Constrained Dynamical Systems and Their Quantization

Transcription

1 Constrained Dynamical Systems and Their Quantization Graduate lecture by Y. Kazama University of Tokyo, Komaba constsys-1

2 Contents 1. Constraints and flows in the phase space 2. Systems with multiple constraints 3. Dirac s theory of constrained systems 4. More general formulation of Batalin, Fradkin and Vilkovisky 5. Application to non-abelian gauge theory constsys-2

3 References Dirac formalism P.A.M. Dirac, Lectures on Quantum Mechanics (1964, Belfer Graduate School of Science, Yeshiva University) Batalin-Fradkin-Vilkovisky formalism Fradkin and Vilkovisky, Quantization of relativistic systems with constrains, PL55B (75)224 Batalin and Vilkovisky, Relativistic S-Matrix of dynamical systems with boson and fermion constraints, PL69B (77) 309 Fradkin and Vilkovisky, Quantization of relativistic systems with constrains: Equivalence of canonical and covariant formalisms in quantum theory of gravitational field, CERN TH-2332 (77) constsys-3

4 Batalin-Vilkovisky formalism (Anti-field formalism) Batalin and Vilkovisky, Gauge algebra and quantization, PL 102B(81) 27 Batalin and Vilkovisky, Quantiation of gauge theories with linearly dependent generators, PRD 28 (83) 2567 Reviews A.J. Hanson, T. Regge, C. Teitelboim, Constrained Hamiltonian Systems, RX-748, PRINT (IAS,PRINCETON), pp. M. Henneaux, Hamiltonian form of the path integral for theories with a gauge freedom, Phys. Rep. 126 (85) 1 66 J. Gomis, J. Paris and S. Samuel, Antibrackets, Antifields and Gauge-Theory Quantization, Phys.Rept.259:1-145,1995. (hep-th/ ) constsys-4

5 Book M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, Princeton University Press 1992 constsys-5

6 1 Constraints and flows in the phase space 1.1 Flows generated by constraints and the geometric meaning of the Poisson bracket The set up: We consider a dynamical system described in a 2n-dimensional phase space M: M = 2n-dimensional phase space x µ = x µ (q i, p i ), i = 1 n: a set of local coordinates of M Suppose the system is subject to a constraint defined by the equation f(x) = c = constant (1) This defines a (2n 1)-dimensional hypersurface in M, to be denoted by constsys-6

7 Σ f. A flow on Σ f : Consider an infinitesimal move on the surface Σ f. Then, we have f(x + dx) f(x) = µ fdx µ = 0 (2) Since dx µ is tangential to Σ f, µ f is a vector normal to Σ f. As we continue this development along the surface, we should get a flow on Σ f of the form x µ (u) with u a parameter of the flow. x µ (u) f(x) = constant Σ f To generate such a flow, we must make sure that dx µ is always normal to µ f. This can be guranteed if there exists an antisymmetric tensor ω µν (x) on M. Then we can set constsys-7

8 dx µ = ω µν ν f du (3) This obviously satisfies the requirement dx µ µ f = 0. The vector field which generates this flow is defined as X ω f (ωµν ν f) µ (4) so that (3) can be written as dx µ = (X ω f xµ )du (5) For a fixed ω µν, we will often omit the superscript ω and write X f for short. We will call this flow F f. constsys-8

9 Variation along the flow and the Poisson bracket: Now consider how an arbitrary function g(x) varies along the flow F f. It is given by g(x + dx) = g(x) + dux f g(x) = g(x) + duω µν ν f µ g = g(x) + du µ gω µν ν f = g(x) + du{g, f}, (6) where we have defined the Poisson bracket 1 {g, f} X f g = µ gω µν ν f = {f, g} = X g f. (7) 1 This should actually be called pre-poisson bracket. For true Poisson bracket, we need the requirements for ω µν to be discussed below. constsys-9

10 (Strictly speaking, for this to be called a Poisson bracket, ω µν must satisfy certain properties to be discussed below.) In any case, the Poisson bracket {g, f} describes the infinitesimal change of g along a flow F f on Σ f. The basic formula is dg du = {g, f} (8) Requirements on ω µν : (1): As we have seen, once ω µν is given, to a function f(x) we can associate a unique flow F f on the surface Σ f. constsys-10

11 We wish to achieve the converse, namely we want a flow to determine the family of surfaces f(x) = c on which it exists. That is, we want to be able to solve dx µ du = ωµν ν f (9) for f(x) up to a constant. Obviously the condition is that ω µν is nondegenerate (invertible). When ω µν is non-degenerate, we will denote its inverse by ω µν : Then it is natural to define the following 2-form ω µν ω νρ = δ µ ρ. (10) ω 1 2 ω µνdx µ dx ν, (11) ω µν = ω( µ, ν ) (12) constsys-11

12 The Poisson bracket can then be written as {f, g} = ω(x g, X f ) (13) since ω(x g, X f ) = ω(ω µν ν g µ, ω αβ β f α ) = ω µν ν gω µα ω αβ β f = µ fω µν ν g (14) (2): To be relevant to physical problems, ω µν should not be arbitrary. We require that there exists a coordinate system (q i, p i ) such that ω takes the standard form ω = n dp i dq i. (15) i=1 constsys-12

13 This amounts to ω pi q j = δ ij, ω qj p i = δ ij (16) ω p iq j = δ ij, ω q jp i = δ ij (17) Thus, in this coordinate system, the Poisson bracket takes the familiar form: {f, g} = µ fω µν ν g = ( f g g ) f q i p i i q i p i (18) In particular, { q i, p j } = δ i j (19) It is obvious that in this coordinate system ω is closed i.e. dω = 0. Since this is a coordinate independent concept, we must demand that it be true in any coordinate system. As we shall see shortly, this will be quite essential for the Poisson bracket to have constsys-13

14 the desired properties. In tensor notation, the closedness reads dω = 1 2 ρω µν dx ρ dx µ dx ν, (20) 0 = ρ ω µν + cyclic permutations. (21) A non-degenerate and closed 2-form ω is called a symplectic form or a symplectic structure. An even-dimensional manifold endowed with a symplectic form is called a symplectic manifold. constsys-14

15 Properties of the Poisson Bracket: We now prove the following basic properties of the Poisson brackets: (1) {f, g} = {f, x µ } µ g (22) (2) {f, g} = {g, f} X g f = X f g (23) (3) {f, gh} = {f, g} h + g {f, h} (24) (4) X {g,f} = [X f, X g ] (25) (5) {f, {g, h}} + cyclic = 0 (26) (1) and (2) are obvious from the definition. (3) is easy to prove using the vector field notation. Once (4) is proved, then (5) is automatic. So, the only non-trivial property to be proved is (4). Proof of (4): We will give a basis independent derivation below and leave the proof in terms of components as an exercise. constsys-15

