8.324 Relativistic Quantum Field Theory II MIT OpenCourseWare Lecture Notes Hong Liu, Fall 200 Lecture 2 3: GENERAL ASPECTS OF QUANTUM ELECTRODYNAMICS 3.: RENORMALIZED LAGRANGIAN Consider the Lagrangian of quantum electrodynamics in terms of the bare quantities: We use the convention: L F B = F ψ B (γ µ ( µ ie B A B B i µ ) m B )ψ B. 4 () {γ µ, γ β } = 2η, (2) γ 0 2 =, γ 0 = γ 0, γ i = γ i, ψ = ψ γ 0, {/k} = k µ γ µ, k/ 2 = k 2. where, in four dimensions, (η ) = diag(,,, ). We expect to find the mass and field renormalizations. Note: we will omit the B signifying bare quantities in what follows. ψ : + +... A µ : + +... (3) along with vertex corrections e : + +.... (4) We will look at how to introduce renormalized quantities. 3..: Fermion self-energy We have that and S αβ (x) = 0 T (ψ α (x)ψ β (0) 0, (5) S αβ (k) = α β + α β + α β +... γ P I = Σ α = = S 0 (k) + S 0 ( Σ)S 0 + S 0 ( Σ)S 0 ( Σ)S 0 +... S 0 + ΣS 0, (6)
where we have omitted the spinor indices in the second and third lines: these are to be read as matrix equations. Hence, we have that S = S 0 + Σ. (7) Recall that and so and we have for the fully interacting two-point function S 0 =, Σ = +..., (8) ik/ + m B S = (ik/ + m B ) + Σ, (9) S =. (0) ik/ + m B iϵ Σ Note that Σ = Σ(k/), since it can only depend on k/ and k 2. Even though it is a function of matrix, we can treat it as an ordinary function. The physical mass is defined by k/ = im, so that We note that, again, Σ will be divergent. Near the pole, we have that m + m B Σ(im) = 0. () Z 2 S ik/ + m iϵ with Z2 = + i dσ dk/. The relations between the bare and physical quantities are given by k/=im where δm = Σ(im) and ψ is the renormalized field. 3..2: Photon self-energy (2) m B = m + δm, ψ B = Z 2 ψ (3) Similarly, we have that and D (x) = 0 T (A B µ (x)a B β (0) 0, (4) P I = iπ D (k) = µ β + µ β + µ β +... γ δ Hence, Recall that = D 0 (k) + D 0 (iπ)d 0 + D 0 (iπ)d 0 (iπ)d 0 +... = D 0 iπd 0 (5) (id) = (id 0 ) Π. (6) [ ] 0 = k 2 η ( ) kµ k β iϵ k 2 = [P T + P L ], k 2 iϵ id where we have defined the transverse projector P T η kµ k2, and the longitudinal projector P L kµ k2. Note that it is not a coincidence that the propagator can be built from these two tensors, η and k µ k β : they are the only two two-tensors allowed by symmetry. These projectors satisfy the properties β k P T = P µ P L = P µ P L P T βα T β, P L βα L α, P T βα = 0. (7) β k 2
Note: T and L are just labels here, and the placing of these indices does not carry meaning. Hence, we have that [ ] (id 0 ) = k 2 P T + PL. (8) We may also expand Π as Therefore, Π = P T f T (k 2 ) + P L f L (k 2 ) k µ k β = η f T + k 2 (f L f T ). (9) k 2 (id) = P T (k 2 f T ) + P L ( f L ), (20) and we have for the full interacting photon two-point function, [ D = i PT + P k 2 L f k 2 T η f L ]. (2) We observe that if f T,L (k 2 = 0) 0, a mass will be generated for the photon. Because Π comes from PI diagrams, it should not be singular at k 2 = 0, and so f L f T = O(k 2 ), as k 0. We will show that gauge invariance ensures that no mass is generated from the loop corrections. 