etc., etc. Consequently, the Euler Lagrange equations for the Φ and Φ fields may be written in a manifestly covariant form as L Φ = m 2 Φ, (S.

Transcription

1 PHY 396 K. Solutions for problem set #3. Problem 1a: Let s start with the scalar fields Φx and Φ x. Similar to the EM covariant derivatives, the non-abelian covariant derivatives may be integrated by parts using µ Φ Φ = D µ Φ Φ = D µ Φ Φ + Φ D µ Φ, µ Φ D µ Φ = D µ Φ D µ Φ + Φ D µ D µ Φ, S.1 µ D µ Φ Φ = D µ D µ Φ Φ + D µ Φ D µ Φ, etc., etc. Consequently, the Euler Lagrange equations for the Φ and Φ fields may be written in a manifestly covariant form as Φ, DΦ D µ D µ Φ Φ, DΦ Φ Φ, DΦ = 0, D µ D µ Φ Φ, DΦ Φ = 0. S.2 For the Lagrangian 13 at hand, D µ Φ = Dµ Φ, D µ Φ = Dµ Φ, Φ = m2 Φ, Φ = m 2 Φ, S.3 so the covariant Euler Lagrange equations for the scalar fields are D µ D µ + m 2 Φ x = 0 and D µ D µ + m 2 Φx = 0. S.4 Now consider the vector fields A a µx. In the previous homework set#2, problem 2e, 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 [ ] 1 δ 2g 2 tr F µνxf µν x = 1 g 2 δa a νx D µ F µν a x + a total derivative. a S.5 The remaining terms in the Lagrangian 1 depend on the vector fields through the covariant 1

2 derivatives D µ Φ and d µ Φ. Thus δd ν Φx = a δa a νx i 2 λa Φx, δd ν Φ x = a δa a νx i 2 Φx λ a, S.6 and therefore δ D ν Φ D ν Φ m 2 Φ Φ = a δa a νx J aν x S.7 where J aν x = i 2 Dν Φ xλ a Φx + i 2 Φ xλ a D ν Φx. S.8 Combining eqs. S.5 and S.7 and integrating d 4 x, we obtain the action variation δs = d 4 x a δa a νx 1 g 2 D µf µν a x J aν x. S.9 Since this variation must vanish for any δa a νx, the vector fields must satisfy D µ F µν a x = g 2 J aν x. S.10 Or in matrix notations, D µ F µν x = g 2 J ν x. Q.E.D. Problem 1b: Taking D ν of both sides of eq. 2, we have g 2 D ν J ν = D ν D µ F µν = 1 2 [D µ, D ν ]F µν = i 2 [F µν, F µν ] = 0. S.11 where the second equality follows from F µν = F νµ, the second from [D µ, D ν ]F = if µν F for any adjoint field F x cf. homework set#2, problem 2, and the third from the fact that F µν x commutes with itself. Q.E.D. 2

3 Problem 1c: First, let s rewrite the non-abelian currents S.8 in matrix notations. Defining J ν a Jaν λ a 2 as in eq. 2 and using the normalization trλa λ b = 2δ ab of the Gell Mann matrices λ a, we have J aν = trλ a J ν = λ a j i J ν j i S.12 implicit summation over i, j = 1,..., N. On the other hand, eqs. S.8 amount to J aν = i 2 Dν Φ i λ a j i Φ j + i 2 Φ j λ a j i Dν Φ j S.13 again, implicit summation over i, j = 1,..., N. Comparing the factors multiplying the λ a j i in these two formulæ, we immediately see that J ν i j = i 2 Φ j D ν Φ i + i 2 Dν Φ j Φ i S.14 In matrix notations, this amounts to J ν = i 2 Φ Dν Φ + i 2 Dν Φ Φ S.15 where denotes the matrix product of a column vector and a raw vector. Note the order of the factors: Φ Φ is a matrix, unlike Φ Φ which is just a single number. The covariant derivative of a matrix product satisfies the usual Leibniz rule D µ Φ Φ µ Φ Φ + i[a µ, Φ Φ ] = D µ Φ Φ + Φ D µ Φ. S.16 Applying this rule to the matrix products in eq. S.15, we find D ν J ν = 2 i D νd ν Φ Φ + 2 i D νφ D ν Φ = 2 i D 2 Φ Φ + D ν Φ D ν Φ + 2 D i ν Φ D ν Φ + Φ D 2 Φ S.17 = i 2 D2 Φ Φ + i 2 Φ D2 Φ. But for scalar fields satisfying their equations of motion S.4, the D 2 Φ D ν D ν Φ is equal to 3

4 m 2 Φ and likewise D 2 Φ = m 2 Φ. Consequently, eq. S.17 becomes D ν J ν = + i 2 m2 Φ Φ i 2 m2 Φ Φ = 0. In other words, when the scalar fields satisfy their equations of motion, the non-abelian current is covariantly conserved. Q.E.D. PS: In general, a non-abelian gauge symmetry may act non-trivially on all kinds of scalar, spinor, etc., fields; collectively all those fields except the gauge fields themselves are called matter. The matter fields always interact with the gauge fields because the kinetic-energy terms in their Lagrangians involve covariant derivatives D µ. Consequently, the gauge fields satisfy D µ F µν = g 2 J ν where now all the matter fields contribute to the non-abelian current J ν, and the explicit formula for this current could be rather complicated. However, when all the matter fields satisfy their equations of motion, the current always ends up being covariantly conserved. Problem 2a: In 3D notations, the Lagrangian 3 for the massive vector field is L = 2 1 E 2 B m2 A 2 0 A 2 J 0 A 0 + J A = 1 2 Ȧ A A m2 A 2 0 A 2 J 0 A 0 + J A. S.18 Note that only the first term on the last line contains any time derivatives at all, and it does not contain the A 0 but only the Ȧ. Consequently, /A 0 = 0 and the scalar potential A 0 x does not have a canonical conjugate field. On the other hand, the vector potential Ax does have a canonical conjugate, namely δl δȧx = Ȧ = Ȧx A 0 x = Ex. S.19 x 4