16 (1) First it is easy to prove that for any vector field Y, we have the identity ( ) ω(y, X f ) = Y (f) (27) Indeed, writing Y = Y α α, ω(y, X f ) = ω(y α α, ω µν ν f µ ) = Y α ω αµ ω µν ν f = Y α α f = Y (f) (28) (2) Next we recall the basis-independent definition of the exterior derivative operator d. On a p-form ω, dω(x 1, X 2,..., X p+1 ) p+1 ( ( 1) 1+i X i ω(x1,..., ˇX i,..., X p+1 ) ) i=1 + ( 1) i+j ω([x i, X j ], X 1,..., ˇX i,..., ˇX j,..., X p+1 ) i<j (29) constsys-16

17 where ˇX i means that it be deleted. For p = 0, i.e. when ω is just a function, one defines For p = 1 the formula takes the form dω(x) = X(ω) (30) dω(x, Y ) = X(ω(Y )) Y (ω(x)) ω([x, Y ]) (31) Let us check this by comparing it to the usual component calculation. Set ω = ω ν dx ν, X = X α α, Y = Y β β. Then the component calculation goes as follows: dω = µ ω ν dx µ dx ν = 1 2 ( µω ν ν ω µ )dx µ dx ν (32) dω(x, Y ) = 1 2 ( µω ν ν ω µ )dx µ dx ν (X α α, Y β β ) = 1 2 ( µω ν ν ω µ )(X µ Y ν X ν Y µ ) (33) constsys-17

18 On the other hand, we have X(ω(Y )) = X α α (ω ν Y ν ) = X α α ω ν Y ν + X α ω ν α Y ν (34) Y (ω(x)) = Y β β ω ν X ν Y β ω ν β X ν (35) ω([x, Y ]) = ω((x α α Y β ) β (Y β β X α ) α ) = X α α Y β ω β + Y β β X α ω α (36) So the last term of the RHS of (31) removes the derivatives on X α and Y β and we get RHS of (31) = µ ω ν (X µ Y ν Y µ X ν ) = LHS of (31) (3) Now apply this to our 2-form ω in the following way and impose the closedconstsys-18

19 ness condition: 0 = dω(y, X f, X g ) = Y (ω(x f, X g )) X f (ω(y, X g )) + X g (ω(y, X f )) ω([y, X f ], X g ) ω([x f, X g ], Y ) + ω([y, X g ], X f ) (37) where Y is an arbitrary vector field. The first term can be written in the following two ways (i) Y (ω(x f, X g )) = Y ({g, f}) = ω(y, X {g,f} ) = ω(x {g,f}, Y ) (ii) Y (ω(x f, X g )) = Y (X f (g)) = Y (X g (f)) = 1 2 (Y (X f(g)) Y (X g (f))) Forming 2 (ii) (i) we have Y (ω(x f, X g )) = Y (X f (g)) Y (X g (f)) + ω(x {g,f}, Y ) (38) constsys-19

20 The rest of the terms can be rewritten in the following way X f (ω(y, X g )) = X f (Y (g)) (39) X g (ω(y, X f )) = X g (Y (f)) (40) ω([y, X f ], X g ) = [Y, X f ] (g) (41) ω([y, X g ], X f ) = [Y, X g ] (f) (42) Combining all the terms, many of the terms cancel and we are left with 0 = ω(x {g,f}, Y ) ω([x f, X g ], Y ) = ω(x {g,f} [X f, X g ], Y ) (43) Since Y is arbitrary and ω is non-degenerate, we must have X {g,f} [X f, X g ] = 0, which proves the assertion. // Exercise: Give a proof in terms of components. constsys-20

21 1.2 Hamiltonian vector field, gauge symmetry and gauge-fixing Hamiltonian vector field and the equation of motion: Consider now the Hamiltonian function H(x) of a dynamical system. For a conservative system, H(x) stays constant, and we can apply the above general consideration with f(x) = H(x). Thus, a flow is generated by the associated Hamiltonian vector field X H on the surface Σ H. When talking about a flow generated by X H, we normally use t as the parameter. Using the general formula (8), we obtain the equation of motion for any physical quantity g(x) dg dt = X Hg = {g, H} (44) constsys-21

22 Compatibility of H and the constraint: Now consider the situation where we have a constraint φ(x) = 0, which is different from the Hamiltonian constraint. Actually, we must consider the set of functions of the form Φ {χ(x) χ(x) = α(x)φ(x)} (45) where α(x) is an arbitrary function non-vanishing and non-singular on the constraint surface Σ φ. Then, any member of Φ vanishes on the surface Σ φ and generates essentially the same flow as φ. Now when the system develops according to the Hamiotonian H(x), it should not leave Σ φ. Otherwise, the imposition of the constraint would be incompatible with the Hamiltonian. So for compatibility, we must require that after an infinitesimal time the change constsys-22

23 of the constraint function should vanish on Σ φ. This is expressed by dφ = {φ, H} Φ (46) dt When this holds, we say that {φ, H} is weakly zero and denote it by {φ, H} 0. Since the surface Σ φ is generated by any member of the set Φ, we must also have χ Φ {χ, H} 0. (47) Indeed this is guaranteed: Since χ can be written as χ = α(x)φ(x), we get {αφ, H} = {α, H} φ + α {φ, H} 0. (48) Similarly, it is easy to show that χ 1 0, χ 2 0 {χ 1, χ 2 } 0 (49) constsys-23

24 (Proof: Σ φ. ) χ i 0 χ i = α i φ. Then, {α 1 χ 1, α 2 χ 2 } clearly vanishes on Equivalence relation on Σ φ and gauge symmetry: When we impose a constraint φ(x) = 0, the dimension of the phase space drops by 1. However, from the point of view of (p, q) conjugate pair, the physical degrees of freedom should drop by 2, not just by 1. Indeed this can happen due to the following mechanism. Let ψ(x) Φ. Then ψ generates a flow on Σ φ given by dx µ = {x µ, ψ} du. (50) The important point is that flows generated by any memeber of Φ are actually the same. Let χ = αψ be another member. Then the flow generated by χ, constsys-24

25 with a parameter v, is described by the equation dx µ = {x µ, χ} dv = ({x µ, α} ψ + α {x µ, ψ})dv = {x µ, ψ} αdv (51) This shows that, apart from the redefinition of the parameter u, the trajectory is the same. Furthermore, since {ψ, H} Φ, Φ as a set is invariant under time development. Thus, the shape of the flow does not change in time t. Thus, all the points on a flow generated by Φ describe the same physical state from the point of view of the Hamiltonian dynamics and they should be identified. Hence, the bonafide physical phase space P is the quotient space P = Σ Φ /, dim P = 2n 2. (52) where denotes the equivalence relation just described. Thus Φ generates gauge transformations on the dynamical variable x µ in the manner constsys-25

26 δ α x µ = α(x) {x µ, φ} Idea of Gauge Fixing: Let us call our constraint φ 1 (x). Each point in P is described by a representative of Σ φ 1. To pick one out, one may try to intersect Σ φ 1 by another hypersurface generated by an additional constraint φ 2 (x) = 0. To guarantee that this picks out a single point on the flow generated by φ 1, we must demand that the value of φ 2 (x) changes monotonically along the flow constsys-26

