Lessons from perturbative unitarity in graviton scattering - PowerPoint PPT Presentation

Lessons from perturbative unitarity in graviton scattering amplitudes Yu-tin Huang National Taiwan University Nima Arkani-Hamed, Tzu-Chen Huang, Ellis Ye Yuan, Warren Siegel Strings and Fields 2016 YITP

The cartoon story of string theory, The presence of world-sheet → high energy softness, infinite excitations (solution to UV completion) There is much more to the story than the world-sheet

Two apparently unrelated developments: • Constraints from/of perturbative completion: • Positivity: For any effective field theory, L = φ ∇ φ + a i O i ( φ ) The existence of unitary, Lorentz invariant UV completion → a i > 0 for certain operators Adams, Arkani-Hamed, Dubovsky, Nicolis, Rattazzi • Causality: For gravitation interactions, any perturbative correction to R produce time advancement interactions. It’s cure require infinite number of higher spin states Camanho, Edelstein, Maldacena, Zhiboedov • World-sheet based field theory amplitudes • Witten’s twistor string theory: Topological B-model in CP 3 | 4 . • “Scattering equations” Cachazo, He, Yuan k i · k a � Scattering Equations : E a = = 0 σ i − σ a i � = a Massless kinematics parameterizes the moduli space of n -punctured Riemann spheres

The “physical” origin of string theory is likely to have nothing to do with the world-sheet In this talk I will pursuit this line of thought by answering • What constraints does perturbative unitarity imposes on the S-matrix? For massive scalar, see Caron-Huot, Komargodski, Sever, Zhiboedov • Are these “world-sheet” field theory a limit of string theory ?

We already have a well known example Longitudinal W scattering: A ( W + W − → W + W − ) ∼ − s / v 2 − t / v 2 , v ∼ 246 GeV Partial wave expansion violates unitarity √ s < 1 . 8 TeV m 2 � s t � s − m 2 + v 2 t − m 2 UV completion via scalar, m < 870 GeV m 2 s + t v 2 s − m 2 UV completion via vector

What is perturbative completion? • The UV degrees of freedom appears while the theory is still weakly coupled • The S-matrix only have poles, no branch cuts • The new degrees of freedom (glue balls and mesons in large N YM) • The high energy fixed angle scattering is improved → all interactions are EFT in some dimensions

What can we expect? • As s → ∞ , for t < 0 causality requires M ( s , t ) | s →∞ ∼ s 2 + α ( t ) , α ( t ) < 0 • At low energies we just have Einstein gravity, 4 ) = � 12 � 4 [ 34 ] 4 M ( h − 1 , h − 2 , h + 3 , h + stu We expect M ( s , t ) = � 12 � 4 [ 34 ] 4 f ( s , t , m i ) , f ( s , t , m i ) | s →∞ ∼ s a with a < − 2 • Then, for fixed t ∗ � dv f ( s , t ∗ ) v − s f ( v , t ∗ ) =  i ( t + 2 m 2  r [ t ] s = m 2 i ) r [ t ] s = 0 � = i ) +  ( s − m 2 i )( s + t + m 2 s ( − s − t ) t i at large s � 1 � t 2 + α + β t 2 r [ t ] s = m 2 i − → 0 Must have infinite higher spin !

General solution: n ( s , t ) M ( s , t ) = � 12 � 4 [ 34 ] 4 f ( s , t , m i ) = � 12 � 4 [ 34 ] 4 � ∞ i = 1 ( s − m 2 i )( t − m 2 i )( u − m 2 i ) But locality requires � � f ( s , t , m i ) → � � s = m 2 i absence of singularity All double poles must have no residue s = a , t = b → n ( a , b ) = 0 s = a , u = b → n ( a , − a − b ) = 0 s = a , t = b → n ( − a − b , b ) = 0

General solution: n ( s , t ) M ( s , t ) = � 12 � 4 [ 34 ] 4 f ( s , t , m i ) = � 12 � 4 [ 34 ] 4 � ∞ i = 1 ( s − m 2 i )( t − m 2 i )( u − m 2 i ) Let { i , j } ( s + m 2 i + m 2 j )( t + m 2 i + m 2 j )( u + m 2 i + m 2 � j ) n ( s , t ) ∼ i ( s − m 2 i )( t − m 2 i )( u − m 2 stu � i ) We’re done, this is string theory!

Massless residues controlled by the interaction of three massless particles ← highly constrained! One only has R , R 2 φ , R 3 . This implies that the massless residue, for s = 0, must be 1 ( ) t t R R 3 3 R R 2 2 On the other hand the massless residue of our ansatz is � { i , j } ( m 2 i + m 2 j )( t + m 2 i + m 2 j )( − t + m 2 i + m 2 j ) M ( s , t ) | s = 0 ∼ � 12 � 4 [ 34 ] 4 � ∞ i = 1 ( m 2 i )( t − m 2 i )( t + m 2 i ) We must have for any two pair of { i , j } there exists an m 2 k such that m 2 i + m 2 j = m 2 k i = 1 m 2 α ′ { 1 , 2 , 3 , · · · }

