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

Null Hypersurface Quantization and Asympotic Symmetries of Maxwell theory

LING YAN HUNG, Fudan University In collaboration with Arpan Bhattacharyya, Yikun Jiang, arXiv: 1708.05606

East Asia Joint Workshop on Fields and Strings 2017, KEK Theory workshop 2017

• Overview: quantization at asymptotic infinity via the

Schwinger brackets and subtleties

• Introducing an IR regulator
• Some example applications
• Future direction

Outline

Overview

• quantization at future null-infinity

ds2 = −du2 − 2dudr + 2r2γz¯

zdzd¯

z J + u = t − r γz¯

z = 2 (1+z¯ z)2

Future null infinity located at r = ∞

based on Frolov ; Strominger et al arXiv: 1407.37892

Overview

• The Schwinger bracket
• Writing the action in first order form

W[φ, φµ] = 1

2

R Kµν(φµ∂νφ − φµφν) − φ∂µ(Kµνφν)) δW = GΣ2 − GΣ1 GΣ = 1

2

R

Σ gµν(φµδφ − φδφµ)dΣν

φµ = ∂µφ

Kµν = √−ggµν

Overview

• The Schwinger bracket

GΣ = 1

2

R

Σ gµν(φµδφ − φδφµ)dΣν

[φ, GΣ] = i

2δφ,

[φµ, GΣ] = i

2δφµ

In a space-like surface, this is equivalent to the usual canonical quantization. On a null surface, are not independent

φ, φµ

Overview

• Write

GN = 1 2 Z

ΣN

√g(∂µφδφ − φ∂µφ) dud2x = Z

ΣN

∂µ ˜ φδ ˜ φ dud2x − Gb Gb = 1

2

R d2x ˜ φ δ ˜ φ

• u1

u0

removed by taking

δ ˜ φ

• u1,u0 = 0

Maxwell theory

• If we do Maxwell theory, the corresponding statement is:

GΣ = R

ΣN dudx2(δAz∂uA¯ z + δA¯ z∂uAz) + Gb

Gb = − 1

2

R dz(AzδA¯

z + A¯ zδAzd¯

z)

• u=+∞

u=−∞

With boundary conditions:

Az = Az(u, z, ¯ z) + O(r−1), Au = 1

rAu(u, z, ¯

z) + O(r−2) r → ∞

( in retarded radial gauge )

Ar = 0

Radiation flux

R

J + FuzF z u

is non-zero and finite

Maxwell theory

• If we set the boundary terms to zero

[Az,¯

z, GΣ] = i 2δAz,¯ z

One recovers the known brackets

[Az(u, z, ¯ z, A ¯

w(u, w, ¯

w)] = − i

4Θ(u − u0)δ2(z − w)

[Az(u, z, ¯ z), Aw(u0, w, ¯ w)] = [A¯

z, A ¯ w] = 0

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

✏Az = @z✏(z, ¯ z) Q = R

J + dud2z ✏ [@u(@zA¯ z + @¯ zAz)]

Modification of brackets : imposing

Fz¯

z|J +, u→±∞ = 0

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

Az(u → ±∞) = ∂zφ±

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

δAz,¯

z(u = +∞) + δAz,¯ z(u = −∞) = 0

IR regularization

• This was first considered by Pasterski 

• Introduce a cutoff for u.

− T

2 ≤ u ≤ T 2

Fuz(¯

z)(u = − T 2 ) = Fuz(¯ z)(u = + T 2 )

Fuz(¯

z)(u) = P∞ m=−∞ αm(¯

αm)ei(2πmu)/T

Periodic boundary conditions - that the boundary field values remain unchanged

IR regularization

Fuz(¯

z)(u) = P∞ m=−∞ αm(¯

αm)ei(2πmu)/T Az(u) = d0 + α0u + P1

m=6=0 T 2πmiαmei(2πmu)/T

A¯

z(u) = ¯

d0 + ¯ α0u + P1

m=6=0 T 2πmi ¯

αmei(2πmu)/T αn, ¯ αn, d0, ¯ d0

are functions of

z, ¯ z

Schwinger brackets revisited

Substitute the expansion into GΣ Now there is no natural distinction between bulk and boundary terms Apply the Schwinger procedure mode by mode

[d0, GΣ] = i

2δd0,

[αn, GΣ] = i

2δαn

There are more equations than unknowns.

