Table Of ContentAnalytical solutions for two heteronuclear atoms in a ring trap
Xing Chen,1 Liming Guan,1 and Shu Chen1,∗
1Institute of Physics, Chinese Academy of Sciences, Beijing 100190, China
We consider two heteronuclear atoms interacting with a short-range δ potential and confined in
a ring trap. By taking the Bethe-ansatz-type wavefunction and considering the periodic boundary
condition properly, we derive analytical solutions for the heteronuclear system. The eigen-energies
represented in terms of quasi-momentums can then be determined by solving a set of coupled
1 equations. Wepresentanumberofresults,whichdisplaydifferentfeaturesfromthecaseofidentical
1 atoms. Our result can be reduced to the well-known Lieb-Liniger solution when two interacting
0
atoms havethe same masses.
2
n PACSnumbers: 34.10.+x,34.50-s,03.75.Hh
a
J
I. INTRODUCTION dimensional ring trap with periodic geometry. The pe-
6
riodic geometry has been experimentally realized as a
2
circularwaveguideortoroidaltrap[21,22]. Forthetwo-
Motivatedbyrecentexperimentalstudiesonheteronu-
] atomsystemwithδ-potentialinteraction,itisconvenient
s clearatoms,mixturesofatomicgaswithdifferentmasses
totreattheinteractiontermbyseparatingcenter-of-mass
a
haverecentlyattractedlotsofattention. Stablemixtures
g andrelativemotion. However,itisnotconvenienttoap-
of heteronuclear atoms, for examples, 40K and 6Li, 40K
- ply the periodic boundary condition by using separated
t and 87Rb, and 6Li and 23Na, have already been real-
n center of mass and relative coordinates. When the peri-
ized in experiments [1–6]. The heteronuclear quantum
a odic boundary condition is taken into account, the rel-
u gasmixtureoffersawiderangeofpossibilitiesforexplor-
ative motion is actually coupled to the center-of -mass
q ing novel phases and new quantum effects. Significant
motion. In the present work, we analytically solve the
t. efforts have been made on research topics such as su- problemofinteractingtwoheteronuclearatomsbytaking
a perfluid properties and pairing mechanisms of heteronu-
m the Bethe-ansatz-type wavefunction and considering the
clear fermions [7–9] and heteronuclear molecule forma-
periodicboundaryconditionproperly. Whilethemomen-
- tion [3, 10–12].
d tumofcenter-of-massmotionisagoodquantumnumber,
n The mass ratio between heteronuclear atoms adds a the relative momentum is coupled to the center-of-mass
o new degree of freedom for exploring physical effects as- momentumduetotheperiodicboundarycondition. Our
c sociated with the mass imbalance. In general, heteronu- results are consistent with the exact Bethe-ansatz solu-
[
clear atomic systems are expected to behave quite dif- tions of the Lieb-Linger model when two atoms have the
1 ferently than the equal-mass counterpart [7–9, 13, 14]. same mass [18]. With the analytical solution, we obtain
v Togaindeepinsightintopropertiesofheteronuclearsys- the energy spectrum, density distribution and momen-
4
tems, someanalyticalsolutionsareespecially important. tum distribution for different mass ratio. Our results in-
7
While most of theoretical works on the many-body het- dicate that they display different features from the iden-
0
5 eronuclearsystemsarewithinthemean-fieldtheory,few- tical two-atom system.
. body systemscanbe treatedmuchmorepreciselywithin The paper is organized as follows: In Sec.II, we intro-
1
nonperturbative microscopic frameworks [14]. Particu- duce the model and present a detailed derivation of the
0
1 larly,thetwo-bodyproblemisvitaltounderstandtheba- analytical solution of the interacting heteronuclear sys-
1 sic physicalphenomenonfromthe beginning ofquantum tem. In the limit of equal mass, our analytical result
: mechanics. The two-atom system composed of identical is compared to the Bethe-ansatz solution of two-particle
v
atoms is exactly solved as one can separate the center- Lieb-Liniger model. In Sec.III, we present the energy
i
X of -mass and relative motions and thus the two-particle spectrum of heteronuclear atoms. In Sec.IV, we show
r system turns to a one-body problem [15–19]. The prob- the density distribution of relative coordinate and the
a lem with different atoms becomes complicated and gen- momentum distribution of heteronuclear atoms. A sum-
erally is hard to get analytical solutions as it does not mary is given in the last section.
separate in center-of-mass and relative coordinates [12].