27 F φ 1. This is condition is expressed by { φ 2, φ 1} = dφ2 0 du never crosses zero (53) This procedure of introducing such an additional constraint φ 2 (x) is called gauge fixing. 1.3 Flow in the physical space and the Dirac bracket The Poisson bracket {g, f} describes an infinitesimal flow of g generated by the function f. (In this context, the function f(x) need not be a geuine constraint of the dynamical system.) Suppose we have genuine constraints φ a = 0, (a = 1, 2) as above. We may still consider the flow {g, f} generated by f. But in general the flow gets out of the physical space P = Σ Φ /. We now explain a convenient way of generating a flow in the physical phase space. constsys-27

28 It would be very nice if we can invent a new bracket which generates a flow that stays in the physical space. This is achieved by the so-called the Dirac bracket, to be deonoted by {g, f} D. First we have the following property: f(x) generates a flow on P {φ a, f} = 0, (a = 1, 2). Proof: This follows immediately from dφ/du = {φ a, f}. If the flow stays on P, then cleary the conditions φ a = 0 does not change and we have {φ a, f} = 0. Conversely, {φ a, f} = 0 means that along the flow dφ a /du = 0 and hence φ a = 0 conditions are preserved. constsys-28

29 Next, the desired bracket should have the following properties: 1. For f(x) satisfying {φ a, f} = 0, we want {g, f} D = {g, f}. (54) 2. When {φ a, f} = 0, the bracket should automatically drop the part of the movement which gets out of the constraint surface. Specifically, we want {φ a, f} D = 0. (55) 3. The bracket should satisfy all the basic properties of the Poisson bracket. The bracket satisfying all these requirements is given by constsys-29

30 { {g, f} D {g, f} {g, φ a } C ab φ b, f }, (56) C ab { φ a, φ b} anti-symmeric, non-degenerate C ab = inverse of C ab (57) The second term subtracts precisely the portion which gets out of the physical space. We will see this more explicitly in the next section. In the meantime, let us check that the properties 1,2,3 are met. Property 1 is obvious. constsys-30

31 Property 2: {φ a, f} D {φ a, f} { φ a, φ b} }{{} C ab C bc {φ c, f} = {φ a, f} {φ a, f} = 0 (58) Excercise: Check the property 3 Since {φ a, f} D = 0 for any function f(x), we can set φ a strongly (identically) to zero when we use the Dirac bracket. constsys-31

32 Meaning of the Dirac bracket and the physical Hamiltonian H P : We now wish to make the meaning of the Dirac bracket clearer and construct the physical Hamiltonian which directly acts on the physical phase space P. This is achieved by employing a convenient coordinate system, to be described below. First we recall that the Poisson bracket is basis independent as seen from the expression {f, g} = ω(x g, X f ). So we may take a special coordinate system y µ (x) where two of the new coordinates coincide with the constraints: x µ y µ (x) {, (59) a a = 1, 2 µ = (60) i i = 3, 4,..., 2n y a (x) = φ a (x) (61) constsys-32

33 Further, we choose y i such that ω ai y = { y a, y i} = { φ a, y i} = 0. This should be possible since this condition simply says that y i (x) generates a flow on the physical space P. Indeed if { y a, y i} = ω ia y 0, then redefine yi as Then, (omitting the subscript y on ω ia ) ỹ i y i ω ib y C bcy c (62) {ỹi, y a} = ω ia ω ib C bc C ca = 0 (63) In this coordinate system ω µν takes the form ( ω ω µν ab 0 = 0 ω ij Now we call U = unphysical space spanned by {y a } P = physical spaces spanned by {y i } ) (64) constsys-33

34 Because of the block diagonal form of ω µν, we have {g, f} = {g, f} U + {g, f} P. (65) It is easy to prove the identity {g, f} = {g, y µ } ω µν {y ν, f} (66) { = {g, φ a } C ab φ b, f } + { g, y i} { ω ij y j, f } (67) Thus we can identify {g, f} P = { g, y i} { ω ij y j, f } (68) { {g, f} U = {g, φ a } C ab φ b, f } (69) Setting g = φ a in this identity, we find {φ a, f} P = { φ a, y i} { ω ij y j, f } { = ω ai ω ij y j, f } = 0 (70) {φ a, f} U = { φ a, φ b} ω bc {φ c, f} = { φ a, φ b} C bc {φ c, f} = 0 (71) constsys-34

35 So {φ a, f} P has precisely the property of the Dirac bracket. In other words, the Dirac bracket is nothing but the bracket in the physical space. Namely {g, f} D = {g, f} {g, f} U = {g, f} {g, φ a } C ab { φ b, f } (72) which is exactly the definition introduced before. If we adopt this special coordinate system, it is clear that the physical Hamiltonian is H P = H(y 1 = y 2 = 0, y i ). (73) and the symplectic structure to be used is ω P 1 2 ω ijdy i dy j (74) We then get dy i dt = { y i, H P } ωp (75) constsys-35

36 But the same result can be obtained in any coordinate system if we use the Dirac bracket and set φ a = 0 strongly. Indeed dy a dt dy i = {y a, H} D = {φ a, H} D = 0 (76) dt = { y i, H } D y a =0 = { y i, y j} ω ij { y j, H } y a =0 = { y i, H P } ωp (77) Thus we always get correct equations of motion by using the Dirac bracket. constsys-36

37 2 Systems with multiple constraints 2.1 Multiple constraints and their algebra From now on, we fix the symplectic structure and take the basis x µ = (q i, p i ), i = 1, 2,..., n such that ω is of the standard form ω = n i=1 dp i dq i. Suppose we have m independent bosonic constraints 2 T α (x) = 0, α = 1, 2,..., m (78) The constrained surface will then be 2n m dimensional. In order for any flow generated by these constraints to remain on this surface, all the Poisson bracket among them must be weakly zero. Also, for these constraints to be consistent with the time evolution, theirpoisson 2 The case where fermionic constraints are present can also be handled, but for simplicity, we shall not do that here. In what follows, we use the BFV convention and use subscripts for the constraint indices. constsys-37

38 bracket with the Hamiltonian, which we write H 0, must vanish weakly as well. In other words, T α s and H 0 must satisfy an involutive algebra of the form {T α, T β } = T γ U γ αβ, (79) {H 0, T α } = T β V β α, (80) U γ αβ and V α β are in general functions of (qi, p i ) Hence the above need not be a Lie algebra. (In fact for gravity this situation occurs.) We shall call this algebra the algebra of constraints. In Dirac s terminology T α s are called the first class constraints. constsys-38

39 2.2 Analysis using the Action Integral Invariance of the action under the transformations generated by the constraints: The action for the constrained system defined above can be written in two ways: (i) S [q, p] = dt ( ) p i q i H T 0 (81) =0 (ii) S [q, p, λ] = dt ( ) p i q i H 0 + λ α T α (82) where in the second expression λ α are the Lagrange multipliers. (1) First let us check that the action (i) is invariant under the gauge transformations generated by the constraints. Denoting by ɛ α (p, q) the infinitesimal local parameters, the transformations are constsys-39

