Quantum phases in resonantly-interacting gasesA collision of two particles is well described by...

Quantum phases in resonantly-interacting gases

Erik Tillema

Institute for Theoretical Physics, University of Utrecht, Netherlands

supervisor: Prof. dr. H.T.C. Stoofdrs. D.B.M. Dickerscheid

June 26, 2007

Contents

1 Introduction 3

2 Scattering theory 5

2.1 Interacting spinless particles . . . . . . . . . . . . . . . . . . . 5

2.2 Relation with bound state . . . . . . . . . . . . . . . . . . . . 11

3 Feshbach resonance 15

4 Atoms in an optical lattice 24

4.1 Optical lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.2 Atom light interaction . . . . . . . . . . . . . . . . . . . . . . 26

5 Feshbach resonance in an optical lattice 31

6 The generalized Hubbard model 37

7 Calculations and results 43

8 Conclusion 44

2

1 Introduction

The year 2005 was the Einstein year because it was exactly 100 years afterthe experimental confirmation of Einstein’s most famous theory, his theoryof relativity. Recently Einstein’s work has been in the news again, becauseRowdy Boeijink, a student from Utrecht Univerisity, discovered an originalmanuscript of Einstein. But this time it didn’t involve relativity but hisprediction of a special state of matter that occurs at temperatures close toabsolute zero temperature, called a Bose-Einstein condensate.

A Bose-Einstein condensate is a state of matter formed by bosons cooled totemperatures very near to absolute zero. Under such supercooled conditions,a large fraction of the particles collapse into the lowest quantum state, atwhich quantum effects become apparent on a macroscopic scale. In order toconfine the particles in the condensate, it is common to use a magnetic fieldminimum.

Ever since the experimentalists Eric Cornell and Carl Wieman confirmedthe existence of this state of matter in 1995, a lot of research has beendone in the area of ultra-low temperature physics. Bose-Einstein conden-sates have proved to be useful in exploring a wide range of questions infundamental physics, including superfluidity and quantized vortices. It islikely to be some time before any practical applications are developed, butatomic lithography, coherent matter wave optics and quantum computingare among the application areas proposed.

In this thesis we also focus on gases at ultra-low temperature, but underspecial circumstances. Firstly, we will mainly consider a mixture of bosonsand fermions.

Secondly, we will only look at values of the magnetic field strength whereparticles interact resonantly, or at least one of the possible interacting par-ticle pairs. This phenomenon is known as a Feshbach resonance.

Thirdly, we will use an optical trapping lattice in our system since veryunstable molecules will be formed, which need to be confined. So we’ll haveto take into account the effect the laser light has on the system.

Combining the atom-atom interaction and the effect of the optical latticewith a Hubbard model, we will give a total many-body Hamiltonian for this

3

system. From this, we will use Bogoliubov theory to obtain the equation ofstate for the system and numerically solve this.

4

2 Scattering theory

We start off with some important concepts of scattering theory. We will alsogive an example of so-called single channel scattering.

2.1 Interacting spinless particles

When we have a gas of particles at low temperature and density, one of theimportant things is how two particles of a gas interact when they get closeto each other. The chance that three or more particles get in each othersinteraction range at the same time is so small compared to two particles,that we won’t even consider this case.

A collision of two particles is well described by scattering theory, so we willintroduce some important concepts here.

Suppose that two spinless particles of mass m move towards each other withrelative momentum 2~~k. In the coordinate system at rest with respect tothe center of mass of the two particles, this situation is described by theHamiltonian

H =−~2~∇2

2µ+ V (~r), (2.1)

with reduced mass µ = m/2 and interaction potential V . Since we assumedthe relative momentum to be 2~~k, we are looking for solutions of the Hamil-tonian with energy E = 2~

2k2

2m = ~2k2

2µ , where we defined k = |~k|.Note that because V is independent of time, we are looking for stationarysolutions of the time-dependent Schrodinger equation of the form

Ψ(~r, t) = ψ(~r)e−iEt~ . (2.2)

For atoms and molecules the interaction potential has a very small range,and as a result V will vanish for large values of |~r|. This means that wecan look at this system as two free particles with relative momentum 2~~k,which interact in a small region and come out of the interaction region astwo free particles again but with relative momentum 2~~k′. Since energy isconserved, |~k′| must be equal to |~k|.

5

Without loss of generality we can assume ~k to be in the direction of thez-axis of our system, so the whole situation can be shown schematically asin figure 2.1.

Figure 2.1: Schematic representation of two-particle scattering in the centerof mass reference frame. The region where the interaction takes place isindicated by the black circle.

From this we see that the solution ψ of equation 2.1 must be independentof φ in terms of the spherical coordinates (r, φ, θ): ψ(~r) = ψ(r, θ).

Outside of the interaction region the particles move freely, so for large r thewave function must have the form of the superposition of an incoming planewave with momentum ~k and a set of outgoing plane waves,

limr→∞ψ(r, θ) = N (ψin + ψout) , (2.3)

ψin(~r) = ei~k·~r, (2.4)

ψout(r, θ) = f(k, θ)eikr

r, (2.5)

where N is a normalization constant.

One can check that ψout satisfies the Hamiltonian of the free particle withmass µ and has eigenenergy ~2k2

2µ . So indeed ψout represents the possible out-

6

going plane waves and the function f determines how the outgoing particlesare distributed.

An important notice is that f completely determines ψ far from the inter-action region, and that f must depend on the interaction potential V .

In order to find the solution to equation 2.1, or equivalently, to find f(k, θ)we follow a method called the method of partial waves (see for exampleBransden and Joachain (2003)). Therefore we expand the wave function ina series of Legendre polynomials as

ψ(k, r, θ) =∞∑

l=0

Rl(k, r)Pl(cos θ), (2.6)

and by combining this with equation 2.1 we get[

d2

dr2− l(l + 1)

r2− U(r) + k2

]ul(k, r) = 0, (2.7)

where we defined ul(k, r) = rRl(k, r) and U(r) = 2µV (r)~2 .

For large r, we know that ψ must have the form of equation 2.3, so weexpand

ei~k·~r =∞∑

l=0

(2l + 1)il

krsin

(kr − lπ

2

)Pl(cos θ), (2.8)

f(k, θ) =∞∑

l=0

fl(k)Pl(cos θ), (2.9)

limr→∞ψ(r, θ) = N

∞∑

l=0

[(2l + 1)il

krsin

(kr − lπ

2

)+ fl(k)

eikr

r

]Pl(cos θ)

.

(2.10)

As is shown in Bransden and Joachain (2003) for example, it turns out that

fl(k) =2l + 12ik

(e2iδl(k) − 1

), (2.11)

for some set of functions δl.

In our case, the gas is at low temperature so the kinetic energy of theparticles will be low. Hence also the value of k will be small. Now by taking

7

a good look at equation 2.7 we see that as l increases the centrifugal barrierterm l(l + 1)/r2 becomes more important than the potential term U(r), sothat for sufficiently large l, U(r) can be neglected and the correspondingpartial-wave amplitude fl is negligible. In our case of ultra-low energy, eventhe l = 1 centrifugal barrier term will be much greater than U(r). Weconclude that for small k only the l = 0 terms have to be taken into accountin the sums over l. This principle is called s-wave scattering.

Now, we define a constant called the s-wave scattering length (or just scat-tering length in short) a by

a = − limk↓0

δ0(k)k

. (2.12)

One can check that f0(0) = −a and because we only take the l = 0 part ofthe sum over l, we conclude that for small k the scattering length completelydetermines the eigenfunction far from the interaction region.

