Atom-Atom Interactions in Ultracold Gases Claude Cohen-Tannoudji

Atom-atom interactions in ultracold gases Claude Cohen-Tannoudji

To cite this version:

Claude Cohen-Tannoudji. Atom-atom interactions in ultracold gases. DEA. Institut Henri Poincaré, 25 et 27 Avril 2007, 2007. cel-00346023

HAL Id: cel-00346023 https://cel.archives-ouvertes.fr/cel-00346023 Submitted on 12 Dec 2008

HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Atom-Atom Interactions in Ultracold Quantum Gases

Claude Cohen-Tannoudji

Lectures on Quantum Gases Institut Henri Poincaré, Paris, 25 April 2007

Collège de France 1 Lecture 1 (25 April 2007) Quantum description of elastic collisions between ultracold atoms The basic ingredients for a mean-field description of gaseous Bose Einstein condensates

Lecture 2 (27 April 2007) Quantum theory of Feshbach resonances How to manipulate atom-atom interactions in a ultracold quantum gas

2 A few general references

1 – L.Landau and E.Lifshitz, Quantum Mechanics, Pergamon, Oxford (1977) 2 – A.Messiah, Quantum Mechanics, North Holland, Amsterdam (1961) 3 – C.Cohen-Tannoudji, B.Diu and F.Laloë, Quantum Mechanics, Wiley, New York (1977) 4 – C.Joachin, Quantum collision theory, North Holland, Amsterdam (1983) 5 – J.Dalibard, in Bose Einstein Condensation in Atomic Gases, edited by M.Inguscio, S.Stringari and C.Wieman, International School of Physics Enrico Fermi, IOS Press, Amsterdam, (1999) 6 – Y. Castin, in ’Coherent atomic matter waves’, Lecture Notes of Les Houches Summer School, edited by R. Kaiser, C. Westbrook, and F. David, EDP Sciences and Springer-Verlag (2001) 7 – C.Cohen-Tannoudji, Cours au Collège de France, Année 1998-1999 http://www.phys.ens.fr/cours/college-de-france/ 8 – C.Cohen-Tannoudji, Compléments de mécanique quantique, Cours de 3ème cycle, Notes de cours rédigées par S.Haroche http://www.phys.ens.fr/cours/notes-de-cours/cct-dea/index.html/ 9 – T.Köhler, K.Goral, P.Julienne, Rev.Mod.Phys. 78, 1311-1361 (2006) 3 Outline of lecture 1 1 - Introduction 2 - Scattering by a potential. A brief reminder • Integral equation for the wave function • Asymptotic behavior. Scattering amplitude • Born approximation 3 - Central potential. Partial wave expansion • Case of a free particle • Effect of the potential. Phase shifts • S-Matrix in the angular momentum representation 4 - Low energy limit • Scattering length a • Long range effective interactions and sign of a 5 - Model used for the potential. Pseudo-potential • Motivation • Determination of the pseudo-potential • Scattering and bound states of the pseudo-potential • Pseudo-potential and Born approximation 4 Interactions between ultracold atoms At low densities, 2-body interactions are predominant and can be described in terms of collisions. We will focus here on elastic collisions (although inelastic collisions and 3-body collisions are also important because they limit the achievable spatial densities of atoms). Collisions are essential for reaching thermal equilibrium At very low temperatures, mean-field descriptions of degenerate quantum gases depend only on a very small number of collisional parameters. For example, the shape and the dynamics of Bose Einstein condensates depend only on the scattering length Possibility to control atom-atom interactions with Feshbach resonances. This explains the increasing importance of ultracold atomic gases as simple models for a better understanding of quantum many body systems Purpose of these lectures: Present a brief review of the concepts of atomic and molecular physics which are needed for a quantitative description of interactions in ultracold atomic gases. 5 Notation Two atoms, with mass m, interacting with a 2-body interaction (GG) potential Vr12− r In lecture 1, we ignore the spins degrees of freedom. They will be taken into account in lecture 2. Hamiltonian pp22(GG) H =++−12Vr r (1.1) 22mm 12 Change of variables GGGG/ G G RrrGG=+()122 Ppp =+1 2Center of mass variables Mmm=+=2 mTotal mass GGG12 G G G/ rrr=− p = pp − 2 Relative variables 12/ ( 1 2)/ μ =+=mm() m m m 2 Reduced mass 12G 1/ 2 / 22G G HP=+22 M( pμ ) + Vr( ) (1.2) G HCM Hrel Hamiltonian of a free Hamiltonian of a “fictitious” particle particle with mass M with mass μ, moving in V(r) 6 Finite range potential G Simple case where Vr( ) = 0 for r> b b is called the range of the potential

One can extend the results obtained in this simple case to potentials decreasing fast enough with r at large distances. For example, for the Van der Waals interactions between atoms 6 decreasing as C6 / r for large r, one can define an effective range / 14 ⎛⎞2μC 6 (1.3) bVdW = ⎜⎟2 ⎝⎠= See for example Ref. 5

7 Outline of lecture 1 1 - Introduction 2 - Scattering by a potential. A brief reminder • Integral equation for the wave function • Asymptotic behavior. Scattering amplitude • Born approximation 3 - Central potential. Partial wave expansion • Case of a free particle • Effect of the potential. Phase shifts • S-Matrix in the angular momentum representation 4 - Low energy limit • Scattering length a • Long range effective interactions and sign of a 5 - Model used for the potential. Pseudo-potential • Motivation • Determination of the pseudo-potential • Scattering and bound states of the pseudo-potential 8 Scattering by a potential. A brief reminder Shrödinger equation for the relative particle (with E>0) ⎡⎤==22G GG k 2 ⎢⎥−Δ+Vr()ψψ() r = E () r E = (1.4) ⎣ 22μ ⎦ μ G 2μ GG ⎡⎤Δ+kr2 ψψ() = Vrr() () ⎣⎦ =2 Green function of Δ + k2 G G ⎡⎤2 (1.5) ⎣Δ+kGr⎦ ( )() =δ r The boundary conditions for G will be chosen later on (1.6) Integral equation for the solution of Shrödinger equation G : ϕ0 (r ) Solution of the equation without the right member 2 G ⎡⎤Δ+krϕ =0 (1.7) ⎣⎦0 () G G GGGG ψ rrrGrrVrr=+ϕψd3 ′ −′′′(1.8) () 0 ( ) ∫ ( ) ( ) ( ) 9 Choice of boundary conditions

We choose for ϕ0 a plane. wave .with wave vector k GGG G GG G / ikr ikκ r (1.9) ϕ0 ()rk==ee κ =k and we choose, for the Green function G, boundary conditions corresponding to an outgoing spherical wave (see Ref.2, Chap.XIX) GG G G 1 eikr− r′ Gr()−=− r′ GG (1.10) + 4π rr− ′ We thus get the following solution for Schrödinger equation GG ikr− r′ ( ) G .G ( ) ( ) ++GGikr 1 3 e G ψ GGrrV=−ed′ GG r′′ψ r(1.11) kk4π ∫ rr− ′ If V has a finite range b, the integral over r’ is restricted to a finite range and we can write: G GGG G G If rbrrrrn ,.− ′ −=′ with nrr / G G ikr + GGik. r Ge ψκG ()rfkn e − (, ,) k r (1.12) GG G G 2μ 3 -.ikn r′ G + G fk(,,)κψ n=− derVr′ ()()′′G r 4π =2 ∫ k 10 Scattering state with an outgoing spherical wave Asymptotic behavior for large r ( ) + G The state ψ G r is a solution of the Schrodinger equation behaving k exp G .G for large r as the sum of an incoming plane wave ()ikr and of ( ,GG, )exp( )/ an outgoing spherical wave fkκ n ikr r Scattering amplitude f ( , , ) - G .G ( ) ( ) GG 2μ 3 ik′′ r GG+ fkκψ n = − d r′ e Vr′′G r (1.13) 2 ∫ k G4π = GG/ We have put kknk′ ==r r ( ,G ,G ) fkκ nis the amplitude of the outgoing spherical wave in the G GG/ direction of kknkrr′ ==GG. It depends only on k and on the polar angles θϕ and of kk′ with respect to Differential cross section Comparing the fluxes along k and k’, one gets : / ( ,G ,G )2 ddσκΩ= f kn (1.14) 11 Born approximation

In the scattering amplitude, the potential V appears explicitly

( , , ) - G .G ( ) ( ) GG 2μ 3 ik′′ r GG+ f knκψ=− der′ Vrr′′G (1.15) =2 ∫ k 4π ( ) + G G ′ To lowest order in V, one can thus replace ψ k r by the zeroth exp( G .G ) order solution of the Schrodinger equation ikr ( ). ( ,G ,GG) 2μ GG G ( ) f knκ =− de3r′ ik− k′′ r Vr′ (1.16) 4π =2 ∫ This is the Born approximation

In this approximation, the scattering amplitude is proportional to the spatial Fourier transform of the potential

12 Low energy limit The presence of V(r’) in the scattering amplitude - . ( , , ) G G ( ) ( ) G G 2μ 3 ik′′ r GG+ f knκψ=− der′ Vrr′′G (1.17) 4π =2 ∫ k restricts the integral over r’ to a finite range r’< b G If kb 1 , one can replace e−ik′. r′ by 1. The scattering amplitude

