Excitation Spectrum of Trapped Bose-Einstein Condensates Benjamin - PowerPoint PPT Presentation

Excitation Spectrum of Trapped Bose-Einstein Condensates Benjamin Schlein, University of Zurich From Many Body Problems to Random Matrices Banff, August 5, 2019 Joint works with Boccato, Brennecke, Cenatiempo, Schraven 1

Introduction Bose-Einstein condensates: in the last two decades, BEC have become accessible to experiments. Goal: understand low-energy properties of trapped condensates, starting from microscopic description. 2

Gross-Pitaevskii regime : N bosons in Λ = [0; 1] 3 , interacting through potential with effective range of order N − 1 , as N → ∞ . Range of interaction much shorter than typical distance among particles: collisions rare, dilute gas . Hamilton operator: has form N N � � N 2 V ( N ( x i − x j )) , on L 2 s (Λ N ) H N = − ∆ x j + j =1 i<j V ≥ 0 with compact support . 3

Scattering length: defined by zero-energy scattering equation � � − ∆ + 1 2 V ( x ) f ( x ) = 0 , with f ( x ) → 1 as | x | → ∞ f ( x ) = 1 − a 0 ⇒ | x | , for large | x | Equivalently, � 8 π a 0 = V ( x ) f ( x ) dx By scaling , � � − ∆ + 1 2 N 2 V ( Nx ) f ( Nx ) = 0 Rescaled potential has scattering length a 0 /N . 4

Ground state energy : [Lieb-Yngvason ’98] proved that E N = 4 π a 0 N + o ( N ) BEC: [Lieb-Seiringer ’02, ’06] showed that ψ N ∈ L 2 s (Λ N ) with � ψ N , H N ψ N � ≤ 4 π a 0 N + o ( N ) exhibits BEC, i.e. reduced density matrix � γ N ( x ; y ) = dx 2 . . . dx N ψ N ( x, x 2 , . . . , x N ) ψ N ( y, x 2 , . . . , x N ) is such that N →∞ � ϕ 0 , γ N ϕ 0 � = 1 lim with ϕ 0 ( x ) = 1 for all x ∈ Λ. Warning: this does not mean that ψ N ≃ ϕ ⊗ N . In fact 0 � = ( N − 1) � ϕ ⊗ N , H N ϕ ⊗ N � V (0) ≫ 4 π a 0 N 0 0 2 Correlations are important!! 5

Main results Theorem [Boccato, Brennecke, Cenatiempo, S., ’17] : There exists C > 0 such that | E N − 4 π a 0 N | ≤ C uniformly in N . Furthermore, if ψ N ∈ L 2 s (Λ N ) such that � ψ N , H N ψ N � ≤ 4 π a 0 N + ζ we have 1 − � ϕ 0 , γ N ϕ 0 � ≤ C ( ζ + 1) N Interpretation: in low-energy states, condensation holds with optimal rate, with bounded number of excitations. Question: Is it possible to resolve order one contributions to the ground state energy? 6

Theorem [Boccato, Brennecke, Cenatiempo, S., ’18] : Let Λ ∗ + = 2 π Z 3 \{ 0 } . Then E N = 4 π a 0 ( N − 1) + e Λ a 2 0 � � � | p | 4 + 16 π a 0 p 2 − (8 π a 0 ) 2 � − 1 p 2 + 8 π a 0 − + O ( N − 1 / 4 ) 2 p 2 2 p ∈ Λ ∗ + where � cos( | p | ) e Λ = 2 − lim p 2 M →∞ p ∈ Z 3 \{ 0 } : | p 1 | , | p 2 | , | p 3 |≤ M Moreover, for the ground state, we have the BEC depletion   � p 2 + 8 π a 0 − | p | 4 + 16 π a 0 p 2 � 1 −� ϕ 0 , γ N ϕ 0 � = 1    + O ( N − 9 / 8 ) �  | p | 4 + 16 π a 0 p 2 N 2 p ∈ Λ ∗ + 7