To get a feeling for the meaning of the scattering length, let us look at asimple example where we take a square well for the interaction potential:

V (r, φ, θ) =

V0 if r < R

0 if r ≥ R, (2.13)

for some R > 0 and V0 < 0. Plugging this potential into equation 2.7 andtaking only s-wave interaction into account, we obtain for l = 0

[d2

dr2+ k2 − 2µV0

~2

]u<(k, r) = 0, (2.14)

[d2

dr2+ k2

]u>(k, r) = 0, (2.15)

where

u0(k, r) =

u<(k, r) if r < R

u>(k, r) if r ≥ R. (2.16)

Solving this yields

u<(k, r) = Aeik<r + Be−ik<r, (2.17)

u>(k, r) = Ceikr + De−ikr, (2.18)

8

with k< =√

k2 − 2µV0

~2 ∈ R. Demanding that the wave function is welldefined at the origin, u<(k, r = 0) must vanish and we obtain A = −B.Also, from the r → ∞ limit in the s-wave part of the wave function (seeequation 2.10) we get

ψ(k, r) = N

[sin(kr)

kr+ f0(k)

eikr

r

]

= N

[1

2ikr

(eikr − e−ikr

)+ f0(k)

eikr

r

]. (2.19)

Comparing this with equation 2.18 and using equation 2.11 we see that

D

C=

−N 12kri

N 12kri + N f0(k)

r

, (2.20)

−C

D= 1 + 2kif0(k)

= e2iδ0(k). (2.21)

This only tells us that if we calculate C and D, we know δ0, and this givesus a from its definition, which in turn gives us ψ.

So we calculate C and D, by demanding that the wave function is smoothlyjoint at r = R:

u<(k, R) = u>(k, R), (2.22)∂u<

∂r(k, R) =

∂u>

∂r(k, R), (2.23)

which yields

A

D

[eik<R − e−ik<R

]= −e2iδ0(k)eikR + e−ikR, (2.24)

A

D

[k<eik<R + k<e−ik<R

]= −e2iδ0(k)keikR − ke−ikR. (2.25)

9

Diving these two equations gives

2i sin (k<R)2k< cos (k<R)

=eiδ0(k)

[−eiδ0(k)eikR + e−iδ0(k)e−ikR]

eiδ0(k)k[−eiδ0(k)eikR − e−iδ0(k)e−ikR

]

=−2i sin (δ0(k) + kR)−2k cos (δ0(k) + kR)

, (2.26)

1k<

tan(k<R

)=

1k

tan (δ0(k) + kR) , (2.27)

δ0(k) = −kR + arctan[

k

k<tan

(k<R

)]. (2.28)

From this we can determine the scattering length, using

limk↓0

tan(k<R

)= tan

(√−2µV0

~2R

), (2.29)

limk↓0

arctan(kx) = kx,, (2.30)

and plugging this in the definition of a,

a = R− 1√−2µV0

~2tan

(√−2µV0

~2R

)

= R

(1− tan γ

γ

), (2.31)

with γ = R√−2µV0

~2 , a dimensionless constant.

In figure 2.2 we see how the scattering length depends on γ. Clearly, when-ever γ = (n + 1

2)π with n ∈ Z, we have a resonance of the scattering length.

We have done a lot of calculations, so let us sum it all up. We have a Hamil-tonian to solve, with the restriction that far from the interaction region weknow the form of the solution. We find the exact eigenfunction correspond-ing to low values of the eigenenergy (through A,B,C and D). B = −A, C/Dis calculated and A/D is calculated. D can be obtained from demanding theeigenfunction to be normalized. And we see that far from the interactionregion the wave function is

limr→∞ψ(r, θ) = N

[ei~k·~r − a

eikr

r

]. (2.32)

10

2 4 6 8 10Γ

-10

-7.5

-5

-2.5

2.5

5

7.5

10

aR

Figure 2.2: Scattering length in units of the potential range as a function of

the parameter γ = R√−2µV0

~2

From this equation we see the meaning of the scattering length: small ameans hardly any particles are scattered and almost all particles will stayundeflected, because of the relatively small contribution of the sphericalwave part in the wave function. However when a is large, a lot of particleswill scatter. Note that the scattered particles are isotropically distributed.

Although for much more complicated interaction potentials the exact eigen-function can no longer be calculated, we will still know the wave functionfor large r by just measuring the scattering length.

An important notice is that the scattering length depends strongly on V0, aswe can see in figure 2.2. It also turns out that there is an important relationbetween the scattering length and the eigenenergy of the least bound statein the system. We will discuss this now.

2.2 Relation with bound state

In order to find the least bound state, we solve the time independent Schrodingerequation Hψb(~r) = Ebψb(~r) with V0 < Eb < 0. Expanding the wave func-tion in spherical harmonics ψb(r, φ, θ) =

∑∞l=0

∑lm=−l Rb,El(r)Ylm(φ, θ) and

11

considering only the s-wave part we get[−~2

2µ

(d2

dr2+

2r

d

dr

)+ V (r)

]Rb,E0(r) = EbRb,E0(r). (2.33)

By using ub(r) = rRb,E0(r) as usual, we obtain[

d2

dr2+

2µ

~2(Eb − V (r))

]ub(r) = 0, (2.34)

and we find the solution

ub(r) =

u<

b (r) if r < R

u>b (r) if r ≥ R

, (2.35)

where

u<b (r) = Aeik<

b r + Be−ik<b r, (2.36)

u>b (r) = Cek>

b r + De−k>b r, (2.37)

moreover, k<b =

√2µ~2 (Eb − V0) ∈ R and k>

b =√−2µ~2 Eb ∈ R. From de-

manding that the wave function is normalizable we obtain B = −A andC = 0, and from demanding the wave function to be smoothly connected atr = R we get k>

b = −k<b cot

(k<

b R), or

√−2µ

~2Eb = −

√2µ

~2(Eb − V0) cot

(R

√2µ

~2(Eb − V0)

). (2.38)

One can prove that this equation has n solutions for Eb in the region V0 <Eb < 0, when (n − 1

2)π < γ < (n + 12)π. Differently put, when V0 is such

that γ is slightly larger than (n + 12)π for certain n, then a new loosely

bound state is added to the system. See figure 2.3 for the eigenenergies ofthe bound states.

This new bound state has energy |Eb| ¿ |V0|, so we can estimate from

12

2 4 6 8 10 12 14Γ

-1

-0.8

-0.6

-0.4

-0.2

EbÈV0È

Figure 2.3: Bound state energy in units of the potential depth as a functionof the parameter γ

equation 2.38 that√−2µ

~2Eb ' −

√2µ

~2(−V0) cot

(R

√2µ

~2(−V0)

)

= −γ cot(γ)R

=[−R

tan(γ)γ

]−1

= (a−R)−1 , (2.39)

or Eb = − ~22µ(a−R)2

.

The conclusion is that if we could alter γ by altering V0 and slowly increaseγ, we would see that the scattering length goes through a resonance (therebysatisfying |a| À |R|) precisely when a new bound state is supported by thepotential, and that the energy of this bound state and the scattering lengthare related by

Eb = − ~2

2µa2if |a| À |R|. (2.40)

13

This relation does not only hold for the specific potential that we have cho-sen. As is shown in R.A. Duine (2003), this relation holds for any attractivepotential.

In principle, therefore, if we had experimental access to the value of theenergy of the least bound state of the system we could alter the scatteringlength and thereby the effective interaction of the particles. This is basicallyimpossible to achieve for the system of spinless particles we chose above, butcan be done when particles with an internal state are considered. We willdiscuss this in the next chapter.

14

3 Feshbach resonance

In this chapter we will show that in a system of two particles with internalstates in an magnetic field, we can control how the particles scatter.

Suppose we have two alkali atoms of the same species in a homogeniousmagnetic field with strength B at ultra-low temperature.

Because the energy of the atoms is low, we can expect that their spin-statewill have minimum energy. This means that all electrons will be in thelowest orbits and the most outer electron (called the valence electron) willbe alone in its outer orbit. Therefore we can look at this atom as if it hada nucleus with charge +e, consisting of the real atom nucleus plus all innerelectrons, and a single electron, the valence electron with charge −e. Inshort, we consider the alkali atom to have only one electron and thereforeonly two internal eigenstates, corresponding to the two eigenstates of thespin-operator of the valence electron.