GG 2μ 3 GG+ ′′G ′ fk(,,)κψ n=− 2 d rVr()() r (1.18) 4π = ∫ k G then no longer depends on the directionG of the scattering vector k′. It is spherically symmetric even if Vr() is not. , ( ,G ,G ) When k →→0 f knκ − a . ( ) G G ikr (1.19) + G ikr e a ψ G ra e − →−1 k rr a is a constant, called “scattering length”, which will be discussed in more details later on 13 Another interpretation of the outgoing scattering state Another expression for this state (see refs. 4 and 8) / ++1 2 ψ GG=+ϕψμLim VTpG =2 (1.20) kkε →0 k + E − Ti+ ε + G For εψ non zero but very small, k appears as the state obtained at tt==0 by starting from the free stateϕ G at −∞ and by k / switching on slowly V on a time interval on the order of = ε Ingoing scattering state G G −−ikr r′ G .G −−( GG) ikr 1 3 e ( ) ( G) ψ GGrrV=−ed′ GG r′′ψ r kk4π ∫ rr− ′ (1.21) −−1 ψϕGG=+Lim V ψG kkε →0 k + E − Ti− ε If one starts from such a state at t = 0 and if one switches off V slowly on a time interval on the order of = / ε , one gets G the free stateϕk at t =+∞ 14 S - Matrix Definition ( , ) GG G G SS==ϕϕ ϕLim ϕ Utt, ji kkji kj21 ki (1.22) t1 →−∞ t →+∞ : 2 U evolution operator in interaction representation

− + One can show that S = ψ G ψ G (1.23) ji kkj i Qualitative interpretation V is switched on slowly (time scale ħ/ε) between -∞ and 0, and then switched off slowly (time scale ħ/ε) between 0 and +∞

One starts from ϕi at t = -∞ and one looks for the probability amplitude to be in ϕ at t = +∞ j , From tt= −∞ to = 0 the initial free state ϕ is transformed . i into ψψ+ Since the evolution operator is unitary, and since − i , j −+ transforms intoϕψjj from tt==+∞0 to ψiis the

amplitude to find the system in. the free state ϕ j at t =+∞ if one starts from ϕ at t =−∞ i 15 Outline of lecture 1 1 - Introduction 2 - Scattering by a potential. A brief reminder • Integral equation for the wave function • Asymptotic behavior. Scattering amplitude • Born approximation 3 - Central potential. Partial wave expansion • Case of a free particle • Effect of the potential. Phase shifts • S-Matrix in the angular momentum representation 4 - Low energy limit • Scattering length a • Long range effective interactions and sign of a 5 - Model used for the potential. Pseudo-potential • Motivation • Determination of the pseudo-potential • Scattering and bound states of the pseudo-potential 16 Central potential V depends only on r 1D radial Schrödinger equation One looks for solutions of the form G GGG ϕklm()rRrYn= kl () lm () nrr= / (1.24) If we put ur() Rr()= kl (1.25) kl r with the boundary condition ukl()00= (1.26) one gets for u the following 1D radial equation kl ( ) ⎡⎤d2 ll+ 12μ ( ) ( ) + kV2 −−rur =0 (1.27) ⎢⎥222kl ⎣⎦drr= 1D Schrödinger equation for a particle moving in a potential which is the sum of V and of the centrifugal barrier ( ) 2 = ll+ 1 (1.28) 2μ r 2 17 Case of a free particle (V=0) The solutions of the Schrödinger equation are: ( ) (G ) 2k 2 ( ) (G ) ϕ 0 r = j kr Y n (1.29) klm π l lm where the jl are the( spherical) Bessel functions of order l ( ) kr l ( ) 1 sin( π ) j kr ( )!! j kr kr− l (1.30) llrr→→0 21lk+ ∞r2 For large r we thus have sin( / ) ( ) (GG) 22( ) kr− lπ ϕ 0 rYn klm r →∞ lm π ( /r ) ( / ) (1.31) ikrl−−−ππ22ikrl 2 (G )⎡⎤ee− = Yn⎣⎦ π lm 2ir = outgoing spherical wave + ingoing spherical wave These functions form an orthonormal set (see Appendix) ( ) ( ) ( ) 00 ϕklm′ ′′ ϕδ klm =−kk δδ′ ll′′ mm (1.32) 18 Expansion of a plane wave in free spherical waves Plane wave / . G GG( ) G G G(GG) rk==2πδ−32 eikr k′ k k− k′ (1.33)

The factor (2π) -3/2 is introduced for the orthonormalization One can show that: / . / * ( ) G G ( ) ∞=+ml( ) ( ) ( ) ( ) −−32ikr 32 l G G 224πππei= ∑∑ YlmnY lm κj l kr lml==−0 * ( ) ∞=+ml( ) ( ) ( ) 1 l GG0 = iYκϕ r (1.34) k ∑∑ lm klm lml==−0 G G / with κ = kk G The transformation from the orthonormal basis {k } to t ( ) 0 he orthonormal basis {}ϕklm is given by the matrix ( ) G ( ) *( ) 0 1 G ϕδklm′′ ′ kkk=−′ Ylm′′ κ (1.35) k 19 Effect of a potential. Phase shifts

We come back to the Schrödinger equation with V≠0. Consider, for r large, an incoming wave exp[-i(kr-lπ/2)]. Since the reflection coefficient of V is 1 (conservation of the norm), the reflected outgoing wave has the same modulus and has just accumulated a phase shift with respect to the V=0 case. The superposition of the 2 waves is thus a shifted sinusoid. We conclude that there is a set of solutions of the Schrödinger equation with V≠0 which behave for large r as: sin / ( ) (GG) 2 ( ) ⎡kr−+ lπδ2 k ⎤ ϕ rYn ⎣ l ⎦ (1.36) klm r →∞ π lm r One can show that these functions are orthonormalized (see Appendix) ( )

ϕk′ l′′ mϕδ klm =−kk δδ′ ll′′ mm (1.37) They don’t form a basis if there are also bound states in the potential V. 20 Partial wave expansion of the outgoing scattering state

Consider the linear superposition of the states ϕklm with the same coefficients as (those) appearing in the expansion (1.34) of the plane 0 waveexp( on) .the ϕklm , each state being multiplied by the phase factor iδ We will show that such/ a state is nothing but the outgoing l ( ) +−32 state ψπG (multiplied by 2 for having an orthonormalized state) k ∞=+ml( ) *( ) + 1 l G iδ ψκJG = iY e l ϕ k ∑∑ lm klm k lml==−0 ∞=+ml ( ) G (1.38) 0 iδl = ∑∑ ϕϕklm k e klm lml==−0 Before demonstrating this identity, let us discuss its physical

meaning.The outgoing scattering state is obtained by( switching) on slowly V on the free state. Each spherical wave ϕ 0 of the G klm expansion of k is transformed intoϕklm , but in addition it acquires iδ a phase factor el which depends on l and which thus varies. from one spherical wave of the expansion to another one 21 Demonstration For large r, the linear superposition introduced in (1.38) behaves as: / / ikrl−+π 22δ −−ikrlπ 2 ∞=ml+ ( ) *( ) ( ) ( l ) ( ) 12l GGee− (1.39) ∑ ∑ iYlmlκ Ym n kilml==e0xp − / π exp 2 r/ 2iδ Using ikrl−+=π 22δ ikrl −π 2 ×⎡ 1+ e l − 1 ⎤ ()()l ⎣ ()⎦ we get: (1.40) / sin( / ) 2iδ −ilπ 2 ∞=+ml( ) *( ) ( )⎡ ()eel − 1 ikr ⎤ 12l GG⎢ kr− lπ 2 e ⎥ ∑ ∑ iYlm κ Ylm n + krlml0 π ⎢ 2ir⎥ =−= ⎣ ⎦ (1.41) The contribution of the first term of the bracket is nothing but the asymptotic expansion of the plane wave in spherical waves. The second term gives an outgoing spherical wave

/ G .G ikr ( )−32⎡ ikr ( , GG, ) e ⎤ 2πκ⎢e + fk n ⎥ (1.42) ⎣ r ⎦ This demonstrates that the state given in (1.38) is an outgoing scattering state and gives in addition the expression of the amplitude f 22 Scattering amplitude in terms of the phase shifts ( ,GG, ) ∞=+ml sin (G ) *(G ) 4π i δl fkκδ n = ∑∑e llmY nY lmκ k lml==−0 ∞ ( ) sin (cos ) (1.43) 1 i δl =+∑ 21lPe δθll (cos )k l =0

where Pl θ is a Legendre polynomial and where θ G G . 2 is the angle between nf and κ Integrating over the G polar angl es of κ gives the scattering cross section ( ) ∞ ( ) ( ) 4π ( )sin ( ) σ kk==σσ kl21+2 δk(1.44) ∑ ll2 l l =0 k Scattering of 2 identical particles Quantum interference between 2 different paths ( ) ( ) ( ) θπ - θ fffkkkθ →+( )θεπθ − / ε =+11 − for bosons (fermions) π 2 sin ( ) ( )2 σ =+2πθθθεπθd ff−(1.45) total ∫0 kk 23 Partial wave expansion of the ingoing scattering state

− ψ G is given by a linear superposition of the states ϕ analogous k klm + to the one introduced for ψ G each state being now multiplied by exp( )k exp( ). the phase factor −iiδδ instead of * ll( ) ∞=+ml( ) ( ) ∞=+ml G − 1 l G −−iiδδ0 ψκJG ==iY eellϕϕϕk k k ∑∑ lm klm ∑∑ klm klm lml==−00lml==− (1.46) The demonstration of this identity is similar to the one given above for the outgoing scattering state