5 Problem 2b: In terms of the Hamiltonian and Lagrangian densities, eq. 4 means H = Ȧ E L. 4 Expressing all fields in terms A, E, and A 0, we have Ȧ = E A 0, Ȧ E = E 2 + E A 0, S.20 L = 1 2 E 2 A m2 A 2 0 A 2 A 0 J 0 A J, and consequently, H = 1 2 E2 + E A m2 A A 0 J A m2 A 2 A J. S.21 Taking the d 3 x integral of this density and integrating by parts the E A 0 term, we arrive at the Hamiltonian 5. Q.E.D. Problem 2c: Evaluating the derivatives of H in eq. 6 gives us δh δa 0 x H A 0 H i i A 0 = m2 A 0 + J 0 i E i. S.22 If the scalar field A 0 had a canonical conjugate π 0 x, t, its time derivative π 0 /t would be given by the right hand side of eq. S.22. But the a 0 x, t does not have a canonical conjugate, so instead of a Hamilton equation of motion we have a time-independent constraint 6, namely m 2 A 0 = J 0 E. S.23 In the massless EM case, a similar constraint gives rise to the Gauss Law E = J 0. But the massive vector field does not obey the Gauss Law; instead, eq. S.23 gives us a formula for the scalar potential A 0 in terms of E and J 0. 5

6 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 δe i x H E i H j j E i = Ei + i A 0, which leads to Hamilton equation t Ax, t = Ex, t A 0x, t. S.24 Similarly, in the second eq. 7 we have δh δa i x H A i H j j A i = m2 A i J i j ɛ jik A k and hence Hamilton equation t Ex, t = m2 A J + A. S.25 Problem 2d: In 3D notations, the Euler Lagrange field equations 8 or µ F µν m 2 A ν = J ν become E m 2 A 0 = J 0, S.26 Ė + B + m 2 A = J, S.27 where E def = Ȧ A 0, 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 A 0, 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

7 Problem 3: Let start with the [Â, Ĥ] commutator. In light of eq. 11, we have [Âi x, Ĥ] = [ d 3 y  i 1 x, 2 Ê ] 2m 2 Ĵ0 Ê Â m2  2 Ĵ  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, 1 2Ê2 y] = 1 2 {Êj y, [Âi x, Êj y]} = 1 2 {Êj y, iδ ij δ 3 x y} S.31 = iδ 3 x y Êi y, while for the second term we have [Âi x, Ĵ0 y Êy] = 0 y j [Âi x, Êj y] = +iδ ij y j δ3 x y S.32 and hence [  i 1 ] 2 x, Ĵ0 2m 2 y Êy = 1 Ĵ0 m 2 y Êy +iδ ij y j δ3 x y = Â0y i y i δ3 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] = d 3 y iδ 3 x y Êi y + Â0y i y i δ3 x y integrating by parts = d 3 y iδ 3 x y Ê i y + = i Ê i x + x i Â0x. y i Â0y S.34 7

8 In other words, [Âx, Ĥ] = iêx i Â0x and consequently in the Heisenberg picture, t Âx, t = i [Âx, Ĥ] = Êx, t Â0x, t. S.35 Clearly, this Heisenberg equation is the quantum equivalent of the classical Hamilton equation S.24. Now consider the [Ê, Ĥ] commutator. Similarly to eq.s.30, we have [Êi x, t, Ĥ] = d 3 [Êi y x, t, 12 Ê m2 Â ˆB ] m2 Â 2 Ĵ Â y, t S.36 where m 2 Â 0 = Ĵ0 Ê according to eq. 10 and ˆB def = Â. At equal times the Êi x operator commutes with all the Êj y and hence with Ê 2 y, Â0y, and Â2 0 y; this eliminates the first two terms in the Hamiltonian density from the commutator S.36. For the remaining three terms we have [Êi x, Ĵ Ây] [Êi ] = Ĵj y x, Âj y = Ĵj y +iδ ij δ 3 x y [Êi x, 1 2 m2 Ây] [Êi x, ˆB j y] [Êi x, 1 2 ˆB 2 y] = iδ 3 x y Ĵi y, = +im 2 δ 3 x y Âi y, = ɛ jkl [Êi ] y k x, Âl y = ɛ jkl y k +iδ il δ 3 x y = +iɛ jki y k δ3 x y, = ˆB j y +iɛ jki y k δ3 x y = iɛ jki y k ˆB j y δ 3 x y + a total derivative = +i ˆB i y δ 3 x y + a total derivative. S.37 Thus [Êx, Ĥy] = iδ 3 x y ˆBy + m 2 Ây Ĵy + a total derivative, S.38 8

9 hence [Êx, Ĥ] = d 3 y iδ 3 x y ˆBy + m 2 Ây Ĵy = i ˆBx + m 2 Âx Ĵx, and therefore in the Heisenberg picture + a total derivative S.39 ] [Êx, t Êx, t = i t, Ĥ = + ˆBx + m 2 Âx Ĵx. S.40 Again, this Heisenberg equation is the quantum equivalent of the classical Hamilton equation S.25. 9