Consider the lagrangian density
L=−21Zφ∂μφ∂μφ−21Zmm2φ2+Yφ−21Zχ∂μχ∂μχ−21ZMM2χ2+61Zggμ~ε/2φ3+21Zhhμ~ε/2φχ2(28.42)
in d=6−ε dimensions, where φ and χ are real scalar fields, and Y is adjusted to make ⟨0∣φ(x)∣0⟩=0. (Why is no such term needed for χ?)
a) Compute the one-loop contributions to each of the Z's in the MS renormalization scheme.
b) The bare couplings are related to the renormalized ones via
g0h0=Zφ−3/2Zggμ~ε/2,=Zφ−1Zχ−1/2Zhhμ~ε/2.(28.43)
Define
G(g,h,ε)H(g,h,ε)=n=1∑∞Gn(g,h)ε−n≡ln(Zφ−3/2Zg),=n=1∑∞Hn(g,h)ε−n≡ln(Zφ−1Zχ−1/2Zh).(28.45)
By requiring g0 and h0 to be independent of μ, and by assuming that dg/dμ and dh/dμ are finite as ε→0, show that
μdμdgμdμdh=−21εg+21g(g∂g∂G1+h∂h∂G1),=−21εh+21h(g∂g∂H1+h∂h∂H1).(28.47)
c) Use your results from part (a) to compute the beta functions βg(g,h)≡limε→0μdg/dμ and βh(g,h)≡limε→0μdh/dμ. You should find terms of order g3, gh2, and h3 in βg, and terms of order g2h, gh2, and h3 in βh.
d) Without loss of generality, we can choose g to be positive; h can then be positive or negative, and the difference is physically significant. (You should understand why this is true.) For what numerical range(s) of h/g are βg and βh/h both negative? Why is this an interesting question?