If we start from thisG state and if we switch off V slowly, it transforms( ) 0 into the free state k . Each wave ϕϕklm is transformed intoklm , but −iδ in addition its phase factor changes from el to 1 which corresponds +iδ to acquiring a phase factor e l , Finally, when we go from tt= −∞ to =( ) +∞ switching on ( and) then 00 switching off V slowly, we start fromϕϕklm and we end in klm acquiring ++iiδδ +δ2 i a global phase factor ell×= e e l 24 S – Matrix in the angular momentum representation G G −+ G GG′ G SkSkkk′′==ψ kψ k (1.47) +− We use the expansion of ψψGJ and G in spherical waves ′ ∞=+ml ( ) G kk ∞=+ml′′ ( ) G + 0 iδ − 0 −iδ ψϕϕJG = k e l ψϕϕJG = k′ e l k ∑∑ klm klm k′ ∑∑ klm′ ′′ klm′′′ lml==−0 lml′′′==−0 (1.48) G G This gives a first expression of Skk′ ∞=+∞=+ml m′′ l G ( ) ( ) G −+ 00+iiδδ S GG==ψψ G G kk′ ϕ ϕ ϕee ϕll kk′′ k k ∑∑∑∑ klm′′ ′ klm ′′ ′ klm klm lmllml==−==−00′′′ (GG) (1.49) δδδkk− ′ ll′′ mm G G On the other hand, a change of basis gives for Skk′ G GG∞=+∞=+ml m′′ l ( ) ( ) ( ) ( ) G 00 00 S GG ==kSk′′kϕϕ ϕϕS k kk′ ∑∑∑∑ klm′′ ′ klm ′′ ′ klm klm ′′′ lmllml==−==−00 , (1.50) GG Comparing the 2 expressions obtained for Skk′ we get ( ) ( ) ( ) 00+2i δl (1.51) ϕklm′ ′′ Skϕδδδ klm =−e k′ ll′′ mm which shows that theS -matrix is diagonale xinp (the angular) momentum

representation, with diagonal elements 2iδl clearly showing the unitarity of S 25 Outline of lecture 1 1 - Introduction 2 - Scattering by a potential. A brief reminder • Integral equation for the wave function • Asymptotic behavior. Scattering amplitude • Born approximation 3 - Central potential. Partial wave expansion • Case of a free particle • Effect of the potential. Phase shifts • S-Matrix in the angular momentum representation 4 - Low energy limit • Scattering length a • Long range effective interactions and sign of a 5 - Model used for the potential. Pseudo-potential • Motivation • Determination of the pseudo-potential • Scattering and bound states of the pseudo-potential 26 Central potential. Low energy limit

Suppose first V=0. The centrifugal barrier in the 1D Schrödinger equation prevents the particle from approaching near the region r=0 ( ) 222 ( ) ll+ 1 ==k 2 = ll+ 1 = 2 2 2μrl 2μ 2μ r ( ) / kr=+ ll 1 =22k 2μ l r ( ) rl rlll =+1 dB If the range b of the potential is small enough, i.e. if

b dB (1.52) a particle with l ≠ 0 cannot feel the potential . Only lV= 0 wave will feel "s -wave scattering" ⎡⎤d2 2μ ( ) ( ) + kVrur2 −=0 (1.53) ⎢⎥22k 0 ⎣⎦dr = 27 Scattering length sin ( ) For ruuk large enough, =+ varies as ⎡⎤rkδ kk00sin ( ) ⎣⎦. Let vk be the function ⎡⎤rkuextending for all r k ⎣⎦+ δ0 k

Let Pv be the intersection point of k with the r -axis which is the closest from the origin.By definition, the scattering length a is the limit of the abscissa of Pk when → 0 (see figure)

Expansion of vk in powers of rk near r= 0 ( ) k sin ( ) sin ( ) cos ( ) v r=+→⎡⎤ krδδδ k k+ kr k k ⎣ 00⎦ 0 kr →0 (1.54) 1 tan ( ) Abscissa of P : − δ k tkan ( 0 ) lim − δ k π ( ) π 0 (1.55) ak=−≤δ0 ≤ + k →0 k 2228 Scattering length (continued) Limit k=0 ⎡⎤d2 2μ ( ) ( ) − Vr u r = 0 ⎢⎥22 0 ⎣⎦dr = (1.56) Far from r=o, the solution of the S.E. is a straight line and ( )

vr0 ∝ r− a (1.57) The abscissa of Q is equal to a

Scattering cross section sin ( ) ( ) 4π ( )sin ( ) ( ) 2 δ k kl21 2 k k4 0 (1.58) σ ll= 2 +⇒=δσπl=0 2 k ( ) ( ) k δσka − k ⇒= ka4π2 (1.59) 0 k →0 (l =0) 2 (1.60) For identical bosons σ l =0 k = 8π a

29 Scattering length for square potentials Square potential barriers Square barrier of / 22 ( ) height Vk00= = 2μ Vr and width b V For( rb) >=(and) k0 0 ( ) ( ) u r = v r ∝−ra vr0 ∝ r− a 0 0 ur0 For( rb) < and( ) k= 0 2 ur0′′ = k00 ur 0 a b r The curvature of( u0 is )

positive and u0 r = 00= We conclude that the scattering length is always positive and smaller than the range b of the potential 0 ≤≤ab

When V0 →∞(hard sphere potential) ab→ 30 Scattering length for square potentials (continued)

Square potential wells / 22 ( ) Vk00=−= 2μ Vr ( ) vr0 ∝− r a For( r) >=b (and) k 0 u r = v r ∝−r a ( ) 0 0 ur 0 For( rb) < and( k) = 0 ′′ 2 a 0 b r ur0 =−k00 ur The curvature of( u0 is ) V0 negative and u0 r = 00=

If V0 is small enough so that there is no bound state in the potential

well, the curvature of urb0 for < is small and a is negative When Vu increases, the curvature of for r< b increases in 00. absolute value and aa→−∞ Then switches suddenly to +∞ and decreases. This divergence of a corresponds to the appearance of the first bound state in the potential well

31 Square potential wells (continued)

Variations of a with k0 figure taken from Ref.5

When the depth of the potential well increases,( divergences) /

of aVk occur for all values of 00 such that b=+212nπ corresponding to the appearances of successive bound states in the potential well. These divergences of a which goes from −∞ to +∞ are called "zero-energy" resonances

32 Long range effective interactions and sign of a The scattering length determines how the long range behavior of the wave functions is modified by the interactions. To understand how the sign of a is related to the sign of the effective long range interactions, it will be useful to consider the particle enclosed in a spherical box with radius R, so that we have the boundary condition ( )

uR0 = 0 (1.61) leading to a discrete energy spectrum In the absence of interactions (V=0), the normalized eigenstates and the eigenvalues of the 1D Schrödinger equation are: sin( / ) ( )( ) 1 Nrππ R =222N , ,... ψ 0 rE==N=12 NN22πμRr 2 sin( / ) R (1.62) Nrπ R Figure corresponding to N=3 0 Rr 33 V = 0 R

V ≠ 0 a > 0 a R The dotted line is the sinusoid outside the range of the potential It has a shorter wavelength than for V=0, and thus a larger wave number k. The kinetic energy in this region, which is also the total energy, is larger

V ≠ 0 a < 0 a R The dotted line is the sinusoid outside the range of the potential It has a longer wavelength than for V=0, and thus a smaller wave number k. The kinetic energy in this region, which is also the total energy, is smaller 34 Correction to the energy to first order in a

a R

( ) / 0 For the state ψ N , we have R = N λ 2 Because of the interactions, these N half wavelengths occupy now a length/ R − a so that ( )/ λλ=→=−22RN/ ′ RaN/ /( ) kkR=→==22πλ ′′πλ k Ra− / / /( ) ⎛⎞2a 2 22′′222 EkNN=→==−=21μ EEkkERRNNa EN ⎜⎟+ ⎝⎠R 2aN=22π 2 Finally, we have δ E =−=EE′ E = a (1.63) NNNR Nμ R3 Long range effective interactions are - repulsive if a > 0 - attractive if a < 0 35 Outline of lecture 1 1 - Introduction 2 - Scattering by a potential. A brief reminder • Integral equation for the wave function • Asymptotic behavior. Scattering amplitude • Born approximation 3 - Central potential. Partial wave expansion • Case of a free particle • Effect of the potential. Phase shifts • S-Matrix in the angular momentum representation 4 - Low energy limit • Scattering length a • Long range effective interactions and sign of a 5 - Model used for the potential. Pseudo-potential • Motivation • Determination of the pseudo-potential • Scattering and bound states of the pseudo-potential 36 Model used for the potential V(r) Why not using the exact potential? The interaction potential is very difficult to calculate exactly. A small error in V can introduce a very large error on the scattering length deduced from this potential. Mean field description of ultracold quantum gases require in general a first order treatment of the effect of V (Born approximation). But Born approximation cannot be in general applied to the exact potential Approach followed here The motivation here is not to calculate the scattering length. This parameter is supposed known experimentally. We are interested in the derivation of the macroscopic properties of the gas from a mean field description using a single parameter which is a. The key idea is to replace the exact potential by a “pseudo-potential” simpler to use than the exact one and obeying 2 conditions: - It has the same scattering length as the exact potential - It can be treated with Born approximation so that mean field descriptions of its effects are possible 37 Determination of the pseudo-potential Derivation “à la” H.Bethe and R.Peierls (Y.Castin, private communication) We add to the 3D Schrödinger equation of a free particle (V=0) a term proportional to a delta function ==22(GG) ( ) k 2( G) −ΔψδrCr + = ψ r (1.64) 22μμ To determine the coefficient C , we impose to the solution of this equation to coincide with the extension to all r of the asymptotic sin ( ) / ( ) / behavior ⎡⎤kr k r of the true wave function u r r ⎣⎦+ δ00 In particular, for k small enough, one should have: (G ) ( ) / ψ rBar r − (1.65) ( r →)0 ( / ) ( ), Inserting (1.65) into 1.64 and using Δ=−14rrπ δ we get an equation containing a delta function multiplied by a coefficient 4π =2 CaB− 2μ which must vanish. This gives the coefficient C appearing in (1.64) 44ππ==22 CgBgaa==where =(1.66) 2μ m 38 Determination of the pseudo-potential (continued) It will be more convenient to express C=gB, not in terms of the coefficient B appearing in the wave function ψ = B(r-a)/r of equation (1.65), but in terms of the wave function ψ itself. We use for that ⎡ d ⎤ Br= ψ ⎢ ⎥ (1.67) ⎣dr ⎦r =0 Equation (1) can be rewritten as: ==22G GGk 2 −Δψψψ()rV + () r = () r (1.68) 22μμpseudo GGd G where Vrgrrrpseudo ψδ()= ()[] ψ () (1.69) dr ( / ) Vr is called the pseudo-potential. The term ⎡⎤ d dr regularizes pseudo (G ) ⎣⎦/ the action. of δ rr when it acts on functions behaving as 1 near rr==00For functions which are regular in , V has the (G ) : pseudo same effect as grδ (GG) ( )/ ( ) ( ) ( ) (G) ψψrurr=≠with u00⇒= V rgur′ 0δ (GG) pseudo ( ) (G ) (G) (1.70) ψψrrregular in =⇒00 Vrgrpseudo =ψδ 39 Scattering states of the pseudo-potential We are looking for solutions of equation (1.68) with E > 0 , Forl ≠ 0 the centrifugal, barrier prevents the (particleG ) ,from approaching r ==00 and one can show that ψ 0 so that, (G ) . according to (1.70), Vrψ = 0 Vgives only s-scattering Gpseudo ( )/ pseudo ( ) . and one can write ψ (r ) =≠ur rwhere u00can be ( ) ( ) ( )00( ) ur⎡⎤ u00 ur− u ( ) ( ) 1 d2u Δ=Δ+0000 =−40πδur +0(1.71) ⎢⎥0 2 rr⎣⎦ r rdr The Schrödinger equation for u becomes: ( )0 ( ) ==2 ⎡⎤( ) (GG) ur′′ ( ) ( ) 22k ur 00′ (1.72) −−⎢⎥40πδur00 + + gruδ 0 = 22μμ⎢⎥rr ⎣⎦(G ) Cancelling the term proportional to δ r and the term independant (G ), of δ r we get 2 equations: ( ) u 0 μ ( ) ( ) 0 ( ) =−ga =− urkur′′ +2 =0 (1.73) ′ 2π =2 00 u0 0 40 Scattering states of the pseudo-potential (continued)

