NGB and their parameters

Transcription

1 NGB and their parameters Gradient expansion: parameters of the NGB s Masses of the NGB s The role of the chemical potential for scalar fields: BE condensation Dispersion relations for the gluons 1

2 Gradient expansion: NGB s parameters Recall from HDET L dv iv D = χ χ π A AB AB B D * 4 AB iv D AB A = ( α,i) G Γ = ψ C ψ, = V Γ AB AT B CD AB CDAB V = (3 δ δ δ δ ) δ δ ( αi)( βj)( γk)( δ) αδ βγ αγ βδ ik j

3 Condensate in CFL: 0 ψ ψ 0 =- 0 ψ ψ 0 ε ε α β α β αβi il jl ir jr iji Changing basis: 1 G = V Γ = ε ε CD αβi AB CDAB iji 1 ψ = λ ψ 9 α α i ( A) i A= 1 ( ) A B i j αβi ψψ = ( λa) α( λb) β ε εiji = Tr A I B I λ ελ ε I A AB = AδAB A1,,8 = = A 9 A = A = 9 = 3

4 Free action L D dv iv = χ χ 4 A A A π A iv Propagator S δ V AB A AB = V V A A V Coupling to the U(1) NGB: ψ e ψ, ψ e ψ i( α+β) i( α β) L L R R α i β i U e U, V e V σ U = e, V = e i σ/f i τ/f σ= f ϕ, τ= f θ σ τ τ Invariant couplings UV ψ, UV L ψ R 4

5 Consider now the case of the U(1) B NGB. The invariant Lagrangian is: L iv U dv A A A D = χ χ 4π U A iv At the lowest order in σ L σ iσ σ 0 + dv f f = χ χ 4π iσ σ + 0 fσ fσ A σ σ A A generates 3-linear and 4-linear couplings 5

6 Generating functional: i A( ) Z[ σ ] = DχDχ e χ σχ iσ σ A( σ ) = S + Γ + Γ 1 A A fσ fσ Γ 0 =, Γ 1 = V 1 S0 A = A V Z[ σ ] = (det[a( σ )]) = e 1 Tr[log A( σ )] 1/ i S eff [ σ ] = Tr[logA( σ)] 6

7 1 iσ σ itr[log A( σ )] = itr log S 1+ S Γ 0 + S Γ 1 = fσ fσ = Γ + Γ n 1 1 ( 1) iσ σ itr logs i is i 0 is i 1 n= 1 n fσ fσ At the lowest order: i is(y, x)i σ(x) is(x, y)i σ(y) Seff = dxdytr i 0 i 0 4 Γ Γ + fσ f σ i is(x,x) σ (x) + dxtr i Γ fσ 1 n 7

8 Feynman rules For each fermionic internal line iδ V AB A isab = iδ ABS(p) = V V A A V For each vertex a term il int For each internal momentum not constrained by momentum conservation: +δ + 4πµ µ d = d d ( π) 4π Factor x(-1) from Fermi statistics and spin. A factor 1/ from replica trick. A statistical factor if needed. δ 8

9 + 1 dv µ il = il (p) + il (p) = A eff I II 3 4π A π fσ V ( + p) σv σ+ V ( + p) σv σ Aσ σ d D A( + p)d A( ) D A( ) D ( ) = V V + iε A A Goldstone theorem: L (0) + L (0) = 0 I II Expanding in p/ : 9

10 L (x) = (V ) σ(x)(v ) σ(x) 9µ 1 dv eff π fσ 4π dv 3 VV µ ν = 1 4π µ ( ) ( ) eff = σ 0 σ σ π f σ L (x) v 1 9µ v σ =, fσ = 3 π CFL 10

11 For the V NGB same result in CFL, whereas in SC 1 4µ v τ =, fτ = SC 3 π With an analogous calculation: 8 µ (1 8log) 1 a a L = ( Π ) v Π π eff 0 36 F a = 1 µ v, F F 3 36π 1 (1 8log) = = T = 11

