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 + 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 CP3|4. • “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 √ 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 → ∞, for t < 0 causality requires 2+α(t) M(s, t)|s→∞ ∼ 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 )|s→∞ ∼ s with a < −2 • Then, for fixed t∗
Z dv f (s, t∗) = f (v, t∗) v − s
2 r[t] 2 (t + 2m ) X s=m i r[t]s=0 = i + (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 Q∞ 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 Q∞ 2 2 2 i=1(s − mi )(t − mi )(u − mi ) Let Q 2 2 2 2 2 2 {i,j}(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 {i,j}(m + m )(t + m + m )(−t + m + m ) M(s, t)| ∼ h12i4[34]4 i j i j i j s=0 Q∞ 2 2 2 i=1(mi )(t − mi )(t + mi )
2 2 2 2 We must have for any two pair of {i, j} there exists an mk such that mi + mj = mk
1 m2 = {1, 2, 3, ···} i α0 1 m2 = {1, 2, 3, ···} i α0 We thus find a simple solution:
∞ 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 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)
|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 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)
|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):
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 → ∞ 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) = {String} + σ3∆
• But s3 stu = (1 − x2) 4 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 + 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. To seek the world-sheet for field theory
Unlearn the world-sheet from string theory We know how to get field-theory:
• α0 → 0 World-line formalism Bern, Kosower, Nair
0
no world sheet • α0 → ∞ Tension-less limit, not a field theory Amati, Ciafaloni, Veneziano, Gross, Mende
Yet s t ∂z (s log(z) + t log(1 − z)) = 0 → + = 0 Scattering Eq. z 1 − z Zeros of flat space string theory amplitudes encode informations of the spectrum:
Consider the following Ansatz for string theory Yang-Mills amplitude
Y (u + ai + aj ) ← Unitarity A(s, t) = (s − a )(t − a ) i i i
The zeros are necessary for the cancelation of double poles (unitarity zeros) Caron-Huot, Komargodski, Sever, Zhiboedov
All pairs of double poles must be canceled, controlled high energy behaviour imposes constraint on the specturm Massless interactions imposes stronger constraint:
4 F Y (u + ai + aj ) ← Unitarity AYM (s, t) = st (s − a )(t − a ) i i i
The only allowed local three point interaction comes from F 2, F 4 2 2 α Res[A(s, t)]| = = h12i [34] + βt s 0 t
On the other hand
2 2 h12i [34] Y (−t + ai + aj ) Res[A(s, t)]| = = s 0 t (−a )(t − a ) i i i
To recover the correct massless residue:
• β = 0 ai + aj ∈ ak , → ak = positive integers! • β 6= 0 2 2 h12i [34] Y (u + ai + aj ) α u − β s (s − a )(t − a ) t (s + 1) i i i For the second term the absence of t = 0, s = 1 double pole renders the zero (u + 1) useless (dangerous), unless we have a tachyon since s = −1, t = 2 → u = −1 • β = 0 SuperString
h12i2[34]2 Γ[1 − s]Γ[1 − t] AYM (s, t) = st Γ[1 + u]
• β 6= 0 Bosonic String
h12i2[34]2 Γ[1 − s]Γ[1 − t] α u AYM (s, t) = − β s Γ[1 + u] t (s + 1)
Moving on to closed strings (gravity) Consider gravity
4 4 R R Y (s + ai + aj )(t + ai + aj )(u + ai + aj ) → stu stu (s − a )(t − a )(u − a ) i i i i
Again massless residues fixes ai ∈ postive integers
∞ Y (s + i)(t + i)(u + i) Γ[−s]Γ[−t]Γ[−u] M(s, t) = h12i4[34]4 = h12i4[34]4 (s − i)(t − i)(u − i) Γ[1 + u]Γ[1 + t]Γ[1 + s] i=0
Staring at zeros and poles we see that closed strings can be factorized!
∞ ∞ Y (s + i)(t + i)(u + i) Y (u + i) (s + i) = (t + i)(t − i) (s − i)(t − i)(u − i) (s − i)(t − i) (s − i)(u − i) i=0 i=0
This is nothing but the KLT relation from worldsheet monodromy relations
M(s, t) = sin(πt)AYM (s, t)AYM (t, u) Studying the zeros of string theory amplitudes retain the information that the world sheet
The positions of the zeros will show us a path to field theory Consider the zeros of closed superstring in the KLT representation
M (s,t) = A (s,t)A (u,t)
(I) A (s,t) (II) A (u,t) u u poles s zeros s t
t
There are zeros in all unphysical channels Consider the zeros of closed superstring in the KLT representation
Flip the signature on one of the open strings
(I) A (s,t) (II) A (u,t) u u
flip s s
t t
All massive poles cancel!