40 expressed as δq i = { } q i, T α ɛ α, (83) δp i = {p i, T α } ɛ α. (84) If the point (p i, q i ) is on the constrained surface, it stays on the surface under these variations because they are generated by the constraints. The action changes by δs = dt ( ) δp i q i + p i δ q i δh T 0. (85) =0 Now for the last term, δh 0 = {H 0, T α } ɛ α 0 ( on the T = 0 surface) (86) constsys-40

41 due to the constraint algebra. As for the first and the second terms, δp i q i = {p i, T α } ɛ α q i = T α q i qi ɛ α (87) p i δ q i = ṗ i { q i, T α } ɛ α = T α p i ṗ i ɛ α, (88) δp i q i + p i δ q i = T α ɛ α = d dt (T αɛ α ) + T α }{{} 0 ɛ α (89) (In the second equation above, we used the integration by parts under dt. ) So for the variations that vanish at the initial and the final time this vanishes upon integration on the T = 0 surface. Thus S is invariant. Since the classical trajectories are obtained by extremizing the action, this shows that various trajectories which differ by the variations generated by the constraints all satisfy the equations of motion. (2) Next consider the variation of S [q, p, λ]. This time, we cannot use the equations T α = 0. constsys-41

42 Using the previous results and the integration by parts, we get δp i q i + p i δ q i = T α ɛ α = T α ɛ α, (90) δh 0 = {H 0, T α } ɛ α = T α V α β ɛβ, (91) δ(λ α T α ) = δλ α T α + λ α {T α, T β } ɛ β = δλ α T α + λ α T γ U γ αβ ɛβ. (92) So δs [q, p, λ] vanishes if we define the variation of λ α as It should be noted that δλ α = ɛ α + V α ν ɛν λ β U α βγ ɛγ. (93) in the case of the usual gauge theory this is precisely of the form of the familiar gauge transformation. (for λ α A a 0 with V α ν = 0). constsys-42

43 Gauge Fixing: Just like in the case of a single constraint, we add m additional constraints to pick out the true physical phase space: Θ α (p i, q i ) = 0 α = 1, 2,..., m. (94) Since the new surface must intersect the original constraint sufrace we need to require δθ α = {Θ α, T β } ɛ β 0 α, ɛ β. (95) so that on any tranjectory the value of Θ α must be changing. In other words, the matrix {Θ α, T β } need not have zero eigenvalue. Thus we demand det {Θ α, T β } 0 (96) Note that Θ α s do not form any involutive algebra. constsys-43

44 Let us collect all the 2m constraints and define ( ) Tα T a =, (97) Θ β C ab = {T a, T b }. (98) Then by looking at the structure of C ab one can show that det C 0. Excercise Prove this fact. To enforce the additional constraints, we introduce m Lagrange multipliers λ β for Θ β. The action can be written as S [ q, p, λ, λ ] = dt ( p i q i H 0 + λ α T α + λ α Θ α) (99) = dt ( ) p i q i H 0 + ξ a T a, (100) ( ) λ where ξ a α = (101) λ β constsys-44

45 The dimension of the physical space is 2(n m). Equations of motion and emergence of the Dirac bracket: Variations with respect to q, p and ξ lead to the following equations of motion: δp i : 0 = q i H 0 p i + T a p i ξ a, (102) δq i : 0 = ṗ i H 0 + T a q i q i ξa, (103) δξ a : 0 = T a (p, q). (104) Since the last condition should not change in time, we have 0 = dt a dt = ṗ T a i + q i T a p i q. (105) i constsys-45

46 Substituting the expressions for ṗ i and q i, this becomes ( 0 = H 0 + T ) ( b Ta H0 q i q i ξb + T ) b ξ b Ta p i p i p i q i = {H 0, T a } {T a, T b } ξ b = {H 0, T a } C ab ξ b. (106) Since the matrix C ab is invertible we can solve for ξ a uniquely and get ξ a = C ab {T b, H 0 }. (107) Putting this back into the equations of motion, we find q i = { q i, H 0 } { q i, T a } C ab {T b, H 0 } = { q i, H 0 }D, (108) ṗ i = {p i, H 0 } {p i, T a } C ab {T b, H 0 } = {p i, H 0 } D. (109) Therefore, the equations of motion are generated precisely by the Dirac bracket. constsys-46

47 3 Dirac s theory of constrained systems Up until now we have been considering constrained Hamiltonian systems. We now study how constraints arise in Lagrangian formulation. Questions: When do we get constraints? How to find constraints systematically? Dirac s theory gives the answer. 3.1 Singular Lagrangian and primary constraints Singular Lagrangian: constsys-47

48 Consider a Lagrangian of the form L = 1 2 qi a ij (q) q j + q i b i (q) V (q) (110) a ij = a ji symmetric (111) The momentum conjugate to q i is p i = a ij (q) q j + b i (q), (112) or a ij (q) q j = p i b i (q). (113) When the matrix a ij (q) is singular, then q cannot be solved in terms of p uniquely. Such a Lagrangian is called singular. In this case, a ij (q) has zero eigenvalues. Let va i (a = 1, 2,..., m 1) be zero eigenvectors. Then, 0 = v i a a ij(q) = v i a (p i b i (q)) φ a (p, q), (114) and we have so called the primary constraints. They arise entirely from the definition of the momenta and therefore may not be compatible with the equations of motion. constsys-48

49 Special but important case of a singular Lagrangian: First order system: It may sometimes happen that a ij = 0 so that there are no quadratic kinetic term. In this case, we get the primary constraints of the form p i b i (q) = 0 (115) This occurs for the Dirac equation, which therefore is a constrained system. We will describe how it should be properly quantized later. Canonical Hamiltonian: Lte us construct the Hamiltonian in the usual way: H can = p i q i L(q, q) (116) It is called the canonical Hamiltonian. As is well known, H can is a function constsys-49

50 only of p and q and does not depend on q. To check this, take the variation: δh can = δp i q i + p i δ q i L L q iδqi qi q iδ = q i δp i L. (117) q iδqi Hence δ q i does not appear in δh can and we have H can / q i = Compatibility of the primary constraints and the Hamiltonian Previously, we assumed the compatibility of the constraints φ a with the Hamiltonian H, which is expressed as {φ a, H} 0. (118) If it is not automatically met, we have to impose these conditions for a consistent theory. This in general produces further constraints. constsys-50

51 To perform this analysis, we must note that because of the existence of the primary constraints the following Hamiltonian, called the total Hamiltonian, is as good as the canonical one: H T = H can + λ a (p, q)φ a (p, q) (119) In fact, its variation δh T is identical to δh can because φ a = δφ a = 0 3 and the solutions of the equations of motions derived from H T for any λ a extremize the action. The equations of motion read, after setting φ a = 0, q i = H can p i ṗ i = H can q i + λ a φ a p i, (120) λ a φ a q i. (121) Remembering that Poisson brackets are computed without enforcing constraints, 3 Since φ a = 0 should be enforced, its variation must also be zero. In other words, the variation must be such that δφ a = 0 must hold. constsys-51

