On the infimum of the excitation spectrum of a homogeneous Bose gas. H.D. Cornean, J. Dereziński,

Transcription

1 On the infimum of the excitation spectrum of a homogeneous Bose gas H.D. Cornean, J. Dereziński, P. Ziń. 1

2 Homogeneous Bose gas n bosons on R d interacting with a 2-body potential v are ( described by the Hilbert space L 2 s (R d ) n) and the Hamiltonian H n = 1 2 i + v(x i x j ) i=1 commuting with total momentum 1 i<j n P n := n i=1 i xi. We assume that the potential v decays fast at infinity, v(x) = v( x) and ˆv(k) 0. 2

3 System in a box I We are interested in the properties at fixed density ρ. Therefore, we enclose the system in Λ = [0,L] d. We assume that the number of particles equals n = ρv, where V = L d is the volume. The Hilbert space is L 2 s(λ n ). We replace the potential by its periodization v L (x) = 1 e ik x ˆv(k). V k 2π L Zd 3

4 System in a box II The Hamiltonian, with periodic boundary conditions, equals H L,n = n i=1 1 2 L i + 1 i<j n v L (x i x j ) and has the ground state energy E L,n := inf sph L,n. The total momentum P L,n := n i=1 i L x i has the spectrum spp L,n = 2π L Zd. 4

5 Infimum of the excitation spectrum For k 2π L Zd we set ǫ L,n (k) := inf sph L,n (k) E L,n. Let k R d. For L, keeping n V (informally) ǫ ρ (k) := lim L ǫ L,n (k). = ρ > 0, we set 5

6 Boost operator Define the Boost operator in the direction of the first coordinate: ( ) i2π n U 1 := exp x i,1. L We easily compute U 1 i=1 U1 P L,n 1 U 1 = P L,n 1 + 2πρL d 1, ( H L,n 1 ) L,n 2ρLd(P 1 ) 2 U 1 = H L,n 1 L,n 2ρLd(P 1 ) 2. Therefore, in particular sph L,n (m2πρl d 1 ê 1 ) = sph L,n (0) + m2 (2π) 2 ρl d

7 Excitation spectrum of free Bose gas in finite volume L (k) - 2 n L 2 n L k 7

8 Excitation spectrum of interacting Bose gas in finite volume L (k) - 2 n L 2 n L k 8

9 1-dimensional case Theorem. In dimension d = 1 we have ǫ(k + 2πρ) = ǫ(k). Proof. If Φ then (H L,n E)Φ = 0, (P L,n k)φ = 0, (H L,n E)UΦ = 1 L (2πk + 2π2 ρ)uφ, (P L,n k 2πρ)UΦ = 0. Then we let L. 9

10 Excitation spectrum of 1-dimensional interacting Bose gas (k) k 10

11 Conjecture. Conjecture about the infimum of the excitation spectrum (1) The map R d k ǫ ρ (k) R + is continuous. (2) Let k R d. If L j, n j ρ, k L d s 2π L j j Z d, then we have that ǫ L j,n j (k j ) ǫ ρ (k). (3) If d 2, then there exists c cr > 0 such that ǫ ρ (k) > c cr k. (4) For some c ph > 0, ǫ ρ (k) c ph k for small k. (5) k ǫ ρ (k) is subadditive, that is ǫ ρ (k 1 ) + ǫ ρ (k 2 ) ǫ ρ (k 1 + k 2 ). 11

12 (1) and (2) can be interpreted a kind of a spectral thermodynamic limit. If (1) and (2) are true around k = 0, then there is no gap in the excitation spectrum. (It is easy to show that ǫ(0) = 0). (3) implies the superfluidity of the Bose gas. More precisely, a drop of Bose gas will travel without friction as long as its speed is less than c cr. (4) suggests that speed of sound for at low energies is well defined and equals c ph. If one can superimpose (almost) independent elementary excitations, then (5) is true. 12

13 The above conjecture has important physical consequences. In particular, it implies the superfluidity of the Bose gas at zero temperature. The above conjecture seems plausible. It is suggested by the arguments going back to Bogoliubov, Hugenholz, Pines, Bieliaev, as well as Bijls and Feynman. Nobody has an idea how to prove it rigorously. 13

14 Subadditive functions We say that R d k ǫ(k) R is subadditive iff ǫ(k 1 + k 2 ) ǫ(k 1 ) + ǫ(k 2 ), k 1,k 2 R d. Let R d k ω(k) R be another function. We define the subbadditive hull of ω to be ǫ(k) := inf{ω(k 1 )+ +ω(k n ) : k 1 + +k n = k, n = 1, 2,...}. Clearly, the subadditive hull is always subadditive. 14

