Quantization of Non-abelian Gauge Theories: BRST symmetry

Transcription

1 Quantization of Non-abelian Gauge Theories: BRST symmetry Zhiguang Xiao May 9, 2018

2 :Becchi-Rouet-Stora-Tyutin The gauge fixed Faddeev-Popov Lagrangian is not invariant under a general gauge transformation, though the path integral is invariant. L FB = 1 4 (Fi µν) 2 + i ψd/ψ m ψψ c a c a gf abc c a µ (A b µc c ) 1 2ξ ( µ A a µ) 2 There is a residue symmetry: To see this, we introduce an auxiliary field B: No kinetic term of B, and can be integrated out to obtain the L FB. L BRST = 1 4 (Fi µν) 2 + i ψd/ψ m ψψ c a µ D ac µ c c + ξ 2 (Ba ) 2 + B a µ A a µ Path integral: [db][d ψ][dψ][da µ][dc][d c] exp{i d 4 xl BRST } ξ 2 (Ba ) 2 + B a µ A a µ = ξ 2 (Ba + 1 ξ µ A a µ) 2 1 2ξ ( µ A a µ) 2 EOM of B a : B a = 1 ξ ( µaaµ )

3 L BRST L YM c a µ D ac µ c c + ξ 2 (Ba ) 2 + B a µ A a µ The L BRST has a Global BRST symmetry: infinitesimal parameter ϵ, an anticommuting grassman number, independent of x. BRST transformation and classical BRST operator Q: (iϵ, c a, i c a are real grassman variables, Hermite conjuate for grassman number (χψ) = ψ χ ) δa a µ =ϵd ac µ c c ϵqa a µ, δψ =igϵc a t a ψ ϵqψ, δ ψ = igϵc a ψt a ϵq ψ, δc a = 1 2 gϵfabc c b c c ϵqc a, δ c a =ϵb a ϵq c a, δb a =0 ϵqb a. Q =D ac µ c c δ (y) δa a µ(y) + igca t a ijψα(y) j δ δψα(y) i igca ψj α (y)t a δ ji δ ψ α(y) i 1 2 gfabc c b c c δ (y) δc a (y) + δ Ba (y) δ c a (y) (repeated indices are summed, including continuous indices y.) Ex. (1) Q 2 = 0.

4 L BRST L YM c a µ D ac µ c c + ξ 2 (Ba ) 2 + B a µ A a µ The transformation of A µ and ψ is just a gauge transformation with parameter :α a (x) = gϵc a (x), L YM is invariant. δ(b a µ A a µ) = B a µ (ϵd ac µ c c ) δ( c( µ D ac µ c c )) = ϵb a }{{} δ c ( µ (D ac µ c c ) c a µ D ac µ 1 c 2 gϵfcb c b c c ) }{{} δc c = + gϵ c a µ ( ) g c a µ ((ϵd be µ c e )f abc c c ) }{{} δa b f abc ( ) = 1 2 µ(fab c c b c c ) 1 2 gad µf adc f cb c c b c c + ( µc b )f abc c c + ga d µf bde c e f abc c c =ga d µ( 1 2 fadc f cb c + f cdb f acc )c b c c =ga d µ( 1 2 fadc f cb c 1 2 fb dc f cc a 1 2 fc dc f cab )c b c c = 0

5 It can be proved that Q 2 = 0, Q carry ghost number 1. Grassman odd operator. The L BRST can be recast into L BRST = L YM[A, ψ] + QΨ; Ψ = c a µ A a µ + ξ 2 ca B a L YM[A, ψ] is the gauge invariant Yang-Mills Lagrangian. In general, for gauge fixing G a [A µ] = f a (A µ) ω a (x), Ψ = c a f a [A µ] + ξ 2 ca B a. QΨ = B a f a [A µ] c a δf a [A µ] δa b µ(y) Dbc c c (y) + ξ 2 Ba B a (repeated indices are summed, including continuous indices y.) Now the invariance of L BRST is BRST invariant can be seen easily: (1)Since the BRST trans for A µ and ψ is just a special gauge transformation with parameter :α a (x) = gϵc a (x), L YM is invariant. δ BRST L YM = ϵql YM = 0 (2) Since Q 2 = 0, δ BRST QΨ = Q 2 Ψ = 0.