12 Dispersion relation for the NGB s E =± 1 3 p Different way of computing: a b ab J µ Π = ifδ p µ, pµ = p, p 3 Current conservation: 1 3 p p= E p = 0 1

13 Masses of the NGB s QCD mass term: ψ Mψ + L R h.c. ψ e g ψ g, ψ e g ψ g i( α+β) T i( α β) T L c L L R c R R g SU(3), g SU(3), e U(1), e U(1) iα iβ c c L,R L,R B A i β L R M e g Mg X g Xg e, Y g Yg e T i( α+β ) T i( α β ) c L c R T 4i T Σ= (Y X) = e θ Σ Σ e g Σg 4i β L R 13

14 X 6i( φ+θ ) 6i( φ θ) Y d = det(x) = e, d = det(y) = e 6i( α+β) 6i( α β) X X Y Y d e d, d e d 6iβ 1iβ det(m) e det(m), det( Σ) e det( Σ) masses dxd = det( Σ ) Y L = c det[m]tr[m Σ+ h.c. c' det( Σ)Tr[(M Σ ) ] ( 1 ) ( ) c" Tr[M Σ ]Tr[M Σ] ( ) 14

15 Calculation of the coefficients from QCD Mass insertion in QCD Effective 4-fermi Contribution to the vacuum energy 3 c =,c' = 0,c" = 0 π 15

16 Why only c? From second order terms in M: ( ψ M ψ ) = ( ψ M ψ )( ψ M ψ ) i j α k l β L R Lα i Rj Lβ k Rl ε ε X M M ε ε Y ε ε M M Σ = ikm γ j l αβδ p* ikm j l p αβγ m i k jlp δ jlp i k m =ε ε MMM (M ) Σ det(m)tr[m Σ] ikm j l a 1 b p 1 jlp i k m a b There is a possible other contribution to NGB masses. 16

17 Consider: a µνa L QCD =ψ (id +µγ0) ψ ψlmψr ψrmψl GµνG Solving for ψ as in HDET,L 1 ψ = γ ψ +γ ψ µ ( i D M ),L 0 +,L 0 +,R ( L R,M M ) 1 4 like chemical potential 1 L D = ψ+,l(iv D) ψ+,l ψ +,L (D ) + MM ψ +,L + µ

18 Consider fermions at finite density: L ( ν i g ν0 ) = ψγ i ψ ν µ as a gauge field A 0 Invariant under: ψ ψ µ µ+α i α(t) e, (t) Define: Invariance under: 1 1 XL = MM, XR = M M µ µ ψ L(t) ψ, ψ R(t) ψ, +,L +,L +,R +,R X L(t)X L (t) + il(t) L (t), L L 0 X R(t)X R (t) + ir(t) R (t) R R 0 18

19 The same symmetry should hold at the level of the effective theory for the CFL phase (NGB s), implying that T MM M M 0Σ Σ= 0 0Σ+ iσ i Σ µ µ T The generic term in the derivative expansion of the NGB effective lagrangian has the form L n m p 0 + imm / µ M q r NGB F Σ Σ F 19

20 L n m p 0 + imm / µ M q r NGB F Σ Σ F Compare the two contribution to quark masses: kinetic term mass insertion F F m 1 m µ F µ 4 4 m 1 m F F F Same order of magnitude for m since F µ 0

21 The role of the chemical potential for scalar fields: Bose-Einstein condensation A conserved current may be coupled to the a gauge field. Chemical potential is coupled to a conserved charge. The chemical potential must enter as the fourth component of a gauge field. 1

22 Complex scalar field: L= + iµ ϕ iµ ϕ ϕ ϕ m ϕϕ λ ϕϕ = ( ) ( ) ( )( ) ( ) 0 0 µ µ ( ) ( m ϕ ) + i ( ) 0 0ϕϕ = ϕ ϕ µ ϕϕ λ ϕ µ ϕ ϕ negative mass term breaks C Mass spectrum: p (m µ ) + µ Qp0 = 0 (Q =± 1) (E +µ Q) = m + p For µ < m mp,p = µ+ m