Schwinger brackets revisited

There is a set of consistent solutions. Obtain a set of brackets:

[αm(z, ¯ z), ¯ αn(w, ¯ w)] = − mπ

T 2 δm+n,0δ2(z − w)

[ ¯ d0(z, ¯ z), αm(w, ¯ w)] = [d0(z, ¯ z), ¯ αm(w, ¯ w)] =

i 2T (−1)mδ2(z − w)

[ ¯ d0(z, ¯ z), α0(w, ¯ w)] = [d0(z, ¯ z), ¯ α0(w, ¯ w)] = i

T δ2(z − w)

Applications

Test 1: Large gauge transformation

Q = T R dzd¯ z (¯ ↵0@z✏ + ↵0@¯

z✏)

[Az,¯

z(u, z, ¯

z), Q] = i@z✏ = iAz,¯

z

Curiously, it gives the correct large gauge transformation without the factor of 2 issue before imposing the constraints on Fz¯

z

Applications

Test 1: Large gauge transformation

Q = T R dzd¯ z (¯ ↵0@z✏ + ↵0@¯

z✏)

[Az,¯

z(u, z, ¯

z), Q] = i@z✏ = iAz,¯

z

Now with an IR regularisation, we can impose constraints systematically.

Fz¯

z = 0

Imposing at the boundary points and obtain the corresponding Dirac brackets

[F, G]D = [F, G] − [F, φa]Cab[φb, G], Cab = [φa, φb]

Applications

Test 1: Large gauge transformation

[F, G]D = [F, G] − [F, φa]Cab[φb, G], Cab = [φa, φb]

φ1 = ∂z ¯ α0 − ∂¯

zα0

φ2 = ∂z( ¯ d0 + P

m6=0 (1)mT i2πm ¯

αm) − ∂¯

z(d0 + P m6=0 (1)mT i2πm αm)

C12 = −C21 = i

T (∂z∂ ¯ w + ∂¯ z∂w)δ2(z − w)

Applications

Test 1: Large gauge transformation

[α0(z, ¯ z), ¯ d0(w, ¯ w)]D = − i

2T δ2(z − w)

[d0(z, ¯ z), αm(w, ¯ w)]D = −

i 4πT (−1)m (z−w)2

[d0(z, ¯ z), ¯ αm(w, ¯ w)]D = 0

It does not change the commutation between Q and A!

Applications

Test 1: Large gauge transformation To make comparison with Strominger et al:

∂zφ+ = d0 + T

2 α0 + P m6=0 T (1)m i2πm αm

∂¯

zφ = ¯

d0 − T

2 ¯

α0 + P

m6=0 T (1)m i2πm ¯

αm [φ+(z, ¯ z), φ−(w, ¯ w)] =

i 4πln|z − w|2

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

S = − 1

8

R d4x r2z¯

z[F 2 + G2], 1 2✏µνρσF ρσ = Gµν

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.

S = Z d4x [−i/2(Gz¯

zFru − Gu¯ zFrz + GuzFr¯ z −

Fz¯

zGru + Fu¯ zGrz − FuzGr¯ z) +

1 2( F 2

z¯ z

r2γz¯

z

+ G2

z¯ z

r2γz¯

z

) + FuzFu¯

z +

GuzGu¯

z − iFuzGu¯ z + iFu¯ zGuz]

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(Cz − iAz) = 0 ∂u(C¯

z + iA¯ z) = 0

Cz = iAz + f, C¯

z = −iA¯ z + ¯

f

Applications

Test 2: Electromagnetic duality The electromagnetic duality symmetry transformation reduces to:

δAz = θCz, δA¯

z = θC¯ z

δCz = −θAz, δC¯

z = −θA¯ z

Applications

Test 2: Electromagnetic duality The Noether charge is given by

Q = iθ Z d2z [¯ α0d0T − α0 ¯ d0T + X

n6=0

T 2 iπnαn¯ αn − X

m6=0

i(−1)mT 2 2πm (α0¯ αm − ¯ α0αm)] +θ 2 Z d2z[T(f ¯ α0 + ¯ fα0)].

Applications

Test 2: Electromagnetic duality

[αn, Q] = −θαn, [¯ αn, Q] = +θ¯ αn

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