i = 1 m 2 α ′ { 1 , 2 , 3 , · · · } We thus find a simple solution: ∞ M ( s , t ) = � 12 � 4 [ 34 ] 4 ( s + i )( t + i )( u + i ) ( s − i )( t − i )( u − i ) = � 12 � 4 [ 34 ] 4 Γ[ 1 − t ]Γ[ 1 − s ]Γ[ 1 − u ] � stu Γ[ 1 + t ]Γ[ 1 + s ]Γ[ 1 + u ] i = 1 This is nothing but the closed superstring amplitude! In fact this is the universal piece in all perturbative string completion: � − 1 � Γ[ 1 − s ]Γ[ 1 − u ]Γ[ 1 − t ] f ( s , t ) = Super Γ[ 1 + s ]Γ[ 1 + u ]Γ[ 1 + t ] stu � − 1 Γ[ 1 − s ]Γ[ 1 − u ]Γ[ 1 − t ] 1 � Heterotic f ( s , t ) = stu + Γ[ 1 + s ]Γ[ 1 + u ]Γ[ 1 + t ] s ( 1 + s ) � − 1 � Γ[ 1 − s ]Γ[ 1 − u ]Γ[ 1 − t ] 2 tu f ( s , t ) = stu + s ( 1 + s ) − Bosonic s ( 1 + s ) 2 Γ[ 1 + s ]Γ[ 1 + u ]Γ[ 1 + t ] Each additional term corresponds to the presence of R 2 φ , R 3 .

How unique is the answer ? Is the four-point amplitude sufficient ?

Unitarity requires the exchange coefficient of each spin is positive: p 4 p 1 p 2 p 3 The residue at s = m 2 is a function of t = − s 2 ( 1 − cos θ ) , i.e. r ( cos θ ) Spin decomposition corresponds to expanding r ( cos θ ) on Gegenbauer polynomial basis: � c ℓ C α r ( cos θ ) = ℓ ( cos θ ) ℓ where α = ( D − 2 ) / 2 ∞ 1 � x ℓ C α ( 1 − 2 x cos θ + x 2 ) α = ℓ ( cos θ ) ℓ = 0 c ℓ > 0 are known as positive functions

What do we expect of r ( x ) ( x = cos θ )? • Permutation invariance: → f ( x ) = f ( − x ) , since t = − s u = − s 2 ( 1 − x ) , 2 ( 1 + x ) • If r ( x ) is positive in D , it is also positive in D ′ < D • If r ( x ) is positive (corollary of Schoenberg Theorem) | r ( x ) | < r ( 1 ) If r ( x ) has root at x = 1, it must be a negative function • If we rescale r ( x ) → r ( ax ) with a > 1, then r ( ax ) − r ( x ) = positive function since � n d n 1 i = 1 ( a − 1 + i ) ( 1 − 2 tx + t 2 ) α = 2 n t n dx n ( 1 − 2 tx + t 2 ) α + n

What do we expect of r ( x ) ( x = cos θ )? • Permutation invariance: → f ( x ) = f ( − x ) , since t = − s u = − s 2 ( 1 − x ) , 2 ( 1 + x ) • If r ( x ) is positive in D , it is also positive in D ′ < D • If r ( x ) is positive (corollary of Schoenberg Theorem) | r ( x ) | < r ( 1 ) If r ( x ) has root at x = 1, it must be a negative function • If we rescale r ( x ) → r ( ax ) with a > 1, then r ( ax ) − r ( x ) = positive function • If r ( x ) is positive, the roots of r ( x ) on the real axes must lie − 1 < x < 1 !

Consider the open string residue (whose square gives closed string): x 2 − ( n − 2 ) 2 � x 2 − 1 � � x 2 − 9 � � � ( t + 1 )( t + 2 ) · · · ( t + n − 1 ) → · · · n 2 n 2 n 2 No-ghost theorem tells us that this must be a positive function! In D = 3 this is just a Fourier transform � = c a cos a θ, c a > 0 a There is no world-sheet less proof of this amazing fact! String theory residues have the following properties • As n → ∞ it’s a boundary positive function • For n = 3: � x 2 − 1 � 0 = c 2 ( x 2 − 1 ↔ c 2 C D 2 + c 0 C D D ) + c 0 9 The critical dimension is D = 9 • Observe primeness at the critical dimensions, “minimal building block”

Can these properties explain why string theory is the unique answer? • A general solution: f ( σ 3 , σ 2 ) A 4 = A Tree 4 � i ( s − a i )( t − a i )( u − a i ) σ 3 = stu , σ 2 = ( s 2 + t 2 + u 2 ) Massless residue fixes the purely σ 2 part: f ( σ 3 , σ 2 ) = { String } + σ 3 ∆ • But stu = s 3 4 ( 1 − x 2 ) So anything with an stu factor cannot be positive!!!! • As n → ∞ { String } → near negative!

Does this mean perturbative string is the only solution? Not yet Consider the following deformation: Γ[ − s ]Γ[ − t ]Γ[ − u ] � stu � 1 + ǫ Γ[ 1 + s ]Γ[ 1 + t ]Γ[ 1 + u ] ( s + 1 )( t + 1 )( u + 1 ) It’s residue at s = n � 1 − ǫ + 1 � � 4 ǫ ( n − 1 ) � Res super n ( 1 + ( 1 − ǫ ) n ) Res bos n Res bos Res n ( x ) = ( x ) + n n − 4 n which is positive for 0 ≤ ǫ ≤ 1

Does this mean perturbative string is the only solution? Not yet Consider the following deformation: Γ[ − s ]Γ[ − t ]Γ[ − u ] � stu � 1 + ǫ Γ[ 1 + s ]Γ[ 1 + t ]Γ[ 1 + u ] ( s + 1 )( t + 1 )( u + 1 ) It’s residue at s = n � 1 − ǫ + 1 � � 4 ǫ ( n − 1 ) � Res super n ( 1 + ( 1 − ǫ ) n ) Res bos n Res bos Res n ( x ) = ( x ) + n n − 4 n which is positive for 0 ≤ ǫ ≤ 1

Lessons: • String theory satisfies positivity in a very non-trivial way • Improved high energy behaviour appears to be in tension with positivity • However, with massless scattering, deformations inheriting the string magic exists One expects further constraints from massive interactions