Theorem [Boccato, Brennecke, Cenatiempo, S., ’18] : The spectrum of H N − E N below a threshold ζ > 0 consists of eigen- values � � | p | 4 + 16 π a 0 p 2 + O ( N − 1 / 4 (1 + ζ 3 )) n p p ∈ Λ ∗ + where n p ∈ N for all p ∈ Λ ∗ + . Interpretation: every excitation with momentum p ∈ Λ ∗ + “costs” � | p | 4 + 16 π a 0 p 2 . energy ε ( p ) = Remark: excitation spectrum is crucial to understand the low- energy properties of Bose gas. The linear dependence of ε ( p ) on | p | for small p can be used to explain the emergence of superfluidity . 8

Previous works Mathematically simpler models described by N N � � − ∆ x j + 1 H β N 3 β V ( N β ( x i − x j )) N = N j =1 i<j for β ∈ [0; 1). In mean field regime , β = 0, excitation spectrum determined in [Seiringer, ’11], [Grech-Seiringner, ’13], [Lewin-Nam-Serfaty- Solovej, ’14], [Derezinski-Napiorkowski, ’14], [Pizzo, ’16]. � | p | 4 + 2 � V ( p ) p 2 . Dispersion of excitations given by ε mf ( p ) = For intermediate regimes , β ∈ (0; 1) (and V small enough) excitations spectrum determined in [BBCS, ’17]. � | p | 4 + 2 � V (0) p 2 . Dispersion of excitations given by ε β ( p ) = For Gross-Pitaevskii regime , β = 1, and V small, excitations spectrum determined in [BBCS, ’18]. 9

Extension to BEC in external potentials Consider N bosons in R 3 , with Hamilton operator N N � � � � N 2 V ( N ( x i − x j )) − ∆ x j + V ext ( x j ) H N ( V ext ) = + j =1 i<j with V ext a trapping potential. [ Lieb-Seiringer-Yngvason, ’00 ] proved that E N E GP ( ϕ ) lim N = min N →∞ ϕ ∈ L 2 ( R 3 ): � ϕ � =1 with the Gross-Pitaevskii energy functional � � |∇ ϕ | 2 + V ext | ϕ | 2 + 4 π a 0 | ϕ | 4 � E GP ( ϕ ) = dx R 3 [ Lieb-Seiringer, ’02 ]: ground state exhibits BEC into minimizer ϕ GP of Gross-Pitaevskii functional, ie. N →∞ � ϕ GP , γ N ϕ GP � = 1 lim 10

Theorem [Brennecke-S.-Schraven, in progress] : Optimal BEC: if ψ N ∈ L 2 s ( R 3 N ) with � ψ N , H N ( V ext ) ψ N � ≤ E N ( V ext ) + ζ then 1 − � ϕ GP , γ N ϕ GP � ≤ C ( ζ + 1) N Excitation spectrum: let h GP = − ∆ + V ext + 8 π a 0 | ϕ GP | 2 and ε 0 = inf σ ( h GP ). Let D = h GP − ε 0 and � D 1 / 2 ( D + 16 π a 0 | ϕ GP | 2 ) D 1 / 2 � 1 / 2 E = Spectrum of H N ( V ext ) − E N ( V ext ) below threshold ζ > 0 consists of eigenvalues having the form � n i e i + o (1) where e i are eigenvalues of E and n i ∈ N . i ∈ N 11

Dynamics generated by change of external fields First results by [ Erd˝ os-S.-Yau, ’06, ’08 ], and by [ Pickl, ’10 ]. Theorem [Brennecke-S., ’16]: let ψ N ∈ L 2 s ( R 3 N ) with reduced density matrix γ N such that a N = 1 − � ϕ GP , γ N ϕ GP � → 0 � � � � � N − 1 � ψ N , H N ( V ext ) ψ N � − E GP ( ϕ GP ) b N = � → 0 Let N N � � N 2 V ( N ( x i − x j )) on L 2 s ( R 3 N ) − ∆ x j + H N = j =1 i<j and ψ N,t = e − iH N t ψ N solve many-body Schr¨ odinger equation . Then � a N + b N + N − 1 � 1 − � ϕ t , γ N,t ϕ t � ≤ C exp( c exp( c | t | )) where ϕ t solves time-dependent Gross-Pitaevskii equation i∂ t ϕ t = − ∆ ϕ t + 8 π a 0 | ϕ t | 2 ϕ t , with ϕ t =0 = ϕ GP . 12

