Holographic Wilson Loops and Topological Insulators Andy OBannon - PowerPoint PPT Presentation

Jumping coupling θ L = − 1 4 g 2 tr F µ ν F µ ν + 32 π 2 ǫ µ νρσ tr F µ ν F ρσ − 1 1 2 g 2 tr( D µ Φ i D µ Φ i ) + 4 g 2 tr([ Φ i , Φ j ][ Φ i , Φ j ]) g + g ( z ) g − z z = 0

Jumping coupling θ L = − 1 4 g 2 tr F µ ν F µ ν + 32 π 2 ǫ µ νρσ tr F µ ν F ρσ − 1 1 2 g 2 tr( D µ Φ i D µ Φ i ) + 4 g 2 tr([ Φ i , Φ j ][ Φ i , Φ j ]) Breaks all SUSY Preserves SO(3,2) x SO(6) “Conformal Interface”

� � � � � � � � � � � � � � � � � � Dielectric Interfaces � � L = 1 E 2 − 1 ε � � B 2 8 π µ ε ∝ 1 /g 2 Image charges! � � � �

SL (2 , R ) τ = C 0 + ie − 2 φ τ → a τ + b a, b, c, d ∈ R c τ + d ab − cd = 1 Jumping axion Jumping dilaton

SL (2 , R ) τ = θ 2 π + i 2 π g 2 τ → a τ + b a, b, c, d ∈ R c τ + d ab − cd = 1 Jumping coupling Jumping θ - angle Topological Insulator

Outline: • Motivation: Topological Insulators • Holographic Conformal Interfaces • Holographic Wilson Loops • Static Quark Potential • Future Directions

Wilson Loops in N = 4 SYM �� �� W R [ C ] = 1 x µ + Φ i θ i | ˙ � tr R P exp iA µ ˙ x | ds N c C of SU ( N c ) R = N c T C =

Wilson Loops in N = 4 SYM �� �� W R [ C ] = 1 x µ + Φ i θ i | ˙ � tr R P exp iA µ ˙ x | ds N c C of SU ( N c ) R = N c T C = L

Wilson Loops in N = 4 SYM �� �� W R [ C ] = 1 x µ + Φ i θ i | ˙ � tr R P exp iA µ ˙ x | ds N c C 1 V ( L ) = − lim T ln � W [ C ] � T →∞ V ( L ) = f ( λ ) L

Wilson Loops in N = 4 SYM λ ≪ 1 Perturbatively � W [ C ] � = 1 − N c � � x µ ( s ) ˙ d ˜ s ( ˙ x ν (˜ s ) � PA µ ( x ( s )) A ν ( x ( s )) � ds 2 C C s ) | θ i θ j � P Φ i ( x ( s )) Φ j ( x (˜ � − | ˙ x ( s ) || ˙ x (˜ s )) � + . . .

Wilson Loops in N = 4 SYM Sum “ladder” diagrams � + · · · 1 + +

Wilson Loops in N = 4 SYM λ ≫ 1 Holographically r = ∞ C Maldacena r = 0 b. Rey and Yee 1998

Wilson Loops in N = 4 SYM � 1 � d 2 σ S NG = − − det g 2 πα ′ S NG | solution = A � W [ C ] � ∝ e − A 1 A V ( L ) = − lim T ln � W [ C ] � = lim T T →∞ T →∞

Wilson Loops in N = 4 SYM S NG | solution C diverges at r = ∞ b. Infinite self-energy

Wilson Loops in N = 4 SYM Drukker, Gross, Ooguri 1999 Legendre transform � � � A = S NG − d σ P r r � � ∂ AdS

Wilson Loops in N = 4 SYM “Straight string” Legendre transform cancels the divergence! A = 0 � W [ C ] � = 1 ⇒ due to SUSY

N = 4 SYM jumping g Wilson Loops from AdS 5 C b. Legendre Subtracting Subtracting transform = = self-energy straight string

Holography Maldacena Rey and Yee 1998 √ 4 π 2 2 λ V ( L ) = − λ ≫ 1 Γ (1 / 4) 4 L Ladder Diagrams Erickson Semenoff Zarembo 2000  2 λ − 1 λ ≪ 1 L , 4 π   V ( L ) = √ 2 λ  − 1 λ ≫ 1 . L ,  π

Wilson Loops in N = 4 SYM Conformal Interface z z X 3 X 3 L left right z left z right X X 3 3 D

z X 3 L D V ( L, D ) = f ( λ , D/L ) L

Wilson Loops in N = 4 SYM Conformal Interface Perturbatively � PA µ ( x ( s )) A ν ( x ( s )) � acquires image terms � P Φ i ( x ( s )) Φ j ( x (˜ s ) � unchanged Clark, Freedman, Karch, Schnabl 2004

Wilson Loops in N = 4 SYM Conformal Interface Holographically 4 2 -1 1 2 3 4 5 -2 -4

Outline: • Motivation: Topological Insulators • Holographic Conformal Interfaces • Holographic Wilson Loops • Static Quark Potential • Future Directions

Electromagnetism g II g I + z − z ˜ Q Q Q = Qg 2 II − g 2 ˜ I g 2 II + g 2 I

Electromagnetism g II g I + z − z ˜ Q Q attracted to side with SMALLER coupling Q

Electromagnetism g 2 I > g 2 Q = +1 II V g 2 I / 4 π g 2 g 2 II I z

N = 4 SYM jumping g g II g I + z − z ˜ Q Q C � W [ C ] � with straight-line

N = 4 SYM jumping g Straight string in Janus � � � A = S NG − d σ P r r � � ∂ AdS (!) A � = 0 A V = lim T � = 0 T →∞ Interaction energy with image charge

