Scattering Amplitudes LECTURE 2 Jaroslav Trnka Center for Quantum - PowerPoint PPT Presentation

Scattering Amplitudes LECTURE 2 Jaroslav Trnka Center for Quantum Mathematics and Physics (QMAP), UC Davis ICTP Summer School, June 2017

Review of Lecture 1

Spinor helicity variables ✤ Standard SO(3,1) notation for momentum p µ = ( p 0 , p 1 , p 2 , p 3 ) p j ∈ R p 2 = p 2 0 + p 2 1 + p 2 2 − p 2 ✤ Matrix representation 3 ✓ p 0 + ip 1 ◆ p 2 + p 3 p ab = σ µ ab p µ = p 2 − p 3 p 0 − ip 1 p 2 = det( p ab ) = 0 On-shell: Rank ( p ab ) = 1

Spinor helicity variables ✤ Rewrite the four component momentum a λ 1 a e p µ 1 = σ µ λ 1˙ a ˙ a ✤ Little group scaling λ → t λ p → p λ → 1 e e λ t ✤ Invariants b e a e h 12 i ⌘ ✏ ab � 1 a � 2 b [12] ≡ ✏ ˙ � 1˙ � 2˙ a ˙ b s 12 = h 12 i [12]

Three point amplitudes ✤ Three point kinematics p 2 1 = p 2 2 = p 2 p 1 + p 2 + p 3 = 0 3 = 0 ✤ Two solutions: h 12 i = h 23 i = h 13 i = 0 [12] = [23] = [13] = 0 λ 1 ∼ e e λ 2 ∼ e λ 1 ∼ λ 2 ∼ λ 3 λ 3 ( − − +) (+ + − ) No solution for real momenta ✓ [12] 3 ✓ h 12 i 3 ◆ S ◆ S E.g. spin-S amplitudes [23][31] h 23 ih 31 i

T ree-level amplitudes ✤ Single function: locality and unitarity constraints 1 M − P 2 =0 M L P 2 M R − − → ✤ On-shell constructibility: amplitude fixed by poles ✤ Consistency of four point amplitude: only spins ≤ 2

Recursion relations

T ree level amplitudes ✤ Tree-level amplitude is a rational function of kinematics momenta N X A = (Feyn . diag) = polarization vectors j P 2 Q j Feynman propagators X P j = p k ✤ Only poles, no branch cuts k ✤ Gauge invariant object: use spinor helicity variables

Reconstruction of the amplitude ✤ Amplitude on-shell constructible: fixed only from factorizations: try to reconstruct it 1 “Integrate the relation” M − P 2 =0 M L P 2 M R − − → 1 ✤ First guess: X M = M L P 2 M R P

Reconstruction of the amplitude ✤ Amplitude on-shell constructible: fixed only from factorizations: try to reconstruct it 1 “Integrate the relation” M − P 2 =0 M L P 2 M R − − → 1 ✤ First guess: X WRONG M = M L P 2 M R P Overlapping factorization channels ✤ Solution: shift external momenta

Momentum shift ✤ Let us shift two external momenta e λ 1 → e λ 1 → λ 1 − zλ 2 λ 1 λ 2 → e e λ 2 + z e λ 2 → λ 2 λ 1 ✤ Momentum is conserved, stays on-shell ( λ 1 − zλ 2 ) e λ 1 + λ 2 ( e λ 2 + z e λ 1 ) = λ 1 e λ 1 + λ 2 e λ 2 ✤ This corresponds to shifting p 1 , p 2 , ✏ 1 , ✏ 2

Shifted amplitude ✤ On-shell tree-level amplitude with shifted kinematics A n ( z ) = A (ˆ p 1 ( z ) , ˆ p 2 ( z ) , p 3 , . . . , p n ) N ( z ) ✤ Analytic structure A n ( z ) = Q j P j ( z ) 2 P j ( z ) = P j − z λ 2 e p 1 ∈ P j ✤ Location of poles: λ 1 if P j ( z ) = P j + zλ 2 e p 2 ∈ P j λ 1 if P j ( z ) = P j otherwise

Shifted amplitude ✤ On the pole if p 1 ∈ P j P j ( z ) 2 = P 2 j � 2 z h 1 | P j | 2] = 0 P 2 j z = 2 h 1 | P j | 2] ⌘ z j ✤ Shifted amplitude: location of poles N ( z ) A n ( z ) = Q j P j ( z ) 2

