Problem 1(a): As discussed in class, Euler Lagrange equations for charged fields can be written in a manifestly covariant form as L (D µ φ) L

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1 PHY 396 K. Solutions for problem set #. Problem 1a: As discussed in class, Euler Lagrange equations for charged fields can be written in a manifestly covariant form as D µ D µ φ φ = 0. S.1 In particularly, for φ = Φ, we have D µ Φ = Dµ Φ, Φ = m Φ, which gives us D µ D µ Φ + m Φ = 0. S. Likewise, for φ = Φ we have D µ Φ = Dµ Φ, Φ = m Φ, and therefore D µ D µ Φ + m Φ = 0. S.3 As for the vector fields A ν, the Lagrangian 1 depends on µ A ν only through F µν, which gives us the usual Maxwell equation µ F µν = J ν where J ν A ν. S.4 To obtain the current J ν, we notice that the covariant derivatives of the charged fields Φ and 1

2 Φ depend on the gauge field: D µ Φ A ν = iqδ ν µφ, D µ Φ A ν = iqδ ν µφ. S.5 Consequently, J ν = D ν Φ iqφ = iq ΦD ν Φ Φ D ν Φ. D ν Φ iqφ S.6 Note that all derivatives on the last line here are gauge-covariant, which makes the current J ν gauge invariant. In a non-covariant form, J ν = iqφ D ν Φ iqφ D ν Φ q Φ Φ A ν. S.7 To prove the conservation of this current, we use the Leibniz rule for covariant derivatives, D ν XY = XD ν Y + Y D ν X. This gives us µ Φ D µ Φ = D µ Φ D µ Φ = D µ Φ D µ Φ + Φ D µ D µ Φ, µ ΦD µ Φ = D µ ΦD µ Φ = D µ ΦD µ Φ + Φ D µ D µ Φ, S.8 and hence in light of eq. S.6 for the current, νj ν = iq D ν ΦD ν Φ + ΦD ν D ν Φ = iqφ D Φ iqφ D Φ by equations of motion = iqφ m Φ iqφ m Φ = 0. + iq D ν Φ D ν Φ + Φ D ν D ν Φ S.9 Q.E.D.

3 Problem 1b: According to the Noether theorem, where T µν Noether = µ A λ ν A λ + µ Φ ν Φ + = T µν µν Noether EM + T Noether matter µ Φ ν Φ g µν L S.10 similar to the free EM fields, and T µν Noether EM = F µλ ν A λ gµν F κλ F κλ S.11 T µν Noether matter = Dµ Φ ν Φ + D µ Φ ν Φ g µν D λ Φ D λ Φ m Φ Φ. S.1 Both terms on the second line of eq. S.10 lack µ ν symmetry and gauge invariance and thus need λ K λµν corrections for some K λµν = K µλν. We would like to show that the same K λµν = F µλ A ν we used to improve the free electromagnetic stress-energy tensor will now improve both the T µν µν EM and T mat at the same time! Indeed, to improve the scalar fields stress-energy tensor we need T µν matter T µν µν true matter T Noether matter = D µ Φ D ν Φ ν Φ + D µ ΦD ν Φ ν Φ = D µ Φ iqa ν Φ + D µ Φ iqa ν Φ S.13 = A ν iqφ D µ Φ iqφd µ Φ = A ν J µ, while the improvement of the EM stress-energy requires cf. previous homework 1. T µν EM = F µλ F ν λ ν A λ = +F µλ λa ν = λ F λµ A ν + A ν J µ. S.14 Altogether, to symmetrize the whole stress-energy tensor, we need Q.E.D. T µν tot T µν µν true total T Noether total = λ F µλ A ν K λµν. 3

4 Problem 1c: Because the fields Φx and Φ x have opposite electric charges, their product is neutral and therefore µ Φ Φ = D µ Φ Φ = D µ Φ Φ + Φ D µ Φ. Similarly, µ D µ Φ D ν Φ = D µ D µ Φ D ν Φ + D µ Φ D µ D ν Φ = m Φ D ν Φ + D µ Φ D ν D µ Φ + iqf µν Φ S.15 where we have applied the field equation D µ D µ + m Φ x = 0 to the first term on the right hand side and used [D µ, D ν ]Φ = iqf µν Φ to expand the second term. Likewise, µ D µ Φ D ν Φ = D µ D µ Φ D ν Φ + D µ Φ D µ D ν Φ = m Φ D ν Φ + D µ Φ D ν D µ Φ iqf µν Φ S.16 and µ [ g ] µν D λ Φ D λ Φ m Φ Φ = ν D λ Φ D λ Φ + m ν Φ Φ = D ν D µ Φ D µ Φ D µ Φ D ν D µ Φ + m Φ D ν Φ + m Φ D ν Φ. Together, the left hand sides of eqs. S.15, S.16 and S.17 comprise µ T µν mat S.17 cf. eq. 7. On the other hand, combining the right hand sides of these three equations results in massive cancellation of all terms except those containing the gauge field strength tensor F µν. Thus, µ T µν mat = D µ Φ iqf µν Φ + D µ Φ iqf µν Φ = F µν iqφ D µ Φ iqφ D µ Φ S.18 = F µν J ν. And since in the previous homework we have shown µ T µν EM = F µν J ν, it follows that the total stress tensor 4 is conserved, µ T µν = 0. Q.E.D. 4

5 Problem a: Thanks to unitarity of the Ux matrix, the derivative of U x U 1 x is µ U x = U x µ Ux U x. S.19 Consequently, the ordinary derivative µ Φx of the adjoint field transforms under the local Ux symmetry into µ Φ x µ UxΦxU x = µ UΦU + U µ ΦU + UΦ µ U = U µ UU S.0 = U µ ΦU + [ µ UU, UΦU ]. At the same time, the vector field matrix A µ x transforms as discussed in class, A µx = UxA µ xu x + i µ UxU x. S.1 Given these two formulae, the covariant derivative 1 transforms to [D µ Φx] = µ Φ x + i [ A µx, Φ x ] = µ UΦU + i [ UA µ U, UΦU ] [ µ UU, UΦU ] = U µ ΦU + [ µ UU, UΦU ] + iu[a µ, Φ]U [ µ UU, UΦU ] = U µ Φ + i[a µ, Φ] U Ux D µ Φx U x. In other words, the covariant derivative defined by eq. 1 is indeed covariant. Q.E.D. S. Problem b: D µ D ν Φ = D µ ν Φ + i[a ν, Φ] = µ ν Φ + i[a ν, Φ] + i [ A µ, ν Φ + i[a ν, Φ] ] Similarly, = µ ν Φ + i[ µ A ν, Φ] + i[a ν, µ Φ] + i[a µ, ν Φ] [A µ, [A ν, Φ]]. S.3 D ν D µ Φ = ν µ Φ + i[ ν A µ, Φ] + i[a µ, ν Φ] + i[a ν, µ Φ] [A ν, [A µ, Φ]]. S.4 Three out of five terms on the right hand sides of these formulæ are identical and hence cancel 5

6 out of the difference D µ D ν Φ D ν D µ Φ. The remaining terms comprise the commutator [D µ, D ν ]Φ = i[ µ A ν, Φ] i[ ν A µ, Φ] [A µ, [A ν, Φ]] + [A ν, [A µ, Φ]] = i[ µ A ν ν A µ, Φ] [[A µ, A ν ], Φ] S.5 i[f µν, Φ]. Problem c: First, let us evaluate D λ F µν = λ µ A ν ν A µ + i[a µ, A ν ] + i [ A λ, µ A ν ν A µ + i[a µ, A ν ] ] = λ µ A ν λ ν A µ + i [A µ, λ A ν ] [A ν, λ A µ ] + i [A λ, µ A ν ] [A λ, ν A µ ] [A λ, [A µ, A ν ]]. S.6 For each group of terms here, summing over cyclic permutation of the Lorentz indices λ µ ν λ produces a zero: λ µ A ν λ ν A µ + µ ν A λ µ λ A ν + ν λ A µ ν µ A λ = 0, [Aµ, λ A ν ] [A ν, λ A µ ] + [A ν, µ A λ ] [A λ, µ A ν ] + [A λ, ν A µ ] [A µ, ν A λ ] = 0, [Aλ, µ A ν ] [A λ, ν A µ ] + [A µ, ν A λ ] [A µ, λ A ν ] + [A ν, λ A µ ] [A ν, µ A λ ] = 0, and consequently [A λ, [A µ, A ν ]] + [A µ, [A ν, A λ ]] + [A ν, [A λ, A µ ]] = 0, S.7 D λ F µν + D µ F νλ + D ν F λµ = 0. 9 Alternative Solution: Suppose φx is a fundamental multiplet of fields, i.e. a column vector of N complex fields transforming according to ψ x = Uxψx. Then [D µ, D ν ]φx = if µν x φx and hence [D λ, [D µ, D ν ]]φx = i[d λ, F µ,ν x ]φx = id λ F µν x φx, S.8 6

7 where the second equality follow from the Leibniz rule D λ F µν φ = D λ F µν φ + F µν D λ φ. S.9 Consequently, for any φx, i D λ F µν x + D µ F νλ x + D ν F λµ x φx = S.30 = [D λ [D µ, D ν ]]φx + [D µ, [D ν, D λ ]]φx + [D ν, [D λ, D µ ]]φx = 0 because of Jacobi identity for the commutators: for any linear operators A, B, and C, [A, [B, C]] + [B, [A, C]] + [C, [A, B]] = 0 and in particular [D λ, [D µ, D ν ]] + [D µ, [D ν, D λ ]] + [D ν, [D λ, D µ ]] = 0. S.31 And since eq. S.30 holds for any φx, it implies D λ F µν + D µ F νλ + D ν F λµ = 0. 9 Problem d: δ F µν = µ A ν ν A µ + i[a µ, A ν ] = µ δa ν ν δa µ + i[δa µ, A ν ] + i[a µ, δa ν ] = D µ δa ν D ν δa µ S.3 Problem e: Working out the Euler Lagrange equations for the vector fields A µ x = a Aa µx λa is rather painful and it s easier to directly evaluate the action variation δs. In light of the previous question d, we have δ tr F µν F µν = tr F µν δf µν = tr F µν D µ δa ν D ν δa µ = 4 tr F µν D µ δa ν = 4 µ tr F µν δa ν 4 tr δa ν D µ F µν = total derivative a δa a ν D µ F µν a S.33 7

8 where the last equality follows from δa ν = a δa a µ λ a, D µf µν = b D µ F µν b λb λ a, and tr λ b = δab. S.34 Consequently, the Yang Mills action S = 1 g d 4 x tr F µν F µν S.35 varies by δs = 1 g d 4 x a δa a νx D µ F µν a x, S.36 and demanding that δs = 0 for any δa a µx gives us the Yang Mills field equations D µ F µν x = 0. S.37 8