N = 4 SYM jumping g V � � 2 √ λ I λ II 2 π e 2 g 2 g 2 z

Electromagnetism g II g I + z − z D D ˜ Q 1 Q 1 = +1 L ˜ Q 2 Q 2 = − 1 Q = Qg 2 II − g 2 ˜ I g 2 II + g 2 I

Electromagnetism g II g I + z − z D D ˜ Q 1 Q 1 = +1 L ˜ Q 2 Q 2 = − 1 � � + Q 1 ˜ + Q 2 ˜ Q 1 ˜ V = g 2 Q 1 Q 2 Q 1 Q 2 Q 2 I + L 2 + 4 D 2 √ L 2 D 2 D 4 π

Electromagnetism g II g I + z − z D D ˜ Q 1 Q 1 = +1 L ˜ Q 2 Q 2 = − 1 Again attracted to side with SMALLER coupling

Electromagnetism V L g 2 I / 4 π II = 1 g 2 g 2 2 g 2 3 I I z 1 L � 2 � 1 1 2 � 1 � 3 � 5

Electromagnetism V L I / 4 π − (images) g 2 II = 1 g 2 g 2 2 g 2 1.0 I I 0.5 z � 10 � 5 5 10 L � 0.5 � 1.0 � 1.5 � 2.0

N = 4 SYM jumping g g II g I + z − z D D ˜ Q 1 Q 1 = +1 L ˜ Q 2 Q 2 = − 1 C � W [ C ] � with rectangular

N = 4 SYM jumping g V L � � 2 √ λ I λ II 2 π g 2 e 2 g 2 5 z � 4 � 2 2 4 L � 5

N = 4 SYM jumping g V L − (images) 2 √ λ I λ II � � 2 π e 2 g 2 g 2 � 1.0 � 1.2 � 1.4 � 1.6 � 1.8 � 2.0 z � 2.2 L � 4 � 2 2 4

Electromagnetism θ I θ II + z − z ( ˜ Q e , ˜ Q m ) Q − Q g 4 ( θ II − θ I ) 2 − 4 π Q g 2 ( θ II − θ I ) ˜ ˜ Q m = Q e = 16 π 2 + g 4 ( θ II − θ I ) 2 16 π 2 + g 4 ( θ II − θ I ) 2

Electromagnetism θ I θ II + z − z ( ˜ Q e , ˜ Q m ) Q − Q g 4 ( θ II − θ I ) 2 − 4 π Q g 2 ( θ II − θ I ) ˜ ˜ Q m = Q e = 16 π 2 + g 4 ( θ II − θ I ) 2 16 π 2 + g 4 ( θ II − θ I ) 2

Electromagnetism θ I θ II + z − z ( ˜ Q e , ˜ Q m ) Q Interface always attractive!

Electromagnetism V g 2 / 4 π z

N = 4 SYM jumping θ θ I θ II + z − z ( ˜ Q e , ˜ Q m ) Q − Q g 4 ( θ II − θ I ) 2 − 4 π Q g 2 ( θ II − θ I ) ˜ ˜ Q m = Q e = 16 π 2 + g 4 ( θ II − θ I ) 2 16 π 2 + g 4 ( θ II − θ I ) 2

N = 4 SYM jumping θ θ I θ II + z − z ( ˜ Q e , ˜ Q m ) Q ⎷ − Q g 4 ( θ II − θ I ) 2 − 4 π Q g 2 ( θ II − θ I ) ˜ ˜ Q m = Q e = 16 π 2 + g 4 ( θ II − θ I ) 2 16 π 2 + g 4 ( θ II − θ I ) 2

N = 4 SYM jumping θ V √ 2 λ / 2 π z

Electromagnetism θ I θ II + z − z D D ( ˜ 1 , ˜ 1 ) Q 1 = +1 Q e Q m L ( ˜ 2 , ˜ 2 ) Q e Q m Q 2 = − 1 Interface always attractive!

Electromagnetism V L g 2 / 4 π z � 1 � 0.5 0.5 1 L � 10 g 2 ∆ θ = 10 2 � 20 � 30

Electromagnetism V L g 2 / 4 π − (images) z � 15 � 10 � 5 5 10 15 L � 0.2 � 0.4 g 2 ∆ θ = 10 2 � 0.6 � 0.8 � 1

N = 4 SYM jumping θ θ I θ II + z − z D D ( ˜ 1 , ˜ 1 ) Q 1 = +1 Q e Q m L ( ˜ 2 , ˜ 2 ) Q e Q m Q 2 = − 1 − Q g 4 ( θ II − θ I ) 2 − 4 π Q g 2 ( θ II − θ I ) ˜ ˜ Q m = Q e = 16 π 2 + g 4 ( θ II − θ I ) 2 16 π 2 + g 4 ( θ II − θ I ) 2

N = 4 SYM jumping θ V L θ = π θ = 0 √ � 2 λ 2 π z � 1.45 L � 1.50 � 1.55 � 1.60 � 4 � 2 2 4

N = 4 SYM jumping θ V L − (images) √ 2 λ / 2 π � 1.405 θ = 0 θ = π � 1.410 � 1.415 � 1.420 � 1.425 z � 1.430 L � 4 � 2 2 4

N = 4 SYM jumping θ V L − (images) √ 2 λ / 2 π θ = 0 θ = π � 1.4056 � 1.4058 � 1.4060 � 1.4062 � 1.4064 � 1.4066 z � 1.4068 � 0.2 � 0.1 0.1 0.2 L