Exercise 1. Two-loop self-energy in massless φ

QFT II Problem Set 11.

FS 2019 Prof. M. Grazzini https://www.physik.uzh.ch/en/teaching/PHY552/FS2019.html Due by: 20/5/2019

Exercise 1. Two-loop self-energy in massless φ

-theory in six dimensions

Last week you did the one-loop renormalization of the φ

-theory and also had a quick look at the divergence structure at two-loops. You shall now explicitly calculate the two-loop self-energy diagrams, shown in figure 1 and show, that the non-local divergences cancel. For simplicity we will work in the massless theory, m = 0.

p p

k k

k+p l k+l

(i)

p p

l k

k+p l+p

k−l

(ii)

Figure 1: The two diagrams contributing to the self-energy at two-loop.

(a) The contribution of the left diagram is given by Π

(p

) = g

Z d

k (2π)

i

−1

(k

)

(k + p)

Π

(k

) , (1) where

Π

(k

) = g

2

Z d

k (2π)

i

1

k

(k + l)

= g

2(4π)

Γ

2 − D

2 Z

0

dx [−x(1 − x)k

]

(2) is the one-loop self-energy you know from last week, with m = 0. Show that evaluation of the Feynman integral yields

Π

(k

) = g

2(4π)

Γ 2 −

Γ

− 1

Γ(D − 2) (−k

)

. (3) (b) Insert the result for Π

(k

) into Π

(p

), introduce Feynman parameters and perform the

integrations.

(c) Express the result in terms of ε = (6−D)/2 and expand in ε. Keep only the parts divergent as ε → 0. You should find

Π

(p

) = g

(4π)

p

1 2(3!)

− 1 2ε

+ 1

ε

ln −p

4πµ

+ γ − 43 12

+ finite . (4) (d) Explain why at two-loop order we also have to include the following diagram containing

the one-loop counter term:

Π

(p

) ≡

p p

k k

k+p

(5)

1

which corresponds to Π

(p

) with Π

(k

) replaced by B

k

, where B

k

= − g

(4π)

k

2(3!)

1

ε = Z

− 1 + O g

(6) has been derived on the last sheet. (Note that, since we are working in a massless theory, we do not need the mass renormalization.)

(e) Calculate Π

(p

) and expand in ε, neglecting finite terms of O ε

. (f) Show that this term exactly cancels the non-local divergence

∼ 1 ε ln

−p

4πµ

(7) in equation (4).

(g) The contribution of the right diagram is given by Π

(p

) = g

2

Z d

l (2π)

i

1

l

(l + p)

Λ

(p, l) (8) where

Λ

(p, l) = −g

Z d

k (2π)

i

1

k

(k + p)

(k − l)

(9)

= g

(4π)

Γ

3 − D 2

Z

dx Z

0

dy

−x(1 − x)p

− y(1 − y)l

− 2xypl

(10) is the one-loop vertex correction known from the last sheet, with m = 0. The evaluation of this integral is more involved than the previous one, so we just give the result:

Π

(p

) = − g

4(4π)

(−p

)

Γ(5 − D)Γ

− 2

Γ

− 5

× 2

3 − D + B

3 − D, 3D 2 − 5

− B

3 − D, 3 − D 2

. (11) Expand this result in ε, keep only the divergent parts.

(h) You should find that the result again contains a non-local divergence. What diagrams are we still missing to cancel it? Call the contribution of those diagrams Π

(p

). Calculate the divergent part of Π

(p

) and show that the non-local divergence is exactly cancelled.

(i) Determine the two-loop counter term needed to remove the remaining overall divergence.

Hint. U seful formulae:

• B(x, y) = Γ(x)Γ(y) Γ(x + y) =

Z

dt t

(1 − t)

•

Z d

l

(2π)

1

(l

+ X)

= (4π)

Γ ξ −

Γ(ξ) X

2