Null Hypersurface Quantization and Asympotic Symmetries of Maxwell - PowerPoint PPT Presentation

East Asia Joint Workshop on Fields and Strings 2017, KEK Theory workshop 2017 Null Hypersurface Quantization and Asympotic Symmetries of Maxwell theory LING YAN HUNG, Fudan University In collaboration with Arpan Bhattacharyya, Yikun Jiang, arXiv: 1708.05606

Outline • Overview: quantization at asymptotic infinity via the Schwinger brackets and subtleties • Introducing an IR regulator • Some example applications • Future direction

Overview based on Frolov ; Strominger et al arXiv: 1407.37892 • quantization at future null-infinity J + ds 2 = − du 2 − 2 dudr + 2 r 2 γ z ¯ z dzd ¯ z 2 u = t − r γ z ¯ z = (1+ z ¯ z ) 2 Future null infinity located at r = ∞

Overview • The Schwinger bracket • Writing the action in first order form W [ φ , φ µ ] = 1 R K µ ν ( φ µ ∂ ν φ − φ µ φ ν ) − φ∂ µ ( K µ ν φ ν )) 2 φ µ = ∂ µ φ K µ ν = √− gg µ ν δ W = G Σ 2 − G Σ 1 G Σ = 1 R Σ g µ ν ( φ µ δφ − φδφ µ ) d Σ ν 2

Overview • The Schwinger bracket G Σ = 1 R Σ g µ ν ( φ µ δφ − φδφ µ ) d Σ ν 2 [ φ , G Σ ] = i [ φ µ , G Σ ] = i 2 δφ , 2 δφ µ In a space-like surface, this is equivalent to the usual canonical quantization. φ , φ µ On a null surface, are not independent

Overview • Write 1 Z √ g ( ∂ µ φδφ − φ∂ µ φ ) dud 2 x = G N 2 Σ N Z ∂ µ ˜ φδ ˜ φ dud 2 x − G b = Σ N � u 1 d 2 x ˜ φ δ ˜ δ ˜ G b = 1 � � R φ φ u 1 ,u 0 = 0 removed by taking � 2 u 0

Maxwell theory • If we do Maxwell theory, the corresponding statement is: Σ N dudx 2 ( δ A z ∂ u A ¯ R G Σ = z + δ A ¯ z ∂ u A z ) + G b � u =+ ∞ � G b = − 1 R dz ( A z δ A ¯ z + A ¯ z δ A z d ¯ z ) 2 u = −∞ A r = 0 With boundary conditions: ( in retarded radial gauge ) r → ∞ A u = 1 z ) + O ( r − 1 ) , A z = A z ( u, z, ¯ r A u ( u, z, ¯ z ) + O ( r − 2) R J + F uz F z Radiation flux is non-zero and finite u

Maxwell theory • If we set the boundary terms to zero z , G Σ ] = i [ A z, ¯ 2 δ A z, ¯ z One recovers the known brackets w )] = − i 4 Θ ( u − u 0 ) δ 2 ( z − w ) [ A z ( u, z, ¯ w ( u, w, ¯ z, A ¯ z ) , A w ( u 0 , w, ¯ [ A z ( u, z, ¯ w )] = [ A ¯ w ] = 0 z , A ¯

Subtleties! • It is observed by Strominger et al that the brackets do not reproduce the expected commutator of the charge Q that generates large gauge transformation � ✏ A z = @ z ✏ ( z, ¯ z ) J + dud 2 z ✏ [ @ u ( @ z A ¯ R Q = z + @ ¯ z A z )] Modification of brackets : imposing z | J + , u → ± ∞ = 0 F z ¯ trick: new brackets obtained by taking limits of the previous brackets

Subtleties! trick: new brackets obtained by taking limits of the previous brackets and introducing a pure gauge mode A z ( u → ± ∞ ) = ∂ z φ ± 1) Q is a charge that depends on the boundary values of A. It’s not obvious what we should do for other charges that do not localize at the boundary 2) Puzzled that the boundary conditions consistent with ignoring Gb is not natural z ( u = + ∞ ) + δ A z, ¯ z ( u = −∞ ) = 0 δ A z, ¯

IR regularization • This was first considered by Pasterski 
 • Introduce a cuto ff for u. − T 2 ≤ u ≤ T 2 z ) ( u = − T z ) ( u = + T 2 ) = F uz (¯ 2 ) F uz (¯ α m ) e i (2 π mu ) /T z ) ( u ) = P ∞ m = −∞ α m (¯ F uz (¯ Periodic boundary conditions - that the boundary field values remain unchanged

IR regularization α m ) e i (2 π mu ) /T z ) ( u ) = P ∞ m = −∞ α m (¯ F uz (¯ A z ( u ) = d 0 + α 0 u + P 1 T 2 π mi α m e i (2 π mu ) /T m = 6 =0 z ( u ) = ¯ α 0 u + P 1 T α m e i (2 π mu ) /T d 0 + ¯ 2 π mi ¯ A ¯ m = 6 =0 α n , d 0 , ¯ are functions of z, ¯ α n , ¯ d 0 z

Schwinger brackets revisited Substitute the expansion into G Σ Now there is no natural distinction between bulk and boundary terms Apply the Schwinger procedure mode by mode [ d 0 , G Σ ] = i [ α n , G Σ ] = i 2 δ d 0 , 2 δα n There are more equations than unknowns.

