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 CFT d | r i = � ( r ) | � i , x = r 2 or the four point function on the real line with x = ¯ G 4 ( x ) = h � | � (1) � ( x , ¯ x ) | � i / h r | r i .

The problem Consider the state in a Euclidean CFT d | r i = � ( r ) | � i , x = r 2 or the four point function on the real line with x = ¯ G 4 ( x ) = h � | � (1) � ( x , ¯ x ) | � i / h r | r i . Using � ⇥ � OPE, one can decompose X | r i = C �� O V O ( r ) |O i O X C 2 �� O x � 2 ∆ φ F ∆ , ` ( x ) / X C 2 �� O h O| V † G 4 ( x ) = O ( r ) V O ( r ) |O i O O

The problem Suppose we want to understand X | r i = C �� O V O ( r ) |O i . O

The problem Suppose we want to understand X | r i = C �� O V O ( r ) |O i . O Consider the CDF = 1 � G ∆ ∗ F ( ∆ ⇤ , x ) = h r | P ∆ O < ∆ ∗ | r i ( x ) 4 G 4 ( x ) , h r | r i where X G ∆ ∗ C 2 �� O x � 2 ∆ φ F ∆ , ` ( x ) ( x ) = 4 O , ∆ O > ∆ ∗ [Pappadopulo,Rychkov,Espin,Rattazzi ’12; Rychkov,Yvernay ’15]

A simpler problem Ignore conformal symmetry, use only scaling symmetry ! “scaling blocks”, Z 1 �� O x ∆ O � 2 ∆ φ = X C 2 x ∆ � 2 ∆ φ g ( s ) ( ∆ ) d ∆ . G 4 ( x ) = 0 O

A simpler problem Ignore conformal symmetry, use only scaling symmetry ! “scaling blocks”, Z 1 �� O x ∆ O � 2 ∆ φ = X C 2 x ∆ � 2 ∆ φ g ( s ) ( ∆ ) d ∆ . G 4 ( x ) = 0 O From [PRER ’12], in a given theory, for su ffi ciently large ∆ ⇤ 1 Γ (2 ∆ � + 1) ∆ 2 ∆ φ G ∆ ∗ x ∆ ∗ � 2 ∆ φ . ( x ) . 4 ⇤ Can we obtain more information on the structure of F ( ∆ ⇤ , x )?

Crossing symmetry Using a di ff erent channel for OPE expansion one finds G 4 ( x ) = G 4 (1 � x ) , and so @ n G 4 ( x ) = ( � @ ) n G 4 (1 � x ) .

Crossing symmetry Using a di ff erent channel for OPE expansion one finds G 4 ( x ) = G 4 (1 � x ) , and so @ n G 4 ( x ) = ( � @ ) n G 4 (1 � x ) . At x = 1 / 2 one obtains Z 1 [ ∆ � 2 ∆ � ] (2 k +1) � ( s ) 1 / 2 ( ∆ ) d ∆ = 0 , 0 [ ↵ ] ( n ) = x � ↵ + n @ n x ↵ = ↵ ( ↵ � 1) . . . ( ↵ � n + 1) � ( s ) x ( ∆ ) = x ∆ � 2 ∆ φ g ( s ) ( ∆ )

Crossing symmetry Suppose ∆ � � 1 in Z 1 [ ∆ � 2 ∆ � ] (2 k +1) � ( s ) 1 / 2 ( ∆ ) d ∆ = 0 , 0 Then for k ⌧ p ∆ � approximate [ ∆ � 2 ∆ � ] (2 k +1) ' ( ∆ � 2 ∆ � ) 2 k +1 , Z 1 w 2 k +1 � ( s ) 1 / 2 ( w + 2 ∆ � ) dw ' 0 . � 2 ∆ φ 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 ∆ � � ∆ ). � ( s ) R Normalize 1 / 2 ( ∆ ) d ∆ = 1. R ∆ ∗ � ( s ) Then F ( ∆ ⇤ , 1 / 2) = 1 / 2 ( ∆ ) d ∆ is antisymmetric up to a 0 constant.

Reflection symmetry Suppose the symmetry is exact, � ( s ) 1 / 2 ( ∆ ) = � ( s ) 1 / 2 (4 ∆ � � ∆ ). � ( s ) R Normalize 1 / 2 ( ∆ ) d ∆ = 1. R ∆ ∗ � ( s ) Then F ( ∆ ⇤ , 1 / 2) = 1 / 2 ( ∆ ) d ∆ is antisymmetric up to a 0 constant. ℱ 1 1 2 Δ 2 Δ ϕ 4 Δ ϕ

Reflection symmetry Suppose the symmetry is exact, � ( s ) 1 / 2 ( ∆ ) = � ( s ) 1 / 2 (4 ∆ � � ∆ ). � ( s ) R Normalize 1 / 2 ( ∆ ) d ∆ = 1. R ∆ ∗ � ( s ) Then F ( ∆ ⇤ , 1 / 2) = 1 / 2 ( ∆ ) d ∆ is antisymmetric up to a 0 constant. ℱ 1 1 2 Δ 2 Δ ϕ 4 Δ ϕ