52 we can write this as q i = { q i, H can } + λ a { q i, φ a }, (122) ṗ i = {p i, H can } + λ a {p i, φ a }. (123) In the sense of weak equality, this can be written also as q i { q i, H T }, (124) ṗ i {p i, H T }. (125) Compatibility with H T : Now we check the compatibility, i.e., the invariance of the constraint surface, call it Σ 1, as the system evolves according to H T. This reads ( ) 0 {φ a, H T } = {φ a, H can } + λ b {φ a, φ b }. (126) Let the rank of {φ a, φ b } on Σ 1 be r 1. Then by forming appropriate linear constsys-52

53 combinations of {φ a }, we can bring it to the form ( ) Cαβ 0 {φ a, φ b } =, (127) 0 0 φ a = (φ α, φ A ), (128) α = 1, 2,..., r 1, A = r 1 + 1,..., m 1 (129) (i) For the subspace corresponding to the subscript α, we can use the inverse of C αβ to solve for the mulitiplier from ( ): λ α = C αβ {φ β, H can } (130) (C αβ = (C 1 ) αβ ) (131) (ii) For the subspace corresponding to the subscript A, ( ) demands the compatibility {φ A, H can } 0. (132) LHS can contain parts which cannot be written as a linear combination of primary constraints. constsys-53

54 Then we must impose certain number, m 2 of secondary constraints. Constraint surface now becomes a more restricted one Σ 1+2. At this stage, primary and the secondary constraints must be regarded on the same footing and we re-set φ a = all the constraints a = 1, 2,..., m 1 + m 2,(133) and repeat the analysis again with of course new H T. Continue until no new constraints are generated. For a system with a finite degrees of freedom, this process obviously terminates after a finite number of steps. In the case of field theory, where we have infinite degrees of freedom, it may require infinite steps. However, we only need finite steps if the Poisson brackets always contain δ( x y) so that we only get local constraints. constsys-54

55 We then end up with m constraints φ a : a + 1, 2,..., m (134) m = m 1 + m 2 + 2n (135) (2n = dim. of the original phase space) r = r 1 + r 2 + m. (136) Using (130), the total Hamiltonian will be of the form H T = H can φ α C αβ {φ β, H can } + φ A λ A (137) It is easy to check that for all the constraints, Poisson bracket with H T produces a linear combination of constraints and hence weakly vanish: {φ a, H T } = φ b V b a 0. (138) constsys-55

56 3.3 1st and 2nd class functions and quantization procedure It is important to introduce the notion of the first and the second class functions (constraints). A function R(q, p) is classified as either 1st class or 2nd class according to: 1st class {R, φ a } 0 for all constraints φ a {R, φ a } = φ b r b a (139) 2nd class otherwise (140) Then we have Theorem: R, S : 1st class = {R, S} : 1st class (141) Proof is easy using the Jacobi identity for Poisson bracket. Thus the 1st class constraints form a closed (involutive) algebra under Poisson bracket operation. constsys-56

57 Quantization with the second class constraints: There are several methods: (1) Solve the second class constraints explicitly and eliminate unphysical coordinates. However, in general this is difficult and spoils manifest symmetries. (2) Use the Dirac bracket (as discussed in Chapter 2) and replace it with the quantum bracket in the following way: quantum bracket = [ ] { q i, p j i q i, p j (142) }D (3) Use path-integral formalism as already discussed. Quantization with the first class constraints: Since 1st class constraints generate gauge transformations, we must either fix the gauge or select gauge invariant physical states by imposing these constraints. constsys-57

58 More explicitly, 1. Add gauge-fixing constraints to make them second class and then use the methods above. 2. Replace the usual Poisson bracket with quantum bracket (with appropriate i factor) and then impose them on physical states. φ A Ψ = 0 (143) In this approach, operator ordering is an important problem. For compatibility, we must have [φ a, φ b ] Ψ = 0 (144) and for this purpose, one must find an ordering such that [φ a, φ b ] = U c ab φ c (145) holds i.e. φ c appears to the right of Uab c. constsys-58

59 When there does not exist any such ordering, then the system becoms inconsistent and it is said to possess quantum commutator anomaly. 3.4 Application to abelian gauge theory We now apply the Dirac s theory to the Maxwell field and see how it works. Analysis of constraints: Lagrangian: L = 1 4 d 3 xf µν F µν (146) Momentum Π µ conjugate to A µ : F µν = µ A ν ν A µ (147) constsys-59

60 Vary with respect to A µ : Therefore we get δl = 1 2 d 3 xf µν δf µν d 3 xf µ0 δf µ0 d 3 xf µ0 δ A µ (148) ( ) Π µ = F µ0 (149) Equal time Poisson bracket: {A µ (x), Π ν (y)} = δ ν µ δ(x y) (150) where x means x and the δ-function is the 3-dimensional one. Now from ( ) above, we get the following primary constraint since F 00 = 0: Π 0 = 0 (151) constsys-60

61 Form of H can : H can = = Rewrite ( ): d 3 xπ µ A µ,0 L d 3 x F i0 A i,0 }{{} ( ) F ij F ij F i0 F i0 (152) ( ) = F i0 (A i,0 A 0,i ) + F i0 A 0,i = F i0 F i0 + F i0 A 0,i (153) Putting this back in and using the definition of the momentum we get ( ) 1 H can = d 3 x 4 F ij F ij Πi Π i A 0 i Π i (154) Compatibility of Π 0 = 0 with H can : We must demand { } Π 0, H can 0. This immediately gives the Gauss law constraint as a secondary constraint: constsys-61

62 G i Π i = 0 (155) Compatibility of G = 0 with H can : Prepare some formulas {A 0 (x), G(y)} = 0 (156) { Π 0 (x), G(y) } = 0 (157) {A i (x), G(y)} = δ δπ i (x) G(y) = y i δ(x y) (158) { Π i (x), G(y) } = δ G(y) = 0 δa i (x) (159) {G(x), G(y)} = 0 (160) Note that the second equation tells us that the multiplier λ in the term λg in H T is not determined. constsys-62

63 H T is given by H T = H can + λg (161) Taking the Poisson bracket with Π 0, we get { } { } Π 0, H T = Π 0, H { can + λ Π 0, G } (162) so that λ is not determined. The 3rd equation expresses the gauge transformation of A i. Indeed, introducing the gauge parameter Λ(y), we have { } A i (x), G(y)Λ(y) = dyλ(y) y i δ(x y) = dy i Λ(y)δ(x y) = i Λ(x) (163) constsys-63

64 Now let us compute {H can, G(y)} making use of these formulas. { } 1 {H can, G(y)} = d 3 x 4 F jk F jk (x) Πi Π i (x), G(y) = d 3 x 1 2 F jk (x) {F jk (x), G(y)} (164) But {F jk (x), G(y)} = { j A k (x) k A j (x), G(y)} = x j {A k(x), G(y)} x k {A j(x), G(y)} = x j y k δ(x y) x k y j δ(x y) = x j y k δ(x y) + x k x j δ(x y) = 0 (165) Thus we have {H can, G(y)} = 0 identically and no new constraints are generated. So actually H can is already H T. constsys-64