15 Quadratic Hamiltonians Consider the Hamiltonian H = ω(k)a k a kdk, R d and the total momentum P = ka ka k dk. R d We will call the function ω appearing in H the elementary excitation spectrum. Clearly, the infimum of the energy-momentum spectrum of H is equal to the subadditive hull of ω. 15

16 Excitation spectrum of free Bose gas energy (k) (k) k 16

17 Hypothethic excitation spectrum of interacting Bose gas with no rotons energy (k) (k) k 17

18 Hypothethic excitation spectrum of interacting Bose gas with rotons energy (k) (k) k 18

19 Criterion for subadditivity Theorem. (1) Let f be an increasing concave function with f(0) 0. Then f( k ) is subadditive. (2) Let ǫ 0 be subadditive and ǫ 0 ω. Let ǫ be the subadditive hull of ω. Then ǫ 0 ǫ. It often happens that the subadditive hull of a function is equal to zero everywhere. This is the case e.g. when ω(k) = ck 2, which corresponds to the free Bose gas. But not for superfluid systems: 19

20 Subadditive hulls with phononic shape and positive critical velocity Corollary. Suppose that c cr,c s > 0 and ω satisfies 1. ω(k) c cr k ; 2. lim k 0 ω(k) k = c s. Let ǫ be the subadditive hull of ω. Then ǫ also satisfies 1. ǫ(k) c cr k ; 2. lim k 0 ǫ(k) k = c s. 20

21 Landau s argument for the superfluidity I We add to H the perturbation u travelling at a speed w: i d dt Ψ t = ( n H + λ u(x i wt) ) Ψ t. i=1 We go to the moving frame: Ψ w t (x 1,...,x n ) := Ψ t (x 1 wt,...,x n wt). We obtain a Schrödinger equation with a time-independent Hamiltonian i d dt Ψw t = ( n H wp + λ u(x i ) ) Ψ w t. i=1 Is the ground state HΨ gr = EΨ gr stable against the travelling perturbation? 21

22 Bose gas travelling slower than critical velocity energy k 22

23 Bose gas travelling faster than critical velocity energy k 23

24 Travelling Bose gas in finite volume energy - 2 n L 2 n L k 24

25 Stability Define the global critical velocity c L,n cr := inf k ǫ L,n (k) k If w < c L,n cr, then the ground state of H L,n remains the ground state of the tilted Hamiltonian, hence it is stable. For the free Bose gas we have c L,n cr c L,n cr π L = π L > 0. In general,. Hence the global critical velocity is very small and vanishes in the thermodynamic limit. 25

26 Metastability Define the restricted critical velocity below the momentum R as { } ǫ c L,n L,n cr,r := inf (k) k 0, k < R. k We expect that for repulsive potentials c ρ cr,r := lim L cl,n cr,r, exists and, in dimension d 2, lim inf R cρ cr,r > 0. n V = ρ, We expect metastability against a travelling perturbation travelling at a smaller speed. 26

27 2nd quantized grand-canonical approach I Consider the symmetric Fock space Γ s (L 2 (Λ)). For a chemical potential µ > 0, we define the grand-canonical Hamiltonian H L = a x ( 1 2 x µ)a x dx + 1 a 2 xa yv L (x y)a y a x dxdy, = n=0(h n,l µn). 27

28 2nd quantized grand-canonical approach II In the momentum representation it equals H L = k + 1 2V ( 1 2 k2 µ)a ka k k 1,k 2,k 3,k 4 δ(k 1 + k 2 k 3 k 4 )ˆv(k 2 k 3 )a k 1 a k 2 a k3 a k4. The momentum operator equals P L := k ka k a k. If E L is the ground state energy of H L, then one can get the corresponding density by µ E L = V ρ. 28

29 Infimum of the excitation spectrum a rigorous definition I We define the ground state energy in the box E L = inf sph L. For k 2π L Zd, we define the infimum of the excitation spectrum in the box ǫ L (k) := inf sph L (k) E L. 29

30 Infimum of the excitation spectrum a rigorous definition II For k R d we define the infimum of the excitation spectrum in the thermodynamic limit ( ǫ(k) := sup δ>0 lim inf L ( inf ǫ L (k L) k L 2π L Zd, k k L <δ )). Proposition. At zero total momentum, the excitation spectrum has a global minimum where it equals zero: ǫ L (0) = ǫ(0) = 0. 30

31 Conjecture. Conjecture about the infimum of the excitation spectrum (1) The map R d k ǫ(k) R + is continuous. (2) Let k R d. If L j, k j 2π L j Z d, k j k, then ǫ L j (k j ) ǫ(k). (3) If d 2, then there exists c cr > 0 such that ǫ(k) > c cr k. (4) For some c ph > 0 such that ǫ(k) c ph k for small k. (5) k ǫ(k) is subadditive. 31