Recently, the scattering of heteronuclear atom pairs in a
one-dimensionalopticallatticehasbeenstudiedbyusing
II. THE MODEL AND ITS SOLUTIONS
the method of Green’s function [20].
In this article, we study the problem of two unequal-
We consider an interacting two-atom system in a 1D
mass atoms with δ-potential interaction in a one-
ring trap described by the Hamiltonian
¯h2 ∂2 ¯h2 ∂2
Hˆ =− − +gδ(x −x ), (1)
∗Electronicaddress: [email protected] 2m ∂x2 2m ∂x2 1 2
1 1 2 2
2
where g is the interaction strength and the general case wherek =Km /(m +m )+k,k =Km /(m +m )−
1 1 1 2 2 2 1 2
with different masses m and m is considered. The ge- k,k′ =Km /(m +m )−k,andk′ =Km /(m +m )+
1 2 1 1 1 2 2 2 1 2
ometry of the ring trap enforces the periodic boundary k. The periodic boundary condition then enforces
condition
A+eik1L = A−, (9)
Ψ(x1+L,x2)=Ψ(x1,x2+L)=Ψ(x1,x2) (2) B+eik1′L = B−, (10)
to the eigenfunction Ψ(x1,x2) of Hamiltonian Hˆ. In- A−eik2L = A+, (11)
txro−duxci,nXg r=ela(mtivexa+ndmcexnt)e/r(-mof-m+amss)c,ooorrdalitneartneastixvel=y B−eik2′L = B+. (12)
1 2 1 1 2 2 1 2
x = X + m2 x, x = X − m1 x, we can rewrite
1 m1+m2 2 m1+m2 By multiplying Eqs. (9) and (11) or Eqs (10)and (12),
the Hamiltonian Eq.(1) as the separated form we get eiKL =1, which implies
Hˆ = HˆX(X)+Hˆx(x) KL=2πn (13)
¯h2 ∂2 ¯h2 ∂2
= − − +gδ(x), (3) with n the integer. Combining Eqs. (9) (10) and (6), we
2M ∂X2 2µ∂x2
can represent B+, A−, B− in terms of A+. Substituting
where µ = m m /(m +m ) and M = m +m . The them into Eq. (7), we can derive the following relation
1 2 1 2 1 2
wavefunctions of the Hamiltonian can take the form of
Ψ(X,x) = eiKXϕ(x), where eiKX is the eigenstate of ¯h2 sin(kL)
k = . (14)
HˆX(X) and the eigenstate of Hˆx(x) has the form of µg cos K m1L −cos(kL)
m1+m2
(cid:16) (cid:17)
ϕ(x) = (A eikx+B e−ikx)θ(x)
+ +
From Eq.(14), it is clear that k is coupled with the K
+(A−eikx+B−e−ikx)θ(−x), (4) whereasK isuniquelydeterminedbyEq.(13). Bysolving
where A± and B± are coefficients to be determined. It Eqs. (13) and (14), we can get solutions of k and K and
is straightforwardto get the eigenenergy given by therefore eigenvalues of the system by Eq.(5).
