The Hamiltonian formulation of gauge theories

Transcription

1 The Hamiltonian formulation of gauge theories I [p, q] = dt p i q i H(p, q) " # q i =[q i, H] ṗ i i i, H] 1. Symplectic geometry, Hamilton-Jacobi theory, The first (general) quantization method [p i, q j ]=i~ j i, H(p, q) 3. The Energy E = H(p, q) functional is built-in in the formalism.

2 Symmetries Symmetries are displacements sq i such that, for all q i (t), I [q i + s q i ] I [q i ]= dt d dt B s(q i, sq i ) (1) On-shell variations have the property that, for all q i (t), I [qos i + q i ] I [qos] i = dt d dt (p i q i ) (2) I In (1), q i is arbitrary while q i is restricted. I In (2), q i is restricted while q i is arbitrary. Replacing q i = qos i in (1), and q i = s q i in (2), the left hand sides are equal. Subtracting yields Noether theorem, 0= d B s (q i, sq ) q i sq i

3 Hamiltonian & inverse Noether Theorem Let Q(p, q) be a conserved charge, [Q, H] = 0. (Poisson brackets) Then, the following transformations define a symmetry: q i = [q i, i p i = [p i, i Indeed, the variation of R dt(p q H) is a total derivative: 1. (p i q i )= p i q i ṗ i q i + d dt (p i q i q i i + d dt (p i q i )= dq dt + d dt (p i q i ) 2. i q i p @p i =[H, Q] = i

4 Symmetries Noether Symmetries dq(p, q) dt =0, q i =[q i, Q] p i =[p i, Q] We shall now explore gauge symmetries with the properties: Gauge Symmetries (q, p) =0, q i =[q i, ] p i =[p i, ] I It may happen that A µ µ is generated by a non-zero charge Q. If so, we do not call it gauge. I We reserve the word gauge for transformations generated by constraints.

5 Important gauge theories are 1. Yang-Mills theories (including QED) I YM [A a µ]= 1 Tr(F µ F µ ), A a 4 µ(x) =D a µ (x) 2. Einstein Gravity (and its generalizations, f (R) gravity, Gauss-Bonnet, Chern-Simons...) pg(r I GR [g µ ]= 2 ), gµ (x) = g µ, +,µg +, g µ 3. The string worldsheet action phh I [X µ, h X X µ, X µ X µ 4. Chern-Simons I [A] = Tr AdA + 23 A3

6 Examples of Lagrangians with a gauge symmetry in particle mechanics are: 1. The parameterized non-relativistic point particle (its constraint is Schroedinger equation), 1 I [q( ), t( )] = 2ṫ q2 ṫv(q) d 2. The relativistic point particle (its constraint is Klein-Gordon s equation) I [x µ ]= dx m r µ dx µ d d d

7 Gauge symmetry implies a Hamiltonian constraint Chern-Simons : I [A µ ]= Tr AdA + 23 A3 h i = Tr AȦ A 0F ( A a µ = D µ a ) Yang-Mills : I [A a µ]= 1 4 = apple E i aȧa i F µ a F a µ, ( A a µ = D a ) 1 8 (~ E 2 + ~ B 2 )+A a 0 r ~E a pgr, Gravity : I [g] = ( gµ = L g µ ) = ijġ ij NH? N i H i

8 General structure of a gauge theory in Hamiltonian form: I [p i, q j, ]= dt p i q i H 0 (p, q) (p, q) The -equations of motion are constraints (p, q) =0. There is a gauge symmetry if are zero at all i q i ṗ i =[, H 0 + ]=0. for all (t). This happens if and only if, [H 0, ] = C [, ] = f The constraints are said to satisfy a a first class algebra. Max Bañados PUC-Chile

9 General proof of gauge invariance Consider a Hamiltonian action of the form, I [p i, q j, ]= dt p i q i H 0 (p, q) (p, q). If the constraints are first class, then the following transformation is a gauge symmetry of the action, q i (t) = [q i, ] (t) p i (t) = [p i, ] (t) (t) = (t) C (t) f (t) where (t) isafullyarbitraryfunction of time. I In QED, the Lagrange multiplier is A 0.Recallthat A µ µ thus A 0 =, asexpected.thisisanabelian theory with C =0=f.

10 Proof: For any function A(p, q) of the canonical variables: A(q i, p i p i q i ].Inparticular, i H 0 = [H 0, ] = C = [, ] = f The variation of the kinetic term is (p i q i ) = p i q i ṗ i q i + total derivative = [p i, ] q i ṗ i [q i, i q i ṗ i = = + total derivative ) The variations of H 0,, p i q i all give terms proportional to the constraints. Max Bañados PUC-Chile

11 Putting all together, the variation of the full action becomes: I = (p i q i ) H 0 = + C + f +. We can choose to cancel everything making the action invariant. = 0 up to total derivatives.

12 Gauge theories, equivalent classes of solutions Why two configurations that di er by a deformation generated by a constraint are physically indistinguishable? ṗ i = [p i, H 0 ]+[p i, ] q i = [q i, H 0 ]+[q i, ] = 0 I Given the fields at time t, the equations fixed them at time t + t, only up to a gauge transformation. These theories seem inconsistent...? Nop. A clever interpretation is available: Only combinations that do not see are physical. [ must be unobservable.] We must mod out by the set of all gauge transformations

13 I The electric ( ~ E) and magnetic ( ~ B)fields(F µ )inqed. I Curvature invariants, g µ R µ, R µ R µ,... in gravity. I Wilson loops PeH A in Yang-Mills theory.

14 Comment 1. A gauge theory can also have Noether symmetries: 8 gauge : A I [A] = 1 µ µ (x >< ) F µ F µ 4 Lorentz : A >: µ = µa, µ = µ Translations : A µ A µ Exercise: The (classical) global group of QED (and Yang-Mills) is much larger, A µ = F µ (x), ( µ, +,µ = 1 2 µ ) (3) 1. Prove that (3) is a symmetry of the Maxwell action 2. Prove that (3) contains Lorentz, Translations, but also Dilatations, and Special Conformal Transformations (last two broken in QM) 3. Compute the translational Noether current ( µ = a µ ) J µ = F µ 1 F 4 F µ F a, T µ : {z } T µ Max Bañados PUC-Chile Symmetric Gauge invariant

15 Comment 2: Is the scalar field action phh I [X ]= µ X, ) 1 p µ hh X =0 h on a curved, but fixed background h µ, gauge invariant? No. There are no constraints, no Lagrange multipliers. No gauge symmetry. If the metric is dynamical (string worldsheet action) 8 1 phh p phh >< I [X, h µ ]= µ X =0 µ X ) >: µ X 2 h µ X =0 apple (Virasoro constraints) the action is gauge invariant. Varying h µ yields constraints, and h 0µ are the Lagrange multipliers. Max Bañados PUC-Chile

16 Number of degree of freedom Gauge theories and degrees of freedom How many independent initial conditions does a gauge theory have? F(p,q)=0 Gauge orbits 2N {p i, q i } s with first order equations: 2N initial conditions. I g constraints (p, q) = 0 on initial conditions. I Two initial conditions related by a gauge are the same. Number of degrees of freedom 1 (2N g g) 2 = N g.

17 Examples I 4d Gravity: g ij = 6 functions - 4 symmetries = 2 I QED : A i = 3 functions - 1 symmetry = 2 I d-dimensional Gravity: g ij = symmetries = d(d 3) 2. (d 1)d 2 functions - d I d-dimensional Yang-Mills: A a i =(d symmetries = N(d 2) 1)N fields - N

18 Old Dirac quantization condition Quantize q i, p j. What is the role of the constraints ˆ = (ˆq, ˆp)? For example, in particle quantum mechanics, rotations are generated by ~ L = ~r ~p. A rotated state is i = i~ ~L i In a gauge theory, the symmetry is generated by ˆ and the rotated state will be i = ˆ i But gauge transformations are not observable. States must be invariant. Dirac imposed, i =0 ) ˆ i =0

19 Examples of Dirac quantization: Free relativistic particle I [X µ ( )] = m d r dx µ d dx d µ p µ = m Ẋ µ qẋ µ Ẋ µ apple Invariant under! 0 = f ( ) and it follows directly that We now quantize ˆp µ µ p µ p µ + m 2 =0, and Dirac condition becomes: ( + m 2 ) =0, Kein-Gordon equation

20 Parametrized non-relativistic particle (time t( ) as a canonical variable) m d~r 2 I [~r(t)] = dt V (~r)! 2 dt I [~r( ), t( )] = ~p ~ṙ = ~r p t Quantize = m 2 ~r 2 ṫ 2 d m 2 V (~r). ~r 2 ṫ ) ṫv(~r)!, apple Invariant under! 0 = f ( ) = p t + 1 2m ~p2 + V (~r) =0 p t = ~p = and the Dirac condition becomes: = 2m r2 + V, Schroedinger equation

21 The string worldsheet action...infinitely many degrees of freedom, it requires a detailed analysis: 8 1 phh p phh >< µ X =0 I [X, h µ ]= µ X ) >: µ X 2 h µ X =0 apple (Virasoro constraints) µ X 1 2 h µ X : ) ˆL n, ˆ L n Dirac improved condition becomes L n i =0, n > 0

22 Are gauge and Noether symmetries really disconnected? I Global symmetries, generated by non-zero charges Q n. Their action change states. Q n (ˆp, ˆq) i = 0 i. I Gauge symmetries, generated by constraints = 0. Their action do not change physical states. (ˆp, ˆq) i = 0. However, as we shall see, some gauge symmetries on manifolds with boundary are generated by a combination! G( )= d 3 x a a + Q[ ] And they cannot be disentangle. We conclude:

23 Gauge Farm All gauge transformations have the same interpretation, but some have more interpretation than others.

24 AdS 3 /CFT 2 The coordinate transformation (see previous Lecture) that maps ds 2 = e 2 dzd z + d c T (z)dz2 into ds 2 = e 2 0 dz 0 d z 0 + d c T 0 (z 0 )dz 2 with T 0 (z 0 )=T 0 f falls precisely in this category. 2 c 12 {f, z0 }. I Metrics with di erent values of T (z) do represent di erent states. I This explains how AdS 3 which naively has no degrees of freedom can be dual to a CFT 2 with infinitely many degrees of freedom.

25 Stop talking and calculate! We shall exhibit the Regge-Teitelboim e ect with the example of Chern-Simons theory which is: I simple I yet not trivial (like QCD) I modern and fun mathematically I lots of applications I...I understand it well See you tomorrow.