The solution of the second equation can be written: ( ) sin ur=+ kr δ 00( ) (1.74) Inserting this solution into the first equation gives: tan

δ0 = − ka (1.75) On the other hand, the s-wave scattering amplitude is equal to: ( ) sin 1 i δ fk= e 0 δ (1.76) 0 k 0 Using equation (1.75) giving tanδ0 finally gives after simple algebra: ( ) a fk0 =− (1.77) . 1 + ika

Vapseudo is proportional to A first order treatment of Vpseudo thus gives the correct result for) .the scattering amplitude in the zero energy limit (ka = 0 This shows that Born

approximation can be used with Vpseudo for ultracold atoms

The 2 conditions imposed above on Vpseudo are thus fulfilled 41 The unitary limit

From the expression of the scattering amplitude obtained above, we deduce the scattering amplitude for identical bosons ( ) 8π a2 σ k = 1 + ka22 (1.78) which is valid for all k. ( ) The low energy limit ka 1 gives the well known result: 2 σπ (ka ) 8 (1.79) ka 1 There is another interesting limit, corresponding to high energy, ( ) or strong interaction ka 1 leading to result independent of a : 8π σ (k ) (1.80) ka 1 k 2 This is the so called “unitary limit”

42 Bound state of the pseudo-potential

The calculation is the same as for the/ scattering states, except that/ we replace the positive energy ==22k 22μ by a negative one - 2κμ 2 The 2 equations derived from the Schrodinger equation are now: ( ) u 0 ( ) ( ) 0 ( ) =−au′′ ru−κ 2 r =0 (1.81) ′ 00 u0 0 The solution of the second equation (finite for r→∞) is: ( ) −κ r ur0 = e (1.82) which inserted into the first equation gives: / κ = 1 a (1.83) The pseudo-potential thus has a bound state with an energy =2 E =− (1.84) 2μa2 and a wave function: exp⎛ r ⎞ ⎜ − ⎟ (1.85) ⎝ a ⎠ 43 Energy shifts produced by the pseudo-potential

We come back to the problem of a particle in a box of radius R. We have calculated above the energy shifts of the discrete energy levels of this particle produced by a potential characterized by a scattering length a. To first order in a, we found: =22π N 2 δ E = a (1.86) N μ R3 This result was deduced directly from the modification induced by the interaction on the asymptotic behavior of the wave functions and not from a (perturbative) (treatment) of V. We show now that: 00 (1.87) δψEVNN= pseudo ψ N to first order in Vpseudo which is another evidence for the fact that the effect of the pseudo potential can be calculated perturbatively, which is not the case for the real potential. For example, a hard core potential (V=∞ for r< a) cannot obviously be treated perturbatively, but its scattering length is a , and using a pseudo-potential with scattering length a allows perturbative calculations. 44 Demonstration

The unperturbed normalized eigenfunctions of the particle in the spherical box are: sin( / ) ()( ) 0 1 Nrπ R ψ N r = (1.47) ()( ) 2π R r ψ 0 r is regular in r = 0 and N () ( ) 11/ 0 (1.88) ψπN 0 = N 2π R32 so that ( )( ) (G ) ( )( ) 00 (1.89) Vrgrpseudoψδψ N = N 0

We deduce that 22 22 2 ( )( )2 NNππ= δψE ===gg0 0 a(1.90) NN 2πμRR33

which coincides with the result obtained above in (1.63).

45 Conclusion

Elastic collisions between ultracold atoms are entirely characterized by a single number, the scattering length Effective long distance interactions are attractive if a<0 and repulsive if a>0 Giving the same scattering length as the real potential, the pseudo-potential gives the good asymptotic behavior for the wave function describing the relative motion of 2 atoms, and thus correctly describes their long distance interactions In a dilute gas, atoms are far apart. The pseudo-potential is proportional to a and can be treated perturbatively. A first order treatment of the pseudo-potential is the basis of mean field description of Bose Einstein condensates where each atom moves in the mean field produced by all other atoms. Next step: can one change the scattering length?

46 Atom-Atom Interactions in Ultracold Quantum Gases

Claude Cohen-Tannoudji

Lectures on Quantum Gases Institut Henri Poincaré, Paris, 27 April 2007

Collège de France 1 Lecture 1 Quantum description of elastic collisions between ultracold atoms The basic ingredients for a mean-field description of gaseous Bose Einstein condensates

Lecture 2 Quantum theory of Feshbach resonances How to manipulate atom-atom interactions in a quantum ultracold gas

2 A few general references

Ultracold atoms with interactions manipulated by Feshbach resonances become a very attractive system for getting a better understanding of quantum many body systems

5 Purpose of this lecture

- Provide a physical interpretation of Feshbach resonances in terms of a resonant coupling of the state of a colliding pair of atoms to a metastable bound state belonging to another collision channel - Present a simple two-channel model allowing one to get analytical predictions for the scattering states and the bound states of the two colliding atoms near a Feshbach resonance • How does the scattering length behave near a resonance? • When can we expect broad resonances or narrow resonances? • Are there bound states near the resonances? What are their binding energies and wave functions? - In addition to their interest for ultracold atoms, Feshbach resonances are a very interesting example of resonant effect in collision processes deserving to be studied for themselves This lecture will closely follow the presentation of Ref.9: T.Köhler, K.Goral, P.Julienne, Rev.Mod.Phys. 78, 1311-1361 (2006) See also the references therein 6 Microscopic atom-atom interactions

Case of two identical alkali atoms G ,G Unpaired electrons for each atom with spins SS GG, 12 Nuclear spins II 12, ; ,

Hyperfine states fm1122ff fm Born Oppenheimer potentials (2 atoms fixed at a distance r) 2 potential curves:

VT(r) VT(r) for the triplet state S=1 V (r) for the singlet state S=0 : S S quantumGGG number for the total spin SSS=+ ( ) ( )12( )

VS(r) Vr= VS rPSTT+ V rP (2.1)

PS :Projector on S = 0 states

r PST :Projector on = 1 states 7 Microscopic atom-atom interactions (continued) Electronic interactions ( ) ( ) ( )

Vrel =+ VrPSS VrP TT 13( ) ( ) 1( ) ( ) G .G (2.2) =++Vr Vr⎡⎤ Vr − Vr SS 44ST2=2 ⎣⎦ TS12 This interaction depends on the electronic spins because of Pauli principle (electrostatic interaction between antisymmetrized states). It is called also “exchange interaction” Does not depend on the orientation in space of the molecular axis (line joining the nuclei of the 2 atoms)

Magnetic spin-spin interactions Vss Dipole-dipole interactions between the 2 electronic spin magnetic moments. Depends on the orientation in space of the molecular axis Interaction Hamiltonian V V V (2.3) int ss = el + Vel is much larger than Vss 8 Outline of lecture 2 1 - Introduction 2 - Collision channels • Spin degrees of freedom. • Coupled channel equations • Strong couplings and weak couplings between channels 3 - Qualitative interpretation of Feshbach resonances 4 - Two-channel model • Two-channel Hamiltonian • What we want to calculate 5 - Scattering states of the 2-channel Hamiltonian • Calculation of the outgoing scattering states • Asymptotic behavior. Scattering length • Feshbach resonance 5 - Bound states of the 2-channel Hamiltonian • Calculation of the energy of the bound state • Calculation of the wave function 9 Channels Two atoms entering a collision in a s-wave (A = 0) and in well defined hyperfine and Zeeman states. This defines the “entrance channel” α defined by the set of quantum numbers: : , , , ,

α {fm1122ff fm A = 0} The eigenstates of the total Hamiltonian with eigenvalues E can be written: (G ) (2.4) ψαψ= ∑ α r α where ψα(r) is the wave function in channel α whose radial part is of the form: ( , )

Fα rE r Because the interaction has off diagonal elements between different channels, the Fα do not evolve independently from each other 10 Coupled channel equations

The coupled equations of motion of the Fα are of the form: 2 ∂ ( , ) 2μ ( , ) FrE+ ⎡⎤ Eδ −= V FrE 0 (2.5) 22αα∑ ⎣⎦βαββ ∂r = β ( ) 2 ⎡⎤AA+ 1 = int ( ) , , (2.6) VEαβ =++⎢⎥fm E fm 2 δ αβ +Vr ⎣⎦if122f 2μr Solving numerically these coupled differential equations gives the

asymptotic behavior of Fα for large r from which one can determine the phase shift δ0 and the scattering length in channel α. Importance of symmetry considerations

The symmetries of Vel(r) and Vss determine if 2 channels can be coupled by the interaction. In particular, if 2 channels can be

coupled by Vel, the Feshbach resonance which can appear due to this coupling will be broad because Vel is large. If the symmetries are such that only Vss can couple the 2 channels, the Feshbach resonance will be narrow. 11 Examples of symmetry considerations

If the magnetic field B0 is the only external field, the projection M of the total angular momentum along the z-axis of B0 is conserved.