32 Minimization among coherent states For α C, we define the displacement or Weyl operator of the zeroth mode: W α := e αa 0 +αa 0. Set Ω α := W α Ω. Note that P L Ω α = 0. The expectation of the Hamiltonian in those coherent states equals (Ω α H L Ω α ) = µ α 2 + ˆv(0) 2V α 4, and is minimized for α = e iτ V µ ˆv(0). 32

33 Translation of the zero mode We apply the Bogoliubov translation to the zero mode of H L by W(α). This means making the substitution a 0 = ã 0 + α, a 0 = ã 0 + α, a k = ã k, a k = ã k, k 0. Note that ã k = Wαa k W α, ã k = Wαa kw α, and thus the operators with and without tildes satisfy the same commutation relations. We drop the tildes. 33

34 Translated Hamiltonian H L := V µ2 2ˆv(0) + ( 1 2 k2 + ˆv(k) µ ) a k ˆv(0) a k k + µ ( ˆv(k) e i2τ a k a k + e i2τ a 2ˆv(0) ka k) k + ˆv(k) µ (e ˆv(0)V iτ a k+k a k a k + e iτ a ka k a k+k ) k,k + δ(k 1 + k 2 k 3 k )ˆv(k 2 k 3 ) 4 a k 2V 1 a k 2 a k3 a k4. k 1,k 2,k 3,k 4 Let H L bg denote the first 3 lines of the above expression. 34

35 Translated Hamiltonian with a coupling constant If we (temporarily) replace the potential v(x) with λv(x), where λ is a (small) positive constant, the translated Hamiltonian can be rewritten as H λ,l = λ 1 H L 1 + H L 0 + λh L λh L 1, Thus the 3rd and 4th terms are in some sense small, which suggests dropping them. 35

36 Bogoliubov rotation We want to analyze Hbg L. To this end, for nonzero k we substitute a k = c kb k s kb k, a k = c k b k s k b k, with c k = 1 + s k 2, so that [b k,b k ] = δ k,k, [b k,b k ] = 0. For the zero mode we introduce p 0 = 1 2 (a 0 + a 0 ) and x 0 = i 2 (a 0 a 0 ). 36

37 Bogoliubov Hamiltonian after rotation H L bg = µp k ωbg (k)b kb k + E L bg, where the elementary excitation spectrum of Hbg L is 1 ω bg (k) = 2 k2 ( 1 2 k2 + 2µˆv(k) ˆv(0) ), and its ground state energy equals E L bg = µ 2 V 2ˆv(0) k 1 2 ( ( 1 2 k2 + µˆv(k) ) ) ωbg (k) ˆv(0). 37

38 Zeroth mode The number α has an arbitrary phase. Thus we broke the symmetry when translating the Hamiltonian. This is related to the fact that the zero mode is not a harmonic oscillator it has continuous spectrum. The zeroth mode can be interpreted as a kind of a Goldstone boson. 38

39 Infimum of excitation spectrum in the Bogoliubov approximation I The infimum of the excitation spectrum of Hbg L by is given ǫ bg (k) := inf{ω bg (k 1 ) + + ω bg (k n ) : k k n = k, n = 1, 2,...}. ω bg (k) and ǫ bg (k) have a phononic shape and a positive critical velocity. 39

40 Infimum of excitation spectrum in the Bogoliubov approximation II Replace the potential v(x) with λv(x), where λ > 0. Let ǫ λ (k) be the grand-canonical IES for the potential λv. Conjecture. Let d 2. Then for a large class of repulsive potentials the Bogoliubov method gives the correct IES in the weak coupling limit: lim λց0 ǫλ (k) = ǫ bg (k). 40

41 Bogoliubov transformations commuting with momentum Let α C and 2π L Zd k θ k C be a sequence with θ k = θ k. Set U θ := k e 1 2 θ ka k a k +1 2 θ ka k a k Then U α,θ := U θ W α is the general form of a Bogoliubov transformation commuting with P. 41

42 Improving the Bogoliubov method I Let Ω denote the vacuum vector. Ψ α,θ := Uα,θ Ω is the general form of a squeezed vector of zero momentum. Vectors Ψ α,θ,k := Uα,θ a kω have momentum k, that means Clearly we have bounds (P L k)ψ α,θ,k = 0. E L (Ψ α,θ H L Ψ α,θ ) E L + ǫ L (k) (Ψ α,θ,k H L Ψ α,θ,k ) 42

43 Improving the Bogoliubov method II After the translation, for all k we make the substitution where Note that a k = c kb k s kb k, a k = c k b k s k b k, c k := cosh θ k, s k := θ k θ k sinh θ k. U θ a k U θ = b k, U θ a ku θ = b k, 43

44 Improving the Bogoliubov method III H L = B L + C L b 0 + CL b O L (k)b 2 kb k Then k + terms higher order in b s. k O L (k)b k b k + k D L (k)b kb k (Ψ α,θ H L Ψ α,θ ) = B L, (Ψ α,θ,kl H L Ψ α,θ,k ) = B L +D L (k). 44