The above equations can be also represented in terms
E = ¯h2 K2+ ¯h2k2. (5) of the quasi-momentum k1 and k2 via K = k1+k2 and
2M 2µ k = (m k −m k )/(m +m ). In terms of k and k ,
2 1 1 2 1 2 1 2
the eigenenergy is given by
Although formally the center-of-mass and relative parts
are separated, we shall show that the relative quasi- ¯h2 ¯h2
momentum k is actually coupled with the total momen- E = k2+ k2. (15)
2m 1 2m 2
tum K due to the periodic boundary condition (2). 1 2
Thecontinuousconditionofthewavefunction(4)gives Inorderto compareto the case withthe samemasses,
the following restriction to the coefficients: we introduce g = h¯2c and rewrite Eq. (14) as
m1
A++B+ =A−+B−. (6) 1 k sin(kL)
1+ = , (16)
Byintegratingtheeigen-equationHˆΨ(X,x)=EΨ(X,x) (cid:18) α(cid:19) c cos KL −cos(kL)
from the negative infinitesimal −ǫ to the positive in- (cid:16)1+α(cid:17)
finitesimal ǫ, we get
whereα=m /m isthemassratio. Whenthemassratio
2 1
¯h2 ∂ ∂ α=1,ourresults(13)and(14)shouldbeconsistentwith
− 2µ(cid:18)∂xϕ(x)x=ǫ− ∂xϕ(x)x=−ǫ(cid:19)+gϕ(0)=0, thewell-knownBethe-ansatzsolutionoftheLieb-Liniger
model [18]. Using the above relation (16) and taking
which leads to α=1, we can get
µ
ik(A+−B+−A−+B−)= ¯h2g(A++B++A−+B−). 2k+ic = eikL−cos(KL/2) .
(7) 2k−ic cos(KL/2)−e−ikL
The Eq.(6) and Eq.(7) can not uniquely determine all
Since cos(KL/2) = cos(πn) = ±1, the above equation
the parameters. Additional relations can be obtained
reduces to
by considering the periodic boundary condition (2). In
ordertoapplytheperiodicboundarycondition,itismore k −k +ic
1 2 =±eikL,
convenientto workin the coordinate space of x1 and x2. k −k −ic
1 2
Intermsofx andx ,thetotalwavefunctionΨ(x ,x )=
1 2 1 2
Ψ(X,x)=eiKXϕ(x) can be also represented as where + (or −) corresponds to n =even (or odd). By
using the relation eikL = ei(k1−k2)L/2 = eik1Le−iKL/2 =
Ψ(x1,x2) ±eik1L, we then have
= (A eik1x1+ik2x2 +B eik1′x1+ik2′x2)θ(x −x )+
+ + 1 2 k −k +ic
(A−eik1x1+ik2x2 +B−eik1′x1+ik2′x2)θ(x2−x1), (8) eik1L = k11−k22−ic, (17)
3
which is just the Bethe-ansatz equation of the two-
particle Lieb-Liniger model [18]. Another Bethe-ansatz 100
equation can be obtained via eik2L = e−ik1L. When
m =m , we have k′ =k and k′ =k . From Eqs. (9), 80 K=0
1 2 1 2 2 1
(10) and (6), we can also get A− = B+ and B− = A+. 60
Therefore the wavefunction also has the same form of
) 40
the Lieb-Linigermodel [18], which is invariantunder the
(
the exchange of x and x . We note that, for the gen- e 20
1 2
eral case with mass imbalance, the wavefunction is not
0
invariant under the exchange of x and x since k′ 6= k
1 2 1 2
and k2′ 6=k1. -20 =1 bound state
=2.175
-40
-60
III. THE ENERGY SPECTRUM -100 -80 -60 -40 -20 0 20 40 60 80 100
Without lossofgenerality,only the caseofα≥1 shall
be considered. For convenience, in the following calcu-
lation we shall take ¯h = 1 and 2m = 1. In the units FIG. 1: (Color online) The dimensionless energy spectrum
1
of energy ¯h2/(2m )=1, the energy expression Eq.(5) or versusγ for K =0, α=2.175 (dashed line) and α=1 (solid
1
Eq.(15) becomes line). Forγ <0, thereexists a boundstate.
