Physics 523, Quantum Field Theory II Homework 9 Due Wednesday, 17 th March 2004

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1 Physics 5 Quantum Field Theory II Homework 9 Due Wednesday 7 th March 004 Jacob Lewis Bourjaily β-functions in Pseudo-Scalar Yukawa Theory Let us consider the massless pseudo-scalar Yukawa theory governed by the renormalized Lagrangian L = µφ + ψi ψ igψγ 5 ψφ λ 4! φ4 + δ φ µ φ ψiδ ψ ψ igδ g ψγ 5 ψφ δ λ 4! φ4 In homework 8 we calculated the divergent parts of the renormalization counterterms δ φ δ ψ δ g and δ λ to -loop order These were shown to be δ φ = g 8π δ λ = M λ π g4 π Λ δ ψ = g Λ π M ; Λ M δ g = g Λ 6π M Using the definitions of B i and A i in Peskin and Schroeder these imply that A φ = γ φ = g 8π B λ = g4 π λ π Therefore we see that A ψ = γ ψ = g π ; B g = g 6π β g = gb g ga ψ ga φ = g g g + g 6π π + g g 8π = 5g 6π ; λ β λ = B λ 4λA φ = π g4 π + 4λ g 8π = λ + 8λg 48g 4 6π While it was supposedly unnecessary the running couplings were computed to be 6π gp = 0 p/m ; λp = λ = g g 45/ g 45/5 Notice that both g and λ generally become weaker at large distances because for typical values of g λ we see that β g and β λ are both positive However if λ << g then β λ will be negative and so λ will grow stronger at larger distances Near small values of g and λ the theory shows interesting interplay between g and λ Also interesting is the characteristic Landau pole in λ suggesting that we should not trust this theory at too large a scale Below is a graph of g versus λ indicating the direction of Renormalization Group flow as the interaction distance grows larger g p Λ p Figure Renormalization Group Flow as a funciton of scale Arrow indicates flow in the direction of larger distances For this plot M was taken to be 0 4 See appendix

2 JACOB LEWIS BOURJAILY Minimal Subtraction Let us define the β-function as it appears in dimensional regularization as βλ = M d λ where it is understood that βλ = lim 0 βλ We notice that the bare coupling is given by λ 0 = M Z λ λ λ where Z λ is given by an expansion series in Z λ λ = + a ν λ ν ν= We are to demonstrate the following a Let us show that Z λ satisfies the identity βλ + λz λ + βλλ dz λ = 0 Noting the general properties of differentiation from elementary analysis we will proceed by direct computation βλ + λ Z λ + βλ λ dz λ = βλ Z λ + λz λ + βλ dz λλ βλ Z λ = λz λ + M dλ 0 M λ0 λ0 = λz λ Mλ 0 M = λz λ M + M Z λ λ = 0 βλ + λz λ + βλλ dz λ = 0 b Let us show that βλ = λ + βλ We have demonstrated in part a above that βλ + λz λ + βλλ dz λ = 0 Dividing this equation by Z λ and rearranging terms and expanding in Z λ we obtain βλ + λ = βλ λ Z λ dz λ = βλ λ Z λ = βλ λ da + da + da + da + a + Now we know that βλ must be regular in as 0 and so we may expand it as a terminating power series βλ = β 0 + β + β + + β n n We notice that βλ = β 0 in this notation Let us consider the limit of For any n > 0 we see that the order of the polynomial on the left hand side has degree n whereas the polynomial on the left hand side has degree n because as da the equation becomes β n n = β n n λ But this is a contradiction Therefore both the right and left hand sides must have degree less than or equal to 0 Furthermore because the left hand side is βλ + λ = β 0 + β + λ must have degree zero we see that β = So expanding βλ as a power series of we obtain βλ = λ + βλ Professor Larsen does note believe this to be necessary However we have been unable to demonstrate the required identity without assuming a terminating power series

3 ci Let us show that βλ = λ da PHYSICS 5: QUANTUM FIELD THEORY II HOMEWORK 9 By rewriting the identity obtained from part a above and expanding in Z λ we see that βλ + λ Z λ = βλ λ dz λ βλ + λ + a + = βλ λ da + We see that because there is no term on the right hand side of order 0 it must be that βλ + λa = 0 which implies that βλ = λa Furthermore by equating the coefficients of we have in general that βλ a n n + λa n+ = βλ λ da n By rearranging terms and using noticing the chain rule of differentiation we see that this implies that λa n+ = βλ λ da n + a n = βλ dλa n This fact will be important to the proof immediately below Now by the result of part b above we know that βλz λ = βλ + λz λ = βλ λ dz λ βλ + a + = βλ + λz λ = βλ λ Equating the coefficients of terms of order βλa = βλ λ da da + on the far left and right sides we see that Now using our result from before that βλ = λa we see directly that βλ = λ da cii Let us show that βλ dλa ν = λ da ν+ By our result in part b above we have that βλ = βλ + λ βλ dz λλ = βλ + λ dz λλ = βλ + λ + βλ + dλa + dλa + Equating the coefficients of on both sides we see that by using the identities shown ν above βλ dλa ν So we see in general that = βλ dλa ν = βλ dλa ν = βλ dλa ν = λ da ν+ βλ dλa ν + λ dλa ν+ + λ da ν+ + λa ν+ + λ da ν+ βλ dλa ν = λ da ν+

4 4 JACOB LEWIS BOURJAILY In the minimal subtraction scheme we define the mass renormalization by m 0 = m Z m where Z m = + b ν ν ν= Similarly we will define the associated β-function β m λ = mγ m λ which is given by β m λ = M dm di Let us show that γ m λ = λ db Because m 0 is a constant we know that dm 0 = 0 Therefore writing m 0 = m Z m we see that this implies β m λ β m λ m0 dm 0 = 0 = Z mm dm + dz m m = Z m m β mλ M 0 = Z m β m λ + mm + dz m m = 0; dz m ; β m λz m = mβλ dz m + b + db = mβλ + + b + = m βλ λ db + We see that the coefficient of the 0 term on the left hand side is β m λ and on the right hand side it is mλ db Therefore because these terms must be equal we see that β m λ = m λ db γ m λ = λ db dii Let us prove that λ dbν+ = γ m λb ν + βλ dbν β m λ + b + Continuing our work from part di above we have that It must be that the coefficients of ν = m βλ λ db + are equal on both sides Therefore we see that β m λb ν = mβλ db ν + mλdb ν+ mγ m λb ν = mβλ db ν + mλdb ν+ γ m λb ν = βλ db ν + λdb ν+ Rearranging terms we see that λ db ν+ = γ mλb ν + βλ db ν

5 PHYSICS 5: QUANTUM FIELD THEORY II HOMEWORK 9 5 Appendix Calculation of the Running Couplings g and λ Let us now solve for the flow of the coupling constants g λ to the Callan-Symanzik equation will satisfy dg d p/m = β g = 5g 6π + Og5 We have in general that solutions This is an ordinary differential equation We see that g = 5 p/m + C 6π and so g 8π p = 5 p/m + C The constant C is found so that gp = M = This yields C = / To find the flow of λ however it will be convenient to introduce a new variable η λ/g We must then solve the equation dη d p/m = β λ g λβ g η g = η 48 g 6π + Og 4 This is again a simple ordinary differential equation We see that this implies dη g η η 48 = d p/m 6π g Note that from our work above 6π d p/m = g 6π d 8π 5g = 5g dg Therefore dη η η 48 = 5g dg And so η 45 η + = 45 g + C 45 5 Solving this equation in terms of η we see that we have 45/5 η = Cg Cg 45/5 = 45/5 45/5 Cg Cg + Cg /5 Cg 45/5 = + 45 C + 45/5 g C g 45/5 λ = g + 45 C + 45/5 g C g 45/5 As before the constant term C is found by requiring that λp = M = C = The constant is then It can be argued that this is a poor choice of C because it requires the reference scale to be non-perturbative Nevertheless it is not a free parameter