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 + aiOi(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

Scattering Equations : Ea =

• i=a

ki · ka σi − σa = 0 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/v2 − t/v2, v ∼ 246GeV Partial wave expansion violates unitarity √s < 1.8TeV m2 v2

• s

s − m2 + t t − m2

• UV completion via scalar, m < 870GeV

m2 v2 s + t 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

M(s, t)|s→∞ ∼ s2+α(t), α(t) < 0

• At low energies we just have Einstein gravity,

M(h−

1 , h− 2 , h+ 3 , h+ 4 ) = 124[34]4

stu We expect M(s, t) = 124[34]4f(s, t, mi), f(s, t, mi)|s→∞ ∼ sa with a < −2

• Then, for fixed t∗

f(s, t∗) =

• dv

v − s f(v, t∗) =  

i

r[t]s=m2

i (t + 2m2

i )

(s − m2

i )(s + t + m2 i ) +

r[t]s=0 s(−s − t)t   at large s r[t]s=m2

i −

1 t2 + α + βt2

• → 0

Must have infinite higher spin !

General solution: M(s, t) = 124[34]4f(s, t, mi) = 124[34]4 n(s, t) ∞

i=1(s − m2 i )(t − m2 i )(u − m2 i )

But locality requires f(s, t, mi)

• s=m2

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: M(s, t) = 124[34]4f(s, t, mi) = 124[34]4 n(s, t) ∞

i=1(s − m2 i )(t − m2 i )(u − m2 i )

Let n(s, t) ∼

• {i,j}(s + m2

i + m2 j )(t + m2 i + m2 j )(u + m2 i + m2 j )

stu

i(s − m2 i )(t − m2 i )(u − m2 i )

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

t

( )

1

t R R R R

2 2 3 3

On the other hand the massless residue of our ansatz is M(s, t)|s=0 ∼ 124[34]4

• {i,j}(m2

i + m2 j )(t + m2 i + m2 j )(−t + m2 i + m2 j )

∞

i=1(m2 i )(t − m2 i )(t + m2 i )

We must have for any two pair of {i, j} there exists an m2

k such that m2 i + m2 j = m2 k

m2

i = 1

α′ {1, 2, 3, · · · }

m2

i = 1

α′ {1, 2, 3, · · · } We thus find a simple solution: M(s, t) = 124[34]4 stu

∞

• i=1

(s + i)(t + i)(u + i) (s − i)(t − i)(u − i) = 124[34]4 Γ[1 − t]Γ[1 − s]Γ[1 − u] Γ[1 + t]Γ[1 + s]Γ[1 + u] This is nothing but the closed superstring amplitude! In fact this is the universal piece in all perturbative string completion: Super f(s, t) = Γ[1−s]Γ[1−u]Γ[1−t] Γ[1+s]Γ[1+u]Γ[1+t] −1 stu

• Heterotic

f(s, t) = Γ[1−s]Γ[1−u]Γ[1−t] Γ[1+s]Γ[1+u]Γ[1+t] −1 stu + 1 s(1+s)

• Bosonic

f(s, t) = Γ[1−s]Γ[1−u]Γ[1−t] Γ[1+s]Γ[1+u]Γ[1+t] −1 stu + 2 s(1+s) − tu 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:

p1 p2 p3 p4

The residue at s = m2 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: r(cos θ) =

• ℓ

cℓCα

ℓ (cos θ)

where α = (D − 2)/2 1 (1 − 2x cos θ + x2)α =

∞

• ℓ=0

xℓCα

ℓ (cos θ)

cℓ > 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 2 (1 − x), u = − s 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 dn dxn 1 (1 − 2tx + t2)α = 2ntn n

i=1(a − 1 + i)

(1 − 2tx + t2)α+n

What do we expect of r(x) (x = cos θ)?

• Permutation invariance: → f(x) = f(−x), since

t = − s 2 (1 − x), u = − s 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): (t+1)(t+2) · · · (t+n−1) →

• x2− 1

n2 x2− 9 n2

• · · ·
• x2− (n − 2)2

n2

• No-ghost theorem tells us that this must be a positive function! In D = 3 this is just a

Fourier transform =

• a

ca cos aθ, ca > 0 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:
• x2 − 1

9

• ↔ c2CD

2 + c0CD 0 = c2(x2 − 1

D ) + c0 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:

A4 = ATree

4

f(σ3, σ2)

• i(s − ai)(t − ai)(u − ai)

σ3 = stu, σ2 = (s2 + t2 + u2) Massless residue fixes the purely σ2 part: f(σ3, σ2) = {String} + σ3∆

• But

stu = s3 4 (1 − x2) 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] Γ[1 + s]Γ[1 + t]Γ[1 + u]

• 1 + ǫ

stu (s + 1)(t + 1)(u + 1)

• It’s residue at s = n

Resn(x) =

• 1 − ǫ + 1

n Ressuper

n

(x) + 4ǫ(n − 1) n(1 + (1 − ǫ)n) Resbos

n Resbos n−4

• 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] Γ[1 + s]Γ[1 + t]Γ[1 + u]