E =(1+α)−1K2+(1+1/α)k2 (18)
whichisobtainedbyinsertingk =iλintoEq.(19). From
or E = k2 + (1/α)k2. Since the total momentum K Eq.(21), one can also check that λ has only solution for
1 2
is a conserved quantity determined by Eq.(13), the en- c′ < 0. The energy of the bound state can be obtained
ergy spectrum of the system E can be characterized by via E =−(1+1/α)λ2.
different K. The ground state corresponds to K = 0. To give a concrete example, we consider a two-atom
Given K = 2πn/L, k can be determined by solving system composed of 40K and 87Rb with the mass ration
Eq.(14) or Eq.(16). For the repulsive case with c > 0, α = 2.175. In Fig.1, we plot the energy spectrum for
Eq. (16) has only real solutions which describe the scat- states with K = 0. The energy spectrum for α = 2.175
teringstates. However,fortheattractivecasewithc<0, is represented by dashed lines. The energy spectrum of
Eq. (16) may have imaginary solutions corresponding to theequal-masssystemisalsogivenforcomparison. Here
the bound states of the attractive system. we use the dimensionless energy e(γ)=E/(Nρ2), where
Firstwe shallconsiderthe casewith the totalmomen- ρ = N/L and N = 2 is the atom number. It is shown
tum K =0. For K =0, Eq.(16) becomes thatthegroundenergyforthesystemofα>1haslower
energy than that of the system with α=1. Actually, all
1 k sin(kL) the other spectrum correspondingto quantum number I
1+ = .
(cid:18) α(cid:19) c 1−cos(kL) for α>1 is below its correspondence for α=1.
Nowweconsiderstateswiththenonzerototalmomen-
With similar steps as the equal-mass case, we can get tum,forexample,K =±2π,whichcorrespondtoexcited
L
states of the system. When K 6= 0, Eq.(16) can not be
k+ic′
eikL = (19) simplified to a similar form as Eq.(19) or Eq.(20) for an
k−ic′ arbitrary α. Nevertheless, formally we have
withc′ =c/ 1+ 1 . Forc>0,khasonlyrealsolutions. k kL KL
α kL=2πI −2arctan −2arctan[tan cot2 ],
It is conveni(cid:0)ent to(cid:1)solve the real solution of k from the c′ 2 2(1+α)
logarithm form of Eq.(19) given by (22)
where I takes integer. When K = 0, it reduces to
′
kL=2π(I −1/2)−2arctan(k/c), (20)
Eq.(20). The relative momentum can be obtained by
numerically solving Eq.(16) or (22) with K = ±2π and
where I takes integer. By numerically solving the k so- L
lution, we can get the energy E = 1+ 1 k2 for states theenergyisgivenbyE =(1+α)−1 2Lπ 2+ 1+ α1 k2.
withK =0. Thegroundstateenerg(cid:0)ycorαre(cid:1)spondstothe The energies for K = 2π/L and K =(cid:0)−2(cid:1)π/L(cid:0)are de(cid:1)gen-
erate. In Fig.2, we display energy spectrums of states
solution of Eq.(20) with I = 1. For c < 0, Eq.(19) has
with K = ±2π/L for α = 2.175 and α = 1. It is shown
imaginary solution k = iλ. The real solution λ can be
that the energy spectrum for α > 1 is always below the
obtained by solving
corresponding spectrum for α=1.
λ+c′ Theeffectofmassratioonenergyspectrumcanbeun-
e−λL = , (21)
λ−c′ derstood from two aspects. First, we can see that coeffi-
4
60
=1 =1
K=0
=1 =2.175
3.0 =5 =1
L =5 =2.175
40 1/ 2.5 =500 =1
f =500 =2.175
) o
e( 20 es 2.0
t
ni 1.5
u
0
x 1.0
=1
-20 =2.175 0.5
-60 -40 -20 0 20 40 60 0.0
-1.0 -0.5 0.0 0.5 1.0
x(unite s of L)
FIG. 2: (Color online) The dimensionless energy spectrum FIG.3: (Coloronline)Theprobabilitydensitydistributionof
versus γ for K =±2π/L, α=2.175 (dashed line) and α=1 relative motion for the ground state with γ = 1 and α = 1
(solid line). (dash dot line), γ = 1 and α = 2.175 (dashed line), γ = 5
and α=1 (short dashed line), γ =5 and α=2.175 (dashed
dotted dot line), γ = 500 and α = 1 (short dashed dotted
line), γ =500 and α=2.175 (red dotted).
