Reflections on conformal spectra Petr Kravchuk with Hyungrok Kim - - PowerPoint PPT Presentation

Reflections on conformal spectra

Petr Kravchuk

with Hyungrok Kim and Hirosi Ooguri

Walter Burke Institute for Theoretical Physics, Caltech

CERN, 14 Dec 2015

Outline

• 1. Convergence bounds for large ∆
• 2. Cardy-like formula for large ∆
• 3. A convergence bound for finite ∆

The problem

Consider the state in a Euclidean CFTd | ri = (r) |i ,

• r the four point function on the real line with x = ¯

x = r2 G4(x) = h| (1)(x, ¯ x) |i / h r| ri .

The problem

Consider the state in a Euclidean CFTd | ri = (r) |i ,

• r the four point function on the real line with x = ¯

x = r2 G4(x) = h| (1)(x, ¯ x) |i / h r| ri . Using ⇥ OPE, one can decompose | ri = X

O

COVO(r) |Oi G4(x) = X

O

C 2

Ox2∆φF∆,`(x) /

X

O

C 2

O hO| V † O(r)VO(r) |Oi

The problem

Suppose we want to understand | ri = X

O

COVO(r) |Oi .

The problem

Suppose we want to understand | ri = X

O

COVO(r) |Oi . Consider the CDF F(∆⇤, x) = h r| P∆O<∆∗ | ri h r| ri = 1 G ∆∗

4

(x) G4(x) , where G ∆∗

4

(x) = X

O,∆O>∆∗

C 2

Ox2∆φF∆,`(x)

[Pappadopulo,Rychkov,Espin,Rattazzi ’12; Rychkov,Yvernay ’15]

A simpler problem

Ignore conformal symmetry, use only scaling symmetry ! “scaling blocks”, G4(x) = X

O

C 2

Ox∆O2∆φ =

Z 1 x∆2∆φg(s)(∆)d∆.

A simpler problem

Ignore conformal symmetry, use only scaling symmetry ! “scaling blocks”, G4(x) = X

O

C 2

Ox∆O2∆φ =

Z 1 x∆2∆φg(s)(∆)d∆. From [PRER ’12], in a given theory, for sufficiently large ∆⇤ G ∆∗

4

(x) . 1 Γ(2∆ + 1)∆2∆φ

⇤

x∆∗2∆φ. Can we obtain more information on the structure of F(∆⇤, x)?

Crossing symmetry

Using a different channel for OPE expansion one finds G4(x) = G4(1 x), and so @nG4(x) = (@)nG4(1 x).

Crossing symmetry

Using a different channel for OPE expansion one finds G4(x) = G4(1 x), and so @nG4(x) = (@)nG4(1 x). At x = 1/2 one obtains Z 1 [∆ 2∆](2k+1)(s)

1/2(∆)d∆ = 0,

[↵](n) = x↵+n@nx↵ = ↵(↵ 1) . . . (↵ n + 1) (s)

x (∆) = x∆2∆φg(s)(∆)

Crossing symmetry

Suppose ∆ 1 in Z 1 [∆ 2∆](2k+1)(s)

1/2(∆)d∆ = 0,

Then for k ⌧ p∆ approximate [∆ 2∆](2k+1) ' (∆ 2∆)2k+1, Z 1

2∆φ

w2k+1(s)

1/2(w + 2∆)dw ' 0.

This suggests that (s)

1/2(w + 2∆) is approximately symmetric

around w = 0.

Reflection symmetry

Suppose the symmetry is exact, (s)

1/2(∆) = (s) 1/2(4∆ ∆).

Reflection symmetry

Suppose the symmetry is exact, (s)

1/2(∆) = (s) 1/2(4∆ ∆).

Normalize R (s)

1/2(∆)d∆ = 1.

Then F(∆⇤, 1/2) = R ∆∗ (s)

1/2(∆)d∆ is antisymmetric up to a

constant.

Reflection symmetry

Suppose the symmetry is exact, (s)

1/2(∆) = (s) 1/2(4∆ ∆).

Normalize R (s)

1/2(∆)d∆ = 1.

Then F(∆⇤, 1/2) = R ∆∗ (s)

1/2(∆)d∆ is antisymmetric up to a

constant.