Schwinger brackets revisited There is a set of consistent solutions. Obtain a set of brackets: [ ¯ w )] = i T δ 2 ( z − w ) d 0 ( z, ¯ z ) , α 0 ( w, ¯ w )] = [ d 0 ( z, ¯ z ) , ¯ α 0 ( w, ¯ w )] = − m π T 2 δ m + n, 0 δ 2 ( z − w ) [ α m ( z, ¯ z ) , ¯ α n ( w, ¯ [ ¯ i 2 T ( − 1) m δ 2 ( z − w ) d 0 ( z, ¯ z ) , α m ( w, ¯ w )] = [ d 0 ( z, ¯ z ) , ¯ α m ( w, ¯ w )] =

Applications Test 1: Large gauge transformation R Q = T dzd ¯ z (¯ ↵ 0 @ z ✏ + ↵ 0 @ ¯ z ✏ ) [ A z, ¯ z ( u, z, ¯ z ) , Q ] = i @ z ✏ = i � A z, ¯ z Curiously, it gives the correct large gauge transformation without the factor of 2 issue before imposing the constraints on F z ¯ z

Applications Test 1: Large gauge transformation R Q = T dzd ¯ z (¯ ↵ 0 @ z ✏ + ↵ 0 @ ¯ z ✏ ) [ A z, ¯ z ( u, z, ¯ z ) , Q ] = i @ z ✏ = i � A z, ¯ z Now with an IR regularisation, we can impose constraints systematically. and obtain the corresponding Imposing at the boundary points F z ¯ z = 0 Dirac brackets [ F, G ] D = [ F, G ] − [ F, φ a ] C ab [ φ b , G ] , C ab = [ φ a , φ b ]

Applications Test 1: Large gauge transformation [ F, G ] D = [ F, G ] − [ F, φ a ] C ab [ φ b , G ] , C ab = [ φ a , φ b ] φ 1 = ∂ z ¯ α 0 − ∂ ¯ z α 0 φ 2 = ∂ z ( ¯ ( � 1) m T ( � 1) m T d 0 + P i 2 π m ¯ α m ) − ∂ ¯ z ( d 0 + P i 2 π m α m ) m 6 =0 m 6 =0 C 12 = − C 21 = i z ∂ w ) δ 2 ( z − w ) T ( ∂ z ∂ ¯ w + ∂ ¯

Applications Test 1: Large gauge transformation w )] D = − ( − 1) m i [ d 0 ( z, ¯ z ) , α m ( w, ¯ 4 π T ( z − w ) 2 w )] D = − i z ) , ¯ 2 T δ 2 ( z − w ) [ α 0 ( z, ¯ d 0 ( w, ¯ w )] D = 0 [ d 0 ( z, ¯ z ) , ¯ α m ( w, ¯ It does not change the commutation between Q and A!

Applications Test 1: Large gauge transformation To make comparison with Strominger et al: T ( � 1) m ∂ z φ + = d 0 + T 2 α 0 + P i 2 π m α m m 6 =0 T ( � 1) m z φ � = ¯ d 0 − T α 0 + P ∂ ¯ α m 2 ¯ i 2 π m ¯ m 6 =0 i 4 π ln | z − w | 2 [ φ + ( z, ¯ z ) , φ − ( w, ¯ w )] = Exact agreement!

Applications Test 2: Electromagnetic duality It was observed that the conserved charge of electromagnetic duality in free Maxwell theory coincides with the helicity operator of photons. D. Zwanziger 
 We wanted to know what it is like at null infinity

Applications Test 2: Electromagnetic duality Strategy: 1) Work with the Schwarz-Sen action that makes the electromagnetic duality explicitly a symmetry of the action 2) Quantize the theory at null infinity and obtain the commutator of the conserved charge

Applications Test 2: Electromagnetic duality z [ F 2 + G 2 ] , 2 ✏ µ νρσ F ρσ = G µ ν S = − 1 1 d 4 x r 2 � z ¯ R 8 This has to be modified to remove extra degrees of freedom, by imposing the duality condition as a constraint. We follow a standard method introduced in Schwarz- Sen. Z d 4 x S = [ − i/ 2( G z ¯ z F ru − G u ¯ z F rz + G uz F r ¯ z − F z ¯ z G ru + F u ¯ z G rz − F uz G r ¯ z ) + 2( F 2 + G 2 1 z ¯ z ¯ z z ) + F uz F u ¯ z + r 2 γ z ¯ r 2 γ z ¯ z z G uz G u ¯ z − iF uz G u ¯ z + iF u ¯ z G uz ]

Applications Test 2: Electromagnetic duality The duality condition shows up as equations of motion of this action. Interestingly, at the asymptotic null infinity, it reduces to ∂ u ( C z − iA z ) = 0 ∂ u ( C ¯ z + iA ¯ z ) = 0 z + ¯ C z = iA z + f, C ¯ z = − iA ¯ f

Applications Test 2: Electromagnetic duality The electromagnetic duality symmetry transformation reduces to: δ A z = θ C z , δ A ¯ z = θ C ¯ z δ C z = − θ A z , δ C ¯ z = − θ A ¯ z

Applications Test 2: Electromagnetic duality The Noether charge is given by Z α 0 d 0 T − α 0 ¯ d 2 z [¯ Q = d 0 T + i θ T 2 i ( − 1) m T 2 X X i π n α n ¯ ( α 0 ¯ α m − ¯ α 0 α m )] α � n − 2 π m n 6 =0 m 6 =0 + θ Z α 0 + ¯ d 2 z [ T ( f ¯ f α 0 )] . 2

Applications Test 2: Electromagnetic duality [ α n , Q ] = − θα n , [¯ α n , Q ] = + θ ¯ α n Q continues to behave like the helicity operator, giving the correct helicity of these modes . Unfortunately, the algebra doesn’t get further extended.

Summary • Demonstrate the virtue of an IR regulator that makes quantization and handling of zero modes more transparent • We demonstrate in a few simple examples how it works • Applied in the discussion of the BMS algebra