Residue theorem N ( z ) ✤ Shifted amplitude A n ( z ) = Q k ( z − z k ) ✤ Let us consider the contour integral Z dz No pole at z → ∞ z A n ( z ) = 0 Residue at ✤ Original amplitude A n = A n ( z = 0) z = 0 ◆ � ✓ A n ( z ) � X ✤ Residue theorem: A n + Res = 0 � z � � k z = z k

Residue theorem ◆ � ✓ A n ( z ) � X A n = − Res � z � � k z = z k P j ( z ) 2 = 0 Residue on the pole ✤ Unitarity of shifted tree-level amplitude 1 A n ( z ) − P j ( z ) 2 =0 A L ( z ) P j ( z ) 2 A R ( z ) − − − − − →

Residue theorem ◆ � ✓ A n ( z ) � X A n = − Res � z � � k z = z k Residue on the pole P j ( z ) 2 = 2 h 1 | P j | 2]( z j � z ) = 0 P 2 j ✤ Unitarity of shifted tree-level amplitude z j = 2 h 1 | P j | 2] 1 A n ( z ) � z = z j A L ( z j ) � � ! 2 h 1 | P j | 2] A R ( z j )

Residue theorem ◆ � ✓ A n ( z ) � X A n = − Res � z � � k z = z k 2 h 1 | P j | 2] A R ( z j ) ⇥ 2 h 1 | P j | 2] 1 = A L ( z j ) 1 A L ( z j ) A R ( z j ) P 2 P 2 j j Final formula P 2 A L ( z j ) 1 X j A n = − A R ( z j ) z j = P 2 2 h 1 | P j | 2] j j

BCFW recursion relations (Britto, Cachazo, Feng, Witten, 2005) A L ( z j ) 1 P 2 X A n = − A R ( z j ) j z j = P 2 2 h 1 | P j | 2] j j Chosen such that internal line is on-shell ˆ 2 2 Sum over all distributions of legs keeping 1,2 on different sides

Comment on applicability ✤ The crucial property is for A n ( z ) → 0 z → ∞ ✤ In Yang-Mills theory this is satisfied if λ 1 → λ 1 − z λ 2 Helicity + λ 2 → e e λ 2 + z e Helicity - λ 1 ✤ Same is true for Einstein gravity, and many others ✤ This means that amplitudes in these theories are fully specified by residues on their poles

Generalizations ✤ In Standard Model and other theories more general recursion relations needed: shift more momenta ✤ Include masses: go back to momenta q 2 = ( p 1 · q ) = ( p 2 · q ) = 0 p 1 → p 1 + zq Shifted momenta on-shell, p 2 → p 2 − zq q completely fixed ✤ Extension to effective field theories (Cheung, Kampf, Novotny, JT, 2015)

Example: amplitudes of gluons

Color decomposition ✤ Sum of Feynman diagrams in Yang-Mills X M = (Color) × (Kinematics) F D Polarization vectors Gauge dependent ✤ Color factors Tr( T a 1 T a 2 T a 3 . . . T a n ) ✤ Decomposition Tr( T σ 1 T σ 2 T σ 3 . . . T σ n ) A (123 . . . n ) X M = σ

Color decomposition ✤ Sum of Feynman diagrams in Yang-Mills X M = (Color) × (Kinematics) F D Polarization vectors Gauge dependent ✤ Color factors Tr( T a 1 T a 2 T a 3 . . . T a n ) ✤ Decomposition Tr( T σ 1 T σ 2 T σ 3 . . . T σ n ) A (123 . . . n ) X M = σ

Color ordered amplitude A (123 . . . n ) Particles are ordered, other orderings: permutations Gauge invariant ✤ This is a key object of our interest ✤ Consider: All particles massless and on-shell All momenta incoming Helicities fixed

Example 1: 4pt amplitude ✤ Let us consider amplitude of gluons A 4 (1 + 2 − 3 − 4 + ) 4 + 3 − Only one term P + P − contributes ˆ 1 + ˆ 2 − [ˆ ˆ h ˆ 14] 3 23 i 3 λ 1 = λ 1 − z λ 2 1 ˆ [ˆ h ˆ λ 2 = e e λ 2 + z e 1 P ][4 P ] 2 P ih 3 P i s 23 λ 1 z takes the value when P is on-shell momentum