2 Δϕ 4 Δϕ

Δ

1 2

1 ℱ

Reflection symmetry

Suppose the symmetry is exact, (s)

1/2(∆) = (s) 1/2(4∆ ∆).

Normalize R (s)

1/2(∆)d∆ = 1.

Then F(∆⇤, 1/2) = R ∆∗ (s)

1/2(∆)d∆ is antisymmetric up to a

constant.

2 Δϕ 4 Δϕ

Δ

1 2

1 ℱ

Reflection symmetry

Suppose the symmetry is exact, (s)

1/2(∆) = (s) 1/2(4∆ ∆).

Normalize R (s)

1/2(∆)d∆ = 1.

Then F(∆⇤, 1/2) = R ∆∗ (s)

1/2(∆)d∆ is antisymmetric up to a

constant.

2 Δϕ 4 Δϕ

Δ

1 2

1 ℱ

G4(x) = x2∆φ2∆φ or G4(x) = x2∆φ + (1 x)2∆φ

Reflection symmetry

For general x the reflection is between (s)

x

and (s)

1x, relating

∆ 2∆ x $ ∆ 2∆ 1 x

Reflection symmetry

For general x the reflection is between (s)

x

and (s)

1x, relating

∆ 2∆ x $ ∆ 2∆ 1 x Let x > 1/2, ∆x = 2∆φ

1x . Then (s) 1x(∆) = 0 for ∆ < 0 )

(s)

x (∆) ' 0 for ∆ ∆x.

Δx

Δ

1 ℱ

Saddle point interpretation

The same relation for general x can be obtained if one assumes that the four-point function is dominated by a saddle point at ∆ = ∆(x), ∆(x) = 2∆ + @ log G4(x) @ log x .

Saddle point interpretation

The same relation for general x can be obtained if one assumes that the four-point function is dominated by a saddle point at ∆ = ∆(x), ∆(x) = 2∆ + @ log G4(x) @ log x . G4(x) = G4(1 x) ) ∆(x) 2∆ x = ∆(1 x) 2∆ 1 x

Threshold bound

Can we compute a bound on the tail which exhibits ∆x threshold?

Threshold bound

Can we compute a bound on the tail which exhibits ∆x threshold? Consider the linear programming formulation [Rattazzi,Rychkov,Tonni,Vichi 2008] (s)

1/2(∆) 0

Z 1 (s)

1/2(∆)d∆ = 1

Z 1 [∆ 2∆](2k+1)(s)

1/2(∆)d∆ = 0

max Z 1

∆∗

(s)

1/2(∆)d∆ =?

Threshold bound

Can we compute a bound on the tail which exhibits ∆x threshold? Consider the linear programming formulation [Rattazzi,Rychkov,Tonni,Vichi 2008] (s)

1/2(∆) 0

Z 1 (s)

1/2(∆)d∆ = 1

Z 1 [∆ 2∆](2k+1)(s)

1/2(∆)d∆ = 0

max Z 1

∆∗

(s)

1/2(∆)d∆ =?

This is of the form A~ x = ~ b, x 0 max ~ c · ~ x =?

Linear programming duality

The linear programming problem of the form A~ x = ~ b, x 0, max ~ c · ~ x =?, is dual to another problem, AT~ y ~ c, min ~ b · ~ y =?

Linear programming duality

The linear programming problem of the form A~ x = ~ b, x 0, max ~ c · ~ x =?, is dual to another problem, AT~ y ~ c, min ~ b · ~ y =? With the property that ~ c · ~ x  ~ b · ~ y.

Linear programming duality

The linear programming problem of the form A~ x = ~ b, x 0, max ~ c · ~ x =?, is dual to another problem, AT~ y ~ c, min ~ b · ~ y =? With the property that ~ c · ~ x  ~ b · ~ y. ~ c · ~ x  ~ y · A~ x = ~ y · ~ b.