6 In the whole fock state space, including unphysical polarization and ghost excitations, we can define a hermitian operator ( quantum operator) Q, From Q 2 ϕ = 0, we have δ BRST ϕ = ϵqϕ = i[ϵq, ϕ] [Q, [Q, ϕ] ±] = Q 2 ϕ = 0 [Q 2, ϕ] = Q 2 ϕ ϕq 2 = 0 for any ϕ. Either Q 2 = 0 or Q 2 = I, Q 2 ghost charge = 2, it can not be I. So Q 2 = 0. We want to identify the physical states in the fock state space, in a Lorentz covariant way. The physical amplitude should be invariant under the change of the gauge fixing function, changing of the gauge, or changing of ξ an arbitrary change of the Ψ, δψ. 0 = δ α ph β ph = i α ph Q δψ β ph = α ph [Q, δψ] + β ph Because δψ is arbitrary, the only possibility is: Q Phy states = 0. For any physical state α ph, we can not physically distingush it from α ph + Q γ, (γ any fock state), since β ph α ph + Qγ = β ph α ph + β ph Q γ = β ph α ph + Qβ ph γ = 0

7 Physical state: { Physical state condition:q Phy states = 0 α ph α ph + Q γ The whole fock space is divided into 3 parts: H 0, H 1, H 2: H 0: α 0, Q α 0 = 0, but α 0 = Q γ, H ph is a subspace of H 0. H 1: α 1, Q α 1 = 0. Unphysical states. H 2: α 2 = Q γ. (1) Q α 2 = 0, (2) zero norm: α 2 α 2 = γ Q Q γ = γ Q 2 γ = 0, (3) orthogonal to H 0, α 0 α 2 = α 0 Q γ = Qα 0 γ = 0. BRST closed states: Q α = 0, Kernel of Q BRST exact states: α = Q γ, Image of Q kernel of Q So the physical state space H ph is defined as H ph = { α : Q α = 0} { β = Q γ } = {Kernel of Q} {Image of Q} = Cohomology of Q

8 (reference Weinberg, The Quantum theory of fields, II) Consider the asymptotic particle state, in the t ±, interactions are adiabaticaly turned off g 0. BRST transformation: [Q, A a µ ] = i µca, [Q, c a µ ] + = 0, [Q, c] + = ib a EOM ===== i ξ µaa,µ A µ (x) = c(x) = d 3 k (2π) 3 2E (aµ k e ik x + a µ k eik x ) d 3 k (2π) 3 2E (c ke ik x + c k eik x ), c(x) = d 3 k (2π) 3 2E ( c ke ik x + c k eik x ) [Q, a µ k ] = kµ c k, [Q, c k ] = 1 ξ k a k, [Q, c k ] = 0

9 [Q, a µ k ] = kµ c k, [Q, c k ] = 1 ξ k a k, [Q, c k ] = 0 The superscript a and subscript k are supressed in the following. A physical state: ψ, satisfies Q ψ = 0. Adding another vector boson, e, ψ = e µa µ ψ. Physical condition: 0 = Q e, ψ = e µk µ c ψ, so k µ e = 0. ( k ϵ + = 0, k ϵ 0. ) for e µ ϵ +µ, k ϵ + = 0, ϵ + µ a µ ψ satisfies Physcal condition. for e µ ϵ µ, k ϵ 0, ϵ µ a µ ψ H 1 is not a physical state. for e µ ϵ Tµ, k ϵ T = 0, ϵ T µa µ ψ satisfies physical condition. Add a ghost: c ψ = Q e, ψ /(e k), (e k 0,) is BRST exact 0, H 2. Q c ψ = 1 ξ kµ a µ ψ, BRST exact, ϵ +µ a µ ψ H 2., e µ + αk µ, ψ = ψ + Qα c ψ e µ, ψ, this is the gauge equivalent condition for the vector boson. Q c ψ = 1 ξ kµ a µ ψ = 0 c ψ is not physical H 1. ϵ T a ψ H 0, c ψ H 1, ϵ a ψ H 1, ϵ + a ψ H 2, c ψ H 2. For each equivalent class of states, we choose ϵ T a ψ to represent the class, H ph can be restricted to states with only transverse polarizations A; tr ϵ T a ψ. So the physical H ph H 0 is ghost free.

10 We can choose the external polarizations contain only the transverse polarizations. Because the hamiltonian H commutes with Q, then QS = SQ. QS A, tr = SQ A, tr = 0 H ph, S A, tr = C; tr + Qexact. The S matrix in the full fock space is (pseudo-)unitary: S S = S S = 1. ( intermediate states include a set of orthogonal independent basis {C} of the whole fock space H 0 + H 1 + H 2). The restricted S-matrix on the transversely polarized states is also Unitary : S A; tr = C; tr + Q χ, S B; tr = D; tr + Q ψ C; tr = X;tr X; tr X; tr S A; tr, D; tr = X;tr X; tr X; tr S B; tr δ(a, B) = A; tr B; tr = A; tr S S B; tr = C; tr + Qχ D; tr + Qψ = C; tr D; tr = A; tr S X; tr X; tr S B; tr X;tr (See Kugo and Ojima, Prog. Theo. Phys., V60, NO for details.)