3..3: Ward identities Consider the path integral for the generating functional: ˆ Z [J µ, η, η] = DA µ DψD ψe is[a µ,λ, λ] (22) where S = S QED + d 4 x J µ A B µ + ηψ B +ψ B η, where we note explicitly these couplings are in terms of bare quantities. L QED = F F iψ (γ µ D µ m)ψ ( µ A µ ) 2. (23) 4 2 We define the generating functional for connected diagrams, W [J µ, η, η ], by Z [J µ, η, η ] = e iw [Jµ,η,η ]. (24) For example, 0 T (ψ α (x)ψ β (y)) 0 = i δ2 W [J µ, η, η ], δη α (x)δη β (y) J=η=η =0 T (A B (x)a B β (y)) = i δ2 W [J µ, η, η ] 0 µ 0. δj µ (x)δj β (y) J=η=η =0 Recall, for infinitesimal gauge transformations, δa µ = µ λ, δψ = ie B λψ, and δ ψ = ie B λ ψ. Consider a change of variables in the path integral: A µ A µ = A µ + δa µ, ψ ψ ψ = ψ + δψ, ψ = ψ + δ ψ. Then we have ˆ ˆ DA µ µdψ Dψ e is[a,λ,λ ] = DAµ DψD ψe is[aµ,λ, λ ], (25) as this is just of a change of the dummy integration variables. Note that the measure is unchanged by this shift: DA µdψ Dψ = DA µ DψD ψ, (26) 3
and the action for the two sets of variables are related by S [ A µ, ψ, ψ ] = S [ A µ, ψ, ψ ] ˆ ˆ d 4 x µ A µ 2 λ + d 4 x J µ µ λ + ie B λ ηψ ie B λ ψη. (27) Hence, we must have ˆ ˆ [ ] ψe is[a,λ, λ ] d 4 x λ(x) DA A µ J µ µ DψD 2 µ µ + ie B ( ηψ ψη) = 0. (28) Since δz δw A µ (x) i δj µ (x) = Z δj µ (x), we have that that is, or, written in momentum-space, If we now write with k µ P L = k β, the Ward identity reduces to and so δz δw ψ(x) i = Z, δη (x) δη (x) δz δw ψ(x) i = Z, δη(x) δη(x) δ 2 W 2 µ δj µ (x)δj β (y) + β δ (4) (x y) = 0, (29) J=η=η =0 i 2 µ D (x y) + β δ (4) (x y) = 0, (30) i k 2 k µ D (k) + k β = 0. (3) D (k) = P T D D L (k 2 T (k 2 ) + P L ), (32) i k 2 k β D L (k 2 ) + k β = 0, (33) D L (k 2 ) = i, (34) k 2 and the longitudinal part of the two-point function is completely determined. Comparing this with (2), we find that f L (k 2 ) = 0, and we thus conclude that Π is purely transverse. That is, from (9), we have that ( ) k µ k β Π = η k 2 f T (k 2 ). (35) For Π (k) to be non-singular at k = 0, we must have f T (k 2 ) = k 2 Π(k 2 ), (36) where Π(0) is non-singular. Hence, for the two-point function in the interacting theory, we have [ ] i P T D = + P L. (37) k 2 iϵ Π(k 2 ) Remarks:. The longitudinal part of D does not receive any loop corrections: it is completely determined by the Ward identities. The physics should not depend on this part. For example, in the Landau gauge, = 0, D is purely transverse. 4
2. Since Π(k 2 ) is non-singular at k 2 = 0, the photon remains massless to all orders. There are exceptions to this: it is not true in quantum electrodynamics in + dimensions, or in theories where an additional Higgs field is introduced. 3. The residue at the k 2 = 0 pole is given by Z = Π(0), and we have that 3 near k 2 = 0. The renormalized field is given by A B µ = Z 3 A µ. id T Z 3 P T (38) k 2 iϵ 5
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