cientsinEq.(18)decreaseasthe massrationαincreases.
On the other hand, while K in Eq.(18) is independent
of the mass ratio, k is related with α via the nonlinear
c′ effectively. In the limit of c → ∞, the wave function
equation (16). It is not easy to analytically analyze the
Ψ(x ,x ) vanishes whenever x =x .
influence of the mass ratio on k from Eq.(16). However, 1 2 1 2
The momentum distribution for atom with mass m
forthe caseofK =0,fromEq.(19)orEq.(20),wenotice 1
can be represented as
that the mass ratio α affects the solution of k by effec-
tively changing the interaction strength via c′ and thus
∗
k increases with the increase in α. n (p )= Ψ (p ,p )Ψ(p ,p )dp ,
1 1 1 2 1 2 2
Z
For the attractive case with c < 0, it is worthy of in-
dicating that there exists a bound state for each set of
where Ψ(p ,p ) is the Fourier transformation of
1 2
statescharacterizedbyK whentheinteractionexceedsa
Ψ(x ,x ). The definition of momentum distribution
1 2
threshold c . The threshold of γ = c /ρ for the appear-
t t t n (p)fortheatomwithmassm issimilar. Fortheequal-
2 2
ance of bound state is given by
mass case, the momentum distribution for atom 1 or 2
should be the same since the wave function Ψ(x ,x ) is
1 1 KL 1 2
γt = 1+ cos −1 . (23) invariantunderthe exchangeofx1 andx2. However,the
2(cid:18) α(cid:19)(cid:20) (cid:18)1+α(cid:19) (cid:21) unequal-mass atomic system will show different features
since the wave functions are not necessary to be sym-
For K =0, we have γ =0. This implies that the atoms
t metrical for exchanging two unequal mass atoms except
form bound state for arbitrary weak attraction γ < 0.
of the case of K =0. For K =0, we find that the wave-
For K = 2π/L, we have γ < 0. In this case, atoms can
t functionisinvariantunderexchangeofx andx evenfor
1 2
form bound state only for γ <γ .
t the unequal-mass case, and thus we have n (p)=n (p).
1 2
In Fig.4, we display the ground state momentum distri-
bution for the unequal-mass system with α = 2.175 and
IV. DENSITY AND MOMENTUM
theequal-masssystem. Anobviouspeakaroundthezero
DISTRIBUTION
momentum position can be observed. With the increase
in the interaction strength, the momentum distribution
To show the influence of the interaction on the wave- tends to spread out widely and the height of the peak
function, we plot the ground state density distribution decreases. There is no obvious difference for momentum
for the relative motion defined as ρ(x) = ϕ∗(x)ϕ(x) in distributionsofsystemswithα=2.175andα=1asthe
Fig.4. The density distribution displays a cusp at x= 0 influence of α can be attributed to the effective interac-
due to the repulsive interaction when two atoms coin- tion c′.