Example 1: 4pt amplitude ✤ Let us consider amplitude of gluons A 4 (1 + 2 − 3 − 4 + ) P 2 = h ˆ 14 i [14] = 0 z = h 14 i h ˆ 14 i = h 14 i � z h 24 i = 0 h 24 i We can now rewrite Shouten identity λ 1 = λ 1 � z λ 2 = λ 1 � h 14 i h 24 i λ 2 = h 12 i ˆ h 24 i λ 4 λ 1 = [12] λ 2 = e e λ 2 + z e e λ 3 Use of momentum [13] conservation

Example 1: 4pt amplitude ✤ Let us consider amplitude of gluons A 4 (1 + 2 − 3 − 4 + ) λ 1 = h 12 i ˆ h 24 i λ 4 Calculate on-shell momentum P ✓ h 12 i ◆ λ 1 e λ 1 + λ 4 e λ 1 + e e P = ˆ λ 4 = λ 4 λ 4 h 24 i λ P = h 23 i λ P = λ 4 e e λ 3 h 24 i

Example 1: 4pt amplitude ✤ Let us consider amplitude of gluons A 4 (1 + 2 − 3 − 4 + ) λ 1 = h 12 i 4 + 3 − ˆ h 24 i λ 4 P + P − λ 2 = [12] ˆ e e λ 3 [13] ˆ 1 + ˆ 2 − [ˆ h ˆ λ P = λ 4 14] 3 23 i 3 1 λ P = h 23 i [ˆ h ˆ 1 P ][4 P ] 2 P ih 3 P i s 23 e e λ 3 h 24 i

Example 1: 4pt amplitude ✤ Let us consider amplitude of gluons A 4 (1 + 2 − 3 − 4 + ) λ 1 = h 12 i 4 + 3 − ˆ h 24 i λ 4 P + P − λ 2 = [12] ˆ e e λ 3 [13] ˆ 1 + ˆ 2 − [ˆ h ˆ λ P = λ 4 14] 3 23 i 3 1 λ P = h 23 i [ˆ h ˆ 1 P ][4 P ] 2 P ih 3 P i s 23 e e λ 3 h 24 i h 23 i 4 = h 12 ih 23 ih 34 ih 41 i

Example 1: 4pt amplitude ✤ Let us consider amplitude of gluons A 4 (1 + 2 − 3 − 4 + ) 4 + 3 − P + P − One gauge invariant object equivalent to ˆ 1 + ˆ 2 − three Feynman diagrams

Example 2: 6pt amplitude ✤ Let us consider and shift legs 3,4 A 6 (1 − 2 − 3 − 4 + 5 + 6 + ) 2 _ _ _ 1 2 _ _ _ _ 2 3 1 3 vs 3 _ 6 + + 220 Feynman + _ _ diagrams _ + + + 1 4 6 + 4 4 5 + + + + 6 5 5 (c) h 1 | 2 + 3 | 4] 3 h 1 | 2 + 3 | 4] = h 12 i [24] + h 13 i [34] [23][34] h 56 ih 61 i s 234 h 5 | 3 + 4 | 2]

Example 2: 6pt amplitude ✤ Let us consider and shift legs 3,4 A 6 (1 − 2 − 3 − 4 + 5 + 6 + ) 2 _ _ _ 1 2 _ _ _ _ 2 3 1 3 3 _ 6 + + + _ _ _ + + + 1 4 6 + 4 4 5 + + + + 6 5 5 (c) h 1 | 2 + 3 | 4] 3 Spurious pole [23][34] h 56 ih 61 i s 234 h 5 | 3 + 4 | 2]

Remark on BCFW ✤ Extremely efficient (3 vs 220 for 6pt, 20 vs 34300 for 8pt) ✤ Terms in BCFW recursion relations Gauge invariant Spurious poles ✤ Amplitude = sum of these terms dictated by unitarity ✤ Note: not all factorization channels are present when 1,2 are on the same side

Unitarity methods

One-loop amplitudes ✤ Sum of Feynman diagrams Z d I j = d 4 ` I j X M 1 − loop = where d I j j Rational ✤ Re-express as basis of canonical integrals Z Z Z d I (4) d I (3) d I (2) M 1 − loop = X X X + R a j b j c j + + j j j j j j Box Triangle Bubble