The total wave function of the system of two alkali atoms must be an eigen-function of the Hamiltonian (acting on the spatial part of the total wavefunction) and the total spin operators |~Stot|2 and Sz,tot (acting on the spinpart of the total wave function). And from atomic physics, we know thatthe eigenfunctions of the spin part of the total wave function are one anti-symmetric singlet state en three symmetric triplet states.

Furthermore, a deeper study of atom-atom interactions, considering electronorbitals and the Pauli exclusion principle, shows that the effective potentialin the Hamiltonian is deeper when the spin eigenstate is a singlet statethan when it’s a triplet state. Therefore we have different eigenenergiesand eigenstates of the spatial part of the system, depending on whether thespin state is a triplet or a singlet. From now on, we will label spatial wavefunctions ψ and potentials V with a T or S indicating we are dealing witha triplet or singlet spin state, respectively.

These two spin-states or internal states are said to be two channels throughwhich the atoms can scatter. We should compare this with the previouschapter, where we considered two spinless particles. Since the spinless par-ticle system has no internal states, we called this system a single channelsystem.

In order to calculate the eigenstate and eigenenergie of the spatial part of

15

our two-channel system, we look at the system from the coordinate systemwhere the center of mass of the two particles is at rest, as usual. Let ~r bethe relative coordinate and µ the reduced mass.

Another important fact when considering this system, is that there is acoupling between the two spin-states. The cause of this coupling is theatomic hyperfine interaction between the spin of the atom nucleus and thespin of the electron, so the coupling is given by

Vc =ahf

~2

(~I1 · ~S1 + ~I2 · ~S2

), (3.1)

where ahf is the hyperfine constant, ~Ij the nucleus spin of particle j and ~Sj

the electron spin of particle j.

The general time-independent Schrodinger equation for a system with twocoupled eigenstates is

(HT(~r) Vc

Vc HS(~r)

)(ψT(~r)ψS(~r)

)= E

(ψT(~r)ψS(~r)

). (3.2)

When we also apply a homogeneous magnetic field with strength B on thewhole system, we get a Zeeman-shift (see for example Bransden and Joachain(2003)) and we have the following Hamiltonian:

Hi(~r) =−~2~∇2

2µ+ Vi(~r) + ciB, (3.3)

where i ∈ T,S, Vi is the effective potential and ci the magnetic moment.

So we get the following matrix for our system(−~2 ~∇2

2µ + VT(~r)− E Vc

Vc−~2 ~∇2

2µ + VS(~r) + ∆cB − E

)(ψT(~r)ψS(~r)

)= 0, (3.4)

with ∆c = cS − cT > 0 and all energy levels are shifted by an amount ofcTB.

As an example, we take the following situation. The triplet and singletpotentials are both a square well:

Vi(r, φ, θ) =

Vi if r < R

0 if r ≥ R, (3.5)

16

for some R > 0 and Vi < 0, i ∈ T, S. We assume that |VS| > |VT| sincethe effective singlet potential is deeper than the triplet potential, and weassume that 0 < Vc ¿ Vi, ∆cB.

Also, we choose B such that ∆cB is of the order of |VS|. This will turn outto be the condition for which the singlet potential supports a bound statewith small energy ES < 0, as we will see shortly.

The triplet state is coupled to the bound singlet state, but actually the boundsinglet state is also coupled to other triplet states with different energies. So,through the singlet state, the triplet state is coupled to the whole continuousspectrum of triplet states. This means that we cannot simply perform quasi-degenerated perturbation theory to obtain the perturbed eigenstates andenergies from the unperturbed eigenstates and energies.

Instead, we follow a different approach where we first solve the unperturbedsinglet and triplet states and then use a perturbation matrix with energyoperators and states rather than eigenenergy values and wave function am-plitudes.

In this way, we obtain the perturbed eigenenergies and the perturbed eigen-states. By comparing the wave function of the perturbed eigenstates withsmall but positive energies with the general scattering wave function, weobtain the scattering length of the system as a function of B.

So, firstly, we calculate the unperturbed triplet state ψT with energy ET =~2k2

2µ . We solve the time independent Schrodinger equation HTψT = ETψT

with ET = ~2k2

2µ and HT = −~2 ~∇2

2µ + VT(~r).

Just like in the single channel case of chapter 2, we get

u<T(r) = ATeik<

T r + BTe−ik<T r, (3.6)

u>T(r) = CTeikr + DTe−ikr, (3.7)

with k<T =

√k2 − 2µVT

~2 . Demanding that the wave function is smooth andsquare integrable, we get the values of AT, BT, CT and DT.

Secondly, we calculate the unperturbed singlet state ψS with energy ES < 0by solving HSψS = ESψS with ES / 0 and HS = −~2 ~∇2

2µ + VS(~r) + ∆cB.

17

Following the same steps we made in section 2.2, we obtain[

d2

dr2− 2µ

~2(VS + ∆cB − ES)

]u<

S (k, r) = 0, (3.8)[

d2

dr2− 2µ

~2(∆cB − ES)

]u>

S (k, r) = 0, (3.9)

and

u<S (r) = ASe

ik<S r −ASe

−ik<S r, (3.10)

u>S (r) = 0 + DSe

−k>S r, (3.11)

where where k<S =

√2µ~2 (ES −∆cB − VS) ∈ R and k>

S =√−2µ~2 (ES −∆cB) ∈

R.

By comparing equation 3.8 and 3.9 with 2.34 we see that if VS supports abound state, ES−∆cB depends on VS as is shown in figure 2.3. Let’s say thatVS supports one bound state −αVS with 0 ≤ α ≤ 1, thus ES−∆cB = −αVS.By choosing B such that ∆cB / αVS we see that we have a solution withES / 0. In other words, by varying B while keeping ∆cB ≤ αVS we cancontrol the energy of the bound state directly. More specifically, we can setthe bound state energy approximately equal to the unbound triplet state.

Notice also that actually we don’t need ES to be negative. If ES > 0, thenfor some B it will still hold that ES−∆cB = −αVS thereby being a solutionto the Schrodinger equation. And the wave function will still be of the form3.11, guaranteeing that its state is bound in the sense that the chance ofdetecting the two atoms at distance r À R is very small. We concludethat the bound singlet state eigenenergy is dependent on the experimentallyaccessible magnetic field strength according to

ES = ∆c(B −B0), (3.12)

where B0 = αVS∆c .

In order to calculate the perturbed eigenstates and eigenenergies, we rewritethe Hamiltonian matrix in a more simple form and we describe the systemin terms of states instead of wave functions,

(HT −E Vc

Vc HS − E

)(|ψT〉|ψS〉

)= 0. (3.13)

18

analysis to estimate the second integral, we obtain

E −ES = g2 2µ

~2

∫ 2π

0dφ

∫ π2

−π2

dθ

∫ ∞

−∞dr r2 sin(θ)

[1r2− 1

r2 − 2µE~2

]− lim

r↓0

∫~k∈R3

d~kei~k·~r

k2

= g2 2µ

~2

4π

∫ ∞

−∞dr

[−2µE

~2

r2 − 2µE~2

]− lim

r↓01

4πr

. (3.18)

Notice that the first integral in equation 3.18 will surely diverge when E ≥ 0,while it converges when E < 0 since

∫∞−∞ dx a2

x2+a2 = πa. So, we continueassuming E < 0,

E −ES = g2 2µ

~2

(4π2

√−2µE

~2− lim

r↓01

4πr

)

= g2

[4π2

(2µ

~2

) 32 √

−E − limr↓0

µ

2π~2r

]. (3.19)

We see that we have a divergence in equation 3.19, but the divergence isindependent of E. Therefore, we will solve this problem by renormalizingES, such that for the renormalized singlet eigenenergy we have

E − ES = 4π2g2

(2µ

~2

) 32 √

−E. (3.20)

As an example, we now choose VS = −10 ~22µR2 , VT = −1 ~2

2µR2 and Vc =

0.1 ~22µR2 . In figure 3.1 we have plotted the solution of equation 3.20. As

expected, we see that there is no perturbed eigenstate with negative energywhen the singlet potential no longer supports a bound state.

