Problem Set 9 Solutions - Quantum Field Theory | PHY 396K, Assignments of Physics

Download Problem Set 9 Solutions - Quantum Field Theory | PHY 396K and more Assignments Physics in PDF only on Docsity! PHY–396 K. Solutions for problem set #9. Problem 1(a): For p = 0, p0 = +m and 6p−m = m(γ0 − 1). Hence, u(p = 0, s) satisfy (γ0 − 1)u = 0, or in the Weyl basis ( −12×2 12×2 12×2 −12×2 ) u = 0 =⇒ u = ( ζ ζ ) (S.1) where ζ is an arbitrary two-component spinor. It’s normalization follows from u†u = 2ζ†ζ, so we want ζ†ζ = Ep = m. In other words, ζ = √ m× ξ — and hence u is as in eq. (2) — where ξ†ξ = 1. There are two independent choices of ξ, normalized to ξ†sξs′ = δs,s′ so that u †(0, s)u(0, s′) = 2mδs,s′ = 2Epδs,s′ . They correspond to two spin states of a p = 0 electron. Indeed, in the Weyl basis where the spin matrices are S = 1 2 ( σ 0 0 σ ) (cf. problem 8.4(a) of the previous homework set), we have u†(0, s)Su(0, s′) = 2m× ξ†s σ 2 ξs′ . (S.2) Problem 1(b): Dirac equation is Lorentz covariant, so we may obtain solutions for all pµ = (+Ep,p) by simply Lorentz-boosting the solutions (2) for pµ = (+m,0), u(p, s) = M u(0, s) = ( ML 0 0 MR )(√ m ξs √ m ξs ) = (√ m ML ξs √ m MR ξs ) (S.3) where M , ML, and MR are respectively Dirac-spinor, LH–Weyl-spinor, and RH–Weyl-spinor rep- resentations or the Lorentz boost (m,0) 7→ (E,p). In problem 8.4.(c) of the previous homework 1 we saw that ML = √ γ × √ 1− βn · σ , MR = √ γ × √ 1 + βn · σ , (S.4) where βn is the velocity vector of the boost (in units of c) and γ = 1/ √ 1− β2. For the boost we are interested in, γ = E m , γβn = p m , (S.5) so m× γ × (1∓ βn · σ) = E ∓ p · σ (S.6) and hence √ m ML = √ E − p · σ , √ m MR = √ E + p · σ . (S.7) Plugging these matrices into eq. (S.3) gives us u(p, s) = (√ E − p · σ ξs √ E + p · σ ξs ) . (3) Q.E .D. Problem 1(c): v(p, s) = γ2u∗(p, s) = ( 0 +σ2 −σ2 0 ) × (√ E − p · σ ξs √ E + p · σ ξs )∗ = ( +σ2 (√ E + p · σ )∗ × ξ∗s −σ2 (√ E − p · σ )∗ × ξ∗s ) = ( +σ2 (√ E + p · σ )∗ σ2 × ηs −σ2 (√ E − p · σ )∗ σ2 × ηs ) where ηs def = σ2 × ξ∗s = ( + √ E − p · σ × ηs − √ E + p · σ × ηs ) because σ2 σ ∗σ2 = −σ =⇒ σ2 (√ E ± p · σ )∗ σ2 = √ E ∓ p · σ . (S.8) This verifies eq. (4) for the negative-energy plane waves. 2 Now consider the Lorentz invariant products ūu and v̄v. The ū and v̄ are given by ū(p, s) = u†(p, s)γ0 = (√ E − p · σ ξs √ E + p · σ ξs )†( 0 12×2 12×2 0 ) = ( ξ†s × √ E + p · σ , ξ†s × √ E − p · σ ) , v̄(p, s) = v†(p, s)γ0 = ( + √ E − p · σ ηs − √ E + p · σ ηs )†( 0 12×2 12×2 0 ) = (−η†s × √ E + p · σ , +η†s × √ E − p · σ ) . (S.19) Consequently, ū(p, s) u(p, s′) = ξ†s ×× √ E + p · σ ×× √ E − p · σ × ξs′ + ξ†s ×× √ E − p · σ ×× √ E + p · σ × ξs′ = 2m× ξ†sξs′ = 2mδs,s′ (S.20) because √ E + p · σ× √ E − p · σ = √ E − p · σ× √ E + p · σ = √ E2 − (p · σ)2 = √ E2 − p2 = m. (S.21) Likewise, v̄(p, s) v(p, s′) = − η†s ×× √ E + p · σ ×× √ E − p · σ × ηs′ − η†s ×× √ E − p · σ ×× √ E + p · σ × ηs′ = − 2m× η†sηs′ = −2mδs,s′ . (S.22) Problem 1(f): In matrix notations (column× row = matrix), we have u(p, s)× u(p, s) = (√ E − pσ ξs √ E + pσ ξs ) × ( ξ†s √ E + pσ , ξ†s √ E − pσ ) = (√ E − pσ ξs × ξ†s √ E + pσ √ E − pσ ξs × ξ†s √ E − pσ √ E + pσ ξs × ξ†s √ E + pσ √ E + pσ ξs × ξ†s √ E − pσ ) . (S.23) 5 Summing over two spin polarizations replaces ξs × ξ†s with ∑ s ξs × ξ † s = 12×2 . Consequently,∑ s u(p, s)× u(p, s) = = (√ E − pσ [∑ s ξs × ξ † s ]√ E + pσ √ E − pσ [∑ s ξs × ξ † s ]√ E − pσ √ E + pσ [∑ s ξs × ξ † s ]√ E + pσ √ E + pσ [∑ s ξs × ξ † s ]√ E − pσ ) = (√ E − pσ × √ E + pσ √ E − pσ × √ E − pσ √ E + pσ × √ E + pσ √ E + pσ × √ E − pσ ) = ( m E − pσ E + pσ m ) = m× 14×4 + E × γ0 − p · ~γ = 6p + m. (S.24) Similarly v(p, s)× v(p, s) = ( + √ E − pσ ηs − √ E + pσ ηs ) × ( −η†s √ E + pσ , +η†s √ E − pσ ) = ( − √ E − pσ ( ηs × η†s )√ E + pσ + √ E − pσ ( ηs × η†s )√ E − pσ + √ E + pσ ( ηs × η†s )√ E + pσ − √ E + pσ ( ηs × η†s )√ E − pσ ) , (S.25) where ηs × η†s = σ2 ( ξ†s × ξs )∗ σ2 and hence ∑ s ηs × η†s = σ2 (∑ s ξ†s × ξs )∗ σ2 = σ21 ∗ 2×2σ2 = 12×2 . (S.26) Therefore, ∑ s v(p, s)× v(p, s) = ( − √ E − pσ × 12×2 × √ E + pσ + √ E − pσ × 12×2 × √ E − pσ + √ E + pσ × 12×2 × √ E + pσ − √ E + pσ × 12×2 × √ E − pσ ) = ( −m E − pσ E + pσ −m ) = −m× 14×4 + E × γ0 − p · ~γ = 6p − m. (S.27) Q.E .D. 6 Problem 1(g): The constant spinors u ≡ u(p, s) and ū′ ≡ ū(p′, s′) satisfy Dirac equations 6pu = mu and ū′6p′ = mū′. Applying both equations to the Dirac “sandwich” ū′γµu, we have ū′γµu = 1 m ū′6p′ × γµu = 1 m ū′γµ×6pu = 1 2m ū′(6p′γµ + γµ6p)u. (S.28) On the right hand side, we evaluate 6p′γµ + γµ6p ≡ p′νγνγµ + pνγµγν = p′ν (gµν + 2iSµν) + pν (gµν − 2iSµν) = (p′ + p)ν × gµν + 2i(p′ − p)ν × Sµν , (S.29) which gives us the Gordon identity ū′γµu = (p′ + p)µ 2m ū′u + i(p′ − p)ν m ū′Sµνu. (3) Q.E .D. Problem 1(h): The negative-frequency spinors v ≡ v(p, s) and v̄′ ≡ v̄(p′, s′) satisfy Dirac equations 6pv = −mv and v̄′6p′ = −mv̄′. Consequently, proceeding exactly as above modulo signs, we have ū′γµv = (p′ − p)µ 2m × ū′v + i(p ′ + p)ν m × ū′Sµνv, v̄′γµu = (−p′ + p)µ 2m × v̄′u + i(−p ′ − p)ν m × v̄′Sµνu, v̄′γµv = (−p′ − p)µ 2m × v̄′v + i(−p ′ + p)ν m × v̄′Sµνv. (S.30) 7 The second equality here follows by transposition of the Dirac “sandwich” Ψ> · · ·Ψ∗, which carries an extra minus sign because the fermionic fields Ψ and Ψ∗ anticommute with each other. The third equality follows from (γ0)> = +γ0, (γ2)> = +γ2, and Ψ† = Ψγ0. Problem 2(e): By inspection, 1c ≡ γ0γ2γ0γ2 = +1. The γ5 matrix is symmetric and commutes with the γ0γ2, hence γc5 = +γ5. Among the four γµ matrices, the γ1 and γ3 are anti-symmetric and commute with the γ0γ2 while the γ0 and γ2 are symmetric but anti-commute with the γ 0γ2; hence, for all four γµ, γ c µ = −γµ. Finally, because of the transposition involved, (γµγν)c = γcνγcµ = +γνγµ, hence (γ[µγν])c = +γ[νγµ] = −γ[µγν]. Likewise, (γ5γµ)c = (γµ)c(γ5)c = −γµγ5 = +γ5γµ. Therefore, according to eq. (S.37), the scalar S, the pseudoscalar P , and the axial vector Aµ are C–even, while the vector Vµ and the tensor Tµν are C–odd. Problem 2(f): As discussed in class, in the Weyl fermion language, the Lagrangian for χ, χ̃ and their hermitian conjugates is obviously invariant under the charge conjugation C : χ ↔ χ̃. In the Dirac-spinor language, this is less obvious, hence this exercise. In the previous question we saw that ΨΨ is C–even, which means that the mass term −mΨΨ is C–invariant. For the rest of the Lagrangian, we have C : Ψ6∂Ψ 7→ −Ψ>γ2γ0γµ∂µγ2Ψ∗ = + ( ∂µΨ † ) ( γ2γ0γµγ2 )> Ψ = + ( ∂µΨ ) γ0γ2 (γµ)> γ0γ2Ψ = − ( ∂µΨ ) γµΨ = +Ψ6∂Ψ − ∂µ ( ΨγµΨ ) . (S.38) The second line here follows by transposition of the Dirac ‘sandwich’ Ψ> · · · ∂µΨ∗, and the sign changes because of Fermi statistics (Ψα and Ψ ∗ β anticommute). The fourth line follows from γ0γ2 (γµ)> γ0γ2 = −γµ (S.39) in the Weyl basis, which in terms follows from γ1 and γ3 being anti-symmetric matrices commuting with the γ0γ2, while γ0 and γ2 are symmetric matrices anticommuting with the γ0γ2. 10 Altogether, we have C : L = Ψ(i6∂ −m)Ψ 7→ L + a total derivative, (S.40) and the Dirac action ∫ d4xL is C–invariant. 11