Thermodynamic limit Consider N bosons in Λ L = [0; L ] 3 , with N, L → ∞ but fixed density ρ = N/L 3 . As ρ → 0, Lee-Huang-Yang predicted � � 1 + 128 E N 0 ) 1 / 2 + o ( ρ 1 / 2 ) 15 √ π ( ρ a 3 lim N = 4 π a 0 ρ N,L →∞ N/L 3 = ρ Leading order known from [ Lieb-Yngvason, ’98 ]. Upper bound to second order in [ Erd˝ os-S.-Yau,’08 ], [ Yau-Yin,’09 ]. [ Fournais-Solovej, ’19 ] got matching lower bound (next talk!). Remark: Gross-Pitaevskii regime corresponds to limit ρ = N − 2 . Still open: prove BEC and determine excitations in thermodynamic limit. 13

Bogoliubov approximation Fock space: define F = � n ≥ 0 L 2 s (Λ n ). Creation and annihilation operators: for p ∈ 2 π Z 3 , introduce a ∗ p , a p creating and annihilating particle with momentum p . Canonical commutation relations: for any p, q ∈ 2 π Z 3 , � � � � � � a p , a ∗ a ∗ p , a ∗ = δ p,q , a p , a q = = 0 q q Number of particles: a ∗ p a p measures number of particles with momentum p , � a ∗ N = p a p = total number of particles operator p ∈ Λ ∗ 14

Hamilton operator: we write � � p a p + 1 p 2 a ∗ V ( r/N ) a ∗ p + r a ∗ � H N = q a p a q + r N p ∈ Λ ∗ p,q,r ∈ Λ ∗ Number substitution: BEC implies that √ a 0 , a ∗ N ≫ 1 = [ a 0 , a ∗ 0 ≃ 0 ] √ Bogoliubov replaced a ∗ 0 , a 0 by factors of N . He found � � H N ≃ ( N − 1) p 2 a ∗ a ∗ � p a p + � V (0) + V (0) p a p 2 p � =0 p � =0 � � � + 1 2 a ∗ p a p + a ∗ p a ∗ � V ( p/N ) − p + a p a − p 2 p � =0 � � � 1 a ∗ p + q a ∗ − p a q + a ∗ � √ + V ( p/N ) q a − p a p + q N p,q � =0 � + 1 V ( r/N ) a ∗ p + r a ∗ � q a p a q + r N p,q,r � =0 15

Diagonalization: neglecting cubic and quartic terms, and using appropriate Bogoliubov transformation � � � � � a ∗ p a ∗ T = exp − p − a p a − p τ p p ∈ Λ ∗ + one finds V 2 ( p/N ) � � T ∗ H N T ≃ ( N − 1) V (0) − 1 � 2 p 2 2 2 p � =0 � � � V (0) 2 � � − 1 p 2 + � | p | 4 + 2 � V (0) p 2 − V (0) − 2 p 2 2 p � =0 � � | p | 4 + 2 � V (0) p 2 a ∗ + p a p p � =0 Born series: for small potentials, scattering length given by 8 π a 0 = � V (0) ∞ n − 1 � ( − 1) n � V (( p j − p j +1 ) /N ) � � � V ( p 1 /N ) � + V ( p n /N ) p 2 p 2 2 n N n 1 n =1 p 1 ,...,p n � =0 j =1 j +1 16

Scattering length: replacing V 2 ( p/N ) � � V (0) − 1 � � V (0) → 8 π a 0 , → 8 π a 0 2 p 2 N p Bogoliubov obtained T ∗ H N T ≃ 4 π a 0 ( N − 1) � � � | p | 4 + 16 π a 0 p 2 − (8 π a 0 ) 2 � − 1 p 2 + 8 π a 0 − 2 p 2 2 p � =0 � � | p | 4 + 16 π a 0 p 2 a ∗ + p a p p � =0 Hence � � � | p | 4 + 16 π a 0 p 2 − (8 π a 0 ) 2 E N = 4 π a 0 ( N − 1) − 1 � p 2 + 8 π a 0 − 2 p 2 2 p � =0 and excitation spectrum consists of � � | p | 4 + 16 π a 0 p 2 , n p ∈ N n p p � =0 Final replacement makes up for missing cubic and quartic terms! 17