In figure 3.2, we show for this example the scattering length of the perturbedunbound state. This state with positive eigenenergy, which we will label with↑, is also called the open channel, because only in this state the particles cancollide and come out of the interaction area unbound. On the other hand,in the perturbed state with negative eigenenergy, which we will label with ↓,the particles are bound and therefore this state is called the closed channel.

As we can see, the scattering length behaves resonantly with it’s peak at∆cB ≈ 4.6 ~2

2µR2 . We can explain this: the singlet potential supports one

20

-0.02 -0.015 -0.01 -0.005

ESÑ2H2ΜR2L

-0.015

-0.0125

-0.01

-0.0075

-0.005

-0.0025

0.0025

EÑ2H2ΜR2L

Figure 3.1: Perturbed eigenenergy of the bound state for two coupled square-well potentials as a function of the magnetic field strength dependent singletstate energy. The depth of the singlet and triplet channel potentials areVS = −10 ~2

2µR2 and VT = −1 ~22µR2 respectively. The hyperfine coupling is

Vc = 0.1 ~22µR2 .

21

Figure 3.2: Scattering length for two coupled square-well potentials as afunction of ∆cB. The depth of the singlet and triplet channel potentials areVS = −10 ~2

2µR2 and VT = −1 ~22µR2 respectively. The hyperfine coupling is

Vc = 0.1 ~22µR2 . The dotted line shows the background scattering length abg.

22

bound state with energy −αVS ≈ −0.46VS when γS =√

10 as can be seenfrom figure 2.3. So when ∆cB = αVS ≈ 4.6 ~2

2µR2 we have ES = 0, theresonance peak.

Also, we see that there is a certain scattering length called the backgroundscattering length abg, indicated by the horizontal dashed line, to which aconverges far from the resonance. We can explain why abg ≈ −0.56R: farfrom the resonance, ψ↑ ≈ ψT and in figure 2.2 we can see that γT =

√1

corresponds to a ≈ −0.56R.

As is shown in R.A. Duine (2003), it turns out that also for this system,close to the resonance, the bound state energy is related to the scatteringlength by ES = − ~2

2µa2 just like in the single channel case.

Also we state here that, near a resonance, the scattering length dependenceon the magnetic field strength can be approximated by

a(B) = abg

(1− ∆B

B −B0

). (3.21)

In our example, abg ≈ −0.56R, B0 ≈ 4.6 ~22µR2∆c

and ∆B ≈ −0.05 ~22µR2∆c

.

To conclude this chapter, we stress that by altering the value of the magneticfield strength, which is experimentally accessible, we can set the scatteringlength to almost any value and therefore have complete control of the scat-tering process.

23

4 Atoms in an optical lattice

Suppose the two atoms in the model described in the previous chapter area boson and a fermion. Under the right conditions, the boson and fermioncan form a molecule together, which in turn will be fermionic.

When B − B0 ¿ 0, the two atoms will (in equilibrium of the many-atommodel) generally be in the ψ↓ eigenstate because of its low eigenenergy. Inthis case, ψ↓ ≈ ψS which is a bound state in the sense that the chance ofdetecting the two particles at distance r À R is very small. Differently put,when B − B0 ¿ 0 almost all atoms will have formed two-atom molecules.Moreover, these molecules will be fermions and at the ultra-low temperatureswe assume that a Fermi cloud of these molecules will form.

However, when B − B0 À 0, the eigenstate with minimum energy of twoatoms will be equal to a triplet state. This is not a bound state in the sensethat the atoms can be at distances r À R. Thus, at temperatures close tothe absolute zero temperature, the bosonic atoms will form a Bose-Einsteincondensate and the fermionic atoms will form a Fermi cloud of fermionicatoms.

When manually increasing the magnetic field strength, somewhere in be-tween of B−B0 ¿ 0 and B−B0 À 0 there must be a change from a Fermicloud of molecules, to a Bose-Einstein condensate of atoms in combinationwith a Fermi cloud of atoms. As it will turn out later, this will be an instanttransition and not a smooth one. In other words, we will have a so-calledquantum phase transition at a certain value of the magnetic field strength.The main goal of the calculations that will follow in this thesis will be todetermine this value.

But before we continue, we have to explain something about the problemsthat arise when an experimentalist would investigate the quantum phasetransition. When experimentally determining the point of the quantumphase transition, one would have to detect at different values of the magneticfield strength whether the system contains a Fermi cloud of molecules. How-ever, observing the molecular Fermi cloud is problematic: the freely movingmolecules will collide with each other, causing the unstable molecules to fallapart into atoms again.

In order to overcome this problem, we will apply a laser light field on thesystem in such a way that a so-called optical lattice is created, in which

24

particles can only be likely to have position at certain points (the latticepoints) rather than be free to move around the entire space. Moreover, thelaser light field will be set to make tunneling between lattice points unlikely.This way, a particle will be more or less ‘stuck’ at the lattice point whereit’s positioned and this is exactly what we want to happen.

By making sure that every lattice point is unlikely to contain more than oneor two particles, we get the situation in which molecules at a lattice pointcan’t fall apart into atoms because there are no particles for the molecule tocollide with. In other words, we have isolated the molecules. The result isthat experimentally detecting the molecular Fermi cloud is now a lot easier.

4.1 Optical lattice

Constructing such an optical lattice is relatively easy by using six lasers,two for each each of the three space dimensions. In each dimension the twolasers will be put towards each other on either side of the space where theexperiment is taking place, creating a standing wave.

As we will discuss in detail later on in this chapter, the interaction betweenthe particles and the laser light will have the effect that particles will belikely to be positioned in the ‘valleys’ of the standing waves1. One standingwave will force the particles to be situated in a series of surfaces with smallthickness perpendicular to the light propagation vectors of the two lasers,one surface for each valley in the standing wave. Likewise, two standingwaves in two dimensions will force the particles to be situated at the crosssection of the surfaces resulting in a two-dimensional array of lines perpen-dicular to all four laser wave vectors. Furthermore, three standing waveswill create a three-dimensional lattice of points (with small thickness). Seefigure 4.1 for a visual representations of the multi-dimensional lattices. Seealso D. van Oosten (2004).

In order to find out how the atom-atom interaction is changed by the opti-cal lattice, we first have to investigate the atom-light interaction. See alsoBransden and Joachain (2003) and D. van Oosten (2004) for more informa-tion on this subject.

1at least, for a blue-detuned system.

25

Figure 4.1: Visual representation of the atomic density in a one- dimen-sional (left), two-dimensional (middle) and three-dimensional (right) opticallattice.

4.2 Atom light interaction

The force that laser light exerts on atoms can be split into two categories,the radiation force and the dipole force. The radiation force is due to theabsorption and emitting of photons, where the outgoing photons are emittedin all direction whereas the incoming photons are all from the same direction(the laser). The result is an average force in the direction of the laserbeam. Since we have opposite laser beams and since we will choose ourlaser parameters such that the radiation force is very small in comparisonto the dipole force, we will neglect the radiation force.

The dipole force is caused by the fact that the ground and excited stateenergies of the atom shift in the presence of light. We will try to make this aunderstandable a little without going into details too much. See for exampleMetcalf and Van der Straten (2006) for more information.

Suppose a particle has a ground state ψg with energy Eg and an excitedstate ψe with energy Ee. Under the influence of light, it turns out thatthe two states |ψg, N fotons〉 and |ψe, N − 1 fotons〉 are coupled where thecoupling strength ~Ω

2 defines the so-called Rabi frequency Ω:

(Ee −E ~Ω

2~Ω2 Eg + ~ω −E

)(ψe

ψg

)= 0. (4.1)

Origin of the coupling is the potential energy of the atomic dipole ~d in the

26

electric field ~E of the laser light

~Ω2

= 〈ψe| − ~d · ~E|ψg〉= −〈ψe|~d|ψg〉 · ~E (4.2)

∝√

I, (4.3)

where we see from the last equation that the Rabi frequency is proportionalto the square root of the light intensity I. Note that the light intensity andtherefore the Rabi frequency is experimentally accessible.