23 At µ = m, m second order phase transition. Formation of a condensate obtained from: V ( ) ( m ) = µ ϕ ϕ+λ ϕ ϕ Charge density µ m ϕϕ = µ> λ m ( ) V µ ρ= = µ ϕϕ = µ m µ λ ( ) Ground state = Bose-Einstein condensate 3

24 v ϕ =, v = µ m λ 1 ϕ (x) = v + h(x) e ( ) i θ (x)/v 1 µ 1 µ L = µ θ θ+ µ h h λv h µ h 0θ p λ v iµ E Mass spectrum det = 0 iµ E p At zero momentum ( ) M M λv 4µ = 0 M = 0 M = 6µ m 4

25 At small momentum E µ m 3µ m NGB p 9µ m E 6µ m + p 6 µ m massive v µ m 1 = µ 3µ m 3 NGB 5

26 Back to CFL. From the structure mp,p = µ+ m md mu c m ± = + (m u + m d)m s, π µ F ms mu c m ± = + (m K u + m s)m d, µ F ms md c m 0 0 = + (m + m )m K,K µ F d s u F = c 3 = π µ (1 8log ) 36π MM First term from chemical potential like kinetic term, the second from mass insertions 6

27 For large values of m s: c m ± = (m u + m d)m s, π F ms c ms c m ± = + m m, m = + m m K K,K µ F µ F s d 0 0 s u and the masses of K + and K 0 are pushed down. For the critical value 1/3 1µ 3 3 s = crit u,d u,d m m 3.03 m, π F m MeV s crit masses vanish 7

28 For larger values of m s these modes become unstable. Signal of condensation. Look for a kaon condensate of the type: iαλ 4 4 Σ = e 4 = 1 + (cosα 1) λ + iλ sin α (In the CFL vacuum, Σ = 1) and substitute inside the effective lagrangian 1 m s cmsm V( α ) = F sin α+ (1 cos α) µ F negative contribution from the chemical potential positive contribution from mass insertion 8

29 Defining m ( 0 ) s cmsm µ eff =, mk = µ F 1 0 V( α ) = F µ eff sin α+ (m K) (1 cos α) with solution ( 0 m ) K 0 eff K eff cos α=, µ > m µ and hypercharge density V n = =µ F 1 ( 0 m ) 4 K Y eff 4 µ eff µ eff 9

30 Mass terms break original SU(3) c+l+r to SU() I xu(1) Y. Kaon condensation breaks this to U(1) 1 1 Q = 3 8, [Q, ] 0 λ λ Σ = 3 Σ π π η ± (, ) (K,K ) (K,K ) ( ) SU() U(1) I Y breaking through the doublet as in the SM Only NGB s from K 0, K + instead of expected 3 (see Chada & Nielsen 1976) 30

31 Chada and Nielsen theorem: The number of NGB s depends on their dispersion relation If E is linear in k, one NGB for any broken symmetry If E is quadratic in k, one NGB for any two broken generators In relativistic case always of type I, in the non-relativistic case both possibilities arise, for instance in the ferromagnet there is a NGB of type II, whereas for the antiferromagnet there are to NGB s of type I 31

32 Dispersion relations for the gluons The bare Meissner mass The heavy field contribution comes from the term ( D ) DD ψ ψ = P ψ ψ µ+ iv D µ+ iv D µν µ ν h+ h+ h+ h+ 1 P = g V V + V V µν µν µ ν ν µ 3

33 Notice that the first quantized hamiltonian is: g H = p ga + ea 0 p + ga0 gv A + A (v A) p Since the zero momentum propagator is the density one gets ( ) 3 dp 1 (p A) g Nf Tr A 3 ( π) p p spin p <µ N A A m A A, f g µ 1 a a 1 a a = BM 6π a a g µ m N 6 π BM = f 33