45 Minimizing the energy in squeezed states I We look for the infimum of the Hamiltonian among the states Ψ α,θ. This means that B attains a minimum. Computing the derivatives with respect to α and α we obtain C = c 0 α B s 0 α B so that the condition: α B = α B = 0 entails C = 0. 45

46 Minimizing the energy in squeezed states II Computing the derivatives with respect to s and s we obtain O(k) = ( 2c k + s ) k 2 Thus sk B = sk B = 0 entails O(k) = 0. c k sk B s2 k c k sk B. 46

47 Fixed point equation I Instead of s k, c k, it is more convenient to use functions S k := 2s k c k, C k := c 2 k + s k 2. We will keep α = α e iτ instead of µ as the parameter of the theory. We can later on express µ in terms of α 2 : µ = ˆv(0) V α 2 + ˆv(0) + ˆv(k ) 2V k ρ = α 2 + k s k 2 V. (C k 1) e i2τ k ˆv(k ) 2V S k, 47

48 D(k) = S k = Fixed point equation II fk 2 g k 2, g k D(k), C k = f k D k, f k : = k2 2 + α 2 ˆv(k) V + ˆv(k k) ˆv(k ) (C k 1) + ˆv(k ) 2V 2V ei2τ S k, k k g k : = α 2 i2τ ˆv(k) e V ˆv(k k) 2V k S k. 48

49 Limit L In the thermodynamic limit one should take α = V κ, where κ has the interpretation of the density of the condensate. Then one could expect that S k will converge to a function depending on k R d in a reasonable class and we can replace 1 1 by V (2π) dk. d In particular, ˆv(0) D(0) = 2V α2 k k ˆv(k) V S k ˆv(0)κ 2(2π) d ˆv(k)S k dk. Thus it seems to imply D(0) > 0, which would mean that we have an energy gap in this approximation. 49

50 Perturbative approach based on the original Bogoliubov method Recall that if we replace the potential v(x) with λv(x), the Hamiltonian, after applying the original Bogoliubov method can be rewritten as H λ,l = λ 1 H L 1 + H L 0 + λh L λh L 1, Unfortunately, perturbation theory is problematic in this set-up becaus of a serious infra-red problem: the unperturbed operator has no ground state. 50

51 Perturbative approach based on the improved Bogoliubov method I After soving the fixed point equation we can write H λ,l = λ 1 H λ,l 1 + H λ,l 0 + λh λ,l λh λ,l 1, where λ 1 H λ,l 1 = B λ,l is the constant term, H λ,l 1 = k Dλ,L (k)b k b k is the quadratic term, H λ,l 1 2 and H λ,l 1 are respectively the third and fourth order parts of H in operators b k and b k. 51

52 Perturbative approach based on the improved Bogoliubov method II Consider the Hamiltonian H λ,δ,l := δ 1 H λ,l 1 + H λ,l 0 + δh λ,l δh λ,l 1, where δ is an additional parameter introduced for bookkeeping reasons, that we use to produce the perturbation expansion. At the end we set δ = λ. 52

53 Perturbative approach of the old literature I 1. Replace the zeroth mode operator a 0 with a c-number α, obtaining the Hamiltonian H λ,l (α). 2. Substitute α = λ 1 κv and split the Hamiltonian as H λ,l (α) = λ 1 H κ,l 1 + H κ,l 0 + λh κ,l λh κ,l Compute perturbatively the ground state energy: E λ,κ,l = n λn E κ,l n. 4. Compute the desired quantity. 5. Minimize (up to the desired order in λ) E κ,l, obtaining κ λ,l as a function of λ,l. 6. Substitute κ λ,l into expression for desired quantity. 53

54 Perturbative approach of the old literature II This approach was used e.g. by Bogoliubov, Beliaev, Hugenholz-Pines, Gavoret-Nozieres. It is OK if we compute intensive quantities. E.g. it gives the correct energy density, as proven by Lieb, Seiringer, Yngvason. It is dubious for finer quantities, such as the infimum excitation spectrum, since we modify the Hamiltonian. 54

55 Various limits Low density limit. (Rigorous results by Lieb, Seiringer, Yngvason). Fix the potential v, fix the density ρ, go to themodynamic limit L, consider the leading behavior of the desired quantity for small ρ. Gross-Pitaevski limit. (Rigorous results by Lieb, Seiringer, Yngvason). Fix the potential v, fix n/l, go to themodynamic limit L. (Very low density). Weak coupling limit. (Adapted to the Bogoliubov method, implicit in Bogoliubov, Hugenholz-Pines, Gavoret-Nozieres, etc.) Fix µ, consider the potential λv, go to themodynamic limit L with a small λ. (Very high density). 55