When solving equation 4.1, we have to take into account that the eigenenergyof the excited state could have a negative imaginary part, to represent thatthe excited state decays. To see this we write Ee = Re(Ee)− i~Γ

2 , R 3 Γ ≥ 0and we see that the solution of the time dependent Schrodinger equationψe(~r, t) = ψ0(~r)e−

iEt~ = ψ0(~r)e−

iRe(E)t~ e−

Γt2 decays in time.

To simplify the calculations a bit, we introduce a variable called the detuningδ = Re(Ee) − (Eg + ~ω) so equation 4.1, after shifting all energy values byan amount of Eg + ~ω, transforms into

(δ − i~Γ

2 −E ~Ω2~Ω

2 −E

)(ψe

ψg

)= 0. (4.4)

We find E by demanding that the determinant of the matrix on the lefthand side of equation 4.4 vanishes,

E =δ − i~Γ

2

2±

√√√√(~Ω2

)2

+

(δ − i~Γ

2

2

)2

. (4.5)

Let’s first consider the situation δ > 0, or equivalently Re(Ee) > Eg + ~ω,also called red-detuned. In that case we have

Ered,+ ≈ δ − i~Γ2

+

(~Ω2

)2

δ − i~Γ2

, (4.6)

Ered,− ≈ −(~Ω

2

)2

δ − i~Γ2

. (4.7)

where we have assumed that∣∣δ − i~Γ

2

∣∣ À ~Ω.

27

Differently said, the eigenenergies are shifted by an amount of

(~Ω2

)2

δ − i~Γ2

=

(~Ω2

)2δ

δ2 +(~Γ

2

)2 + i

(~Ω2

)2 ~Γ2

δ2 +(~Γ

2

)2

≈(~Ω

2

)2

δ+

i~2

(~Ω2

)2 Γδ2

, (4.8)

where we have assumed that |δ| À |~Γ|. Putting it all together we find for

the ground state that Re(Eg) is light shifted by an amount Els,red = −(~Ω2 )2

|δ| .

The ground state in fact also picks up a small decay factor Γg = (~Ω2 )2Γ

δ2 whichwe shall ignore.

Now in our setup of lasers, the Rabi frequency of the light Ω is positiondependent just like the light intensity is position dependent. Let’s say thatone of the pairs of opposite lasers creating a standing light wave is alignedalong the x-axis. Then the Rabi frequency from this laser is described byΩx(~r) = Ωx,0 cos

(2πxλx

)where λx is the wavelength of the laser light and

Ωx,0 a constant. Plugging this in our equation for Els,red and expandingaround the minima gives us

Els,red,x(~r) = −

(~Ωx,0

2

)2

|δ| cos2(

2πx

λx

)

≈

(~Ωx,0

2

)2

|δ|(

2π

λx

)2 [x− n

λx

2

]2

−

(~Ωx,0

2

)2

|δ| , (4.9)

with n ∈ Z. In the approximation above we have used − cos2(y) ≈ (y −nπ)2 − 1 around y = nπ.

Differently said, around the valleys of Els,red,x (which are at x = nλx2 ) the

energy shift behaves like Els,red,x(~r) ≈(~Ωx,0

2

)2

|δ|(

2πλx

)2x2 −

(~Ωx,02

)2

|δ| . Notethat these valleys are ‘in the light’, or differently put, at positions withmaximum light intensity.

Taking the laser duos in the y and z direction into account in the samemanner and setting all laser light wavelengths equal to λ, we find that aroundthe lattice points (which are at (x, y, z) = (nλ

2 ,mλ2 , lλ

2 ), (n,m, l) ∈ Z3) it

28

holds that

Els,red(r, φ, θ) ≈(~Ω0

2

)2

|δ|(

2π

λ

)2

r2 − 3

(~Ω02

)2

|δ| , (4.10)

where we have chosen Ω0 = Ωx,0 = Ωy,0 = Ωz,0.

From now on, we will take the atom light interaction into account by treat-ing the laser influence as a part of the potential. That is, we write theHamiltonian H as H = H0 + Els where H0 is the Hamiltonian for the samesystem but without lasers.

Combining all information we can now give the Hamiltonian for a free par-ticle in ground state with mass m at a single site in an optical lattice

H =−~2~∇2

2m+

(~Ω02

)2

|δ|(

2π

λ

)2

r2

=−~2~∇2

2m+

12mη2r2, (4.11)

where we have introduced the parameter η, defined as

η =~Ω0π

λ

√2

|δ|m , (4.12)

in order to write the Hamiltonian in a way that resembles the harmonicoscillator Hamiltonian. Also, note that we have shifted all energies by an

amount of 3

(~Ω02

)2

|δ| , equal to the position independent second term on theright hand side of equation 4.10.

Had we chosen the so-called blue-detuned situation δ < 0, then we wouldhave found for the ground state that Re(Eg) was light shifted by an amount

Els,blue = (~Ω2 )2

|δ| . Considering position dependency, we now find for theblue-detuned system that

Els,blue(r, φ, θ) ≈(~Ω0

2

)2

|δ|(

2π

λ

)2

r2, (4.13)

around the lattice points, which are at (x, y, z) = (λ4 + nλ

2 , λ4 + mλ

2 , λ4 + lλ

2 ),(n, m, l) ∈ Z3. Notice that in this case the particles are ‘in the dark’, ordifferently put, at positions with minimum light intensity. However, just like

29

in the red-detuned case, we find the same Hamiltonian given by equation4.11, and in this case without having to perform an energy shift.

In extension, the Hamiltonian of a particle A with mass mA in ground stateand a particle B with mass mB also in ground state which have an interactionpotential Vint around the same lattice point in this system, is given by

H =−~2~∇2

A

2mA+

12mAη2

Ar2A +

−~2~∇2B

2mB+

12mBη2

Br2B + Vint. (4.14)

Solving a system with this Hamiltonian analytically becomes much easier ifwe separate the Hamiltonian in a center of mass part and a relative partin the usual way. However, this is only possible when ηA and ηB are equal.Conveniently, this can be accomplished as follows.

From the definition of η we see that we have to achieve |δA|mA = |δB|mB.Remember that the detuning was defined as δ = Re(Ee)− (Eg + ~ω) so wejust choose our laser parameter

ω =Re(Ee,A)− Eg,A − mB

mA(Re(Ee,B)− Eg,B)

~(1− mBmA

). (4.15)

Of course, if particle A and B are of the same species then for every ω wewill have ηA = ηB.

For example, let particle A be a 87Rb atom in hyperfine state |1, 1〉 andand let particle B be a 40K atom in hyperfine state |9/2,−9/2〉 then onewould choose ω = 2.3 · 1015 s−1. Differently put, one would choose the laserwavelength to be 806 nm.

Separating the center of mass part and the relative part of this Hamiltonian,we obtain

H =−~2~∇2

A

2mA+

12mAη2

Ar2A +

−~2~∇2B

2mB+

12mBη2

Br2B + Vint

=−~2~∇2

COM

2M+

12Mη2r2

COM +−~2~∇2

rel

2µ+

12µη2r2

rel + Vint. (4.16)

where M = mA+mB the total mass, µ = mAmBmA+mB

the reduced mass, rCOM =mArA+mBrB

mA+mBthe position of the center of mass, rrel = rA − rB the relative

position and η = ηA = ηB.

30

5 Feshbach resonance in an optical lattice

Now we have all the information to set up the complete Hamiltonian matrixfor the system we actually have in mind. We examine the system in areference frame at rest with respect to the center of mass.

Just like in the previous chapter, we look at a system of two alkali atoms Aand B at ultra-low temperature in a magnetic field with strength B but inthe presence of an optical lattice as described earlier.

Again, the atoms can form either a singlet or a triplet with different in-teraction potentials and we have a coupling between the singlet and tripleteigenstates.

