Solutions for Homework 8 - Advanced Quantum Mechanics II | PHYSICS 523, Assignments of Physics

Download Solutions for Homework 8 - Advanced Quantum Mechanics II | PHYSICS 523 and more Assignments Physics in PDF only on Docsity! Physics 523, Quantum Field Theory II Homework 8 Due Wednesday, 10th March 2004 Jacob Lewis Bourjaily Renormalization of Pseudo-Scalar Yukawa Theory Let us consider the theory generated by the Lagrangian L = 1 2 (∂µφo)2 − 12m 2 φoφ 2 o + ψo(i 6∂ −meo)ψo − igoψoγ5ψoφo. Superficially, this theory will diverge very similarly to quantum electrodynamics because the fields and the coupling constant have the same dimensions as in quantum electrodynamics. Therefore, we see that the superficial divergence is given by D = 4L − 2Pφ − Pe where L represents the number of loops and Pφ and Pe represent the number of pseudo-scalar and fermion propagator particles, respectively. Furthermore, we see that this can be reduced to D = 4−Nφ − 32Ne, (a.1) where Nφ and Ne represent the number of external pseudo-scalar and fermion lines, respectively. We see that this implies that the following diagrams are superficially divergent: a)
D = 4 b)
D = 3 c)
D = 2 d)
D = 1 e)
D = 0 f)
D = 1 g)
D = 0 Although vacuum energy is an extraordinarily interesting problem of physics, we will largely ignore diagram (a) which is quite divergent. We note that because the Lagrangian is invariant under parity transformations φ(t,x) → −φ(t,−x) any diagram with an odd number of external φ’s will give zero. In particular, the divergent diagrams (b) and (d) will be zero. The first divergent diagram we will consider, (c), is clearly ∼ a0Λ2 +a1p2 log Λ where we note that the term proportional to p in the expansion vanishes by parity symmetry. Similarly, we näıvely suspect that the divergence of diagram (f) would be ∼ a0Λ+ 6p log Λ but the term linear in Λ is reduced to me log Λ by the symmetry of the Lagrangian of chirality inversion of ψ together with φ → −φ. The diagrams (e) and (g) are both ∼ log Λ. All together, there are six divergent constants in this theory. We note that because the diagram (e) diverges, we must introduce a counterterm δλ which implies that our original Lagrangian should have included a term λ4!φ 4. We define renormalized fields, φo ≡ Z1/2φ φ and ψo ≡ Z1/22 ψ, where Zφ and Z2 are as would be defined canonically. Using these our Lagrangian can be written as L = 1 2 Zφ(∂µφ)2 − 12Zφm 2 φoφ 2 − Z2ψ(i 6∂ −meo)ψ −−igoZ2Z1/2φ ψγ5ψφ− λ 4! Z2φφ 4. Let us define the counterterms, δmφ ≡ Zφm2φo−m2φ, δme ≡ Z2meo−me, δφ ≡ Zφ−1, δλ ≡ λoZ2φ−λ, δ1 ≡ go g Z2Z 1/2 φ −1, δ2 ≡ Z2−1. Therefore, we may write our renormalized Lagrangian L = 1 2 (∂µφ)2 − 12m 2 φφ 2 + ψ(i 6∂ −me)ψ − igψγ5ψφ− λ4!φ 4 + 1 2 δφ(∂µφ)2 − 12δmφφ 2 + ψ(iδ2 6∂ − δme)ψ − igδ1ψγ5ψφ− δλ 4! φ4. (a.4) 1 2 JACOB LEWIS BOURJAILY Let us compute the pseudo-scalar self-energy diagrams to the one-loop order, keeping only the diver- gent pieces. This corresponds to: −iM2(p2) =
p p k +
p k + p p k +
p p×Using the ‘canonical procedure’ and dropping all but divergent pieces (linear in ²−1) we see that −iM2(p2) = −iλ 2 ∫ ddk (2π)d i k2 −m2φ − g2 ∫ ddk (2π)d Tr [ γ5i(6k+ 6p + me)iγ5(6k + me) ((k + p)2 −m2e)(k2 −m2e) ] + i(p2δφ − δme), = −iλ 2 1 (4π)d/2 Γ ( 1− d2 ) (m2φ)1−d/2 − 4g2 ∫ 1 0 dx ∫ ddk (2π)d `2 − x(1− x)p2 −m2e (`2 −∆)2 + i(p 2δφ − δm2), = −iλ 2 1 (4π)d/2 m2φ (1− d/2) Γ ( 2− d2 ) (m2)2−d/2 − 4g2 ∫ 1 0 dx [ − i (4π)d/2 d 2 Γ ( 1− d2 ) ∆1−d/2 + i (4π)d/2 Γ ( 2− d2 ) ∆2−d/2 ( x(1− x)p2 + m2e ) ] + i(p2δφ − δm2), ∼ i λm 2 φ 32π2 2 ² − 8g2 i (4π)2 2 ² ∫ 1 0 dx ( m2e − x(1− x)p2 ) + 4g2 i (4π)2 2 ² ∫ 1 0 dx ( m2e + x(1− x)p2 ) + i(p2δφ − δm2), = i λm2φ 16π2 1 ² + i g2 4π2 2 ² ( −2m2e + 2 6 p2 + 1 6 p2 + m2e ) + i(p2δφ − δm2), = i ( λm2φ 16π2 + g2p2 4π2 − g 2m2e 2π2 ) 1 ² + i(p2δφ − δm2). Therefore, applying our renormalization conditions, we see that1 ∴ δmφ = ( λm2φ 16π2 − g 2m2e 2π2 ) 1 ² , δφ = − ( g2 4π2 ) 1 ² . (b.1) Similarly, let us compute the fermion self-energy diagrams to one-loop order, keeping only divergent parts. This corresponds to: −iΣ22(6p) =
p k p− k p +
p p×Again, using the ‘canonical procedure’ and dropping all but divergent pieces (linear in ²−1) we see that −iΣ(6p) = g2 ∫ ddk (2π)d [ γ5 i ((p− k)2 −m2φ) i(6k + me) (k2 −m2e) γ5 ] + i(6pδ2 − δme), = −g2 ∫ ddk (2π)d 6k −me (k2 −m2e)((p− k)2 −m2φ) + i(6pδ2 − δme), = −g2 ∫ 1 0 dz ∫ dd` (2π)d 6pz −me (`2 −∆)2 + i(6pδ2 − δm2), ∼ −i g 2 (4π)2 2 ² ∫ 1 0 dz (6pz −me) + i(6pδ2 − δme), = i ( g2 6p 16π2 − g 2me 8π2 ) 1 ² + i 6pδ2 − iδme . Therefore, applying our renormalization conditions, we see that ∴ δme = − ( g2me 8π2 ) 1 ² , δ2 = − ( g2 16π2 ) 1 ² . (b.2) 1For renormalization conditions and Feynman rules please see the Appendix.