34 Gluons self-energy Vertices from iga a J µ µ a Consider first SC for the unbroken gluons: 00 g µ Π ab(p) =δab p, 18π Π (p) = Π (p) +Π (p) = kl kl,self kl ab ab ab kl g µ p 0 kl g µ kl g µ =δabδ 1 + δabδ =δ abδ p 0, 3π 6 3π 18π g µ 18π 0k ab(p) ab Π =δ pp 0 k from m BM 34

35 Bare Meissner mass cancels out the constant contribution from the s.e. All the components of the vacuum polarization have the same wave function renormalization ( ) k L= F F + Π A A = E E B B + E E 4 g µ k = 18π µν a µν a b a a a a a a a µν ab µ ν i i i i i i Dielectric constant ε = k+1, and magnetic permeability λ =1 1 v = ελ g µ 35

36 Broken gluons a Π 00 (0) 0 3m g / - Π ij (0) 0 m g / 8 3m g m g /3 m g µ = 3 π g 36

37 But physical masses depend on the wave function renormalization g µ Rest mass defined as the energy at zero momentum: m =, a = 4,5,6,7 R gµ m R =, a = 8 π The expansion in p/ cannot be trusted, but numerically mr 0.9, a = 4,5,6,7 37

38 In the CFL case one finds: m g µ m D = (1 8log) = g F 36π g µ 11 1 m π D M = log+ = from bare Meissner mass Recall that from the effective lagrangian we got: m =α gf, m =α vgf D T T M S T implying α =α = 1 and fixing all the parameters Ṣ T 38

39 We find: md g 16 R α 1 = + 3α 16π 3 1 µ m, 7 log m 1.70 R Numerically m 1.36 R 39

40 LOFF phase Different quark masses LOFF phase Phonons 40

41 Different quark masses We have seen that for one massless flavors and a massive one (m s ), the condensate may be disrupted for m µ< s The radii of the Fermi spheres are: ms pf = µ m 1 s µ, pf =µ µ As if the two quarks had different chemical potential (m s /µ) 41

42 Simulate the problem with two massless quarks with different chemical potentials: µ = µ + δµ, µ = µ δµ u µ +µ µ µ µ=, δµ= u d u d Can be described by an interaction hamiltonian d H = δµψ σ ψ I 3 Lot of attention in normal SC. 4

43 LOFF: ferromagnetic alloy with paramagnetic impurities. The impurities produce a constant exchange field acting upon the electron spins giving rise to an effective difference in the chemical potentials of the opposite spins. Very difficult experimentally but claims of observations in heavy fermion superconductors (Gloos & al 1993) and in quasi-two dimensional layered organic superconductors (Nam & al. 1999, Manalo & Klein 000) 43

44 H I changes the inverse propagator S V 1 + δµ σ3 0 * = V +δµσ 3 and the gap equation (for spin up and down fermions): = dv µ d ig 4 π π ( π ) ( δµ ) This has two solutions: 0, ) a) : = b : = δµ 44

45 Grand potential: d 0 = 0 dg ρ g Ω H dg = Ω= g g g 0 ρ d 0 Ω Ω 0 = ( = 0) 0 Also: ρ Ω0( δµ ) Ω 0(0) = δµ Favored solution 0 = 0 45

46 Also: ρ Ω Ω ( δµ ) = δµ + 4 ( ) 0 0 First order transition to the normal state at δµ = δµ = 1 0 For constant, Ginzburg-Landau expanding up to 6 46

47 LOFF phase In 1964 Larkin, Ovchinnikov and Fulde, Ferrel, argued the possibility that close to the first order-line a new phase could take place. 1 According LOFF possible condensation with non zero total momentum of the pair p = k+ q iq x p = k+ q ψ(x) ψ(x) = e More generally ψ(x) ψ(x) = m c m e iq x 47 m

48 Non zero total momentum p + p 1 = q E(p) µ E( ± k + q) µ = δµ q Gap equation: q q / q δµ ξ fixed variationally chosen spontaneously µ v F n n 3 u d g d p 1 n u n d 1 = 3 (π) µ ε(p, ) 48

49 n u,d e 1 = ( ε(p, ) ± µ )/ T + 1 For T T 0 3 = g d p 1 (π) ε(p, ) 1 3 blocking region (1 θ( ε µ ) θ( ε + µ )) ε < µ The blocking region reduces the gap: << LOFF BCS 49

50 Possibility of a crystalline structure (Larkin & Ovchinnikov 1964, Bowers & Rajagopal 00) ψ(x) ψ(x) = q i q i = 1.δµ e iq i x see later The q i s define the crystal pointing at its vertices. The LOFF phase has been studied via a Ginzburg-Landau expansion of the grand potential 50

51 Ω = α + β 4 + γ (for regular crystalline structures all the q are equal) The coefficients can be determined microscopically for the different structures (Bowers and Rajagopal (00) ) 51

52 Gap equation General strategy Propagator expansion Insert in the gap equation 5

53 We get the equation α + β 3 + γ 5 + = 0 Ω Which is the same as = 0 with α = β 3 = The first coefficient has universal structure, independent on the crystal. From its analysis one draws the following results γ 5 = 53

54 ρ ΩBCS Ωnormal = ( BCS δµ 4 ) Ω LOFF Ωnormal = 0. 44ρ( δµ δµ ) LOFF 1.15 ( δµ δµ ) δµ δµ 1 = BCS / BCS Small window. Opens up in QCD? (Leibovich, Rajagopal & Shuster 001; Giannakis, Liu & Ren 00) 54

55 Results of Leibovich, Rajagopal & Shuster (001) µ(mev) δµ / / BCS (δµ δµ 1 )/ BCS LOFF

56 Single plane wave Critical line from Ω = Ω 0, = q 0 Along the critical line ( at T = 0, q = 1.δµ ) 56

57 Bowers and Rajagopal (00) Preferred structure: face-centered cube 57

58 Tricritical point General study by Combescot and Mora (00). Favored structure antipodal vectors At T = 0 the antipodal vector leads to a second order phase transition. Another tricritical point. (Matsuo et al. 1998, Casalbuoni and Tonini 003) Change of crystalline structure from tricritical to zero temperature? 58

59 Two-dimensional case (Shimahara 98, Mora & Combescot 03) Analysis close to the critical line Ω a Ω N = 1 ρb a a a a a a Tc T T c iq r (r) = FFe (r) = FFLO cos(q r) (r) = sq[cos(qx) + cos(qy)] iq1 r iq r iq3 r (r) = tr[e + e + e ] (r) = hex[cos(q1 r) + cos(q + cos(q r)] 3 r) q 1 + q + q 3 = 0 59

60 Phonons In the LOFF phase translations and rotations are broken phonons Phonon field through the phase of the condensate (R.C., Gatto, Mannarelli & Nardulli 00): ψ(x) ψ(x) introducing iq x iφ(x) = e e Φ (x) 1 (x) (x) q x φ = Φ f = q 60 x

61 L phonon 1 φ φ φ = φ + v v x y z Coupling phonons to fermions (quasi-particles) trough the gap term (x) ψ T Cψ e iφ(x) ψ T Cψ It is possible to evaluate the parameters of L phonon (R.C., Gatto, Mannarelli & Nardulli 00) + v = 1 1 δµ q v δµ = q

62 Cubic structure (x) = 8 k= 1 e iq k x = i= 1,,3; ε e i i q ε x =± i i i= 1,,3; ε e i iε =± i Φ (i ) (x) Φ (i) (x) = q x i 1 (i) (i) ϕ (x) = Φ (x) q f x i 6

63 ( x) = 0 ( x) = + 4 ( x) = 4 63

64 Φ (i) (i) (x) transforms under the group O h of the cube. Its e.v. ~ x i breaks O(3)xO h ~ O diag h (i) 1 a φ L phonon = φ i= 1,,3 t i= 1,,3 b ( (i)) ( (i) (j) φ ) i c φ i jφ i= 1,,3 i< j= 1,,3 (i) Coupling phonons to fermions (quasi-particles) trough the gap term (x) ψ T Cψ i= 1,,3; ε e i iε =± i Φ (i ) (x) ψ T Cψ 64

65 we get for the coefficients 1 a = b = 0 1 c = 1 δµ q One can evaluate the effective lagrangian for the gluons in tha anisotropic medium. For the cube one finds Isotropic propagation This because the second order invariant for the cube and for the rotation group are the same! 65

66 Compact stellar objects 66

67 Compact stellar objects High density core of a compact star, a good lab for testing QCD at high density. 67

68 Some features of a compact star For simplicity consider a gas of free massless fermions. Grand potential: 3 4 dp µ Ω= NfV ( ε 3 p µ ) θ( µ ε p) = VN ( π ) 1 π Density: Eq. of state: Ω µ ρ= = VN µ 3 π 3 f f 4 µ P= Ω= VNf P= Kρ 1 π 68 4/3

69 For a non-relativistic fermion: 8 m 4 m P= VN f µ, ρ= VN µ 15 π 3 π 3/ 3/ 5/ 3/ f P= Kρ 5/3 More generally assumed P = Kρ γ For high densities inverse beta decay becomes important ep nν At the equilibrium µ +µ =µ e p n 69

70 µ +µ =µ ρ 1/3 +ρ 1/3 =ρ 1/3 e p n e p n From charge neutrality ρ =ρ e p ρ p = ρ n 1 8 Neutron star Radius of a neutron star (Landau 193) 70

71 N fermions in a box of volume V. Number density n N R Position uncertainty Uncertainty principle p F 3 N R n 1/3 R N 1/3 N c R 1/3 1/3 1/3 n EF Gravitational energy per baryon E G GNm R B 71

72 cn E = EG + EF R GNm R 1/3 B E > 0 otherwise not bounded. This condition gives cn R GNm c > 0 N Nmx a = 10 R GmB 1/ 3 B 3/ 57 Maximum mass Chandrasekhar limit Mmax = NmaxnB 1.5M M 1.4M max 7

73 c 1/3 c c EF mc Nmax R R GmB 1/ R mc 1/ 8 c 5 10 cm(m m e) = cm(m= m B n) = Gm Typical neutron star density ρ g /cm 73

74 Neutron stars are a good laboratory to test hadronic matter at high density and zero temperature 74

75 In neutron stars CS can be studied at T = 0 T ns BCS 10 6 (T ns 10 ns ~10 5 K) 7 (1MeV BCS K) Consider the LOFF state. From (MeV) 14 δµ (MeV) δµ 0.75 BCS Orders of magnitude from a crude model: 3 free quarks M u = M d = 0, M s 0 75

76 Weak equilibrium: µ µ µ u d s = µ µ 3 1 = µ + µ 3 1 = µ + µ 3 e e e,,, p p p u F d F s F = µ = µ = u d µ s M s i= u,d,s µ i N i + µ e N e = µ N Electrical neutrality: q µ e Q Q = Ω µ e N = q 0 = i= u,d,s N i ρ 1 Ω i B = = (p F) 3 µ 3π i= u,d,s 1 76

77 ρ n.m. is the saturation nuclear density ~.15x g/cm At the core of the neutron star ρ B ~ g/cm Choosing µ ~ 400 MeV M s = 00 M s = 300 δp F = 5 δp F = 50 ρb 5 ρ n.m. 6 Right ballpark (14-70 MeV) 77

78 Glitches: discontinuity in the period of the pulsars Standard explanation: metallic crust + neutron superfluide inside LOFF region inside the star providing the crystalline structure + superfluid CFL phase dipole emission ( Ω/Ω 10 6 ) 78

79 Conclusions SC almost 100 years old, but still actual Important technological applications Source of inspiration for other physical theories (SM as an example) Deep implications in QCD at very high density: very rich phase structure Possible applications for compact stellar objects Unvaluable theoretical laboratory 79