All this gives us the following Hamiltonian matrix(−~2 ~∇2

2µ + 12µη2r2 + VT(~r)− E Vc

Vc−~2 ~∇2

2µ + 12µη2r2 + VS(~r) + ∆cB − E

)(ψT(~r)ψS(~r)

)= 0.

(5.1)

Here ~r is the relative position, again ∆c = cS − cT > 0 is the difference inmagnetic moment and all energy levels are shifted by an amount of cTB.

Since this system is hard to solve in general, we will consider the special casewhere particle A is a Potassium atom and particle B is a Rubidium atom.Both these atoms are alkali atoms and the Potassium atom is a fermionwhile the Rubidium atom is a boson. From now on, we will use the label Finstead of A for the Potassium atom to make it easier for the reader to seewhether certain variables apply to the fermion or the boson.

Solving the unperturbed triplet Hamiltonian we can neglect the triplet in-teraction potential VT(~r) in comparison to the harmonic oscillator part ofthe triplet Hamiltonian, because |VT| ¿ 3

2~η. Therefore the unperturbedtriplet eigenstates are just the well known three-dimensional harmonic os-cillator states φn for a particle with mass µ and oscillation frequency η andeigenenergy (n + 3

2)~η, n ∈ N.

Just like in chapter 3, we choose the magnetic field strength B such that theunperturbed singlet Hamiltonian has a loosely bound state (i.e. a molecularstate in the sense that detecting the atoms far from each other is very slim).

31

Therefore, we can neglect the contribution of the harmonic oscillator part ofthe Hamiltonian 1

2µη2r2 since |~r| will be very small in the region where thebound state eigenfunction doesn’t vanish. And although we can’t exactlycalculate the bound state eigenfunction because we don’t know VS, we canexpect that just like in chapter 3 the bound state eigenenergy will dependon the magnetic field strength as ES = ∆c(B −B0).

So we just state here that HSψS = ESψS for some bound state function ψS

and some energy ES strongly dependent on B in the same way as before.

Now, we will use the same approach to obtain the perturbed eigenstates andenergies as we did in chapter 3.

Temporarily rewriting the Hamiltonian matrix in the more simple form anddescribing the system in terms of states instead of wave functions,

(HT −E Vc

Vc HS − E

)(|ψT〉|ψS〉

)= 0, (5.2)

we follow the equations after 3.13, and we see again that

E − ES = 〈ψS| (E −HS) |ψS〉= 〈ψS|Vc (E −HT)−1 Vc|ψS〉=

∑n

〈ψS|Vc (E −HT)−1 |φn〉〈φn|Vc|ψS〉

=∑

n

〈ψS|Vc|φn〉〈φn|Vc|ψS〉E − ET,n

, (5.3)

where we have used (E −HT)−1 |φn〉 = 1E−ET,n

|φn〉 since |φn〉 are the eigen-

states of HT with ET,n = (n + 32)~η.

Now, we’re going to use the pseudo-potential approximation in order toestimate Vc. In this approximation, we have 〈~r|Vc|ψS〉 = gδ(~r) where g isdetermined by the shape of the scattering length resonance when scatteringthe two atoms in singlet state. More specifically,

g = ~

√2πabg∆c∆B

µ, (5.4)

with ∆B the resonance width and abg the background scattering length.Both these latter values are determined experimentally by measuring the

32

scattering length in the resonance and fitting the results with the equationa(B) = abg

(1− ∆B

B−B0

), seen earlier in chapter 3.

Plugging this in twice gives us

E − ES =∑

n

∫

~r∈R3

d~r

∫

~r′∈R3

d~r′〈ψS|Vc|~r〉〈~r|φn〉〈φn|~r′〉〈~r′|Vc|ψS〉

E −ET,n

=∑

n

∫

~r∈R3

d~r

∫

~r′∈R3

d~r′g2 δ(~r)〈~r|φn〉〈φn|~r′〉δ(~r′)E − ET,n

= g2∑

n

φn(0)φ∗n(0)E − ET,n

. (5.5)

In order to solve this complex equation, we follow Busch et al. (1998). Firstwe note that only s-wave functions will contribute to the sum since theseare the only ones for which φn(0) doesn’t vanish. From this, we see thateigenfunctions with odd n don’t contribute to the sum. Also we can writethe contributing functions with even values of n as

φ2n(~r) =(

1πl2

) 34[L

( 12)

n (0)]− 1

2

e−r2

2l2 L( 1

2)n

(r2

l2

), (5.6)

ET,2n =(

2n +32

)~η, (5.7)

where l =√

~µη and L

(α)n are the generalized Laguerre polynomials.

Plugging this in equation 5.5 while inserting a limit and using φ2n(0) =

33

(1

πl2

) 34

[L

( 12)

n (0)] 1

2

we obtain

E −ES = g2 limr→0

∑n

φn(0)φ∗n(r)E − ET,n

= g2 limr→0

∑n

φ2n(0)φ∗2n(r)E −ET,2n

= g2

(1

πl2

) 32

limr→0

e−

r2

2l2∑

n

L( 1

2)n

(r2

l2

)

E − (2n + 3

2

)~η

= − g2

2~η

(1

πl2

) 32

limr→0

e−

r2

2l2∑

n

L( 1

2)n

(r2

l2

)

n− ν

, (5.8)

where ν = E2~η − 3

4 . Now we make use of the following equations fromadvanced mathematics

∑n

L( 1

2)n

(r2

l2

)

n− ν= Γ(−ν)U

(−ν,

32,r2

l2

), (5.9)

limr→0

Γ(−ν)U(−ν,

32,r2

l2

)= −√π

[2Γ(−ν)

Γ(−ν − 12)− l

r+O

(r

l

)], (5.10)

where Γ is the well known Gamma function and U the confluent hypergeo-metric function.

Plugging these in equation 5.8, we finally get

E − ES =g2

~ηπl3

[Γ(−ν)

Γ(−ν − 12)− lim

r→0

l

2r

]

= g2

[1

~ηπl3Γ(−ν)

Γ(−ν − 12)− lim

r→0

µ

2π~2r

]. (5.11)

Note that this equation contains a divergence, but it is of exactly the sameform as the divergence we saw when considering the system without theoptical lattice in equation 3.19. In the case with the optical lattice, thedivergence is caused by the use of the pseudo-potential. For now, we will

34

state that by renormalizing ES again, we can solve this problem and wecontinue with a renormalized ES in

E − ES = g2

[1

~ηπl3Γ(−ν)

Γ(−ν − 12)

]. (5.12)

Also note that this equation gives us the perturbed eigenenergies E of thesystem: for every value of ES (or differently put since ES = ∆c(B − B0):for every value of B), we have the equation variable E on the left hand sideand ν = E

2~η − 34 on the right hand side, determining the allowed values for

E.

-30 -20 -10 10 20 30

ESÑΗ

-10

-8

-6

-4

-2

2

4

EÑΗ

Figure 5.1: Perturbed eigenenergies in units of ~η as a function of the mag-netic field strength dependent bound state energy. The horizontal dashedlines correspond to the unperturbed triplet eigenenergies and the ascendingdashed line corresponds to the unperturbed singlet eigenenergy. This figurewas calculated for g2

(~η)2πl3= 11.2.

The perturbed eigenenergies are shown in figure 5.1. As expected, we seethat whenever the unperturbed singlet eigenenergy ES gets close to a tripleteigenenergy value such as 3

2~η, we have an avoided crossing. Note that thenon-s-wave triplet solutions with energies (2n + 5

2)~η, n ∈ N do not comeforth in the figure, again as expected.

Also, we label all perturbed eigenstates with a σ value, σ = 0 for the eigen-state with the lowest eigenenergy line in figure 5.1, σ = 1 for the eigenstatewith the second lowest eigenenergy line, etc.

35

For later use, we now calculate the amplitude of the singlet state wavefunction in the perturbed wave functions. We define

Zσ = |〈ψS | ψσ〉|2 =1

1− ∂~Σ∂E (Eσ)

, (5.13)

where we use the singlet state self energy ~Σ(E) = g2[

1~ηπl3

Γ(−ν(E))

Γ(−ν(E)− 12)

].

