By Srednicki M.
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B1 g˜2 + b2 g˜3 + . . 45) b) Just make everything into a matrix: we then have g˙ i = b2,ijk gj gk + b3,ijkl gj gk gl + . . and gi = g˜i − c2,ijk g˜j g˜k + . . Everything in part (a) still goes through. 2) a) We use the relation ∆(k ˜ 2 ) is the propagator with a cutoff Λ, and circulate in the loop in fig. 1; in this case, ∆(k 2 −1 2 ˜ differentiating ∆(k ) with respect to k (and setting k2 = 0) yields Z(Λ), the coefficient of the kinetic term when the cutoff is Λ. The vertex factor is −Z 3/2 (Λ0 )g(Λ0 ), and the tree-level propagator is ∆(k2 ) = 1/[Z(Λ0 )k2 ].
1 g˙ = (1 + 2c2 g˜ + . )(b1 g2 + b2 g3 + . ) = (1 + 2c2 g˜ + . )[b1 (˜ g − c2 g˜2 )2 + b2 (˜ g − c2 g˜2 )3 + . ] = b1 g˜2 + b2 g˜3 + . . 45) b) Just make everything into a matrix: we then have g˙ i = b2,ijk gj gk + b3,ijkl gj gk gl + . . and gi = g˜i − c2,ijk g˜j g˜k + . . Everything in part (a) still goes through. 2) a) We use the relation ∆(k ˜ 2 ) is the propagator with a cutoff Λ, and circulate in the loop in fig. 1; in this case, ∆(k 2 −1 2 ˜ differentiating ∆(k ) with respect to k (and setting k2 = 0) yields Z(Λ), the coefficient of the kinetic term when the cutoff is Λ.
4) We subtract eq. 43) from eq. 39), divide by α, and drop higher-order terms to get 1 1 1 dx D ln(D0 /m2 ) + ( 16 κA + 12 )k2 + (κB + 12 )m2 √ √ 1 1 1 (3π 3 − 17)k2 + 61 (π 3 − 6)m2 + ( 61 κA + 12 )k2 + (κB + 12 )m2 . 0098. 5) See section 31. From eq. 5), we have Π(k2 ) = λ 1 + 2 16π ε 1 2 + ln(µ/m) m2 − Ak2 − Bm2 + O(λ2 ) . 60) Mark Srednicki Quantum Field Theory: Problem Solutions 35 We see immediately that A = O(λ2 ) , B = 1 λ + 2 16π ε 1 2 + ln(µ/m) + O(λ2 ) . 61) and that Π(k2 ) = 0 to O(λ).
Quantum field theory - solutions manual by Srednicki M.
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