Solutions to Problem Set #3 in Quantum Field Theory, Assignments of Physics

Download Solutions to Problem Set #3 in Quantum Field Theory and more Assignments Physics in PDF only on Docsity! PHY–396 K. Solutions for problem set #3. Problem 1(a): Let’s start with the scalar fields Φ(x) and Φ†(x). Similar to the EM covariant derivatives, the non-abelian covariant derivatives may be integrated by parts using ∂µ(Φ †Φ) = Dµ(Φ †Φ) = (DµΦ †)Φ + Φ†(DµΦ), ∂µ(Φ †DµΦ) = (DµΦ †)(DµΦ) + Φ†(DµD µΦ), ∂µ((D µΦ†)Φ) = (DµD µΦ†)Φ + (DµΦ†)(DµΦ), (S.1) etc., etc. Consequently, the Euler–Lagrange equations for the Φ and Φ† fields may be written in a manifestly covariant form as Dµ ( ∂L(Φ, DΦ) ∂(DµΦ) ) − ∂L(Φ, DΦ) ∂Φ = 0, Dµ ( ∂L(Φ†, DΦ†) ∂(DµΦ†) ) − ∂L(Φ †, DΦ†) ∂Φ† = 0. (S.2) For the Lagrangian (13) at hand, ∂L ∂(DµΦ) = DµΦ†, ∂L ∂(DµΦ†) = DµΦ, ∂L ∂Φ = −m2Φ†, ∂L ∂Φ† = −m2Φ, (S.3) so the covariant Euler–Lagrange equations for the scalar fields are (DµD µ + m2 )Φ†(x) = 0 and (DµD µ + m2 )Φ(x) = 0. (S.4) Now consider the vector fields Aaµ(x). In the previous homework (set#2, problem 2(e)), we saw that an infinitesimal variation of the vector fields changes the Yang–Mills Lagrangian — i.e., the first term on the RHS of eq. (1) — by δ [ −1 2g2 tr (Fµν(x)F µν(x)) ] = 1 g2 ∑ a δAaν(x)× (DµFµν)a(x) + a total derivative. (S.5) The remaining terms in the Lagrangian (1) depend on the vector fields through the covariant 1 derivatives DµΦ and dµΦ †. Thus δ(DνΦ(x)) = ∑ a δAaν(x)× i2 λ aΦ(x), δ(DνΦ †(x)) = ∑ a δAaν(x)× −i2 Φ(x) †λa, (S.6) and therefore δ ( DνΦ†DνΦ − m2Φ†Φ ) = − ∑ a δAaν(x)× Jaν(x) (S.7) where Jaν(x) = − i2 (D νΦ†(x))λaΦ(x) + i2 Φ †(x)λa(DνΦ(x)). (S.8) Combining eqs. (S.5) and (S.7) and integrating ∫ d4x, we obtain the action variation δS = ∫ d4x ∑ a δAaν(x)× ( 1 g2 (DµF µν)a(x) − Jaν(x) ) . (S.9) Since this variation must vanish for any δAaν(x), the vector fields must satisfy (DµF µν)a(x) = g2Jaν(x). (S.10) Or in matrix notations, DµF µν(x) = g2Jν(x). Q.E .D. Problem 1(b): Taking Dν of both sides of eq. (2), we have g2 DνJ ν = Dν(DµF µν) = −12 [Dµ, Dν ]F µν = − i2 [Fµν , F µν ] = 0. (S.11) where the second equality follows from Fµν = −F νµ, the second from [Dµ, Dν ]F = iFµνF for any adjoint field F (x) (cf. homework set#2, problem 2), and the third from the fact that Fµν(x) commutes with itself. Q.E .D. 2 Problem 2(b): In terms of the Hamiltonian and Lagrangian densities, eq. (4) means H = −Ȧ · E − L . (4′) Expressing all fields in terms A, E, and A0, we have Ȧ = −E − ∇A0, −Ȧ · E = E2 + E · ∇A0, L = 12 ( E2 − (∇×A)2 ) + 12m 2(A20 −A2) − (A0J0 −A · J), (S.20) and consequently, H = 12E 2 + E · ∇A0 − 12m 2A20 + A0J0 + 1 2(∇×A) 2 + 12m 2A2 − A · J. (S.21) Taking the ∫ d3x integral of this density and integrating by parts the E · ∇A0 term, we arrive at the Hamiltonian (5). Q.E .D. Problem 2(c): Evaluating the derivatives of H in eq. (6) gives us δH