cide. The relative motion density distribution at x = 0 Next we discuss the momentum distribution for the
decreasesas the interactionstrengthincreases. The den- lowest state with K = ±2π/L. For these cases, unequal
sity distribution at x = 0 for the case of α > 1 is below mass will have significant impact on the wavefunction
thatoftheequal-masscaseasaconsequenceofincreasing andmomentumdistributionasthewavefunctionsarenot
5
the increase in the strength of repulsive interactions,the
K=0 =5 =1
) 1.0 =5 =2.175 =5 K=2 /L
2 ==550000 ==12.175 ) 0.6 =1 ==55 0K0= K-2=2/L/L
/
L 0.8 2 =500 K=-2 /L
of L/ 0.5
es 0.6 of 0.4
t s
(uni 0.4 nite 0.3
) u
n(p 0.2 p)( 0.2
(
n
0.1
0.0
-5 -4 -3 -2 -1 0 1 2 3 4 5
0.0
p(unites of 2 /L) -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
p(unites of 2 /L)
FIG.4: (Coloronline)Singleparticlemomentumdistribution
of the ground state for γ = 5 and α = 1 (dot), γ = 5 and FIG.6: (Coloronline)Singleparticlemomentumdistribution
α= 2.175 (solid line), γ = 500 and α= 1 (square), γ = 500 ofthelowest statewithK =±2π/Landα=1forγ =5and
and α=2.175(dashed line). γ =500.
) (a)1.0 =5 nn21((pp)) KK==22 //LL momentum distributions become broaderand the height
2 0.8 n2(p) K=-2 /L of the main peak shrinks. As a comparison, in Fig.6 we
/
f L 0.6 n1(p) K=-2 /L display the momentum distribution for the lowest states
o 0.4 of the equal-mass systems with K = ±2π/L. Due to
es 0.2 the exchange symmetry of the wave function, the mo-
t
ni 0.0 mentum distributions n1(p) = n2(p) when m1 = m2.
p)(u (b)0.8-5 -4 -3 =5-200-1 0 1 2 nn21((3pp)) KK==422 //LL 5 pFoearkKsa=ro2uπn/dL0,athnedm2πo/mLe,nwtuhmiledtihsteridbiusttrioibnudtiiosnplhayassttwwoo
n(2 0.6 nn21((pp)) KK==--22 //LL peaks around 0 and −2π/L for K = −2π/L. With the
p) 0.4 increase in the strength of the repulsive interaction, the
( height of the peaks also shrinks and the momentum dis-
n1 0.2 tributions become broader.
0.0
-5 -4 -3 -2 -1 0 1 2 3 4 5
p(unites of 2 /L) V. CONCLUSIONS
In summary, we have exactly solved the problem of
FIG.5: (Coloronline)Singleparticlemomentumdistribution
of the lowest state with K = ±2π/L and α = 2.175 for (a) two heteronuclear atoms interacting with a short-range
γ =5 and (b) γ =500. δ potential in a ring trap. For the general case with an
arbitrary mass ration α, the relative momentum is cou-
pled with center-of-mass momentum due to the periodic
invariantundertheexchangeoftwoheteronuclearatoms. boundary condition. By taking the generalized Bethe-
Consequently, the momentum distributions n (p) and ansatz-type wave function, we have presented a detailed
1
n (p) are different when m 6= m . In Fig.5, we display derivation of the analytical solution. While α = 1, our
2 1 2
the momentum distributions for the lowest states of the solutionreducesto the Bethe-ansatzsolutionof the two-
systemswithK =±2π/Landα=2.175. ForK =2π/L, particle Lieb-Liniger model. We have applied our an-
the momentum shows the peak of n (p) is around 0 and alytical results to studying the energy spectrum, den-
1
that of n (p) is around 2π/L. The momentum distribu- sitydistributionandmomentumdistributionsofthehet-
2
tionforthenoninteractingsystemwiththesameK andα eronuclearsystem,whichdifferfromthe caseofidentical
hasasimilarstructureastheloweststatewithK =2π/L atoms.
is composed of the product of single atom states charac- Acknowledgments.– This work is supported by
terized by k = 0 and k = 2π/L. For K = −2π/L, the NSF of China under Grants No.10821403 and
1 2
themomentumdistributionsshowthatthepeakofn (p) No.10974234, and National Program for Basic Research
1
is around 0 and that of n (p) is around −2π/L. With of MOST.
2
6
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