M = mmmff12++A

Only states with the same value of mmmff12+ + A can be coupled by the interaction Hamiltonian The s-wave entrance channel can be coupled to A ≠ 0 channels only by VV because , which depends only on the distance r between the ss el G 2 atoms, commutes with the molecule orbital angular momentum L Consider the various states Mm=++ m m with a fixed value of . ff12A G , , MF They can be also classified by the eigenvalues of 2 F where GGG. , , , , z F =+FF This gives the states ffFMmwith M+ m = M GG12. , , {}12GGGGGGFFAA, Since SS12 and thus VFel commutes with =+++S1212SIIand L

VFel can couple only states with the same value of and A Examples of application of these symmetry considerations to the identification of broad Feshbach resonances will be give later on 12 Outline of lecture 2 1 - Introduction 2 - Collision channels • Spin degrees of freedom. • Coupled channel equations • Strong couplings and weak couplings between channels 3 - Qualitative interpretation of Feshbach resonances 4 - Two-channel model • Two-channel Hamiltonian • What we want to calculate 5 - Scattering states of the 2-channel Hamiltonian • Calculation of the outgoing scattering states • Asymptotic behavior. Scattering length • Feshbach resonance 5 - Bound states of the 2-channel Hamiltonian • Calculation of the energy of the bound state • Calculation of the wave function 13 Open channel and closed channel

The 2 atoms collide with a very small positive energy E in an V channel which is called “open” The energy of the dissociation Closed threshold of the open channel is channel taken as the zero of energy There is another channel above Eres the open channel where E scattering states with energy E cannot exist because E is below 0 r the dissociation threshold of this channel which is called “closed”

Open There is a bound state in the channel closed channel whose energy

Eres is close to the collision energy E in the open channel 14 Physical mechanism of the Feshbach resonance

The incoming state with energy E of the 2 colliding atoms in the open channel is coupled by the interaction to the bound state ϕres in the closed channel. The pair of colliding atoms can make a virtual transition to the bound state and come back to the colliding state. The duration of this virtual transition scales as ħ / I Eres-E I, i.e. as the inverse of the detuning between the collision energy E and the energy Eres of the bound state.

When E is close to Eres, the virtual transition can last a very long time and this enhances the scattering amplitude Analogy with resonant light scattering when an impinging photon of energy hν can be absorbed by an atom which is brought to an excited discrete state with an energy hν0 above the initial atomic state and then reemitted. There is a resonance in the scattering amplitude when ν is close to ν0

15 Sweeping the Feshbach resonance The total magnetic moment of the atoms are not the same in the 2 channels (different spin configurations). The energy difference between the 2 channels can thus be varied by sweeping a magnetic field V

Closed channel

0 r

Open channel 16 Shape resonances ( ) ( ) / Vr ++AA12 =22μr Can appear in a A≠0 channel Metastable where the sum of the potential state Incoming and the centrifugal barrier gives state rise to a potential well

The 2 colliding atoms arrive in 0 r a state with positive energy

In the potential well, there are quasi-bound states with positive energy which can decay by tunnel effect through the potential barrier due to the centrifugal potential. This is why they are metastable If the energy of the incoming state is close to the energy of the metastable state, there is a resonance in the scattering amplitude These resonances are different from the zero-energy resonances studied in this lecture. They explain how scattering in A≠0 waves can become as important as s-wave scattering at low temperatures 17 Outline of lecture 2 1 - Introduction 2 - Collision channels • Spin degrees of freedom. • Coupled channel equations • Strong couplings and weak couplings between channels 3 - Qualitative interpretation of Feshbach resonances 4 - Two-channel model • Two-channel Hamiltonian • What we want to calculate 5 - Scattering states of the 2-channel Hamiltonian • Calculation of the outgoing scattering states • Asymptotic behavior. Scattering length • Feshbach resonance 5 - Bound states of the 2-channel Hamiltonian • Calculation of the energy of the bound state • Calculation of the wave function 18 Two-channel model Only two channels are considered, one open and one closed State of the atomic system (G ) (G ) (2.7) opϕϕop rr+ cl cl The wave function has two components, one in each channel Hamiltonian ( ) =2 H =− Δ+V ⎛⎞H Wr op op op 2μ H = ⎜⎟( ) (2.8) (2.9) 2-channel ⎜⎟Wr H =2 ⎝⎠cl H =− Δ+V cl 2μ cl Resonant bound state in the closed channel ( ) ( ) (2.10) HrErclϕ res = resϕ res Eres =Δ=

The energy Eres of this state, denoted also =Δ, is close to the energy E 0 of the colliding atoms in the open channel 19 What we want to calculate We want to calculate the eigenstates and eigenvalues of H ( ) 2-channel

⎛HWrop ⎞⎛⎞⎛⎞ϕop ϕop ⎜( ) ⎟⎜⎟⎜⎟= E ⎜ ⎟⎜⎟⎜⎟ (2.11) ⎝Wr Hcl ⎠⎝⎠⎝⎠ϕ cl ϕcl (G ) ( ) (GG) ( ) Hopϕ op rWrr+ ϕϕcl = Eop r ( ) (G ) (GG) ( ) (2.12) Wrϕϕϕop r+= Hcl cl r Ecl r Eigenstates with positive eigenvalues E>0 They describe the scattering states of the 2 atoms in the presence of the coupling W. In particular, we are interested in the behavior of

the scattering length when Eres is swept around 0 The 2 components of the scattering state corresponding to an G GG kk incoming wave k are denoted ϕϕop and cl Eigenstates with negative eigenvalues Eb<0 They describe the bound states of the 2 atoms in the presence of W bb Their 2 components are denoted ϕop and ϕcl 20 Single resonance approximation

We will neglect all eigenstates of Hcl other than ϕres Near the resonance we want to study (Eres close to 0), they are too far from E=0 and their contribution is negligible We will use the following expression for the Hamiltonian of the closed channel

(2.13) HEcl= resϕϕ res res

The resolvent operator (or Green function) of Hcl will be thus given by: ( ) 1 ϕϕres res (2.14) Gzcl == zH−−cl zEres

21 Outline of lecture 2 1 - Introduction 2 - Collision channels • Spin degrees of freedom. • Coupled channel equations • Strong couplings and weak couplings between channels 3 - Qualitative interpretation of Feshbach resonances 4 - Two-channel model • Two-channel Hamiltonian • What we want to calculate 5 - Scattering states of the 2-channel Hamiltonian • Calculation of the outgoing scattering states • Asymptotic behavior. Scattering length • Feshbach resonance 5 - Bound states of the 2-channel Hamiltonian • Calculation of the energy of the bound state • Calculation of the wave function 22 Scattering states of the

two-channel Hamiltonian H2-channel

Open channel component of the scattering state of H2-channel The first equation (2.12) can be written G ( ) ( ) G ( ) kkG G (2.15) ()E −=HrWrropϕϕ op cl Its solution is the sum of a solution of the equation without the right- side member and a solution of the full equation with the right-side member considered as a source term. G ( ) G ( ) kk++ + 1 G ϕop =+ϕϕk GEWop cl GEop = (2.16) E −+Hiop ε + In (2.16), G op(E) is a Green function of Hop. The term +iε, where ε is a positive number tending to 0, insures that the second term of (2.16) has the asymptotic behavior of an outgoing scattered state for r→∝.

The first term of (2.16), involving only Hop, is chosen as an outgoing scattering state of H , in order to get the good behavior for r→∝. op G G .G 2 ++( GG) 11⎡⎤ikr ( ) p ϕ GGrV=+/ e ϕ rT=(2.17) kk( )32⎢⎥op 2π ETi−+ε 2μ ⎣⎦23 Scattering states of the

two-channel Hamiltonian H2-channel (continued)

Closed channel component of the scattering state of H2-channel The second equation (2.12) can be written: G G G G kk( ) ( ) ( ) (2.18) ()E −=HrWrrclϕϕ cl op

Its solution can be written in terms of the Green function of Hcl: GG( ) ( ) −1 kk (2.19) ϕcl==GEW cl ϕop GEcl ( E− Hcl ) Using the single resonance approximation (2.14), we get: G k G (GG) ( ) ϕres W ϕop ϕ k rr= ϕ (2.20) cl res EE− G res k The closed channel component ϕcl is thus proportional to ϕres Dressed states and bare states G G The 2 components ϕϕkk and of the scattering states of H op cl 2−channel. are called dressed states because they include the effect of W + The eigenstates ϕϕG and of HH and are called bare states k res op cl 24 Open channel components of the scattering

states of H2-channel in terms of bare states

Inserting (2.20) into (2.16), we get: G k G ( ) ϕ W ϕ k ++ res op G (2.21) ϕϕop =+k ϕGEWop res EE− res G In order to eliminate ϕ k in the right side, we multiply both sides of , op

(2.21) by ϕres W which gives: G k + G ϕϕres W op ϕϕW = res k ( ) (2.22) + EE− res EE−−resϕϕ res WGEW op res Inserting (2.22) into (2.21), we finally get:

G ( ) k ++ WWϕϕres res + ϕ =+ϕϕG GE ( ) G (2.23) op k op + k EE−−resϕϕ res WGEW op res Only the bare states appear in the right side of (2.23). 25 Connection with two-potential scattering Equation (2.23) can be rewritten in a more suggestive way. Il we introduce the effective coupling Veff defined by:

ϕϕres res VW= ( ) W (2.24) eff + EE−−resϕϕ res WGEW op res we get, by inserting (2.24) into (2.23): G k + 1 + G G (2.25) ϕop =+ϕϕkkVeff EH−+op iε Veff acts only, like Vop, inside the open channel space. It describes the effect of virtual transitions to the closed channel subspace. The two- channel scattering problem can thus be reformulated in terms of a single-channel scattering problem (in the open channel), but with a new potential Vtot in this channel, which is the sum of 2 potentials

VVVtot= op+ eff (2.26)

Equation (2.25) then appears as the scattering produced by Veff on waves “distorted” by Vop. (Generalized Lippmann-Schwinger equation) (see for example ref.4, Chapter 17) 26 Asymptotic behavior

of the scattering states of H2-channel Let us come back to (2.23). Only the asymptotic behavior of the open channel component is interesting because the closed channel component, proportional to ϕ vanishes for large r. res G k We expect the asymptotic behavior of ϕop to be of the form: GG(G ) 1 ⎡⎤.G ( ,GGG)eikr / ϕ kirf ( ) / e kr+=knnrr(2.27) op 32⎢⎥ r →∞ 2π ⎣ r ⎦ In the limit k→0, the scattering amplitude becomes spherically symmetric and gives the scattering length we want to calculate ( ,G ) f kn→− a (2.28) k →0 The asymptotic behavior of the first term of (2.23) describes the scattering in the open channel without coupling to the closed channel.

It gives the scattering length aWop in the open channel alone (= 0) . This scattering length is often called the background scattering length.

aa= (2.29) op bg 27 Position of the resonance The second term of (2.23) is the most interesting since it gives the effects due to the coupling W. The scattering amplitude given by its asymptotic behavior becomes large if the denominator of the second term of (2.23) vanishes, i.e. if: ( ) + EE=+resϕ res WGEW op ϕres (2.30) When E is close to 0, the last term of (2.30) is equal to: 2 + ( ) ϕϕW G + res k (2.31) ϕϕresWG op 0 W res = =Δ= 0 ∑G G k −+Eik ε

Its interpretation is clear. It gives the shift ħΔ0 of ϕres due to the second order coupling induced by W between ϕres and the continuum of Hop We thus predict that the scattering amplitude, and then the scattering length, will be maximum (in absolute value), not when Eres is close to 0, but when the shifted energy of ϕres (2.32) EEres= res +Δ= 0 is close to the energy E 0 of the incoming state 28 Remark ( ) + −1 G Strictly speaking, the Green function GEop =−+( E Ek iε ) appearing in (2.30) is equal to: 11⎛⎞ =−−P ⎜⎟iEEπδ()G (2.33) EE−+GG iε ⎜⎟ EE − k kk⎝⎠ where P means principal part. Because of the last term of (2.33), equation (2.31) should also

contain an imaginary term describing the damping of ϕres due to its coupling induced by W with the continuum of Hop. But we are considering here the limit of ultracold collisions E → 0 and the density of states of the continuum of H vanishes near , op E G = 0 which means that the damping of ϕ can be ignored in k . res the limit E → 0

For large values of Eres, the imaginary term of (2.33) can no longer be ignored, and it can be shown that it gives rise to an imaginary term in the scattering amplitude, proportional to k. 29 Variations of EEres and res with B The spin configurations of the two channels have different magnetic moments. The energies of the states in these channels vary differently when a static magnetic field B is applied and scanned. If ξ is the difference of magnetic moments in the 2 channels, the difference between the energies of 2 states belonging to the channels varies linearly with B with a slope ξ. If we take the energy of the dissociation threshold of the open channel as the zero of energy, the energy Eres of ϕres is equal to:

Eres =−ξ (BBres ) (2.34)

Eres is degenerate with the energy of the ultracold collision state when B=Bres In fact, the position of the Feshbach resonance is given, not by the zero of EEres , but by the zero of res (2.35) Eres=+Δ=−EBB res = 0 ξ ( 0 )

This equation gives the correct value, B0, at which we expect a divergence of the scattering length. 30 E

= Δ0

Eres

B0 Bres B

Eres

We suppose here ξ < 0

Since Δ0 is also negative according to (2.31), B0 is smaller than Bres.

31 Contribution of the inter channel coupling W to the scattering length G k Asymptotic behavior of the W-dependent term of ϕop Using (2.30) and (2.32), we can rewrite (when E0) equation (2.23):

G ( ) k + ++WWϕϕres res ϕ =+ϕϕG GE G (2.36) op kkop EE− res To find the contribution of W to the scattering length, we have to find the asymptotic behavior for r large of the wave function of the last term ( ) G ++WWϕϕres res rG E ϕ G = op k EE− res (2.37) ( ) 3 GGG+ WWϕϕres res + d rrGE′′′ r r ϕ G ∫ op k EE− res We need for that to know the asymptotic behavior for r large of the Green function of H op , , + G GG 1 G GErrop ()′ = r r′ (2.38) EH−+op iε 32 Contribution of the inter channel coupling W to the scattering length (continued) One can show (see Appendix) that: , , ikr * / + GG e 2μπ − GGG GErr()′′ −=⎡⎤ϕ G ()r n rr(2.39) op r →∞ r =2 2 ⎣⎦kn −−GG* G Using ⎡′⎤′GGrr and the closure relation for r′ , we get ⎣⎦ϕϕkn ()= kn for the asymptotic behavior of (2.37 ):

ikr − + G G e 2μ ϕkn WWϕϕres res ϕk − 2π 2 (2.40) 2 r =, , EE− res ++ − −+ In the limit kE→→→00ϕϕGGand ϕ G ϕϕ →=GGsince / / k 00kn 0 e±ikr rr→ 1 so that (2.40 ) can be also written, using (2.35 ): 22 ++ ϕϕGGWWϕϕ 12μμ00res 12 res −=22ππ22+ (2.41) 22 rr==0 − Eres ξ ()BB− 0 The coefficient of -1/r in (2.41) gives the contribution of the inter- channel coupling to the scattering length 33 Scattering length The asymptotic behavior of the first term of (2.23) gives the background scattering length. Adding the contribution of the second term we have just calculated, we get for the total scattering length: 2 + ϕϕG W 2μ 0 res ⎡ ΔB ⎤ aa=−21π 2 =a − bg 2 bg ⎢ ⎥ (2.42) = −−ξ ()BB0 ⎣ BB− 0 ⎦ where: 2 + ϕϕG W 2μ 0 res B 2 2 (2.43) Δ=2 π = ξ abg This is the main result of this lecture.

- The scattering length diverges when B = B0 - It changes sign when B is scanned around B0 - It vanishes for B – B0 = ΔB The variations of the scattering length with the static field are represented in the next figure

34 Scattering length versus magnetic field a

B Bres 0 0 B

abg

ΔB

Figure corresponding to two colliding Rb85 atoms each in the state fm==2,f -2 in a s-wave (A = 0). In this case, we have and abg <<00ξ 35 Examples of broad and narrow Feshbach resonances

- Entrance channel : ee

fm1122= 2222,,,,ff=− fm = =−A = mA = 0

Mm=++=−ff12 m mA 4 - Other channels with the same

Mm= −==40A A gg, fh eg,df They are open because they are above the entrance channel. Zeeman and hyperfine levels of Rb85 They have the same negative slope ξ (Figure taken from Ref.9) with respect to ee when B is varied G G G Classification by other quantum numbers (,),ffFM12 ,A == mA 0 ( F =+ f1 f 2 )

If ff12= = 2,, F= 024, (Odd values of F are forbidden for identical bosons) Only FM= 44can give =− ⇒ Channel ee corresponds to (,22), FM= 4=− 4

If ff12==3,F = 0246 ,,, (Odd values of F are forbidden for identical bosons) Only F ==4,64can give M − ⇒ Channel gg and fh give rise to 2 types of states ()33,,F ==− 4M 4and (),, 33 F ==−64M 36 Feshbach resonances associated with gg and fh

In the potential wells of the channels (33) F = 6 or 4, M = - 4, there are vibrational levels v = -1,-2,-3,… staring from the highest one v = -1 The energy level (33) F = 4, M= - 4,v = -3 crosses the energy (~0) of the entrance channel around B=155 G The energy level (33) F = 6, M= - 4,v = -3 crosses E~0) around B=250 G (Figure taken from Ref.9 ) (Lower part of the figure) The 2 levels which cross at B=155 G correspond to the same value of F and can thus be coupled by the strong interaction Vel . This is why the corresponding Feshbach resonance is broad The 2 levels which cross at B=250 G correspond to different values of F and can thus be coupled only by the weak interaction Vss . This is why the corresponding Feshbach resonance is narrow (Upper part of the figure) 37 Outline of lecture 2 1 - Introduction 2 - Collision channels • Spin degrees of freedom. • Coupled channel equations • Strong couplings and weak couplings between channels 3 - Qualitative interpretation of Feshbach resonances 4 - Two-channel model • Two-channel Hamiltonian • What we want to calculate 5 - Scattering states of the 2-channel Hamiltonian • Calculation of the outgoing scattering states • Asymptotic behavior. Scattering length • Feshbach resonance 5 - Bound states of the 2-channel Hamiltonian • Calculation of the energy of the bound state • Calculation of the wave function 38 Bound states of the two-channel

Hamiltonian H2-channel

Are there bound states for H2-channel for B close to B0? How are they related to the bound state ϕres of Hcl? How do their energy Eb and wave function vary with B? We denote such a bound state ( ) ( ) b G b G opϕϕop rr+ cl cl (2.44)

bb ϕop and ϕcl are the components of the bound state in the open channel and the closed channel, respectively, obeying the normalization condition: bb bb (2.45) ϕϕop op ϕϕ+ cl cl = 1 Expressing that the state (2.44) is an eigenstate of the Hamiltonian