δA0(x) ≡ ∂H ∂(A0) − ∇i ∂H ∂(∇iA0) = −m2A0 + J0 − ∇iEi. (S.22) If the scalar field A0 had a canonical conjugate π0(x, t), its time derivative ∂π0/∂t would be given by the right hand side of eq. (S.22). But the a0(x, t) does not have a canonical conjugate, so instead of a Hamilton equation of motion we have a time-independent constraint (6), namely m2A0 = J0 − ∇ · E . (S.23) In the massless EM case, a similar constraint gives rise to the Gauss Law ∇ · E = J0. But the massive vector field does not obey the Gauss Law; instead, eq. (S.23) gives us a formula for the scalar potential A0 in terms of E and J0. 5 However, Hamilton equations for the vector fields A and E are honest equations of motions. Specifically, evaluating the derivatives of H in the first eq. (7), we find δH δEi(x) ≡ ∂H ∂(Ei) − ∇j ∂H ∂(∇jEi) = Ei + ∇iA0, which leads to Hamilton equation ∂ ∂t A(x, t) = −E(x, t) − ∇A0(x, t). (S.24) Similarly, in the second eq. (7) we have δH δAi(x) ≡ ∂H ∂(Ai) − ∇j ∂H ∂(∇jAi) = m2Ai − J i − ∇j(jik(∇×A)k) and hence Hamilton equation ∂ ∂t E(x, t) = m2A − J + ∇× (∇×A). (S.25) Problem 2(d): In 3D notations, the Euler–Lagrange field equations (8) or ∂µF µν −m2Aν = Jν become ∇ · E − m2A0 = J0 , (S.26) −Ė + ∇×B + m2A = J, (S.27) where E def = −Ȧ − ∇A0, (S.28) B def = ∇×A. (S.29) Clearly, eq. (S.26) is equivalent to eq. (S.23) while eq. (S.27) is equivalent to eq. (S.25) (provided B is defined as in eq. (S.29)). Finally, eq. (S.28) is equivalent to eq. (S.24), although their origins differ: In the Lagrangian formalism, eq. (S.28) is the definition of the E field in terms of A0, A and their derivatives, while in the Hamiltonian formalism, E is an independent conjugate field and eq. (S.24) is the dynamical equation of motion for the Ȧ. Q.E .D. 6 Problem 3: Let start with the [Â, Ĥ] commutator. In light of eq. (11), we have [Âi(x), Ĥ] = ∫ d3y [ Âi(x), ( 1 2Ê 2 + 1 2m2 (Ĵ0 −∇·Ê)2 + 12(∇× Â) 2 + 12m 2Â2 − Ĵ·Â ) (y) ] (S.30) where all operators are at the same time t as the Âi(x, t). Since all the Âi(x) operators commute with each other at equal times, the last three terms in the Hamiltonian density do not contribute to the commutator (S.30). But for the first term we have [Âi(x), 12Ê 2(y)] = 12{Ê j(y), [Âi(x), Êj(y)]} = 12{Ê j(y),−iδijδ(3)(x− y)} = −iδ(3)(x− y)× Êi(y), (S.31) while for the second term we have [ Âi(x), ( Ĵ0(y) − ∇·Ê(y) )] = 0 − ∂ ∂yj [Âi(x), Êj(y)] = +iδij ∂ ∂yj δ(3)(x− y) (S.32) and hence[ Âi(x), 1 2m2 ( Ĵ0(y) − ∇·Ê(y) )2] = 1 m2 ( Ĵ0(y) − ∇·Ê(y) ) ×+iδij ∂ ∂yj δ(3)(x− y) = Â0(y)× i ∂ ∂yi δ(3)(x− y) (S.33) where the second equality follows from eq. (10). Plugging these all these commutators into eq. (S.30) and integrating over y, we obtain [Âi(x), Ĥ] = ∫ d3y ( −iδ(3)(x− y)× Êi(y) + Â0(y)× i ∂ ∂yi δ(3)(x− y) + 0 + 0 + 0 ) integrating by parts = ∫ d3y (−i)δ(3)(x− y)× ( Êi(y) + ∂ ∂yi Â0(y) ) = −i ( Êi(x) + ∂ ∂xi Â0(x) ) . (S.34) 7