Reflection symmetry Suppose the symmetry is exact, � ( s ) 1 / 2 ( ∆ ) = � ( s ) 1 / 2 (4 ∆ � � ∆ ). � ( s ) R Normalize 1 / 2 ( ∆ ) d ∆ = 1. R ∆ ∗ � ( s ) Then F ( ∆ ⇤ , 1 / 2) = 1 / 2 ( ∆ ) d ∆ is antisymmetric up to a 0 constant. ℱ 1 1 2 Δ 2 Δ ϕ 4 Δ ϕ G 4 ( x ) = x 2 ∆ φ � 2 ∆ φ or G 4 ( x ) = x � 2 ∆ φ + (1 � x ) � 2 ∆ φ

Reflection symmetry For general x the reflection is between � ( s ) and � ( s ) 1 � x , relating x ∆ � 2 ∆ � $ � ∆ � 2 ∆ � x 1 � x

Reflection symmetry For general x the reflection is between � ( s ) and � ( s ) 1 � x , relating x ∆ � 2 ∆ � $ � ∆ � 2 ∆ � x 1 � x Let x > 1 / 2, ∆ x = 2 ∆ φ 1 � x . Then � ( s ) 1 � x ( ∆ ) = 0 for ∆ < 0 ) � ( s ) x ( ∆ ) ' 0 for ∆ � ∆ x . ℱ 1 Δ Δ 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 G 4 ( 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 G 4 ( x ) . @ log x G 4 ( x ) = G 4 (1 � x ) ) ∆ ( x ) � 2 ∆ � = � ∆ (1 � x ) � 2 ∆ � x 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 0 Z 1 [ ∆ � 2 ∆ � ] (2 k +1) � ( s ) 1 / 2 ( ∆ ) d ∆ = 0 0 Z 1 � ( s ) 1 / 2 ( ∆ ) d ∆ =? max ∆ ∗

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 0 Z 1 [ ∆ � 2 ∆ � ] (2 k +1) � ( s ) 1 / 2 ( ∆ ) d ∆ = 0 0 Z 1 � ( s ) 1 / 2 ( ∆ ) d ∆ =? max ∆ ∗ This is of the form x = ~ A ~ b , x � 0 max ~ c · ~ x =?

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

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

Linear programming duality The linear programming problem of the form x = ~ A ~ b , x � 0 , max ~ c · ~ x =? , is dual to another problem, A T ~ y � ~ c , min ~ b · ~ y =? With the property that x  ~ c · ~ b · ~ y . ~ y · ~ c · ~ x  ~ y · A ~ x = ~ 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 X y k [ ∆ � 2 ∆ � ] (2 k � 1) , Q ( ∆ ) = y 0 + k Q ( ∆ ) � 0 , 8 ∆ � 0 , Q ( ∆ ) � 1 , 8 ∆ � ∆ ⇤ , min y 0 =? ,

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 X y k [ ∆ � 2 ∆ � ] (2 k � 1) , Q ( ∆ ) = y 0 + k Q ( ∆ ) � 0 , 8 ∆ � 0 , Q ( ∆ ) � 1 , 8 ∆ � ∆ ⇤ , min y 0 =? , or alternatively X � k [ ∆ � 2 ∆ � ] (2 k � 1) , Q ( ∆ ) = k Q ( ∆ ) � � 1 , 8 ∆ � 0 , Q ( ∆ ) � Q 0 , 8 ∆ � ∆ ⇤ , 1 Q 0 + 1 =? . min

Large ∆ � X � k [ ∆ � 2 ∆ � ] (2 k � 1) , Q ( ∆ ) = k Q ( ∆ ) � � 1 , 8 ∆ � 0 , Q ( ∆ ) � Q 0 , 8 ∆ � ∆ ⇤ , 1 Q 0 + 1 =? . min

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

Large ∆ � q ( v ) 2 P odd 2 n � 1 , q ( v ) � � 1 , 8 v � � 1 , q ( v ) � q 0 , 8 v � v ⇤ , 1 q 0 + 1 =? min

Large ∆ � q ( v ) 2 P odd 2 n � 1 , q ( v ) � � 1 , 8 v � � 1 , q ( v ) � q 0 , 8 v � v ⇤ , 1 q 0 + 1 =? min For v ⇤ > 1 the solution is given by the Chebyshev polynomial, q ( v ) = T 2 n � 1 ( v ).

Large ∆ � q ( v ) 2 P odd 2 n � 1 , q ( v ) � � 1 , 8 v � � 1 , q ( v ) � q 0 , 8 v � v ⇤ , 1 q 0 + 1 =? min For v ⇤ > 1 the solution is given by the Chebyshev polynomial, q ( v ) = T 2 n � 1 ( v ). 1 1 � F ( ∆ ⇤ , 1 / 2)  ⌘ , ∆ ⇤ � 4 ∆ � ⇣ ∆ ∗ � 2 ∆ φ 1 + T 2 n � 1 2 ∆ φ

Large ∆ � 1 1 � F ( ∆ ⇤ , 1 / 2)  ⌘ , ∆ ⇤ � 4 ∆ � ⇣ ∆ ∗ � 2 ∆ φ 1 + T 2 n � 1 2 ∆ φ

Large ∆ � 1 1 � F ( ∆ ⇤ , 1 / 2)  ⌘ , ∆ ⇤ � 4 ∆ � ⇣ ∆ ∗ � 2 ∆ φ 1 + T 2 n � 1 2 ∆ φ 2 1 � F ( ∆ ⇤ , x )  ⌘ , ∆ ⇤ � ∆ x ⇣ ∆ ∗ � ∆ x / 2 1 + T 2 n � 1 ∆ x / 2