(2.8) with eigenvalue Eb, we get the following 2 equations: (G ) ( ) (GG) ( ) H ϕϕϕbbrWrr+=b E r op( ) op ( ) cl ( ) b op( ) (2.46) b G bbGG Wrϕϕϕop r+= Hcl cl r Eb cl r 39 Bound states of the two-channel

Hamiltonian H2-channel (continued)

To solve equation (2.46), we can use the Green functions of Hop and Hcl without the iε term because E is negative (below the threshold of V ) b ( ) op ϕ b = GEWϕ b op op( b ) cl (2.47) b b ϕcl= GEW cl b ϕop

As above, we can use the single resonance approximation for Gcl: ( ) ϕϕres res (2.48) GEcl b = EEb − res b Inserting (2.48) into ,the second equation (2.47) shows that ϕcl is proportional to ϕ so that we can write: res ( ) ⎛⎞b ϕop 1 ⎛⎞GEWop b ϕres ⎜⎟= ⎜⎟ (2.49) ⎜⎟b N ⎜⎟ϕ ⎝⎠ϕcl b ⎝⎠res where Nb is a normalization factor ( ) NW=+1 ϕϕG2 EW (2.50) bbres op res 40 Implicit equation for the energy Eb Inserting (2.48) into the second equation (2.47) gives:

b 1 b ϕcl = ϕϕres res ϕW op (2.51) EEb − res which, inserted into the first equation (2.47) leads to: ( ) b 1 b ϕop = GEWop b ϕϕres res ϕ W op (2.52) EEb − res As for equation (2.21), we can eliminate the dressed state ϕ b by . op multiplying both sides of this equation at left by ϕ W This gives: ( ) res

EEbb−=resϕ res WGEW op ϕres (2.53) Now, using the identity ( ) 1111 GEop bb==−+E (2.54) Ebb−−HHop op HEHop op we can rewrite (2.53) as: ( ) ( ) ( )

EEbb=+resϕϕϕ res WGW op 00res − Eres WGGEW op op b ϕres) (2.55) 41 Implicit equation for the energy Eb (continued) . The second term of the right side of (2.55) is the shift =Δ of ϕ , , 0 res Adding it to EE we get so that (2.55) can be rewritten: res res ( ) ( ) EEbresresopopres=− Ebbϕ WGGEW0 ϕ (2.56)

To go further, we introduce the spectral decomposition of Gop(z) ++ ( ) GG ( ) ϕϕkk Gz=+d3 k / Gzb (2.57) op ∫ zk− =22 2μ op The .last term of (2.57) gives the contribution of the bound ,states of

HEop Se suppose here that their energy if far below = 0 so that we can ignore this term. Using (2.57), we can then write (2.56) as: 2 + G 2 ϕϕres W k EE=− ()2μ Ekd3 (2.58) bbres ∫ 22 22 ==kk()+ 2μ Eb

This is an implicit equation for Eb that we will try now to solve

42 Calculation of the energy Eb To calculate the integral of (2.58), we introduce the new variable: = k u = (2.59)

2μ Eb which allows one to rewrite, after angular integration, the integral of (2.58) as: 2 + ϕϕW G 14π ∞ res k du (2.60) =3 ∫0 u2 + 1 2μ Eb () 2 . + G Let kW0 be the width of ϕϕres k considered as a function of k

This defines a value uu0 of

= k0 u0 = (2.61) 2μ Eb characterizing the width in u of the numerator of the integral of (2.60).?

Two different limits can then be considered: uu00 1 and 1 43 Calculation of the energy Eb (continued) / 22 First limit uEk0012⇔ b =μ The denominator of the integral of (2.60) varies more rapidlyG withG u than the numerator which can be replaces by its value for k = 0 Equation (2.60) can then be approximated by:

2 ∞ 14π + du ϕϕW G (2.62) =32res 0 ∫0 u + 1 2μ Eb / =π 2 Replacing the integral of (2.58) by (2.62)/ then leads to: 2 32 2 22πμ( ) + EE=+ E ϕϕWG (2.63) bbres =3 res 0 2 + One can then reexpress ϕϕWBG in terms of Δ thanks to (res )0 (2.43) and EBres in terms of ξ − B0 thanks to (2.35) and finally use (2.43) to show that the solution of (2.6 ) is, to a good approximation: =2 E =− (2.64) b 2 2μa 44 Calculation of the energy Eb (continued) / 22 Second limit uEk0012⇔ b =μ The numerator of the integral of (2.60) varies more rapidly with u than the denominator, so that we can neglect the term u2 in the denominator. In fact, this approximation amounts to neglecting =22k compared to

2μ Eb in the denominator of the integral of (2.58) This approximation allows one to transform (2.58) into: 2 + G 2 ϕϕres W k EE=+ ()2μ d3 k b res ∫ 2μ =22k 2 + G ϕϕres W k (2.65) =+Ek d3 / res ∫ =22k 2μ =−Δ==EEBBres = 0 res ξ () −res

We have used the expression (2.31) of ħΔ0 and equation (2.35) 45 Eb

= Δ0 Eres

B B0 res B

Asymptote Eb with a slope ξ Eres

- The bound state of H2-channel appears for B > B0, in the region a>0. - Eb first decreases quadratically with B-B0 and then tends to the unperturbed energy Eres of the bound state ϕres of the closed channel - If B0 is swept through the Feshbach resonance from the region a<0 to the region a>0, a pair of ultracold atoms can be transformed into a molecule 46 Wave function of the bound state Weight of the closed channel component of the bound state b According to (2.49) and (2.50), the relative weight of ϕcl in the

(normalized) wave function of H2-channel is given by: 1 ( ) ϕ bbϕϕϕ==NW221 +GEW cl cl 2 bbres op res (2.66) Nb Using ( ) 11∂ ( ) ( ) GE= ⇒ GE=− =−GE2 op bbbEH−∂ Eop 2 op bbop ()EH− b op (2.67) we can rewrite the second equation (2.66) as: ( ) 2 ∂ NWbb=−1 ϕresG op EWϕres (2.68) ∂Eb The last term of (2.68) can be transformed using (2.53) ( ) EE=+ϕ WGEWϕ (2.69) bbNres res op res =−ξ ()BBres 47 Wave function of the bound state (continued) Taking the derivative of (2.69) with respect to B, we get: ∂E ∂ ( ) ∂E bb=+ξϕWG ϕ E W ∂BE∂∂res op b res B b (2.70) =−1 N 2 This finally gives: /b 1 ∂Eb ∂B (2.71) 2 = Nb ξ The weight of the closed channel component in the wave function of the bound state, for a given value of B, is thus equal to the slope of

the curve giving Eb(B) versus B, divided by the slope ξ of the asymptote of the curve giving Eb(B) versus B (see Figure page 46) Conclusion When the bound state of the 2-channel Hamiltonian appears near

B=B0 in the region a > 0, the slope of the curve Eb(B) is equal to 0 and the weight of the closed channel component in its wave function

is negligible. For larger values of B, near the asymptote of Eb(B), this weight tends to 1 48 Wave function of the bound state (continued) Expression of the wave function of the bound state The previous conclusion means that, near the Feshbach resonance, the coupling with the closed channel can be neglected for calculating the wave function of the bound state and that we can thus look for 2 2 the eigenfunction of Hop with an eigenvalue –ħ /2μa . The asymptotic behavior of this wave function (at distances larger than the range of Vop) can be obtained by solving the 1D radial Schrödinger equation for u (r) with V =0. (0 ) op ==22d2ur ( ) −=0 −ur (2.72) 2μ d ra222μ 0 The 3D wave function of the bound state thus behaves asymptotically as exp( / ) −ra (2.73) r

49 Comparison with quantitative calculations

Note the logarithmic scale of the r-axis

When one gets closer to the Feshbach resonance, the extension of the wave function becomes bigger and the weight of the closed channel component smaller: 4.7 % at B=160 G 0.1 % at B=155.5 G

Figure taken from Ref. 9 50 Conclusion

The coupling between the collision state of 2 ultracold atoms and a bound state of these 2 atoms in another closed collision channel gives rise to resonant variations of the scattering length a when the energy of the bound state is varied around the threshold of the closed channel by sweeping a static magnetic field B.

The scattering length a diverges for the value B0 of B for which the energy of the bound state in the closed channel, perturbed by its coupling with the continuum of collision states in the open channel, coincides with the threshold of the open channel. The scattering length can thus take positive or negative values, very large values. It vanishes for a certain value of B depending on the background scattering length in the open channel. By choosing the value of B, one can thus obtain an attractive gas, a repulsive one, a perfect gas without interactions (a=0), a gas with very strong interactions (a very large, corresponding to the unitary limit). 51 Conclusion (continued) The width of the resonance, given by the distance between the value of B for which a diverges and the value of B for which it vanishes, depends on the strength of the coupling between the 2 channels. The resonance is broad if the 2 channels are coupled by the spin exchange interaction, narrow if they can be coupled only by the magnetic dipole-dipole spin interactions.

Near B=B0, in the region a>0, the two-atom system has a bound state, with a very weak binding energy, equal to ħ2/2μa2. The wave function of this bound state has a very large spatial extent of the order of a. Its closed channel component is negligible compared to the open channel component.