Linear programming duality

Any feasible solution to the dual problem provides an upper bound for the primal problem. In our case the dual problem is Q(∆) = y0 + X

k

yk[∆ 2∆](2k1), Q(∆) 0, 8∆ 0, Q(∆) 1, 8∆ ∆⇤, min y0 =?,

Linear programming duality

Any feasible solution to the dual problem provides an upper bound for the primal problem. In our case the dual problem is Q(∆) = y0 + X

k

yk[∆ 2∆](2k1), Q(∆) 0, 8∆ 0, Q(∆) 1, 8∆ ∆⇤, min y0 =?,

• r alternatively

Q(∆) = X

k

k[∆ 2∆](2k1), Q(∆) 1, 8∆ 0, Q(∆) Q0, 8∆ ∆⇤, min 1 Q0 + 1 =?.

Large ∆

Q(∆) = X

k

k[∆ 2∆](2k1), Q(∆) 1, 8∆ 0, Q(∆) Q0, 8∆ ∆⇤, min 1 Q0 + 1 =?.

Large ∆

Q(∆) = X

k

k[∆ 2∆](2k1), Q(∆) 1, 8∆ 0, Q(∆) Q0, 8∆ ∆⇤, min 1 Q0 + 1 =?. For large ∆ we truncate at k = n ⌧ p∆ to find an approximate truncated version (v = (∆ 2∆0)/2∆) q(v) 2 Podd

n

, q(v) 1, 8v 1, q(v) q0, 8v v⇤, min 1 q0 + 1 =?

Large ∆

q(v) 2 Podd

2n1,

q(v) 1, 8v 1, q(v) q0, 8v v⇤, min 1 q0 + 1 =?

Large ∆

q(v) 2 Podd

2n1,

q(v) 1, 8v 1, q(v) q0, 8v v⇤, min 1 q0 + 1 =? For v⇤ > 1 the solution is given by the Chebyshev polynomial, q(v) = T2n1(v).

Large ∆

q(v) 2 Podd

2n1,

q(v) 1, 8v 1, q(v) q0, 8v v⇤, min 1 q0 + 1 =? For v⇤ > 1 the solution is given by the Chebyshev polynomial, q(v) = T2n1(v). 1 F(∆⇤, 1/2)  1 1 + T2n1 ⇣ ∆∗2∆φ

2∆φ

⌘, ∆⇤ 4∆

Large ∆

1 F(∆⇤, 1/2)  1 1 + T2n1 ⇣ ∆∗2∆φ

2∆φ

⌘, ∆⇤ 4∆

Large ∆

1 F(∆⇤, 1/2)  1 1 + T2n1 ⇣ ∆∗2∆φ

2∆φ

⌘, ∆⇤ 4∆ 1 F(∆⇤, x)  2 1 + T2n1 ⇣

∆∗∆x/2 ∆x/2

⌘, ∆⇤ ∆x

Large ∆

1 F(∆⇤, 1/2)  1 1 + T2n1 ⇣ ∆∗2∆φ

2∆φ

⌘, ∆⇤ 4∆ 1 F(∆⇤, x)  2 1 + T2n1 ⇣

∆∗∆x/2 ∆x/2

⌘, ∆⇤ ∆x GFF with ∆ = 100 and T5 bound

Δ2/3=6Δϕ

Δ

1 ℱ

Other cases

The essential ingredient in the above analysis was the formula @nef (x) = [f 0(x)]nef (x) 1 + O(n21)

• .

We had = ∆ 2∆0 and f (x) = log x. More generally, the same approach works for many other cases when there is a UV-IR crossing-like equation and a large parameter limit. The analysis can be extended to

• 1. Conformal block expansion, (∆x = 2∆/p1 x)
• 2. “Scaling block” expansion in ⇢ coordinate

(∆x = 2∆/p1 x)

• 3. Large space-time dimension limit of conformal block

expansion (∆x is more complicated)

• 4. Large central charge limit of modular-invariant partition

function in CFT2 (∆⌧ = (1 + |⌧|2)c/12)

Cardy-like formula

In [PRER ’12] an asymptotic formula for the OPE coefficients was found, similar in spirit to Cardy formula. At large central charge and under an additional sparse light spectrum condition Cardy formula can be shown to work for operators ∆ ⇠ c [Hartman,Keller,Stoica ’14].

Cardy-like formula

In [PRER ’12] an asymptotic formula for the OPE coefficients was found, similar in spirit to Cardy formula. At large central charge and under an additional sparse light spectrum condition Cardy formula can be shown to work for operators ∆ ⇠ c [Hartman,Keller,Stoica ’14]. Consider a situation in which ∆ is large and the OPE coefficients with light operators are sufficiently small so that for x < 1/2 we have log G4(x) = 2∆ log x + O(1), I.e. essentially the contribution from the identity operator.