Γ(s)Γ(t) Γ(−s)Γ(−u) π sin(πt)A(−s, −t)A(u, t) = sin(πt) = Γ(1 + s + t) Γ(1 − u − s) stu
We obtain the amplitude of maximal supergravity Consider the closed bosonic string in the KLT representation
M (s,t) = A (s,t)A (u,t)
(I) A (s,t) (II) A (u,t) u u
poles s zeros s
t
t
(I) A (s,t) (II) A (u,t) u u
flip s s
t t The spectrum contains massless closed string states and two mysterious massive states (super)gravity from (super)string2
Demonstrated at 4 and 5-pts, conjectured to hold for all closed string theories
n−3 sugra Y open open N=2 Mn = (sin πsi ) An,susy An,susy i=1
n−3 sugra Y open open N=1 Mn = (sin πsi ) An,bosonic An,susy , i=1 w tachyon ghost or massive spin-2 state
n−3 gra Y open open Bosonic Mn = (sin πsi ) An,bosonic An,bosonic , i=1 w tachyon ghost and massive spin-2 state Back to the world sheet
Msugra(s, t) = (sin πt) Aopen(−s, −t)Aopen(t, u) What does this mean on the world sheet ? Z Z Mclosed(s, t) = d2z |1 − z|2s|z|2t = d2z (1 − z)szt (1 − z¯)sz¯t
Z the flip → Msugra(s, t) = d2z (1 − z)−sz−t (1 − z¯)sz¯t
The change of space-time signature of the left handed OPE is equivalent to a change of boundary condition for Greens function
ln zz¯ → ln zz¯ − 2 ln z = ln z¯ − ln z
The integrand is no-longer single valued, how does one define the integral? Back to the world sheet
More precisely, the convergence of the two integrals are in opposite regime:
Z 1 B(α, β) = z−1+α(1 − z)−1+β , Re[α] > 0, Re[β] > 0 0 The world sheet integrand appears not well defined in any regime Z Msugra(s, t) = d2z (1 − z)−sz−t (1 − z¯)sz¯t
There is a natural definition of the contour for integrals (with α + β + γ = 0) Z α β γ dz2(z2 − z1) (z2 − z3) (z2 − z4)
→ = sin(πα) sin(πβ)B(α, β) Back to the world sheet There is a natural definition of the contour for integrals (with α + β + γ = 0) Z α β γ dz2(z2 − z1) (z2 − z3) (z2 − z4)
→ = sin(πα) sin(πβ)B(α, β)
Setting γ is a positive integer the contour collapses and one recovers zero
The poles are at α, β = non-positive integers (α = −s, β = −t, γ = −u) Back to the world sheet
This chiral theory has a well defined in oscillator langauge
The change in the boundary conditions corresponds to Bogoliubov transformation for ¯z modes a¯ → a¯†, a¯† → −a¯ The resulting Verasoro generators become:
1 1 L = α0( p)2 + N − 1, L¯ → −α0( p)2 + N − 1 0 2 0 2 and the constraint becomes 1 L + L¯ → N + N − 2 = 0, L − L¯ → α0p2 + N − N = 0, 0 0 0 0 2 restrict the spectrum to levels and masses
1 (N, N; α0m2) = (1, 1; 0), (2, 0; 1), (0, 2; −1) 4 We now have a world-sheet theory (at finite α0) for field theory amplitudes.
For type II superstring and Heterotic Yang-Mills the result contains only massless states
The theory is the same when α0 → ∞ what is its relation to tensionless strings? Particle limit of tensionless string
Tensionless string is best studied with Hamiltonian
λ H = 0 (α0P2 + (∂ X)2/α0) + λ P · ∂ X 2 1 1 1 As α0 → ∞ the two first class constraints become
2 P = 0, P · ∂1X = 0
Consider the oscillators in the pn xn language √ √ 0 0 α xn α x−n αn = pn − in √ , α˜n = p−n − in √ 2 α0 2 α0
Two inequivalent vacuumCasali, Tourkine
• pn|0i = 0 for all n. Descends from the usual αn|0i =α ˜n|0i = 0 (the usual interpretation of high energy behaviour)
• pn|0i = xn|0i = 0 for n > 0. This would correspond to αn|0i =α ˜−n|0i = 0 for n > 0. The new twisted model for type II superstring and Heterotic Yang-Mills, is a quantum version of tensionless string. Relation to CHY
For superstring the final result is α0 independent, α0 → ∞ He
Z 0 dz2(··· )eα [−s log z−t log(1−z)+s log z¯+t log(1−z¯)]
s t Depending on kinematics, saddle point localizes on z + 1−z = 0, the phase factor cancels and one attains CHY.