• 1 + ǫ

stu (s + 1)(t + 1)(u + 1)

• It’s residue at s = n

Resn(x) =

• 1 − ǫ + 1

n Ressuper

n

(x) + 4ǫ(n − 1) n(1 + (1 − ǫ)n) Resbos

n Resbos n−4

• 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

Evidence: (a) For massless interactions, higher-spin theories are ruled out after considering graviton interactions (b) Consider leading trajectory at level-2 V0(1)V0(2)V0(3)V2(4) = Γ[−1−t/2]Γ[−1−s/2] Γ[u/2] =

• (ǫ · p1p1) − 2(ǫ · p1p2) t + 2

u + (ǫ · p2p2) t(t + 2) u(u + 2)

• .

we find the three independent residue polynomial is given by: :

N−1

• i=1
• x −

N − 2i

• (4 + N)(4 + N2)/N
• :

x

N−1

• i=1
• x −

N − 2i

• (4 + N)(4 + N2)/N
• :

x

N−1

• i=1
• x −

N + 2 − 2i

• (4 + N)(4 + N2)/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:

• α′ → 0 World-line formalism Bern, Kosower, Nair

no world sheet

• α′ → ∞ Tension-less limit, not a field theory Amati, Ciafaloni, Veneziano, Gross, Mende

Yet ∂z (s log(z) + t log(1 − z)) = 0 → s z + t 1 − z = 0 Scattering Eq.

Zeros of flat space string theory amplitudes encode informations of the spectrum: Consider the following Ansatz for string theory Yang-Mills amplitude A(s, t) =

• i

(u + ai + aj) ← Unitarity (s − ai)(t − ai) 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: AYM(s, t) = F 4 st

• i

(u + ai + aj) ← Unitarity (s − ai)(t − ai) The only allowed local three point interaction comes from F 2, F 4 Res[A(s, t)]|s=0 = 122[34]2 α t + βt

• On the other hand

Res[A(s, t)]|s=0 = 122[34]2 t

• i

(−t + ai + aj) (−ai)(t − ai) To recover the correct massless residue:

• β = 0 ai + aj ∈ ak, → ak = positive integers!
• β = 0

122[34]2 s

• i

(u + ai + aj) (s − ai)(t − ai) α t − β u (s + 1)

• 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

AYM(s, t) = 122[34]2 st Γ[1 − s]Γ[1 − t] Γ[1 + u]

• β = 0 Bosonic String

AYM(s, t) = 122[34]2 s Γ[1 − s]Γ[1 − t] Γ[1 + u] α t − β u (s + 1)

• Moving on to closed strings (gravity)

Consider gravity R4 stu → R4 stu

• i

(s + ai + aj)(t + ai + aj)(u + ai + aj) (s − ai)(t − ai)(u − ai) Again massless residues fixes ai ∈ postive integers M(s, t) = 124[34]4

∞

• i=0

(s + i)(t + i)(u + i) (s − i)(t − i)(u − i) = 124[34]4 Γ[−s]Γ[−t]Γ[−u] Γ[1 + u]Γ[1 + t]Γ[1 + s] Staring at zeros and poles we see that closed strings can be factorized!

∞

• i=0

(s + i)(t + i)(u + i) (s − i)(t − i)(u − i) =

∞

• i=0

(u + i) (s − i)(t − i) (s + i) (s − i)(u − i) (t + i)(t − i) 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 poles zeros (I) A (s,t) s t u s t u (II) A (u,t) A (s,t)A (u,t) M (s,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

s t u (I) A (s,t) (II) A (u,t) s t u flip

All massive poles cancel! sin(πt)A(−s, −t)A(u, t) = sin(πt) Γ(s)Γ(t) Γ(1 + s + t) Γ(−s)Γ(−u) Γ(1 − u − s) = π stu We obtain the amplitude of maximal supergravity

Consider the closed bosonic string in the KLT representation

poles zeros (I) A (s,t) s t u (II) A (u,t) A (s,t)A (u,t) M (s,t) = s t u

s t u (I) A (s,t) (II) A (u,t) s t u flip 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=2 Msugra

n

=

n−3

• i=1

(sin πsi) Aopen

n,susyAopen n,susy

N=1 Msugra

n

=

n−3

• i=1

(sin πsi) Aopen

n,bosonicAopen n,susy,

w tachyon ghost or massive spin-2 state Bosonic Mgra

n

=

n−3

• i=1

(sin πsi) Aopen

n,bosonicAopen n,bosonic,

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 ? Mclosed(s, t) =

• d2z |1 − z|2s|z|2t =
• d2z (1 − z)szt(1 − ¯

z)s¯ zt the flip → Msugra(s, t) =

• d2z (1 − z)−sz−t(1 − ¯

z)s¯ zt The change of space-time signature of the left handed OPE is equivalent to a change

• f boundary condition for Greens function

ln z¯ z → ln z¯ z − 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: B(α, β) = 1 z−1+α(1 − z)−1+β, Re[α] > 0, Re[β] > 0 The world sheet integrand appears not well defined in any regime Msugra(s, t) =

