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Lessons from Perturbative Unitarity in Graviton Scattering Amplitudes

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Lessons from Perturbative Unitarity in Graviton Scattering Amplitudes 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 = φrφ + ai Oi (φ) The existence of unitary, Lorentz invariant UV completion ! ai > 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 CP3j4. • “Scattering equations” Cachazo, He, Yuan X ki · ka Scattering Equations : Ea = = 0 σ − σ i6=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 ∼ 246GeV p Partial wave expansion violates unitarity s < 1:8TeV m2 s t + v 2 s − m2 t − m2 UV completion via scalar, m < 870GeV m2 s + t v 2 s − m2 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 ! 1, for t < 0 causality requires 2+α(t) M(s; t)js!1 ∼ s ; α(t) < 0 • At low energies we just have Einstein gravity, h12i4[34]4 M(h−; h−; h+; h+) = 1 2 3 4 stu We expect 4 4 a M(s; t) = h12i [34] f (s; t; mi ); f (s; t; mi )js!1 ∼ s with a < −2 • Then, for fixed t∗ Z dv f (s; t∗) = f (v; t∗) v − s 0 2 1 r[t] 2 (t + 2m ) X s=m i r[t]s=0 = @ i + A (s − m2)(s + t + m2) s(−s − t)t i i i at large s 1 2 r[t]s=m2 − + α + βt ! 0 i t2 Must have infinite higher spin ! General solution: n(s; t) M(s; t) = h12i4[34]4f (s; t; m ) = h12i4[34]4 i Q1 2 2 2 i=1(s − mi )(t − mi )(u − mi ) But locality requires f (s; t; mi ) ! = 2 s mi 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) = h12i4[34]4f (s; t; m ) = h12i4[34]4 i Q1 2 2 2 i=1(s − mi )(t − mi )(u − mi ) Let Q 2 2 2 2 2 2 fi;jg(s + m + m )(t + m + m )(u + m + m ) n(s; t) ∼ i j i j i j Q 2 2 2 stu i (s − mi )(t − mi )(u − mi ) We’re done, this is string theory! Massless residues controlled by the interaction of three massless particles highly constrained! One only has R, R2φ, R3. This implies that the massless residue, for s = 0, must be 1 t ()t R R 3 3 2 2 R R On the other hand the massless residue of our ansatz is Q 2 2 2 2 2 2 fi;jg(m + m )(t + m + m )(−t + m + m ) M(s; t)j ∼ h12i4[34]4 i j i j i j s=0 Q1 2 2 2 i=1(mi )(t − mi )(t + mi ) 2 2 2 2 We must have for any two pair of fi; jg there exists an mk such that mi + mj = mk 1 m2 = f1; 2; 3; · · · g i α0 1 m2 = f1; 2; 3; · · · g i α0 We thus find a simple solution: 1 h12i4[34]4 Y (s + i)(t + i)(u + i) Γ[1 − t]Γ[1 − s]Γ[1 − u] M(s; t) = = h12i4[34]4 stu (s − i)(t − i)(u − i) 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: Γ[1−s]Γ[1−u]Γ[1−t] −1 Super f (s; t) = Γ[1+s]Γ[1+u]Γ[1+t] stu Γ[1−s]Γ[1−u]Γ[1−t] −1 1 Heterotic f (s; t) = + Γ[1+s]Γ[1+u]Γ[1+t] stu s(1+s) Γ[1−s]Γ[1−u]Γ[1−t] −1 2 tu Bosonic f (s; t) = + − Γ[1+s]Γ[1+u]Γ[1+t] stu s(1+s) s(1+s)2 Each additional term corresponds to the presence of R2φ, R3. How unique is the answer ? Is the four-point amplitude sufficient ? Unitarity requires the exchange coefficient of each spin is positive: p 4 p p 1 2 p 3 2 s The residue at s = m is a function of t = − 2 (1 − cos θ), i.e. r(cos θ) Spin decomposition corresponds to expanding r(cos θ) on Gegenbauer polynomial basis: X α r(cos θ) = c`C` (cos θ) ` where α = (D − 2)=2 1 1 X = x`Cα(cos θ) (1 − 2x cos θ + x2)α ` `=0 c` > 0 are known as positive functions What do we expect of r(x) (x = cos θ)? • Permutation invariance: ! f (x) = f (−x), since s s t = − (1 − x); u = − (1 + x) 2 2 • If r(x) is positive in D, it is also positive in D0 < D • If r(x) is positive (corollary of Schoenberg Theorem) jr(x)j < 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 dn 1 Qn (a − 1 + i) = 2ntn i=1 dxn (1 − 2tx + t2)α (1 − 2tx + t2)α+n What do we expect of r(x) (x = cos θ)? • Permutation invariance: ! f (x) = f (−x), since s s t = − (1 − x); u = − (1 + x) 2 2 • If r(x) is positive in D, it is also positive in D0 < D • If r(x) is positive (corollary of Schoenberg Theorem) jr(x)j < 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): 1 9 (n − 2)2 (t+1)(t+2) ··· (t+n−1) ! x2− x2− ··· x2− n2 n2 n2 No-ghost theorem tells us that this must be a positive function! In D = 3 this is just a Fourier transform X = ca cos aθ; ca > 0 a There is no world-sheet less proof of this amazing fact! String theory residues have the following properties • As n ! 1 it’s a boundary positive function • For n = 3: 1 1 x2 − $ c CD + c CD = c (x2 − ) + c 9 2 2 0 0 2 D 0 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: Tree f (σ3; σ2) A4 = A4 Q i (s − ai )(t − ai )(u − ai ) 2 2 2 σ3 = stu; σ2 = (s + t + u ) Massless residue fixes the purely σ2 part: f (σ3; σ2) = fStringg + σ3∆ • But s3 stu = (1 − x2) 4 So anything with an stu factor cannot be positive!!!! • As n ! 1 fStringg ! 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 + Γ[1 + s]Γ[1 + t]Γ[1 + u] (s + 1)(t + 1)(u + 1) It’s residue at s = n 1 super 4(n − 1) bos bos Resn(x) = 1 − + Res (x) + Res Res n n n(1 + (1 − )n) n n−4 which is positive for 0 ≤ ≤ 1 Does this mean perturbative string is the only solution? Not yet Consider the following deformation: Γ[−s]Γ[−t]Γ[−u] stu 1 + Γ[1 + s]Γ[1 + t]Γ[1 + u] (s + 1)(t + 1)(u + 1) It’s residue at s = n 1 super 4(n − 1) bos bos Resn(x) = 1 − + Res (x) + Res Res n n n(1 + (1 − )n) n n−4 which is positive for 0 ≤ ≤ 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 Evidence: (a) For massless interactions, higher-spin theories are ruled out after considering graviton interactions (b) Consider leading trajectory at level-2 Γ[−1−t=2]Γ[−1−s=2] hV (1)V (2)V (3)V (4)i = 0 0 0 2 Γ[u=2] t + 2 t(t + 2) = ( · p p ) − 2( · p p ) + ( · p p ) : 1 1 1 2 u 2 2 u(u + 2) we find the three independent residue polynomial is given by: N−1 ! Y N − 2i : x − p 2 i=1 (4 + N)(4 + N )=N N−1 ! Y N − 2i : x x − p 2 i=1 (4 + N)(4 + N )=N N−1 ! Y N + 2 − 2i : x x − ; p 2 i=1 (4 + N)(4 + N )=N The last term contains zeros closer to 1 then the bosonic string at fixed s = n.
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