A Massive Momentum-Subtraction Scheme Ava Khamseh RBC and UKQCD Collaborations Peter Boyle and Luigi Del Debbio July 26, 2016 1 / 21
RBC & UKQCD Members BNL and RBRC Luchang Jin Plymouth University Mattia Bruno Bob Mawhinney Nicolas Garron Tomomi Ishikawa Greg McGlynn Taku Izubuchi David Murphy University of Southampton Chulwoo Jung Daiqian Zhang Jonathan Flynn Christoph Lehner Vera G¨ ulpers University of Connecticut Meifeng Lin James Harrison Tom Blum Christopher Kelly Andreas J¨ uttner Edinburgh University Hiroshi Ohki Andrew Lawson Peter Boyle Shigemi Ohta (KEK) Edwin Lizarazo Guido Cossu Amarjit Soni Chris Sachrajda Luigi Del Debbio Sergey Syritsyn Francesco Sanfilippo Richard Kenway Matthew Spraggs CERN Ava Khamseh Tobias Tsang Marina Marinkovic Julia Kettle Columbia University Antonin Portelli York University (Toronto) Ziyuan Bai Brian Pendleton Renwick Hudspith Norman Christ Oliver Witzel Xu Feng Azusa Yamaguchi 2 / 21
Motivation for a Mass Dependent Scheme 48 3 × 96 Physical 64 3 × 128 Physical 48 3 × 96 Fine Lattice 1 / a (GeV) 1.73 2.36 2.8 M D l (GeV) 1.42 – 1.68 1.49 – 2.12 1.51– 2.42 l = 1.86480(14) GeV 1 PDG M D ± = 1.86957(16) GeV, M D 0 l massless quarks: am ≪ a µ ≪ π Reduction in lattice artefacts when performing continuum extrapolation in a massive scheme, by potentially removing mass dependent O ( a 2 ) terms 1see talk by Tobi Tsang , Friday, July 29, 13:00, arXiv: hep-lat/1602.04118v1 and hep-lat/1511.09328 3 / 21
The Charm Project Determine the decay constants f D and f D s using cq | D q ( p ) � = f D q p µ � 0 | A µ D q where q = d , s and the axial current A µ cq = ¯ c γ µ γ 5 q . To obtain the decay constant, we need to renormalize the bare axial current. 4 / 21
Kinematics 3 = q 2 = − µ 2 with µ 2 > 0 Symmetric Minkowski momentum p 2 2 = p 2 Γ ( q ) ¯ Γ = ¯ Vertex G a Γ ( p 3 , p 2 ) = � O a ψ ( p 3 ) ψ ( p 2 ) � , fermion bilinear O a ψ Γ τ a ψ Γ spans all the element of the basis of the Clifford algebra, Γ = S , P , V , A , T i Propagator S ( p ) = p − m − Σ( p )+ i ǫ / Amputated vertex function Λ a Γ ( p 2 , p 3 ) = S ( p 3 ) − 1 G a Γ ( p 3 , p 2 ) S ( p 2 ) − 1 At 1-loop − − − − − → 5 / 21
Continuum Ward identities Consider chiral transformations with a regulator that does not break the symmetry, e.g. dim-reg Vector and axial transformations on ¯ ψ i , ψ j in the path integral imply: V = iS ( p 2 ) − 1 − iS ( p 3 ) − 1 Vector WI: q · Λ a P − γ 5 iS ( p 2 ) − 1 − iS ( p 3 ) − 1 γ 5 Axial WI: q · Λ a A = 2 mi Λ a Flavor non-singlet τ a = σ + / 2 6 / 21
Renormalization ψ R = Z 1 / 2 ψ , S R ( p ) = Z q S ( p ) , m R = Z m m q � ¯ R = Z A A µ , � R = Z Γ ¯ A µ V µ R = Z V V µ ψ Γ ψ ψ Γ ψ , Renormalization of Λ Γ : Λ Γ , R = Z Γ Z q Λ Γ In general, Z = Z ( g , a µ, am ) Regulator a Renormalization scale µ Renormalization constants are determined by imposing renormalization conditions. e.g. RI/SMOM 2 . 2arXiv: hep-ph/0901.2599 7 / 21