By sweeping B near B0, one can transform a pair of colliding atoms into a molecule or vice versa. A few problems not considered here: - Influence of the speed at which B is scanned. - Stability of the “Feshbach molecules”. How do inelastic and 3-body collisions limit their lifetime. Bosonic versus fermionic molecules. D.Petrov, C.Salomon, G.Shlyapnikov, Phys.Rev.Lett. 93, 090404 (2004) 52 APPENDIX

For the 2 lectures of Claude Cohen-Tannoudji on “Atom-Atom Interactions in Ultracold Quantum Gases”

1 Purpose of this Appendix 1 – Demonstrate the orthonormalization relation ( ) (A.1) ϕk′ l′′ mϕδ klm =−kk δδ′ ll′′ mm - The wave function 2 ur() ϕ ()rY= kl (,)θϕ (A.2) klm π r lm describes, in the angular momentum representation, a particle of mass μ, with energy E=ħ2k2/2μ, in a central potential V(r) - The radial wave function ukl(r) is a regular solution of ( ) ⎡⎤d2 21μ ( ) ( ) ( ) ( ) 2 + +−kVrurVrVr2 =0 = + (A.3) ⎢⎥22tot kl tot 2 ⎣⎦drr ( ) 2μ

ukl 00= (A.4) which behaves, for r→∞, as: ( ) sin / ( ) ur ⎡ krl−+πδ2 k⎤ (A.5) kl r →∞ ⎣ l ⎦ - There are other (non regular) solutions behaving, for r→∞, as: ( ) exp / ( ) exp( ) ± l ur∓⎡⎤±− ikrlπ 2 =i ± ikr (A.6) kl ⎣⎦( ) r →∞ 2 / ( ) 2 – Calculate the Green function of: H =+pVr2 2μ with outgoing and ingoing asymptotic behavior ( ) ± , 22/ (EHGrr−=−=) ( ′) δ ( rr′) E k2μ (A.7) - Show that: ( ) , 21μ exp( ) * ( , ) ( , ) ( ) ( ) G± rr′′=− ±iδθϕθϕ Y Y′ urur± ()2 ∑ llm lm klkl< > krr′ lm (A.8) where r> (r<) is the largest (smallest) of r and r’ - Introducing the Heaviside function:

θ (rr− ′) =+1 if r > r′ (A.9) = 0 if rr< ′ (A.8) can also be written: ( ) , 21μ exp( ) * ( , ) ( , ) Grr± ()′′=− ±iYδθϕθϕ Y ′× 2 ′ ∑ llm lm krr lm( ) ( ) ( ) ( ) ×−⎡⎤θθr rurur′ ′′±± +− r rurur ′ ⎣⎦()kl kl kl () kl (A.10) 3 – Calculate the asymptotic behavior of these Green functions and demonstrate Equation (2.39) of Lecture 2 3 Wronskian Theorem The calculations presented in this Appendix use the Wronskian theorem (see demonstration in Ref.2 Chapter III-8) - Consider the 1D second order differential equation: ( ) ( ) ( ) yr′′ + Fryr= 0 (A.11) Equation (A.4) is of this type with: ( ) 2μ ( ) F rk=−2 Vr (A.12) 2 tot ( ) ( ) - Let yr and yr be 2 solutions of this equation corresponding to 12( ) ( ),

2 different functions Fr12 and Fr respectively. The wronskian of y and y is by definition: ( 1, ) 2 ( ) ( ) ( ) ( )

W yy12= y 1r y 2′ r − y 2r y 1′ r (A.13) - One can show that: ( , )b ( , ) ( , ) W yy =−⎡W yy⎤⎡W yy ⎤ 12ar⎣ 12⎦⎣=br12 ⎦=a(A.14) b ( ) ( ) ( ) ( ) =−⎡⎤F rFry r y rrd ⎣⎦1212 ∫a 4 Demonstration of (A.1)

We consider 2 different values k1 and k2 of k. According to (A.12): ( ) ( ) 22 F1212rFrkk− =− (A.15) (A.14) then gives the scalar product of y = u and y = u 12kl12k l b ( ) ( ) 1 ( , )b y r y rrd = Wyy (A.16) ∫a 12 22 12a , ( , ) kk12− If we take aWyy==00⎡⎤ because of (A.4) ⎣⎦12ra= ( ), If we take bR= very large compared to the range of Vr we can use the asymptotic behavior (A.5) of uukl and k l ( ) ( ) ( ) (12) ( ) ( ) R 1 ⎡ ⎤ ukl r u k l rd r =−ukl ru k′′ l r uk l ru kl r (A.17) ∫0 12 kk22− ⎣ 12 21⎦rR= 12( ) , ( ) ,

Using (A.15) and putting δllkk1122= δδ δ = we get: sin ( ) ( ) ⎡ kkR+−++πδ δ⎤ R 1 ⎣( 12) 12⎦ ururrkl k l d = −+ ∫0 12 2 kk+ sin 12 (A.18) ⎡⎤ 1 ()kkR12−+−δδ 12 + ⎣⎦ 2 kk12− 5 - When R→∞, the first term of the right side of (A.18) vanishes as a distribution, because it is a rapidly oscillating function of k1+k2 (k1 and k2 being both positive k1+k2 cannot vanish)

- The second term becomes important when k1-k2 is close to zero (we have then δ1-δ2=0) -Using: sin lim 1 Rx ( ) = δ x (A.19) R →∞ π x we get: ∞ ( ) ( ) π ( ) ururrkl k l d =−δ k12 k (A.20) ∫0 12 2 - We then have, according to (A.2):

* * ( ) ( ) 2 ( , ) ( , ) ( ) ( ) dd3rrrϕ ϕ =Ω Yθϕ θϕ Y ururr d ∫∫∫k′′ l m ′ klm π l′′ m lm kl k′ l =δδ π ( ) ll′′ mm =−δ kk′ 2 ( ) (A.21) =−δδδkk′ ll′′ mm which demonstrates (A.1). 6 Demonstration of (A.8)

Let us apply E-H to the right side of (A.8). Using (A.10) and: 22( ) ⎡12∂2L2μ ( )⎤ HV=− Δ+r =− − − Vr(A.22) ⎢ 2222 ⎥ 22μμ⎣r ∂ rr⎦ we get, using (A.12): ( ) , 1 exp( ) * ( , ) ( , ) ± ′′′ ()()EHGrr−=−±∑ iδθϕθϕllm Y Y lm × krr′ lm ⎪⎧⎛⎞( ) ∂2 ( ) ( ) ( ) ( )⎪⎫ ×+Fr⎡θθ r − rurur′ ′′±± +− r rurur ′⎤ ⎨⎜⎟2 ⎣ ()kl kl kl () kl ⎦⎬ ⎩⎭⎪⎝⎠∂r ⎪ To calculate the second line of (A.23), we use: (A.23) ∂∂ θθδ()rr12−=−()() rr21 −= rr 12 − ∂∂rr11 ⎡⎤∂∂( ) ( ) ( )⎡⎤(A.24) ⎢⎥δδδ()rrfr12−=−−+− 1 frrr′ 2() 12 fr 2⎢⎥() rr 12 ⎣⎦∂∂rr1 ⎣⎦1 The second order derivative( of the) second( line )of (A.23)( gives) 3 types of terms:( proportional)/ to θθδrr−−−′′ and r r , to rr ′ and to ∂−δ rr′ ∂ r 7 ( ) ( ) / ( ) - The terms ∝−θ rr′ are multiplied by ⎡⎤Fr+∂∂( 22 r) ur± ( ) ⎣⎦kl which vanishes because ur± is a solution of (A.3). kl ( ) ′ The same argument( ) applies/ for the terms( ) ∝−θ rrwhich are ⎡⎤2 2 multiplied by Fr +∂() ∂rurkl = 0 ⎣⎦( )/ - The terms proportional to ∂δ rr−∂′ rcancel out

- The only terms surviving( in the), second line of (A.23) are those proportional to δ r − r′ which gives for this line: ( ) ( )/ ( ) ( )/ ( ) ±± ⎡⎤ur′ ∂∂−∂∂− ur′′ r ur ′ ur ′′ r δ rr ′ (A.25) ⎣⎦kl ()kl kl ()kl ± - We recognize in the bracket of (A.25) the Wronskian of uukl and kl We can thus use (A.14) with FF= since u and u± correspond . 12 kl kl to the same value of k - Equation. (A.14) shows that the Wronskian is independant of r when

FF12= We can thus calculate it for very large values of r where we ± know the asymptotic behavior (A.5) and (A.6) of uukl and kl 8 - The calculation of the Wronskian appearing in (A.25) is straightforward using (A.5) and (A.6) and gives: ( , ) exp( ) + Wukl u kl =− k∓ iδl (A.26) - Inserting (A.26) into (A.25) and then in (A.23) gives: ( ) , 1 ( ) * ( , ) ( , ) EHGrr−=−± ′ δ rr′′ Yθϕ θ Y ϕ ′ ()()2 ∑ lm lm (A.27) r lm - We can then use the closure relation for the spherical harmonics (see Ref. 3, Complement AVI): * ( , ) ( , ) (cos cos ) ( ) ′ ′′′(A.28) ∑ YYlm θ ϕθϕδlm =− θ θδϕϕ − lm to obtain: ( ) , 1 ( ) ( c o s c o s ) ( ) ()()EHGrr−=−−−± ′ δ rr′′′δθ θδϕϕ r 2 =−δ ( rr′ ) (A.29) which demonstrates (A.8). 9 Asymptotic behavior of G+ For r very large, only the first term of the bracket of (A.10) is non zero and we get: ( ) , *( , ) ( , ) ( ) ( ) 21μ iδ Grr+ ()′′ − e l Yθϕ θ Y ϕ ′ urur′+ r →∞ 2 ′ ∑ lm lm kl kl krr lm (A.30) According to (A.6), we have e ( ) , ( ) * ( , ) ( , ) ( ) ikr 21μ iδ Grr+ ()′′ −−il e l Yθϕ θ Y ϕ ′ ur′ r →∞ 2 ∑ lm lm kl kr′ lm r (A.31) On the other hand, from Eq. (1.46) of lecture 1 and (A.2), we have: * ( ) ( ) ( ) exp( ) ( ) ( ) ′ − 12 l urkl rr′ ′′′ ϕδkn rii=−∑ lY lmnY lm nnn== krπ lm ′ rr′ (A.32) Using (A.32), we can rewrite (A.31) as: ( ) * e , ( ) ikr +− 2μπ Grr()′′ − ⎡⎤ϕ r (A.33) r →∞ 2 2 ⎣⎦kn r which demonstrates Eq. (2.39) of lecture 2. 10