Cardy-like formula

In [PRER ’12] an asymptotic formula for the OPE coefficients was found, similar in spirit to Cardy formula. At large central charge and under an additional sparse light spectrum condition Cardy formula can be shown to work for operators ∆ ⇠ c [Hartman,Keller,Stoica ’14]. Consider a situation in which ∆ is large and the OPE coefficients with light operators are sufficiently small so that for x < 1/2 we have log G4(x) = 2∆ log x + O(1), I.e. essentially the contribution from the identity operator. It then follows from the approximate reflection symmetry that we should expect the dominant contribution for x > 1/2 to come from ∆ = ∆x = 2∆/(1 x). This implies an asymptotic formula for the OPE coefficients for operators of dimension ∆ > ∆1/2 = 4∆.

Cardy-like formula

Using the technology of [HKS ’14] one shows for ∆ > 4∆ ¯ g(s)(∆) = exp  ∆ log ✓ 1 2∆ ∆ ◆ + 2∆ log ✓ ∆ 2∆0 1 ◆ + O(∆↵

)

• ,

where ¯ g(s) is g(s) averaged over interval of size ⇠ ∆↵

with

1/2 < ↵ < 1.

Finite ∆

Chebyshev bound shows only a polynomial decay, but we know that it should be exponential in the end. Can we find a similar bound which would be exponentially small?

Finite ∆

Chebyshev bound shows only a polynomial decay, but we know that it should be exponential in the end. Can we find a similar bound which would be exponentially small? Recall the exact dual problem Q(∆) = X

k

k[∆ 2∆](2k1), Q(∆) 1, 8∆ 0, Q(∆) Q0, 8∆ ∆⇤, min 1 Q0 + 1 =?

Finite ∆

Chebyshev bound shows only a polynomial decay, but we know that it should be exponential in the end. Can we find a similar bound which would be exponentially small? Recall the exact dual problem Q(∆) = X

k

k[∆ 2∆](2k1), Q(∆) 1, 8∆ 0, Q(∆) Q0, 8∆ ∆⇤, min 1 Q0 + 1 =? Take Q(∆) / [∆ 2∆](2k1) for some optimal k. This gives (interpolation weakens the bound) 1 F(∆, 1/2)  1 1 + Γ(∆2∆φ1)Γ(2∆φ)

Γ( ∆+3

2 )Γ( ∆−1 2

)

Finite ∆

1 F(∆, 1/2)  1 1 + Γ(∆2∆φ1)Γ(2∆φ)

Γ( ∆+3

2 )Γ( ∆−1 2

)

• r, defining k(∆) = d(∆ 4∆ 3)/4e, without interpolation

1 F(∆, 1/2)  1 1 [∆2∆φ]2k(∆)+1

[2∆φ]2k(∆)+1

,

Finite ∆

1 F(∆, 1/2)  1 1 + Γ(∆2∆φ1)Γ(2∆φ)

Γ( ∆+3

2 )Γ( ∆−1 2

)

• r, defining k(∆) = d(∆ 4∆ 3)/4e, without interpolation

1 F(∆, 1/2)  1 1 [∆2∆φ]2k(∆)+1

[2∆φ]2k(∆)+1

, Asymptotically, 1 F(∆⇤, 1/2)  p 2⇡ Γ(2∆)∆

2∆φ 1

2

⇤

✓1 2 ◆∆∗

Finite ∆

1 F(∆, 1/2)  1 1 + Γ(∆2∆φ1)Γ(2∆φ)

Γ( ∆+3

2 )Γ( ∆−1 2

)

• r, defining k(∆) = d(∆ 4∆ 3)/4e, without interpolation

1 F(∆, 1/2)  1 1 [∆2∆φ]2k(∆)+1

[2∆φ]2k(∆)+1

, Asymptotically, 1 F(∆⇤, 1/2)  p 2⇡ Γ(2∆)∆

2∆φ 1

2

⇤

✓1 2 ◆∆∗ Compare to G4(1/2) (1 F(∆⇤, 1/2)) . 1 Γ(2∆ + 1)∆2∆φ

⇤

✓1 2 ◆∆∗2∆φ

Finite ∆

T3, finite-∆ bounds and GFF at ∆ = 10

4 Δϕ 5 Δϕ 6 Δϕ

Δ

1 2

1 ℱ

Summary

I Approximate reflection symmetry in spectral decomposition

Summary

I Approximate reflection symmetry in spectral decomposition I Bound on scaling dimensions of dominant operators; e.g. 4∆0

• r 2

p 2∆0 for x = 1/2.