RI/SMOM Conditions �� 1 � iS R ( p ) − 1 � p � lim 12 p 2 Tr p 2 = − µ 2 = 1 � � m R → 0 � � 1 p 2 = − µ 2 − 1 � − iS R ( p ) − 1 �� lim Tr 2 Tr [( q · Λ A , R ) γ 5 ] | sym = 1 � 12 m R m R → 0 1 lim 12 q 2 Tr [( q · Λ V , R ) � q ] | sym = 1 m R → 0 1 lim 12 q 2 Tr [( q · Λ A , R ) γ 5 � q ] | sym = 1 m R → 0 1 lim 12 i Tr [Λ P , R γ 5 ] | sym = 1 m R → 0 1 lim 12Tr [Λ S , R ] | sym = 1 m R → 0 8 / 21
Tree level values RC are consistent with trivial renormalizations at tree level e.g. �� 1 iS R ( p ) − 1 � p � � lim 12 p 2 Tr p 2 = − µ 2 = 1 � � m R → 0 �� 1 � � iS ( p ) − 1 � p 12 p 2 Z − 1 lim q Tr p 2 = − µ 2 = 1 � m R → 0 � � 1 � 12 p 2 Z − 1 lim q Tr [( � p − m ) � p ] p 2 = − µ 2 = 1 � � m R → 0 at tree level Z q = 1, same for all the others. This is a property we wish to preserve in the massive scheme 9 / 21
Z V = 1 in SMOM q · Λ V = iS ( p 2 ) − 1 − iS ( p 3 ) − 1 Bare Vector WI: Rewriting in terms of renormalized quantities using, Λ V , R = Z V S R ( p ) = Z q S ( p ) and Z q Λ V ⇒ Z V q · Λ V , R = i Z q S R ( p 2 ) − 1 − i Z q S R ( p 3 ) − 1 Z q multiplying by � q and taking trace, using �� � iS R ( p ) − 1 � p 1 � lim m R → 0 12 p 2 Tr p 2 = − µ 2 = 1 � 1 lim m R → 0 12 q 2 Tr [( q · Λ V , R ) � q ] | sym = 1 gives Z q Z V = Z q ⇒ Z V = 1 10 / 21
Heavy-Heavy RI/mSMOM Conditions �� 1 � iS R ( p ) − 1 � p � lim 12 p 2 Tr p 2 = − µ 2 = 1 � m R → ¯ � m � � 1 p 2 = − µ 2 − 1 � − iS R ( p ) − 1 �� lim Tr 2 Tr [( q · Λ A , R ) γ 5 ] | sym = 1 � 12 m R m R → ¯ m 1 lim 12 q 2 Tr [( q · Λ V , R ) � q ] | sym = 1 m R → ¯ m 1 lim 12 q 2 Tr [( q · Λ A , R − 2 m R i Λ P , R ) γ 5 � q ] | sym = 1 m R → ¯ m 1 lim 12 i Tr [Λ P , R γ 5 ] | sym = 1 m R → ¯ m �� 12Tr [Λ S , R ] − 1 1 � � lim 6 q 2 Tr 2 im R Λ P , R γ 5 / q = 1 � � m R → ¯ m sym 11 / 21
Check at 1-loop in perturbation theory using dim reg Dimensional Regularization, D = 4 − 2 ǫ � k + m ] γ µ γ µ [ / p 2 − / k + m ]Γ[ / p 3 − / Λ (1) = − ig 2 C 2 ( F ) k 2 [( p 2 − k ) 2 − m 2 ][( p 3 − k ) 2 − m 2 ] Γ k 12 / 21
Results at 1-loop � � � � �� m 2 + µ 2 ǫ − γ E + 1 − m 2 µ 2 − m 4 m 2 Z q = 1 + α 1 4 π C 2 ( F ) µ 4 ln − ln m 2 + µ 2 µ 2 ˜ � + B V γ σ + C V ( p 2 , p 3 ) = α � 1 1 (1) σ i ǫ σραβ γ ρ γ 5 p 3 α p 2 β µ 2 ( p σ 2 � p 2 + p σ Λ 4 π C 2 ( F ) � 3 � p 3 ) A V V µ 2 1 1 � µ 2 ( p σ 2 � p 3 + p σ µ ( p σ 2 + p σ + D V 3 � p 2 ) + E V 3 ) � 1 + 4 m 2 � � � � � � � µ 2 − 1 − m 2 1 + m 2 m 2 1 + 4 m 2 4 1 A V = C 0 + log − µ 2 log � m 2 + µ 2 µ 2 µ 2 3 2 1 + 4 m 2 µ 2 + 1 � � � � � � � � � � � 1 − 4 m 2 µ 2 − 2 m 4 3 − m 2 m 2 m 2 1 − 4 m 2 m 2 1 1 B V = − γ E + − C 0 + 2 µ 2 log + log µ 4 µ 2 m 2 + µ 2 µ 2 µ 2 3 ˜ ǫ � 1 + 4 m 2 � � � � � � � µ 2 − 1 1 − m 2 m 2 + µ 2 1 − 2 m 2 1 + 4 m 2 − 4 log − µ 2 log � µ 2 µ 2 µ 2 ˜ 1 + 4 m 2 µ 2 + 1 13 / 21