65 Coulomb gauge-fixing and the Dirac bracket: Let us denote the two 1st class constraints by φ 1 (x) = Π 0 (x) = 0 (166) φ 3 (x) = G(x) = 0 (167) Let us take the Coulomb gauge by adding the following two additional constraints: Non-vanishing Poisson brackets among them are φ 2 (x) = A 0 (x) = 0 (168) φ 4 (x) = i A i (x) = 0 (169) {φ 1 (x), φ 2 (x)} = δ(x y) (170) {φ 3 (x), φ 4 (x)} = { i Π i (x), j A j (y) } { = x i y j Π i (x), A j (y) } = x i y j ( δ(x y)) = 2 xδ(x y) (171) constsys-65

66 Therefore ( ) A 0 C ab (x, y) = {φ a (x), φ b (y)} = 0 B ( ) 0 δ(x y) where A = δ(x y) 0 ( ) 0 B = x 2 δ(x y) δ(x y) 0 2 x (172) (173) (174) It s inverse is given by ( ) A C ab 1 0 (x, y) = 0 B ( 1 ) A 1 0 δ(x y) = = A (175) δ(x y) 0 ( ) B 1 0 D(x y) = (176) D(x y) 0 where 2 xd(x y) = δ(x y) (177) constsys-66

67 Now we can compute the basic Dirac bracket: { A i (x), Π j (y) } D = δi jδ(x y) d 3 ud 3 v { A i (x), φ a (u) } C ab (u, v) {φ b (v), Π j (y)} = δ i jδ(x y) d 3 ud 3 v { A i (x), φ 3 (u) } C 34 (u, v) {φ 4 (v), Π j (y)} = δ i jδ(x y) d 3 ud 3 v ( ) i x δ(x u)) ( D(u v)) ( yj δ(v y) = δ i j δ(x y) δi x δx j D(x y) ( ) = δ i j i j 2 δ(x y) = transverse δ-function (178) We can check that the gauge condition and the Gauss law are satisfied due to constsys-67

68 the transversality of the δ-function: { A i (x), Π j (y) } D = 0 (179) x i j y { A i (x), Π j (y) } D = 0 (180) 4 More general formulation of Batalin, Fradkin and Vilkovisky 4.1 General gauge fixing, including relativistic gauges A big problem with the formulation using the Dirac bracket: One cannot handle relativistic gauge fixing. Example : Lorentz gauge in QED. The constraint is Θ = µ A µ = A 0 + i A i = 0. (181) This involves a time derivative of a mulitiplier A 0. constsys-68

69 ( Recall that no time derivative of A 0 appears in the Lagrangian and, when the Hamiltonian is formed, A 0 appears as the Lagrange multiplier for the Gauss law constraint.) Thus, in a relativistic formulation, we want to consider a general gauge fixing function of the form Θ α = Θ α (q, p, λ, λ, λ), (182) where λ corresponds to A 0 and we may want to take λ α to be the mulitiplier for Θ α itself. So one writes down the action S [ q, p, λ, λ ] ( ) = dt p i q i H 0 + λ α T α (q, p) + λ α Θ α (q, p, λ, λ, λ). (183) But in this form, in general λ α appear also in Θ α. is no longer a multiplier since it may constsys-69

70 More crucially, λ α and λ α (α = 1 m) can become conjugate to each other. In such a case, the dimension of the phase space becomes 2(n + m) and not the correct value 2(n m). (2n =the original dim. of the phase space. ) We must consider a mechanism to kill 4m excess degrees of freedom. 4.2 Introduction of the ghost system For definiteness, we shall deal with the gauge fixing constraint of the form Θ α = λ α + F α (q, p, λ, λ). (184) The action then becomes S [ q, p, λ, λ ] = ) dt (p i q i + λ α λ α H 0 + λ α T α + λ α F α (185) constsys-70

71 Note that clearly the λ α are momenta conjugate to λ α. Now group various functions in the following way: q A = (q i, λ α ), A = 1, 2,..., n + m (186) p A = (p i, λ α ), (187) G a = (T α, i λ α ), (188) χ a = (iλ α, F α ). (189) Then the action can be written more compactly as S [ ] q A, p A = dt ( p A q A H 0 ig a χ a) (190) To kill the unwanted degrees of freedom, we introduce 2m fermionic ghostanti-ghost conjugate pairs: constsys-71

72 (η a, a ), a = 1, 2,..., 2m, (191) {η a, b } = δ a b, { η a, η b} = { a, b } = 0 (192) Later, we will often use the further decomposition of (η a, a ), η a = (η α, η α ), a = ( α, α ) Graded Poisson bracket: α = 1 m To deal with the fermionic degrees of freedom, we need to generalize the concept of Poisson bracket. The general definition of graded-poisson bracket is {F, G} ( F/ Q { A )( / P A )G ( 1) F G ( G/ Q A )( / P A )F, 0 F is bosonic F = 1 F is fermionic, (193) constsys-72

73 where ( F/ Q A ) represents the right-derivative F representation is Q A. So more explicit {F, G} F G ( 1) F G G F (194) Q A P A Q A P A (Q A, P A ) denotes all the conjugate pairs including ghosts, λ α and λ α. This definition is consistent with the Poisson brackets for η a and b given above. An important property of the graded Poisson bracket is the graded Jacobi identity. It can be written as {A, {B, C}} = {{A, B}, C} + ( 1) A B {B, {A, C}}. (195) Exercise: Check this property. (It is rather non-trivial.) constsys-73

74 4.3 Batalin-Vilkovisky theorem Now we modify our action by adding dt( a η a H) : S = dt ( p A q A + a η a H ), (196) H = H 0 + ig a χ a + H. (197) The problem is to find the appropriate H which makes this system equivalent to the canonical system described solely in terms of the physical degrees of freedom. The basic theorem for solving this problem is the following due to Batalin and Vilkovisky: constsys-74

75 4.3.1 BV Theorem Assume that the set of functions G a are algebraically independent 4 and satisfy the involutive algebra: {G a, G b } = G c U c ab (198) {H 0, G a } = G b V b a (199) Let Ψ(q A, p A, η a, a ) be an arbitrary fermionic function, to be called gauge fermion (fermionic gauge-fixing function). Then the following functional integral is independent of the choice of Ψ: 4 It means that a G aa a = 0 A a = 0. constsys-75

76 Z Ψ = S Ψ = dq dp dη d e is Ψ (200) dt ( p A q A + a η a H Ψ ) (201) H Ψ = H 0 + a V a b ηb + i{ψ, Ω}, (202) Ω G a η a ( 1) a a Ubc a ηc η b. (203) Ω is called the BRST operator. (Be careful about the order of the indices of last two η s.) The definition above is valid for a mixed system of bosonic and fermionic constraints G a. a is 0 if G a is bosonic and is 1 if G a is fermionic. Accordingly, the ghosts are fermionic for the former and bosonic for the latter. constsys-76