What about Bosonic and Heterotic Gravity Conclusion
• Perturbative unitarity imposes stringent constraint on possible UV completions • String theory (in a well defined way ) yields the simplest solution. • Positivity constraints are non-trivially satisfied by the scattering of lowest lying string states • Massive string states require more magic • The structure of string theory zeros reveals a novel projection to field theory results without invoking any limits on α0 • For type II superstring and Heterotic Yang-Mills, this corresponds to a quantum version of tensionless string. And has a direct relation to CHY. • For the Heterotic and Bosonic gravity we have massive states (spin-2), provides a “consistent” coupling between massive and massless spin-2 state. Also a hint of CHY. Sketch of proof to all multiplicity
closed open T 0 open grav YM T YM M = (A ) S(α )A , M = (A ) S0A
Open superstring amplitudes can be casted onto YM tree basis:Mafra, Schlotterer, Stieberger
Aopen = F(α0)AYM
The conjecture amounts to stating that
T 0 0 0 F (α )S(α )[F(α )]flip = S0
The α0-dependent function F can be further separated into Schlotterer, Stieberger
F(α0) = P · M
∞ ∞ X X X := k , := ··· ··· , P f2 P2k M fi1 fi2 fip Mip Mi2 Mi1 k=0 p=0 i1,...,ip + ∈2N +1 where
P2k := F k , M2k+1 := F (ζ2) ζ2k+1 Sketch of proof to all multiplicity
T 0 0 0 F (α )S(α )[F(α )]flip = S0
The α0-dependent function F can be further separated into Schlotterer, Stieberger
F(α0) = P · M
∞ ∞ X X X := k , := ··· ··· , P f2 P2k M fi1 fi2 fip Mip Mi2 Mi1 k=0 p=0 i1,...,ip + ∈2N +1 where
P2k := F k , M2k+1 := F (ζ2) ζ2k+1 Two conjectures:
T T + P SP = S0, M2k+1S0 = S0M2k+1, ∀k ∈ N . explicitly verified up to the order α021 at five points, α09 at six points and α07 at seven points Schlotterer, Stieberger Sketch of proof to all multiplicity
T 0 0 0 F (α )S(α )[F(α )]flip = S0
The α0-dependent function F can be further separated into Schlotterer, Stieberger
F(α0) = P · M
Two conjectures:
T T + P SP = S0, M2k+1S0 = S0M2k+1, ∀k ∈ N .
Given this structure of the open superstring amplitudes, we have now
T 0 0 0 T T F (α )S(α )[F(α )]flip = (M P )flippedSPM = S0. (1)
Each P2k has even degree in terms of the Mandelstam variables, and so P = Pflipped. By applying the two conjectures above we obtain an equivalent statement
MflippedM = 1. (2)
Which can be proven via induction! Relation to CHY
A gauge transformation on the world sheet Siegel
1 √ L = − ∂ X∂ X = ggMN ∂ X∂X 2 L R M N where √ 1 1 −g g01 zL,R = ± √ ((λ1 ± λ0)τ + σ), λ0 = , λ1 = 2 λ0 g11 g11 Consider a single parameter family of gauges
1 β λ = , λ = 0 1 + β 1 1 + β
β → 0 conformal gauge, β → ∞ HSZ gauge
p 1 zL = 1 + βz, zR = (z¯ − βz) p1 + β in the HSZ gauge zL,R ∼ z it becomes chiral Relation to CHY
Taking the HSZ gauge on the Greens function hXXi
z β z¯ z¯ log R = log (1 − ) → zL 1 + β βz βz while the remaining simply becomes chiral:
1 1 1 1 1 z¯ → , → + 2 zL z zR β z βz The only z¯ dependence appears as: Relation to CHY
Taking the HSZ gauge on the Greens function hXXi
z β z¯ z¯ log R = log (1 − ) → zL 1 + β βz βz while the remaining simply becomes chiral:
1 1 1 1 1 z¯ → , → + 2 zL z zR β z βz The only z¯ dependence appears as: