Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Homework 9 - Beta Functions in Pseudo-Scalar Yukawa Theory | PHYSICS 523, Assignments of Physics

Material Type: Assignment; Class: Adv Quantum Mech II; Subject: Physics; University: University of Michigan - Ann Arbor; Term: Unknown 2004;

Typology: Assignments

Pre 2010

Uploaded on 09/17/2009

koofers-user-h9b
koofers-user-h9b 🇺🇸

10 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
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=1
2(µφ)2+ψi 6∂ψ igψ γ5ψφ λ
4!φ4
+1
2δφ(µφ)2ψiδψ6∂ψ ig δgψγ5ψφ δλ
4! φ4.
In homework 8, we calculated the divergent parts of the renormalization counterterms δφ, δψ, δg,and
δλto 1-loop order. These were shown to be
δφ=g2
8π2log Λ2
M2, δψ=g2
32π2log Λ2
M2;
δλ=µ3λ2
32π23g4
2π2log Λ2
M2, δg=g2
16π2log Λ2
M2.
Using the definitions of Biand Aiin Peskin and Schroeder, these imply that
Aφ=γφ=g2
8π2, Aψ=γψ=g2
32π2;
Bλ=3g4
2π23λ2
32π2Bg=g2
16π2.
Therefore, we see that
βg=2gBg2gAψgAφ= 2gg2
16π2+ 2gg2
32π2+gg2
8π2=5g3
16π2;
βλ=2Bλ4λAφ= 2 µ3λ2
32π23g4
2π2+ 4λg2
8π2=3λ2+ 8λg248g4
16π2
While it was supposedly unnecessary, the running couplings were computed to be1,
g(p) = s16π2
110 log p/M ;
λ(p) = λ=g2
3Ã1 + 1454145+149
141 +g2145/5
4145+149
141 g2145/5!.
Notice that both gand λgenerally become weaker at large distances because for typical values
of g, λ we see that βgand βλare both positive. However, if λ << g then βλwill b e negative and so λ
will grow stronger at larger distances. Near small values of gand λthe theory shows interesting interplay
between gand λ. Also interesting is the characteristic Landau pole in λsuggesting that we should not
trust this theory at too large a scale.
Below is a graph of gversus λindicating the direction of Renormalization Group flow as the inter-
action distance grows larger.
100 200 300 400 500 600 700
HpL
2.5
5
7.5
10
12.5
15
17.5
g
HpL
Figure 1. Renormalization Group Flow as a funciton of scale. Arrow indicates flow in
the direction of larger distances. For this plot, Mwas taken to be 104.
1See appendix.
1
pf3
pf4
pf5

Partial preview of the text

Download Homework 9 - Beta Functions in Pseudo-Scalar Yukawa Theory | PHYSICS 523 and more Assignments Physics in PDF only on Docsity!

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

  • 2g

g^2 32 π^2

  • g

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 -ˏHpL

5

10

15

gHpL

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

  • m^2

dZm dM

= 2Zmm

βm(λ) M

  • m^2

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

  • mλ dbν+ dλ

∴ 2 γm(λ)bν = −β(λ)

dbν dλ

  • λ

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

  • O(g^5 ).

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

  • O(g^4 ).

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.