77 Remark: For the case at hand with G a = (T α, i λ α ), the involutive algebra is extended. This extension however is rather trivial since T α and H 0 are assumed to be functions only of (q, p) and hence i λ α is completely inert. Typical example of Ψ: The often used form of Ψ is Ψ = a χ a = i α λ α + α F α (204) Then by computing {Ψ, Ω}, we get constsys-77

78 H Ψ = H 0 + ig a χ a + H = H 0 T α λ α λ α F α + H { H = a V a b ηb + i a {χ a, G b } η b + c χ a U c ab ηb } { } a b χ a, U b cd η c η d (205) (206) Note that in general four-ghost interaction is present. For the usual Yang-Mills theory there are no such terms since the structure constants U b cd are indeed constant and { χ a, U b cd} = 0. constsys-78

79 4.3.2 Proof of the BV Theorem For simplicity, we will deal only with the case where the constraints are all bosonic and hence η a are all fermionic. To prove the BV theorem, we first derive some crucial identities. We will use the following notations: G = G a η a, V a = V a b ηb, (207) U a b = U a bc ηc, U a = 1 2 U bc a ηb η c (208) They are very similar to differential forms, with η a playing the role of dx µ. Note that the b, c indices in U a are contracted oppositely to the ones in the definition of Ω. (This leads to the minus sign below in Ω.) Then we have constsys-79

80 Ω = G a U a, (209) H Ψ = H 0 + a V a + i {Ψ, Ω}. (210) Representation of the gauge algebra: The first set of identities, which follow directly from the definitions above and the constraint algebra, are (1) {G a, G} = G b U b a, (211) (2) {G, G} = 2G a U a, (212) (3) {H 0, G} = G a V a. (213) constsys-80

81 First-level Jacobi identities: Now we prove what we call the first-level Jacobi identities: (Ia) {G, U a } = U a b U b, (214) (Ib) {G, V a } = V a b U b U a b V b + {H 0, U a } (215) These are called the first-level because they follow from the Jacobi identity for the Poisson brackets involving the constraints and H 0 alone. Proof of (Ia): Consider the double Poisson bracket {{G, G}, G}. Since G is fermionic, the graded Jacobi identity tells us {{G, G}, G} = {G, {G, G}} {{G, G}, G} = {{G, G}, G} {{G, G}, G} = 0 (216) constsys-81

82 On the other hand, using (2) we get {{G, G}, G} = 2 {G a U a, G} = 2 {G a, G} U a + 2G a {U a, G} = 2G a ( U a b U b + {U a, G} ). (217) Since G a s are assumed to be algebraically independent, for this to vanish we must have the identity (Ia). // Proof of (Ib): Similarly consider {G, {H 0, G}} and compute this in two ways. One way is to use the Jacobi identity for the Poisson brackets. The other way is to compute it directly using (3). Equating them we easily get (Ib). Excercise: Show this explilcitly. constsys-82

83 Second-level Jacobi identities: Next, we will prove the second-level Jacobi identities: (IIa) (IIb) { U a, U b} = 0, (218) { U a, V b} = 0. (219) These are called the second-level because they are relations between the coefficients of the constraint algebra and are derived using the Poisson bracket Jacobi identities with U or V in one of the slots and the first-level Jacobi identities. Proof of (IIa): We consider the double Poisson bracket {{G, G}, U a }. By using the Jacobi constsys-83

84 identity (note U a is bosonic) we can write this as {{G, G}, U a } = {G, {G, U a }} + {{G, U a }, G} = 2 {G, {G, U a }} = 2 { } { G, U a b U b 2U } a b G, U b, (220) where in the last line we used (Ia). On the other hand, using (2) we get {{G, G}, U a } = 2 { G b U b, U a} { = 2G b U b, U a} + 2 {G b, U a } U b. (221) Equating these, we get ( ) { G, U a b } U b U a b {G, U b} = G b { U b, U a} + {G b, U a } U b (222) To go further, we consider the Jacobi identity 0 = {{G, G}, G a } {{G, G a }, G} + {{G a, G}, G} (223) constsys-84

85 After some calculations, this Jacobi identity becomes G c ( U c ab U b + {U c, G a } U c b U b a + { U c a, G}) = 0. (224) Using the linear independence of G c, we obtain U c ab U b + {U c, G a } U c b U b a + { U c a, G} = 0. (225) Now mulitiply this from right by U a. Then the first term vanishes due to the antisymmetry of U c ab in a b and the result is {U c, G a } U a U c b U b a U a + { U c a, G} U a = 0. (226) Substituting this into ( ), we get { U a, U b} = 0. // Proof of (IIb): It is obtained similarly by considering {{G, G}, V a }. Excercise: Supply the details. constsys-85

86 Proof of 3 fundamental relations: We are now ready to prove the following three important relations: (a) {Ω, Ω} = 0, (227) (b) {{Ψ, Ω}, Ω} = 0, (228) (c) {H 0 + a V a, Ω} = 0. (229) Proofs of these relations are straightforward using the formulas already developed. We only show how (a) is proved as an example. Proof of (a): Since Ω is fermionic, this is a non-trivial relation. Recalling Ω = G a U a, constsys-86

87 we have {Ω, Ω} = { G a U a, G b U b} = {G, G} 2 {G, a U a } + { a U a, b U b} (230) (2) gives Using (Ia), the second term of (230) becomes {G, G} = 2G a U a. (231) 2 {G, a U a } = 2 {G, a } U a + 2 a {G, U a } = 2G a U a + 2 a U a b U b. (232) As for the third term of (230), we use (IIa) and get { a U a, b U b} = { a U a, b } U b b { a U a, U b} = a {U a, b } U b b { a, U b} U a = 2 a {U a, b } U b = 2 a U a b U b. (233) constsys-87

88 Adding all the contributions, we get {Ω, Ω} = 0. // Ψ independence of Z Ψ : Finally we show the Ψ-independence of Z Ψ to finish the proof of the BV theorem. Collectively denote all the variables by ϕ = (q, p, η, ) and make a change of variables corresponding to the BRST transformation ϕ ϕ = ϕ + {ϕ, Ω} µ (234) µ is a time-independent fermionic parameter, which nevertheless may depend functionally on the dynamical variables. It is easy to show that the action is invariant. First ) ( δ dt (p A q A + a η a = dt δp A q A ṗ A δq A ) +δ a η a a δη a, (235) constsys-88

89 where we used integration by parts. Now insert the change of variables: δq A = { q A, Ω } µ = Ω p A µ, (236) δp A = {p A, Ω} µ = Ω qaµ, (237) δη a = {η a, Ω} µ = ( / a )Ωµ ( left derivative ), (238) δ a = { a, Ω} µ = ( Ω/ η a )µ ( right derivative ). (239) Assuming no contributions from the boundary, we get ) δ dt (p A q A + a η a ( = dt Ω ) Ω q Aµ qa ṗ A µ + ( Ω/ η a )µ η a a ( / a )Ωµ p A ( = dt q A Ω ) q + ṗ Ω A A + ( Ω/ η a ) η a + a ( / a )Ω µ p A = dt dω dt µ = 0 (240) constsys-89