• d2z (1 − z)−sz−t(1 − ¯

z)s¯ zt There is a natural definition of the contour for integrals (with α + β + γ = 0)

• dz2(z2 − z1)α(z2 − z3)β(z2 − z4)γ

→ = 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)

• dz2(z2 − z1)α(z2 − z3)β(z2 − z4)γ

→ = 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: L0 = α′( 1 2 p)2 + N − 1, ¯ L0 → −α′( 1 2 p)2 + N − 1 and the constraint becomes L0 + ¯ L0 → N + N − 2 = 0, L0 − ¯ L0 → 1 2α′p2 + N − N = 0, restrict the spectrum to levels and masses (N, N; 1 4 α′m2) = (1, 1; 0), (2, 0; 1), (0, 2; −1)

We now have a world-sheet theory (at finite α′) for field theory amplitudes. For type II superstring and Heterotic Yang-Mills the result contains only massless states The theory is the same when α′ → ∞ what is its relation to tensionless strings?

Particle limit of tensionless string

Tensionless string is best studied with Hamiltonian H = λ0 2 (α′P2 + (∂1X)2/α′) + λ1P · ∂1X As α′ → ∞ the two first class constraints become P2 = 0, P · ∂1X = 0 Consider the oscillators in the pn xn language αn = √ α′ 2 pn − in xn √ α′ , ˜ αn = √ α′ 2 p−n − in x−n √ α′ Two inequivalent vacuumCasali, Tourkine

• pn|0 = 0 for all n. Descends from the usual αn|0 = ˜

αn|0 = 0 (the usual interpretation of high energy behaviour)

• pn|0 = xn|0 = 0 for n > 0. This would correspond to αn|0 = ˜

α−n|0 = 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 α′ independent, α′ → ∞ He

• dz2(· · · )eα′[−s log z−t log(1−z)+s log ¯

z+t log(1−¯ z)]

Depending on kinematics, saddle point localizes on s

z + t 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 α′

• 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

Mclosed = (Aopen)T S(α′)Aopen, Mgrav = (AYM)T S0AYM Open superstring amplitudes can be casted onto YM tree basis:Mafra, Schlotterer, Stieberger Aopen = F(α′)AYM The conjecture amounts to stating that FT (α′)S(α′)[F(α′)]flip = S0 The α′-dependent function F can be further separated into Schlotterer, Stieberger F(α′) = P · M P :=

∞

• k=0

f k

2 P2k,

M :=

∞

• p=0
• i1,...,ip

∈2N++1

fi1fi2 · · · fipMip · · · Mi2Mi1, where P2k := F

• (ζ2)k ,

M2k+1 := F

• ζ2k+1

Sketch of proof to all multiplicity

FT (α′)S(α′)[F(α′)]flip = S0 The α′-dependent function F can be further separated into Schlotterer, Stieberger F(α′) = P · M P :=

∞

• k=0

f k

2 P2k,

M :=

∞

• p=0
• i1,...,ip

∈2N++1

fi1fi2 · · · fipMip · · · Mi2Mi1, where P2k := F

• (ζ2)k ,

M2k+1 := F

• ζ2k+1

Two conjectures: PTSP = S0, MT

2k+1S0 = S0M2k+1,

∀k ∈ N+. explicitly verified up to the order α′21 at five points, α′9 at six points and α′7 at seven points Schlotterer, Stieberger

Sketch of proof to all multiplicity

FT (α′)S(α′)[F(α′)]flip = S0 The α′-dependent function F can be further separated into Schlotterer, Stieberger F(α′) = P · M Two conjectures: PTSP = S0, MT

2k+1S0 = S0M2k+1,

∀k ∈ N+. Given this structure of the open superstring amplitudes, we have now FT (α′)S(α′)[F(α′)]flip = (MTPT)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 L = − 1 2 ∂LX∂RX = √ggMN∂MX∂XN where zL,R = ± 1 2 1 √λ0 ((λ1 ± λ0)τ + σ), λ0 = √−g g11 , λ1 = g01 g11 Consider a single parameter family of gauges λ0 = 1 1 + β , λ1 = β 1 + β β → 0 conformal gauge, β → ∞ HSZ gauge zL =

• 1 + βz,

zR = 1

• 1 + β

(¯ z − βz) in the HSZ gauge zL,R ∼ z it becomes chiral

Relation to CHY

Taking the HSZ gauge on the Greens function XX log zR zL = log

• β

1 + β (1 − ¯ z βz )

• → ¯

z βz while the remaining simply becomes chiral: 1 zL → 1 z , 1 zR → 1 β 1 z + ¯ z βz2

• The only ¯

z dependence appears as:

Relation to CHY

Taking the HSZ gauge on the Greens function XX log zR zL = log

• β

1 + β (1 − ¯ z βz )

• → ¯

z βz while the remaining simply becomes chiral: 1 zL → 1 z , 1 zR → 1 β 1 z + ¯ z βz2

• The only ¯

z dependence appears as: