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Physics 523, Quantum Field Theory II
Homework 9
Due Wednesday, 17th^ March 2004
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)^2 + ψi 6 ∂ψ − igψγ^5 ψφ −
λ 4!
φ^4
δφ(∂μφ)^2 ψiδψ 6 ∂ψ − igδg ψγ^5 ψφ −
δλ 4! φ^4. In homework 8, we calculated the divergent parts of the renormalization counterterms δφ, δψ , δg , and δλ to 1-loop order. These were shown to be
δφ = − g^2 8 π^2
log
Λ^2
M 2
, δψ = − g^2 32 π^2
log
Λ^2
M 2
δλ =
3 λ^2 32 π^2
3 g^4 2 π^2
log
Λ^2
M 2
, δg =
g^2 16 π^2
log
Λ^2
M 2
Using the definitions of Bi and Ai in Peskin and Schroeder, these imply that
Aφ = −γφ = −
g^2 8 π^2 , Aψ = −γψ = −
g^2 32 π^2
Bλ = 3 g^4 2 π^2
3 λ^2 32 π^2
Bg = − g^2 16 π^2
Therefore, we see that
βg = − 2 gBg − 2 gAψ − gAφ = 2g
g^2 16 π^2
g^2 32 π^2
g^2 8 π^2
5 g^3 16 π^2
βλ = − 2 Bλ − 4 λAφ = 2
3 λ^2 32 π^2
3 g^4 2 π^2
g^2 8 π^2
3 λ^2 + 8λg^2 − 48 g^4 16 π^2 While it was supposedly unnecessary, the running couplings were computed to be^1 ,
g(p) =
16 π^2 1 − 10 log p/M
λ(p) = λ =
g^2 3
√ 145+ 141 +^ g
2 √ 145 / 5
− 4
√145+ 141 −^ g
2 √ 145 / 5
Notice that both g and λ generally become weaker at large distances because for typical values of g, λ we see that βg and βλ are both positive. However, if λ << g then βλ 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 inter- action distance grows larger.
100 200 300 400 500 600 700 -ΛHpL
5
10
15
gHpL
Figure 1. 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 10^4. (^1) See appendix. 1
2 JACOB LEWIS BOURJAILY
Minimal Subtraction Let us define the β-function as it appears in dimensional regularization as
β(λ, ≤) = M
d dM λ
λ 0 ,≤
where it is understood that β(λ) = lim≤→ 0 β(λ, ≤). We notice that the bare coupling is given by λ 0 = M ≤Zλ(λ, ≤)λ where Zλ is given by an expansion series in ≤,
Zλ(λ, ≤) = 1 +
ν=
aν (λ) ≤ν^
We are to demonstrate the following. a) Let us show that Zλ satisfies the identity (β(λ, ≤) + ≤λ)Zλ + β(λ≤)λ dZ dλλ = 0. proof: Noting the general properties of differentiation from elementary analysis, we will proceed by direct computation.
(β(λ, ≤) + ≤λ) Zλ + β(λ, ≤)λ
dZλ dλ = β(λ, ≤)Zλ + ≤λZλ + β(λ, ≤)
d(Zλλ) dλ − β(λ, ≤)Zλ,
= ≤λZλ + M
dλ dM
λ 0 ,≤
d(λ 0 M −≤) dλ
= ≤λZλ − ≤M λ 0 M −≤−^1 , = ≤λZλ − ≤M 1+≤M −≤−^1 Zλλ, = 0.
∴ (β(λ, ≤) + ≤λ)Zλ + β(λ≤)λ
dZλ dλ
oπ≤ρ^ ‘ ´ ’ ´≤δ≤ι δ≤ιξαι
b) Let us show that β(λ, ≤) = −≤λ + β(λ). proof: We have demonstrated in part (a) above that (β(λ, ≤) + ≤λ)Zλ + β(λ≤)λ dZ dλλ = 0. Dividing this equation by Zλ and rearranging terms and expanding in Zλ, we obtain
β(λ, ≤) + ≤λ = −β(λ, ≤)
λ Zλ
dZλ dλ
= −β(λ, ≤)
λ Zλ
da 1 dλ
≤^2
da 2 dλ
= −β(λ, ≤)λ
da 1 dλ
≤^2
da 2 dλ
a 1 ≤
Now, we know that β(λ, ≤) must be regular in ≤ as ≤ → 0 and so we may expand it as a (terminating)^2 power series β(λ, ≤) = β 0 + β 1 ≤ + β 2 ≤^2 + · · · + βn≤n. We notice that β(λ) = 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 − 1 because as ≤ → ∞, the equation becomes βn≤n^ = −βn≤nλ (^1) ≤da dλ^1. 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 β(λ, ≤) + ≤λ = β 0 + β 1 ≤ + ≤λ must have degree zero, we see that β 1 = −≤. So, expanding β(λ, ≤) as a power series of ≤, we obtain,
∴ β(λ, ≤) = −≤λ + β(λ). oπ≤ρ‘ ´ ’ ´≤δ≤ι δ≤ιξαι
(^2) 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.