90 Next, the Hamiltonian is invariant since δh Ψ = {H Ψ, Ω} µ = 0. What remains is the transformation property of the functional measure. The Jacobian is given by d ϕ = J dϕ, (241) J = det ϕj ϕ ( i = det 1 + ({ ϕ j, Ω } µ ) ) ϕ i Since it will suffice to consider small µ, we can expand. (242) det (1 + A) = e Tr ln(1+a) = 1 + Tr A +. (243) constsys-90

91 So we get J = 1 + = 1 + q A = 1 + Ω ({ q A, Ω } µ ) + q A ( ) Ω µ p A + p A µ q A µ p A q Ω + ( A p A µ Ω = 1 µ ) Ω + q A p A p A q A = 1 {µ, Ω} ({p A, Ω} µ) + p ( A Ω ) q Aµ + e {µ,ω} = e i(i{µ,ω}). (244) Now let us choose µ = dt (Ψ Ψ), (245) where Ψ is infinitesimally different from Ψ. Then, we have d ϕ = dϕe i dt(i{ψ,ω} i{ψ,ω}). (246) constsys-91

92 Inserting it into the functional integral, we prove Z Ψ = Z Ψ // (247) constsys-92

93 4.4 Proof of unitarity: Equivalence with canonical functional Integral Having proved the BV theorem, we now prove the equivalence with the canonical functional integral, i.e. the functional integral involving only the physical degrees of freedom. The proof proceeds in two steps. First step: The first step is to derive a convenient representation of the physical integral. path Let us make a canonical transformation 5 of the form (p i, q i ) (p i, q i ), i = 1 n (248) where the new coordinates (q i, p i ) i=1 n are split into the 5 Canonical transformation is the one which does not change the standard form of the symplectic structure. Namely, ω = i dp i dq i = i dp i dq i. constsys-93

94 physical part (q u, p u ) u=1 n m and the unphysical part (q α, p α ) α=1 m. Moreover, we take p α to be the gauge-fixing constraints themselves, namely p β = Θβ, β = 1 m (249) Then, since canonical transformation does not change the Poisson bracket structure, we have (recall T α are the constraints) det T { } α q = det T β α, p β = det { T α, Θ β} 0 (250) By assumption, Θ β are chosen so that this holds at every point in the phase space. This then guarantees that using the m constraints T α (p, q ) = 0 one can solve for the m variables q α in terms of the rest. For example, consider the constraints near the point where q α are very small. constsys-94

95 Then 0 = T α (p, q u, q β ) = T α (p, q u, 0) + T α q β(p, q u, 0)q β (251) Since the matrix M αβ = T α q β (p, q u, 0) is invertible, we can solve for q α as q α = (M 1 ) αβ T β (p, q u, 0) (252) Since this can be continued away from the origin of q α, we conclude that we can always solve q α in terms of the other variables: q α = q α (q u, p i ) (253) Now the physical partition function Z is given by Z = (dp i dq i )e i dt(p i q i H 0 ) δ(p β ) δ(q α q α (q u, p u )) (254) constsys-95

96 Note that because of the presence of the δ-functions δ(p β ), we can drop the dependence on p β in the expression of q α, so that p i p u. Finally, we will make use of the following : The Liouville measure is invariant. That is (dp i dq i ) = (dp i dq i ). The last δ-function can be rewritten as δ(q α q α (q u, p u )) δ(p β ) = ( ) Tα δ(t α )det δ(p β ) q β = δ(t α )det { T α, Θ β} δ(θ β ) (255) We then finally obtain a useful form of the physical partition function constsys-96

97 Z = (dpi dq i ) δ(t α ) δ(θ β )det { T α, Θ β} e i dt(p i q i H 0 ) Clearly this form is quite natural. Second Step: We now want to prove that the above form is reproduced from the BV theorem. In the BV theorem, let us take the gauge fixing function to be of the form Θ α = λ α + F α (q, p) (256) It is convenient to divide the ghosts as η a = ( η α, η α ), a = ( α, α ) (257) constsys-97

98 Then the BV action takes the form S Ψ = dt (p i q i + λ ) λα α + α η α + α η α H Ψ (258) where H Ψ = H 0 T α λ α [ λ α F α + H (259) H = a V a b ηb + i a b { χ a, U b cd a {χ a, G b } η b + c χ a U c ab ηb } η c η d ] (260) Recalling the extended definitions of G a, χ a (as in (188) and (189)), Vb a Uab c, we can write each term in H more explicitly: and a V a b ηb = α V α β ηβ (261) a {χ a, G b } η b = α η α + γ {F γ, T β } η β (262) c χ a U c ab ηb = γ iλ α U γ αβ ηβ (263) { } { } a b χ a, U b cd η c η d = α β F α, U β γδ η γ η δ (264) constsys-98

99 We now make the following rescaling by a global parameter ɛ, which will be taken to zero: (i) F α 1 ɛ F α (265) (ii) λα ɛ λ α (266) (iii) α ɛ α (267) The functional measure is invariant since (ii) and (iii) are bosonic and fermionic and the factors of ɛ cancel. The purpose of this rescaling is to get rid of the terms λ α λ α and α η α, which scale like ɛ and vanish as ɛ 0. All the other terms in the action remain unchanged. Now since η α has dissappeared, η α only appears as α η α in (262). Thus upon integration over η α, we get δ( α ). constsys-99

100 This in turn allows us to integrate out α and the action simplifies drastically to S Ψ = dt ( p i q i H 0 + T α λ α + λ α F α i α {F α, T β } η β) (268) Now upon integration over α and η β, we get det { T α, F β}. Thus we precisely get the canonical path integral with the gauge-fixing function Θ α replaced by F α (q, p). q.e.d. Remark: In the application to non-abelian gauge theory, to retain the relativistic form, we will not perform the rescaling and simply integrate out α. The details will be discussed later. 5 Application to non-abelian gauge theory In this chapter, we apply the methods developed previously to the important case of non-abelian gauge theory. constsys-100

101 5.1 Dirac s Method First, we make use of the Dirac s method. The basic procedure is the same as the abelian case already described but the details are more involved. For simplicity, we take the gauge group G to be compact and normalize the group metric to be of the form g ab = δ ab. Thus as far as the group indices are concerned, we need not distinguish upper and lower ones. Where convenient, we will use the notations A B A a B a, (269) (A B) a f abc A b B c. (270) Due to the complete antisymmetry of the structure constant, we have the cyclic identity A (B C) = B (C A) = C (A B). (271) constsys-101

102 Conjugate momenta and the basic Poisson bracket: The Lagrangian is given by L = 1 d 3 x F µν F µν, (272) 4 F µν = A ν,µ A µ,ν + ga µ A ν. (273) We will regard A a µ as our fundamental variables. The conjugate momenta are found by the variation with respect to A a µ = Aa µ,0 : δl = 1 2 d 3 x F µν δf µν = + d 3 x F µ0 δa µ,0 Π µ a = F µ0 a. (274) The equal time (ET) Poisson bracket is defined by } {A aµ (x), Πνb (y) = δ a b δν µ δ( x y). (275) From (274) we get the primary constraints: Π 0 a = F 00 a = 0. (276) constsys-102