Summary

I Approximate reflection symmetry in spectral decomposition I Bound on scaling dimensions of dominant operators; e.g. 4∆0

• r 2

p 2∆0 for x = 1/2.

I A version of Cardy formula for weakly coupled light spectrum

Summary

I Approximate reflection symmetry in spectral decomposition I Bound on scaling dimensions of dominant operators; e.g. 4∆0

• r 2

p 2∆0 for x = 1/2.

I A version of Cardy formula for weakly coupled light spectrum I Non-asymptotic convergence bound

Summary

I Approximate reflection symmetry in spectral decomposition I Bound on scaling dimensions of dominant operators; e.g. 4∆0

• r 2

p 2∆0 for x = 1/2.

I A version of Cardy formula for weakly coupled light spectrum I Non-asymptotic convergence bound

Summary

I Approximate reflection symmetry in spectral decomposition I Bound on scaling dimensions of dominant operators; e.g. 4∆0

• r 2

p 2∆0 for x = 1/2.

I A version of Cardy formula for weakly coupled light spectrum I Non-asymptotic convergence bound

Questions

I What about spin? (Transverse derivatives)

Summary

I Approximate reflection symmetry in spectral decomposition I Bound on scaling dimensions of dominant operators; e.g. 4∆0

• r 2

p 2∆0 for x = 1/2.

I A version of Cardy formula for weakly coupled light spectrum I Non-asymptotic convergence bound

Questions

I What about spin? (Transverse derivatives) I Virasoro symmetry?

Summary

I Approximate reflection symmetry in spectral decomposition I Bound on scaling dimensions of dominant operators; e.g. 4∆0

• r 2

p 2∆0 for x = 1/2.

I A version of Cardy formula for weakly coupled light spectrum I Non-asymptotic convergence bound

Questions

I What about spin? (Transverse derivatives) I Virasoro symmetry? I Can we solve the case of scaling blocks exactly? (e.g. the tail

bound)

Summary

I Approximate reflection symmetry in spectral decomposition I Bound on scaling dimensions of dominant operators; e.g. 4∆0

• r 2

p 2∆0 for x = 1/2.

I A version of Cardy formula for weakly coupled light spectrum I Non-asymptotic convergence bound

Questions

I What about spin? (Transverse derivatives) I Virasoro symmetry? I Can we solve the case of scaling blocks exactly? (e.g. the tail

bound)

I Or at least guess the result?

Conformal block for large ∆

We can find the form of the conformal block on real line x = ¯ x in the limit of large intermediate scaling dimension using the quartic Casimir equation with WKB-like approximation. This gives F∆,`(x) =(1 ⇢2)✏1(4⇢)∆ ⇥ exp  1 ∆ ⇢2 1 ⇢2 (1 + ✏ ✏2)∆2 + ✏(✏ 1)`2 ∆2 `2 + O(∆2)

• where

✏ = d 2 2 , ⇢ = x (1 + p1 x)2

Large spacetime dimension

Unitarity bounds ∆ = 0

• r

∆ d 2 2 ⇠ d 2

Large spacetime dimension

Unitarity bounds ∆ = 0

• r

∆ d 2 2 ⇠ d 2 Reflection symmetry then implies ∆ = 0

• r

d 2  ∆  ∆x

• r

∆ = ∆0

x

Large spacetime dimension

Unitarity bounds ∆ = 0

• r

∆ d 2 2 ⇠ d 2 Reflection symmetry then implies ∆ = 0

• r

d 2  ∆  ∆x

• r

∆ = ∆0

x

For ∆ = d

2 , ∆x = d 2 ) resembles free fied

Saturation of tail bound

2d global conformal blocks at Ising point ∆ = 1/8

2 4 6 8 10 10-5 10-4 0.001 0.010 0.100 1