The Heavy-Light Form Factor

Transcription

1 The Heavy-Light Form Factor Christian Bauer Dan Pirjol, Iain Stewart UC San Diego The Heavy-Light Form Factor Christian Bauer p.1

2 What will I do? Heavy-light form factor as F = f F (Q) + f NF (Q) where 1 f F (Q) = N 0 dz dx f NF (Q) = C k (Q, µ) ζ k (Q, µ) 0 dr + T (z, Q, µ 0 ) J(z, x, r +, Q, µ 0, µ)φ M (x, µ)φ B (r +, µ) p 2 ~ Q 2 B M p 2 ~ Λ2 p 2 ~ QΛ p 2 ~ 2 Λ The Heavy-Light Form Factor Christian Bauer p.2

3 SCET Summary Collinear, soft and usoft degrees of freedom Integrating out off-shell fluctuations Wilson lines W and S Wilson lines ensure gauge invariance Simple power counting from scaling of fields in interaction vertices Leading and Subleading Lagrangians known Decoupling of usoft gluons by field redefinition with Wilson line Y The Heavy-Light Form Factor Christian Bauer p.3

4 Issues in B Ll ν Two different types of contributions "Soft" contribution "Hard" contribution m b v+k Γ p 1 m b v+k Γ p 1 m b v+k Γ p 1 q p 2 q p 2 q p 2 p 1 p π, p 2 Λ QCD p 1 xp π, p 2 (1 x)p π Can hard contribution can be written as? F = T φ B φ π Given in terms of light cone wave functions Can not have 1/x 2 singularities in T Form-factor relations for "soft" contributions Relative size of two terms unknown Charles et al. ( 99), CB, Fleming, Pirjol, Stewart ( 00) The Heavy-Light Form Factor Christian Bauer p.4

5 Definition of the states Can pick any interpolating field Pick B v h v γ 5 q s 0, π n ξ n n/γ 5 ξ n 0 Both valence quarks collinear Asymmetric configuration (endpoint of wave function)? O π end Interaction qa c ξ power suppressed The Heavy-Light Form Factor Christian Bauer p.5

6 Definition of the states Can pick any interpolating field Pick B v h v γ 5 q s 0, π n ξ n n/γ 5 ξ n 0 Both valence quarks collinear Asymmetric configuration (endpoint of wave function)? O π n Interaction qa c ξ power suppressed The Heavy-Light Form Factor Christian Bauer p.5

7 Definition of the states Can pick any interpolating field Pick B v h v γ 5 q s 0, π n ξ n n/γ 5 ξ n 0 Both valence quarks collinear Asymmetric configuration (endpoint of wave function)? O π n Interaction qa c ξ power suppressed The Heavy-Light Form Factor Christian Bauer p.5

8 Definition of the states Can pick any interpolating field Pick B v h v γ 5 q s 0, π n ξ n n/γ 5 ξ n 0 Both valence quarks collinear Asymmetric configuration (endpoint of wave function)? O π n Interaction qa c ξ power suppressed The Heavy-Light Form Factor Christian Bauer p.5

9 Power corrections to SCET SCET is expansion in small parameter λ Subleading terms appear in Lagrangians and external currents For Lagrangians distinguish between L ξξ and L ξq For this analysis need L (0) ξξ, L(1) ξξ, L(0) cg, L (1) cg, L (1) ξq, L(2) ξq Lagrangians don t receive perturbative corrections, match at tree level The Heavy-Light Form Factor Christian Bauer p.6

10 Power corrections to Currents At tree level find J (0) i = ξw Γ i h J (1a) i = ξ n/ 2 ( id/ c ) W 1 P Υα i h J (1b) i = ξ Θ α i ( id/ c ) W 1 m b n/ 2 h Including perturbative corrections J (0) i (ω) = [ ξw ] ω Γ i h J (1a) i (ω) = [ ξ i D c α W ] ω J (1b) i (ω 1, ω 2 ) = 1 m b [ ξw ]ω 1 Θ α i 1 P Υα i h, [ ] W 1 P igbc αw Running of Wilson coefficients gives Sudakov logs By RPI C (0) i (ω) = C (1a) i (ω) ω 2 h Same Sudakov logs in C (0) (ω) and C (1a) (ω) (more later) i i The Heavy-Light Form Factor Christian Bauer p.7

11 Kinematics Revisited soft (Λ,Λ,Λ) (Λ,Λ,Λ) (Λ,1,Λ) 2 (Λ,1,Λ) collinear 2 (Λ,1,Λ) The Heavy-Light Form Factor Christian Bauer p.8

12 Kinematics Revisited soft (Λ,Λ,Λ) (Λ,Λ,Λ) (Λ,1,Λ) 2 (Λ,1,Λ) collinear 2 (Λ,1,Λ) Gluon is off-shell, needs to be integrated out Matches onto four-quark operators What do we know about the resulting matrix element? Can we write it in terms of φ π and φ B? Yes, if O 4q [ ξw ] ω1 [W ξ] ω2 [ hs] κ1 [S h] κ2 The Heavy-Light Form Factor Christian Bauer p.8

13 Kinematics Revisited (Λ,Λ,Λ) usoft (Λ,Λ,Λ) (Λ,1, Λ) (Λ,1, Λ) collinear (Λ,1, Λ) Gluon is off-shell, needs to be integrated out Matches onto four-quark operators What do we know about the resulting matrix element? Analyze the factorization on collinear and (u)soft Study in intermediate SCET I, where we have better understanding of collinear and usoft The Heavy-Light Form Factor Christian Bauer p.8

14 Two step matching procedure 1. Matching QCD SCET I Integrate out fluctuations with p 2 m 2 b 2. Factorization in SCET I Factorize usoft from collinear DOF 3. Matching SCET I SCET II Integrate out fluctuations with p 2 m b Λ QCD 4. Matrix elements in SCET II Take matrix elements with physical external states Identify non-perturbative parameters The Heavy-Light Form Factor Christian Bauer p.9

15 Matching QCD SCET I J L ξq L ξξ All particles propagating DOF match currents and Lagrangians L ξq starts at O(λ) power suppression needed Full theory reproduced by the following T-products T 1 = T [ J (1a), il (1) ξq T 3 = T [ J (0), il (2b) ξq T 5 = T [ J (0), il (1) ξξ, il(1) ξq Contributions start at λ 2 ], T2 =T [ J (1b), il (1) ] ξq, ], T4 =T [ J (0), il (2a) ], ξq ], T6 =T [ J (0), il (1) cg, il (1) ξq The Heavy-Light Form Factor Christian Bauer p.10 ].

16 Factorization in SCET I Factorization between collinear and usoft by field redefinition ξ Y ξ (0), A µ c Y A µ(0) c Y L (0) ξξ (ξ, A c, A us ) L (0) ξξ (ξ(0), A (0) c, 0) No coupling in leading order Lagrangian No non-factorizable pieces from matrix element What about factorization of the operator? The Heavy-Light Form Factor Christian Bauer p.11

17 Factorization in SCET I "Factorizable" pieces "Non-factorizable" pieces L (1) ξξ J (0) [ξ (0) W (0) ]Γ[Y h v ] ξ (0) [Y id/ us Y ] 1 n D c (0) id/ c(0) n/ 2 ξ T F 1 = T [ J (1a), il (1) ξq T F 3 = T [ J (0), il (2b) ξq T5 NF = T [ J (0), il (1) ξξ, il(1) ξq ], T F 2 =T [ J (1b), il (1) ] ξq, ], T NF 4 =T [ J (0), il (2a) ] ξq, ], T NF 6 =T [ J (0), il (1) cg, il (1) ]. ξq The Heavy-Light Form Factor Christian Bauer p.12

18 Matching SCET I SCET II Fluctuations in SCET I collinear particles in SCET I too large for physical pion Need to eliminate fluctuations with p 2 m b Λ QCD Match onto SCET II SCET I SCET II hq ξ ξ ΙΙ ΙΙ J hq O 4q ξ ξ ΙΙ ΙΙ Fields ξ II are contained in Hilbert space of SCET I ξ II (k) = ξ I (λ 4, 1, λ 2 ) The Heavy-Light Form Factor Christian Bauer p.13

19 Matrix elements in SCET II Big question: What do we know about the matrix element π O 4q B? If O 4q [ ξw ] ω1 [W ξ] ω2 [ hs] κ1 [S h] κ2 (Valid for T F i ) π O F 4q B Z 1 0 dz Z 1 0 dx Z 0 dr + T (z, Q, µ 0 )J(z, x, r +, Q, µ 0, µ)φ M (x, µ)φ B (r +, µ) Object Origin perturbative? relevant scale T (...) QCD SCET I matching α s (Q) J(...) SCET I SCET II matching α s ( p QΛ QCD ) µ 0 p QΛ QCD φ M (...) light meson LCDA No µ Λ QCD φ B (...) B-meson LCDA No µ Λ QCD f M φ M (x, µ) = i n p M II(p) ξw Γ δ ( (1 2z) n p + P + ) W ξ 0 No 1/x 2 or 1/r 2 + singularities in T and J if defined this way The Heavy-Light Form Factor Christian Bauer p.14

20 Matrix elements in SCET II Big question: What do we know about the matrix element π O 4q B? If O 4q [ ξw ] ω1 [S] [W ξ] ω2 [ hs] κ1 [S h] κ2 (Valid for T NF i ) π O NF 4q B T J Ψ M Ψ B Object Origin perturbative? relevant scale T (...) QCD SCET I matching α s (Q) J(...) SCET I SCET II matching α s ( p QΛ QCD ) µ 0 p QΛ QCD Ψ M (...) light meson matrix element No µ Λ QCD Ψ B (...) B-meson matrix element No µ Λ QCD The non-perturbative objects Ψ M (...) and Ψ B (...) are NOT both 1-d LCDA s! The Heavy-Light Form Factor Christian Bauer p.15

21 Matrix elements in SCET II Big question: What do we know about the matrix element π O 4q B? Matrix elements of "non-factorizable" T-products give rise to new non-perturbative functions Why not write π O NF 4q B C(Q, µ)ζ( QΛ QCD, µ) Object Origin perturbative? relevant scales C(...) QCD SCET I matching α s (Q) ζ(...) all remaining effects No µ Λ QCD, p QΛ QCD Are there less functions ζ(...) than there are form factors? Yes! FF relations hold for ζ(...) if J (0) in T-product Only two (three) functions required for all FF s The Heavy-Light Form Factor Christian Bauer p.16

22 One last point All T-products except T 1 and T 2 contain leading order current J 0 The resulting contributions all obey FF relations Can move all these contributions into definition of ζ(...) Leads to definition of ζ(...) C(Q, µ)ζ( QΛ QCD, µ) = i=3...6 π T i B Furthermore T 1 and T 2 give factorizable terms with only φ + B (r +) No need for φ B (r +) in convolution The Heavy-Light Form Factor Christian Bauer p.17

23 Final result Near q 2 = 0 is F (Q) = f F (Q) + f NF (Q) 1 f F (Q) = N 0 dz dx 0 dr + T (z, Q, µ 0 ) J(z, x, r +, Q, µ 0, µ)φ M (x, µ)φ + B (r +, µ) ) ( QΛQCD ) f NF (Q) = C k (Q, µ ζ k, µ p 2 ~ Q 2 B M p 2 ~ Λ2 p ~ 2 p 2 2 ~ QΛ Λ The Heavy-Light Form Factor Christian Bauer p.18

24 Relative size of two terms Factorizable and non-factorizable same order in power counting Relative size of the two terms is decided by powers of α s (logarithms) Is there Sudakov suppression of one of the terms? Logarithms of NF piece determined by running of C (0) Running known and contains Sudakov logarithms The Heavy-Light Form Factor Christian Bauer p.19

25 Relative size of two terms Factorizable and non-factorizable same order in power counting Relative size of the two terms is decided by powers of α s (logarithms) Is there Sudakov suppression of one of the terms? Logarithms of NF piece determined by running of C (0) Running known and contains Sudakov logarithms µ d dµ C(µ) = γ(µ)c(µ), γ(µ) = α s(µ) (A + B log µ) The Heavy-Light Form Factor Christian Bauer p.19

26 Relative size of two terms Factorizable and non-factorizable same order in power counting Relative size of the two terms is decided by powers of α s (logarithms) Is there Sudakov suppression of one of the terms? Logarithms of NF piece determined by running of C (0) Running known and contains Sudakov logarithms Logarithms of factorizable piece by running of C (1a,b) Since C (1a) = C (0) only unknown is C (1b) But definitely Sudakov logarithms in factorizable piece No reason to be of different size The Heavy-Light Form Factor Christian Bauer p.19

27 What have we learnt? There are four main points we can show with our analysis Cleanly separated factorizable and non-factorizable contributions Factorizable term contains no 1/x 2 singularities Non-factorizable term obeys the form-factor relations Factorizable and non-factorizable same order in power counting No reason for different size in logarithm The Heavy-Light Form Factor Christian Bauer p.20