4 JACOB LEWIS BOURJAILY
In the minimal subtraction scheme, we define the mass renormalization by m^20 = m^2 Zm where
Zm = 1 +
ν=
bν ≤ν^
Similarly, we will define the associated β-function βm(λ) = mγm(λ) which is given by
βm(λ) = M
dm dM
m 0 ,≤
d.i) Let us show that γm(λ) = λ 2 db dλ^1. proof: Because m^20 is a constant, we know that dm
(^20) dM = 0. Therefore, writing^ m
2 0 =^ m
2 Z
m we see that this implies dm^20 dM
= 0 = 2Zmm
dm dM
dZm dM
= 2Zmm
βm(λ) M
dZm dλ
dλ dM
∴ 0 = 2Zmβm(λ) + mM dλ dM
dZm dλ
∴ 2 βm(λ)Zm = −mβ(λ, ≤)
dZm dλ
2 βm(λ)
b 1 ≤
= −mβ(λ, ≤)
db 1 dλ
2 βm(λ)
b 1 ≤
= −m (β(λ) − ≤λ)
db 1 dλ
We see that the coefficient of the ≤^0 term on the left hand side is 2βm(λ) and on the right hand side it is mλ db dλ^1. Therefore, because these terms must be equal, we see that βm(λ) = m
λ 2
db 1 dλ
∴ γm(λ) =
λ 2
db 1 dλ
oπ≤ρ^ ‘ ´ ’ ´≤δ≤ι δ≤ιξαι
d.ii) Let us prove that λ db dλν+1 = 2γm(λ)bν + β(λ) db dλν. proof: Continuing our work from part (d.i) above, we have that
2 βm(λ)
b 1 ≤
= −m (β(λ) − ≤λ)
db 1 dλ
It must be that the coefficients of (^) ≤^1 ν are equal on both sides. Therefore, we see that
2 βm(λ)bν = −mβ(λ)
dbν dλ
dbν+ dλ
2 mγm(λ)bν = −mβ(λ) dbν dλ
∴ 2 γm(λ)bν = −β(λ)
dbν dλ
dbν+ dλ
Rearranging terms, we see that
∴ λ
dbν+ dλ
= 2γm(λ)bν + β(λ)
dbν dλ
oπ≤ρ^ ‘ ´ ’ ´≤δ≤ι δ≤ιξαι
PHYSICS 523: 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. We have in general that solutions to the Callan-Symanzik equation will satisfy
dg d log p/M
= βg =
5 g^3 16 π^2
This is an ordinary differential equation. We see that
−
g^2
16 π^2 log p/M + C,
and so
∴ g^2 (p) = −
8 π^2 5 log p/M + C
The constant C is found so that g(p = M ) = 1.^3 This yields C = − 1 /2. To find the flow of λ, however, it will be convenient to introduce a new variable η ≡ λ/g^2. We must then solve the equation
dη d log p/M
βλ g^2
λβg g^3
3 η^2 − 2 η − 48
g^2 16 π^2
This is again a simple ordinary differential equation. We see that this implies ∫ dη 3 η^2 − 2 η − 48
g^2 16 π^2 d log p/M.
Note that from our work above, g
2 16 π^2 d^ log^ p/M^ =^
g^2 16 π^2 d
− 8 π
2 5 g^2
= (^51) g dg. Therefore, ∫ dη 3 η^2 − 2 η − 48
5 g
dg.
And so,
log
3 η −
3 η +
log g + C.
Solving this equation in terms of η, we see that we have
η =
Cg^2
√ 145 / 5 (√ 145 − 1 )^ + √145 + 1
3 − 3 Cg^2
√ 145 / 5 ,
1 − Cg^2
√ 145 / 5
3 − 3 Cg^2
√ 145 / 5 +^
Cg^2
√ 145 / 5 √145 + √ 145
3 − 3 Cg^2
√ 145 / 5 ,
C + g^2
√ 145 / 5 C − g^2
√ 145 / 5
∴ λ = g^2 3
C + g^2
√ 145 / 5
C − g^2
√ 145 / 5
As before, the constant term C is found by requiring that λ(p = M ) = 1. The constant is then
C = − 4
√ 145+
(^3) 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.