Several values of Z are plotted in figure 5.2.

-30 -20 -10 10 20 30

ESÑΗ

0.1

0.2

0.3

0.4

0.5

Z

Figure 5.2: Amplitude of the singlet state in the perturbed wave functionas a function of the magnetic field strength dependent bound state energy.This figure was calculated for σ ∈ 0, 1, 2 and g2

(~η)2πl3= 11.2.

Now, we have all the information we need from two-body physics to continueto the next step in finding the quantum phase transition. In order to findthe value of B where system changes from a molecular condensate to aatomic condensate, we finally apply many-body theory making use of allthe calculations in this chapter.

36

6 The generalized Hubbard model

Up until now, we have only looked at two interacting atoms. However, inthe end we have an experiment in mind with a whole gas of atoms and notjust two. In order to correctly describe a gas of many interacting atoms wewill use quantum field theory. More specifically, we will set up a so-calledgeneralized Hubbard model to describe our system. This model is a gener-alized version of the original Hubbard model, which describes electrons in acrystal lattice. For more information on Hubbard models, see for exampleRasetti (1991).

As in general in quantum field theory, we start off in second quantizationnotation by writing a system Hamiltonian together with creation and anni-hilation field operators.

As mentioned earlier, the system we have in mind is a gas of Rubidium andPotassium atoms in an optical lattice and under the presence of a magneticfield. Since we will choose the number of particles low with respect to thenumber of lattice points, we assume that no more than two atoms appear ateach of the lattice points. Since we have ultra-low temperature, we considerjust one possible state for each boson and fermion, the ground state. Andwe consider a number of possible states for each duo of a boson and fermion,indicated with a σ and corresponding to the perturbed eigenstates of thetwo-particle system treated in the previous chapter with lowest eigenenergy.From now on, we will call such a duo a molecule, although we take in mindthat depending on the magnetic field strength and the value of σ, the twoparticles in the duo do not necessarily have to be bounded in their relativedistance. All in all we have the following Hamiltonian:

H =− tB∑

〈i,j〉b†ibj − tF

∑

〈i,j〉f †i fj − tm

∑σ

∑

〈i,j〉c†σ,icσ,j

+UBB

2

∑

i

b†ib†ibibi +

UBF,bg

2

∑

i

b†if†i fibi

+ g′∑

σ

∑

i

√Zσ

(c†σ,ibifi + b†if

†i cσ,i

)

+∑

i

(εB − µB)b†ibi +∑

i

(εF − µF)f †i fi +∑

σ

∑

i

(εσ − µB − µF)c†σ,icσ,i,

(6.1)

where b†i and bi are the creation and annihilation operators of the boson at

37

lattice point i respectively, f †i and fi are the creation and annihilation op-erators of the fermion at lattice point i respectively and c†σ,i and cσ,i are thecreation and annihilation operators of the molecular states at lattice pointi respectively; εB and εF are the boson and fermion ground state energyrespectively and εσ are the molecular energies; µB and µF are the chemicalpotentials of the boson and fermion respectively (not to be confused withthe reduced mass µ); tB, tF and tm are the boson, fermion and moleculehopping parameters respectively; UBB and UBF,bg are the boson-boson andbackground boson-fermion interaction parameters respectively; g′ is the ef-fective atom-molecule coupling; Zσ are the amplitudes of the singlet statewave function in the perturbed wave functions;

∑i denotes a sum over all

lattice points and∑〈i,j〉 denotes a sum over all nearest neighbours in the

lattice.

Since we have a pretty extensive Hamiltonian, we will make some remarkson it. The first three sums describe the hopping of the atoms and moleculesbetween different lattice points. The second three sums describe the on-siteinteraction between particles, that is one sum for the on-site boson-boson in-teraction, one sum for the background on-site boson-fermion interaction andfinally one sum for the resonant on-site boson-fermion interaction. Lastly,the last three sums with particle energies and chemical potentials are addedbecause we are going to perform our calculations in the grand-canonical en-semble. Note that because of the Pauli exclusion principle there is no on-sitefermion-fermion interaction.

The hopping parameters can be expressed in terms of lattice parametersusing the so-called tight binding model. This model assumes that the eigen-functions of particles are very small at distances exceeding the lattice con-stant, or in other words the particles are tightly bound to the lattice points.So, in the tight binding model, a particle (boson, fermion or molecule) ata lattice point is simply in the ground state of a harmonic potential withorbital frequency η. See for example Ashcroft and Mermin (1976) for moreinformation on the tight binding model. We obtain

ti =~η2

[1−

(2π

)2](

λ

4li

)2

e−

(λ4li

)2

, (6.2)

li =

√~

miη, (6.3)

with i ∈ B,F, m, η is the orbital frequency of the laser light induced

38

harmonic potential (see chapter 4), li are the harmonic oscillator lengths, λis the laser light wavelength and mB, mF and mm = mB+mF are the boson,fermion and molecule masses, respectively. In fact, the hopping parametercomes from the overlap of the ground state harmonic oscillator wave functionwith neighbouring lattice points, which are at distance λ

2 .

Also, we already know that the atom ground state energies are simply thelowest harmonic oscillator eigenenergies: εB = εF = 3

2~η.

Now, close to the a resonance of the boson-fermion interaction, we expect theon-site boson-boson interaction and the on-site background boson-fermioninteraction to be much smaller than the resonant interaction. Since we areinvestigating our system near resonance, we will neglect these two sums inthe Hamiltonian from now on.

The effective atom-molecule coupling strength in the optical lattice is givenby

g′ = g1

π3/4l3/2rel

, (6.4)

lrel =

√~µη

=√

l2B + l2F, (6.5)

where the atom-molecule coupling g was given by equation 5.4. See forexample D. van Oosten (2004) for how to use so-called Wannier functionsto calculate effective parameters in an optical lattice.

Our next step is to determine the equation of state for the many-particlesystem. Therefore we do a Fourier transform to momentum space for thecreation and annihilation operators in the Hamiltonian, that is we definecreation and annihilation operators b~k

, b†~k, f~k, f †~k , c

σ,~k, c†

σ,~ksuch that

bi =1√Ns

∑

~k

b~ke−i~k·~ri , b†i =

1√Ns

∑

~k

b†~kei~k·~ri , (6.6)

fi =1√Ns

∑

~k

f~ke−i~k·~ri , f †i =

1√Ns

∑

~k

f †~kei~k·~ri , (6.7)

cσ,i =1√Ns

∑

~k

cσ,~k

e−i~k·~ri , c†σ,i =1√Ns

∑

~k

c†σ,~k

ei~k·~ri , (6.8)

where Ns is the number of lattice sites and ~ri is the coordinate of site i. Formathematical convenience, we have started off with a finite volume V for

39

our system. This implies that the momenta ~~k are discretized, which allowsus to write sums instead of integrals. One could take the continuum limitlater.

Using∑

i ei~k·~ri = Nsδ~k

, one can check that

∑

i

b†ibi =∑

~k

b†~kb~k, (6.9)

−tB∑

〈i,j〉b†ibj =

∑

~k

εB,~k

b†~kb~k, (6.10)

where εB,~k

= −2tB∑3

i=1 cos(ki

λ2

)and this also holds for the fermion and

molecule operators. Also using

∑

i

c†σ,ifibi + f †i b†icσ,i =1√Ns

∑

~k,~k′,~k′′

(c†σ,~k

f~k′b~k′′ + f †~k′b†~k′′

cσ,~k

)δ~k−~k′−~k′′ ,

(6.11)

we obtain the Hamiltonian in momentum space

H =∑

~k

(εB,~k

+ εB − µB

)b†~kb~k

+∑

~k

(εF,~k

+ εF − µF

)f †~kf~k

+∑

σ

∑

~k

(εσ,~k

+ εσ − µB − µF

)c†σ,~k

cσ,~k

+g′√Ns

∑σ

∑

~k,~k′,~k′′

√Zσ

(c†σ,~k

f~k′b~k′′ + f †~k′b†~k′′

cσ,~k

)δ~k−~k′−~k′′ . (6.12)

Now, we continue with the Bogoliubov approach: we replace the boson cre-ation and annihilation operators by an expectation value plus a fluctuationand minimize the energy of the gas with respect to the fluctuations, to sec-ond order. See D. van Oosten (2004) for more information on Bogoliubovapproximation.

