The Lagrangian reads

QFT II Problem Set 10.

FS 2019 Prof. M. Grazzini https://www.physik.uzh.ch/en/teaching/PHY552/FS2019.html Due by: 13/5/2019

Exercise 1. One-loop renormalization of φ ³ -theory in six dimensions

In this exercise you shall perform the one-loop renormalization for φ ³ -theory in six dimensions.

The Lagrangian reads

L = 1

2 (∂ ^µ φ)(∂ µ φ) − m ² 2 φ ²

| {z }

L₀

+ g 3! φ ³

| {z }

L_I

. (1)

By power counting the only divergences we need to renormalize arise in the propagator ∆(p ² ) and the vertex function Γ(p, q), which to one-loop take the following form:

∆(p ² ) = 1

m ² − p ² − Π(p ² ) , (2)

Γ(p, q) = g(1 + Λ(p, q)) , (3)

with

Π(p ² ) = g ² 2

Z d ^D k (2π) ^D i

1 m ² − k ²

1

m ² − (k + p) ² , (self-energy) (4)

Λ(p, q) = g ²

Z d ^D k (2π) ^D i

1 m ² − k ²

1 m ² − (k + p) ²

1

m ² − (k − q) ² . (vertex correction) (5)

p p

k

k+p

(i) self-energy

p+q k−q

q

k

p k+p

(ii) vertex correction

(a) Introduce Feynman parameters for the self-energy integral (4) and perform the momentum integration. You do not need to perform the Feynman integral.

(b) Expand your result around ε = 0, where ε = (6 − D)/2.

(c) Insert your result into the expression for the propagator (2). You should find

∆

(p ² ) =∆

₀ (p ² ) − Π(p ² ) (6)

=m ²

1 + g ² ₀ 2(4π) ³

1

ε − γ + ln 4π + 1

(7)

− p ²

1 + g ² ₀ 12(4π) ³

1

ε − γ + ln 4π + 1

(8)

− g ₀ ² 2(4π) ³

Z 1 0

dx

m ² − x(1 − x)p ²

ln m ² − x(1 − x)p ²

µ ² , (9)

1

where ∆ 0 (p ² ) = 1/(m ² − p ² ) is the free propagator and g 0 = gµ

is the dimensionless coupling. Note that in the expression above the divergences are present only in the coeffi- cients of m ² and p ² , which suggests that we can deal with them by modifying the part of the Lagrangian which generates ∆ 0 (p ² ).

(d) In L ₀ , replace φ and m by the renormalized (i.e. finite) quantities φ _r and m _r according to m ² φ ² = m ² _r φ ² _r (1 + A) , (∂ ^µ φ)(∂ _µ φ) = (∂ ^µ φ _r )(∂ _µ φ _r )(1 + B) . (10) How do A and B enter into ∆ 0 (p ² ) and ∆(p ² )?

(e) Choose A and B such that they absorb the divergences. Note that there are arbitrarily many possibilities.

(f) Now repeat steps (a) and (b) for the vertex correction (5).

(g) Insert your result into the expression for the vertex function (3). You should find Γ(p, q) = g 0

1 + g ₀ ² 2(4π) ³

1

ε − γ + ln 4π

− g ₀ ³ (4π) ³

Z 1 0

dx Z 1−x

0

dy ln K

µ ² (11) with

K = m ² − x(1 − x)q ² − y(1 − y)p ² − 2xyp · q . (12) (h) Redefine the interaction term L _I similar to what you did in (d) by introducing a third

renormalization quantity C and choose it such, that the divergences are absorbed.

(i) Usually one does not modify whole terms of the Lagrangian like we did above, but rather the fields, masses and couplings individually:

φ = p

Z _φ φ _r m = p

Z _m m _r g = Z _g g _r . (13) Can you relate the Z i to A, B and C, neglecting terms of O g ³

? Hint. T he following Euclidean integral might be useful:

Z d ^D l E

(2π) ^D 1

(l ² _E + X) ^ξ = (4π)

Γ ξ − ^D ₂

Γ(ξ) X

(14)

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Exercise 2. Zimmermann forest formula

In the lecture you saw two examples for the BPHZ construction, one diagram contributing to the two-loop vertex function as an example with nested divergency and one diagram contributing to the two-loop self-energy as an example for overlapping divergences.

(i) self-energy

(ii) vertex correction

You shall now explicitly check the results obtained in the lecture by applying Zimmermann’s forest formula to both cases:

R _G = X

U

Y

γ∈U

(−T _γ )I _G , (forest formula) (15)

where I _G is the integrand corresponding to a given Feynman diagram G and R _G the respective renormalized integrand. γ are the renormalisation parts of the diagram, that is the proper superficially divergent subdiagrams of G. Note that G itself can be a renormalization part. T γ

is an operator which, when acting on I _G , generates the counterterm for to the renormalization part γ. The sum over U s is the sum over all the possible forests of the diagram G. A forest is defined as a set of non-overlapping renormalisation parts of G (nested or disjoint) including G itself. A forest can be empty. The (unsubtracted) term I _G corresponds to the empty forest.

For the two examples given in the lecture, find all the forests and check the result obtained in the lecture using Zimmermann’s formula.

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