Results at 1-loop � 1 + 4 m 2 � � � � � � � µ 2 − 1 1 − m 2 m 2 m 2 1 − 2 m 2 1 + 4 m 2 2 C V = − µ 2 log + µ 2 log � µ 2 m 2 + µ 2 µ 2 3 1 + 4 m 2 µ 2 + 1 � � � � � � � � � � � �� m 2 + µ 2 2 − m 2 m 2 1 + m 2 1 − 4 m 2 m 2 1 − 4 m 2 + − 2 C 0 − log + log µ 2 µ 2 µ 2 µ 2 µ 2 µ 2 µ 2 ˜ ˜ � � � � � � �� 1 − 2 m 2 1 + m 2 m 2 m 2 2 D V = (1 + C 0 ) − 2 µ 2 log m 2 + µ 2 µ 2 µ 2 3 satisfies bare WI, and ... Z V = 1! Similarly for Z A and all other identities. In particular no µ dependence for the renormalization constant of Noether currents. 14 / 21
Heavy-Light RI/mSMOM Conditions 1 � � q 1 �� i ζ − 1 S H , R ( p 2 ) − 1 − i ζ S l , R ( p 3 ) − 1 � � �� �� lim 12 q 2 Tr q · Λ V , R − ( M R − m R )Λ S , R � sym = lim 12 q 2 Tr � q m R → 0 m R → 0 M R → ¯ M R → ¯ m m 1 � γ 5 � q 1 �� − i γ 5 ζ − 1 S H , R ( p 2 ) − 1 − i ζ S l , R ( p 3 ) − 1 γ 5 � � �� �� lim 12 q 2 Tr q · Λ A , R − ( M R + m R ) i Λ P , R � sym = lim 12 q 2 Tr γ 5 � q m R → 0 m R → 0 M R → ¯ M R → ¯ m m � � � 1 1 1 � − i ζ − 1 S H , R ( p ) − 1 �� � γ 5 � Λ P , R γ 5 �� �� �� � lim Tr sym = lim Tr p 2= − µ 2 − Tr q · Λ A , R + � � � sym m R → 0 12 i m R → 0 12( M R + m R ) 2 M R → ¯ M R → ¯ m m � � � 1 1 � − i ζ S l , R ( p ) − 1 �� � γ 5 �� �� � Tr p 2= − µ 2 − Tr q · Λ A , R . � � sym 12( M R + m R ) 2 √ Z l where M and m refer to heavy and light quark masses respectively and ζ = √ Z H . 15 / 21
Lattice Regularization Lattice WI for chiral symmetry µ ( x ) ψ ( y ) ¯ ψ ( z ) � = 2 m � P a ( x ) ψ ( y ) ¯ ∇ ∗ µ � A a ψ ( z ) � + contact terms + � X a ( x ) ψ ( y ) ¯ ψ ( z ) � X a explicit chiral symmetry breaking by lattice regulator Reproduces usual continuum result when regulator is removed ⇒ X a ( x ) = aO a 5 ( x ) Renormalize operators, O a 5 ( x ) mixed with lower-dim operators Testa 3 : power divergencies do not contribute to the anomalous dimensions ⇒ A R ,µ = Z A ( g , am ) A µ 5 ( x ) + ¯ m a P a ( x ) + Z A − 1 O a O a ∇ ∗ µ A a 5 R ( x ) = Z 5 µ ( x ) a � �� � ˜ Z a ˜ O ( x ) 3arXiv: hep-th/9803147 16 / 21
Summary Generalised SMOM to non-vanishing fermion mass Derived non-perturbatively, checked at 1-loop in perturbation theory Both for heavy-heavy and heavy-light vertex functions such Z cons V , A = 1 Obtain Z local V , A by taking ratios of vertex function with appropriate projectors Numerical implementation and tests will be performed on renormalizing matrix elements used to obtain decay constants and form factors in semi-leptonics 17 / 21
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