In our case, the simplest way to perform the Bogoliubov approach is to con-sider the system at very negative values of ES (or B − B0 ¿ 0). As wasalready mentioned at the beginning of chapter 4, we then expect no conden-sate bosons: 〈b†0b0〉 = 0. Since the expection value is zero, all operators inthe Hamiltonian can now be considered as fluctuations.

40

So we see that there is no part of the Hamiltonian that is constant (H(0))or linear (H(1)) in the fluctuations. Moreover, the quadratic part of theHamiltonian H(2) is given by

H(2) =∑

~k

(εB,~k

+ εB − µB

)b†~kb~k

+∑

~k

(εF,~k

+ εF − µF

)f †~kf~k

+∑

σ

∑

~k

(εσ,~k

+ εσ − µB − µF

)c†σ,~k

cσ,~k

. (6.13)

Therefore, in the Bogoliubov approach we have an effective HamiltonianHeff = H(0) + H(2) = H(2) given above.

Finally, plugging in the Bose distribution for the bosons and the Fermiondistribution for the fermions and molecules, we obtain the total number ofbosons, fermions and molecules

NB =∑

~k

〈b†~kb~k〉Heff =

∑

~k

1

eβ~ω

B,~k − 1, (6.14)

NF =∑

~k

〈f †~kf~k〉Heff =

∑

~k

1

eβ~ω

F,~k + 1, (6.15)

Nσ =∑

~k

〈c†σ,~k

cσ,~k〉Heff =

∑

~k

1

eβ~ω

σ,~k + 1, (6.16)

where the brackets 〈〉Heff denote the expectation value as calculated withthe effective Hamiltonian, and

~ωB,~k

= εB,~k

+ εB − µB, (6.17)

~ωF,~k

= εF,~k

+ εF − µF, (6.18)

~ωσ,~k

= εσ,~k

+ εσ − µB − µF, (6.19)

and β = 1kBT depends on the temperature T ; kB is the Boltzmann constant.

Macroscopically more interesting than the numbers of particles are theirlattice filling fractions, defined as ni = Ni

Ns, i ∈ B, F, σ. Of course, the

total filling fraction n given by

n = nB + nF + 2∑

σ

nσ, (6.20)

is a macroscopic constant. Also, we can experimentally control the fractionbetween the total number of bosons and fermions in the system, whether in

41

atomic form or part of a molecule. Let’s say for now that we choose thisfraction to be one. That implies that under all circumstances there will beas many bosonic atoms as there are fermionic atoms:

nB = nF. (6.21)

Of course, a different fraction would lead to a different linear equation be-tween nB and nF.

It’s important to realize that equations 6.20 and 6.21 fully determine thestate of the system (therefore, the first is also known as the equation ofstate). They determine the numbers of particles in each state and in ourcase specifically whether we have a Bose-Einstein condensate in our systemor not.

Now, given the experimentally accessible values of n, T and B, also giventhat the boson is a 87Rb atom in hyperfine state |1, 1〉 and the fermion isa 40K atom in hyperfine state |9/2,−9/2〉, lastly given the experimentallyaccessible laser parameters λ (satisfying equation 4.15) and Ω0. Using thetwo macroscopic equations 6.20 and 6.21 we can solve for the two unknownvariables µB and µF. This gives us the value of nB and more generallywhether there is a Bose-Einstein condensate or not. In other words, we cannow calculate the phase diagram of the system: a plot showing the stateof the system depending on n, T and B. This way we can see, for fixedvalues of n and B for example, for which temperature the system changesfrom a fermi cloud of molecules to a state with a Bose-Einstein condensateof atoms.

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7 Calculations and results

In the last chapter we saw how, for a specific case, the phase diagram canbe calculated.

For the case where the boson is a 87Rb atom in hyperfine state |1, 1〉 andthe fermion is a 40K atom in hyperfine state |9/2,−9/2〉, we have performedthese calculations numerically on a computer, using an algorithm in theimperative programming language C++. The result has been plotted infigure 7.1.

Figure 7.1: Phase diagram as a function of the total filling fraction, magneticfield strength and temperature.

From this we see for fixed values of n and B for example, for which tem-perature the system changes from a fermi cloud of molecules to a state witha Bose-Einstein condensate of atoms. Or, one can see for which value of Bthe system state changes when keeping n and T constant.

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8 Conclusion

In summary, we have shown how at ultra-low temperatures two particlescan interact resonantly and how this resonance is related to bound particlestates. We have shown how the behaviour of the particles around the reso-nance is experimentally accessible by using particles with spin in a magneticfield and given an example of how the magnetic field dependent interactioncan be calculated for a specific case. Furthermore, we have investigatedhow the use of an optical lattice influences the interaction of two particlesand calculated for a Bose-Fermi mixture of 40K and 87Rb atoms the exactform of the interaction resonance. Lastly, we have shown how a generalizedHubbard model can be used to calculate how a whole gas of the Bose-Fermimixture under the influence of an optical lattice behaves macroscopicallyaround the resonance. We have used this to calculate a phase diagram forlow filling fractions as a function of the applied magnetic field and the tem-perature, thereby determining the conditions for which a quantum phasetransition takes place in the mixture.

In our final calculations, we have used equal trapping frequencies for bothatomic species by using a red-detuned optical lattice. However, as is shownin D.B.M. Dickerscheid (2006) the investigated quantum phase transitionalso exists without this assumption. We leave it for future investigations toexplore this interesting option.

44

References

B.H. Brandsen and C.J. Joachain (2003). Physics of Atoms and Molecules.Pearson Education Limited 1983, 2003.

B.H. Brandsen and C.J. Joachain (2000). Quantum Mechanics. PearsonEducation Limited 1989, 2000.

J.J. Sakurai (1994). Modern Quantum Mechanics. Addison-Wesley Publish-ing Company, 1994.

R.A. Duine (2003). Atom-Molecule Coherence in Bose Gases.

D. van Oosten (2004). Quantum gases in optical lattices: the atomic Mottinsulator.

D.B.M. Dickerscheid (2006). Quantum phases in optical lattices.

N.W. Ashcroft and N.D. Mermin (1976). Solid State Physics. ThomsonLearning, Toronto, 1976.

H.J. Metcalf and P. van der Straten (2006). Laser Cooling and Trapping(Graduate Texts in Contemporary Physics). Springer, 2006.

M. Rasetti (1991). The Hubbard Model (Recent results). World ScientificPub Co Inc, 1991.

D.B.M. Dickerscheid, U. Al Khawaja, D. van Oosten and H.T.C. Stoof(2005). Feshbach resonances in an optical lattice. Phys. Rev. A 71 (2005)043604 cond-mat/0409416.

D.B.M. Dickerscheid, D. van Oosten, E.J. Tillema and H.T.C. Stoof (2005).Quantum phases in a resonantly-interacting Bose-Fermi mixture. Phys.Rev. Lett. 94 (2005) 230404 cond-mat/0502328.

T. Busch, B.-G. Englert, K. Rzazewski and M. Wilkens (1998). Two coldatoms in a harmonic trap. Foundations of Physics 28, 549 (1998)

S. Inouye, J. Goldwin, M.L. Olsen, C. Ticknor, J.L. Bohn and D.S. Jin(2004). Observation of Heteronuclear Feshbach Resonances in a Bose-Fermi Mixture. Phys. Rev. Lett 93 (2004) 183201 cond-mat/0406208.

R. Grimm, M. Weidemuller and Y.B. Ovchinnikov (2000). Optical dipoletraps for neutral atoms. Advances in Atomic, Molecular and OpticalPhysics Vol. 42, 95-170 (2000) cond-mat/9902072.

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