Fig. 3 but now the chemical poten-tial h = − Λ(cid:142)(cid:45) (cid:45) (cid:45) (cid:45) Ε(cid:142)
FIG. 6: Same figure as in Fig. 4 but now the chemical poten-tial h = − IV. CONCLUSIONS
We have developed an exact solution to the problemof chiral atoms and molecules propagating in one dimen-sions with interactions controlled by a Feshbach reso-nance. The remaining outstanding issue is whether thisHamiltonian can be realized in a realistic cold atom ex- periment.A particular tantalizing question is whether thefermionic analog of Eq. (2) (the one where ˆ a † and ˆ a arefermionic creation and annihilation operators, and wherean additional “spin” index is necessary to make the qubicterm non-zero) is integrable. If so, the development of anexact solution of such a problem would be an interestingdirection of further research. The fermionic version of the3-wave model was shown to be integrable in Ref. [12]. Acknowledgments
The author is grateful to P. B. Wiegmann for the initialsuggestion to study this problem, and to M. J. Ablowitzfor discussing the classical analog of the problem definedby Eq. (2) studied here, as well as to R. Wilson for thediscussions at the early stages of this project, to G. As-trakharchik and M. Hermele for useful advice concern-ing the numerical procedure, to L. Radzihovsky for dis-cussing the proper interpretation of the results, and toV. Gritsev for pointing out Refs. [10, 12]. This work wassupported by the NSF grant DMR-0449521. [1] M. Girardeau, J. Math. Phys. , 516 (1960).[2] T. Kinoshita, T. Wenger, and D. Weiss, Science ,1125 (2004).[3] E. Lieb and W. Liniger, Phys. Rev. , 1605 (1963).[4] E. Timmermans, P. Tommasini, M. Hussein, and A. Ker-man, Phys. Rep. , 199 (1999).[5] M. J. Ablowitz, D. Baldwin, private communication. Seealso: D. Baldwin, W. Hereman, J. Nonlin. Math Phys., , 90 (2006).[6] The author is grateful to M. Pustilnik for helping clarifythis issue.[7] D. E. Sheehy and L. Radzihovsky, Phys. Rev. Lett. ,130401 (2005). [8] V. Gurarie, Phys. Rev. A , 033612 (2006).[9] M. J. Ablowitz and H. Segur, Solitons and the InverseScattering Transform (SIAM, Philadelphia, 1981).[10] M. Wadati and K. Ohkuma, J. Phys. Soc. Jpn. , 1229(1984).[11] V. E. Korepin, N. Bogoliubov, and A. G. Izergin, Quan-tum Inverse Scattering Method and Correlation Func-tions (Cambridge University Press, Cambridge, UK,1993).[12] K. Ohkuma and M. Wadati, J. Phys. Soc. Jpn.53