Calculation of the two-loop self-energy in the λφ4 theory
Renormalized perturbation theory
Let us start by recalling the bare Lagrangian,
λB 4
1
1
L = (∂µ φB )2 − m2B φ2B −
φ .
2
2
4! B
(1)
In the bare perturbation theory, the first two terms are taken to define the free part and thus
the propagator, while the last term is treated as an interaction. However, this splitting is
in principle arbitrary as long as the functional integral with the first part of the Lagrangian
can be performed exactly. Let us redefine the field as well as the mass and the coupling in
terms of the renormalized parameters,
1/2
φB = Zφ φR ,
m2B =
1
(m2 + δm2 ),
Zφ R
λB =
1
(λR + δλ).
Zφ2
(2)
By introducing the three new parameters Zφ , δm2 , δλ, we have expressed the freedom to split
the Lagrangian into the free and interaction parts according to our will. The Lagrangian
hence becomes
1
1
1
1
δλ
λR
L = (∂µ φR )2 − m2R φ2R − φ4R + (Zφ − 1)(∂µ φR )2 − δm2 φ2R − φ4R .
2
2
4! }
|2
{z 2
} | 4!
{z
(3)
interactions
free part
This splitting of the Lagrangian gives rise to the renormalized perturbation theory. Since we
eventually want to express every physical observable as a series in the renormalized coupling
λR , it is convenient to similarly expand the anticipated counter-terms,
Zφ − 1 = z1 λR + z2 λ2R + · · · ,
δm2 = m2R (c1 λR + c2 λ2R + · · · ),
δλ = d2 λ2R + d3 λ3R + · · · .
(4)
Note that the expansion of δλ starts at order λ2R : this is because the leading, tree-level
interaction is proportional to λR and loops only provide higher-order corrections to it. With
these definitions, the momentum-space Feynman rules of the theory take the form
=
i
,
p2 − m2R + i
= −iλR ,
= i(Zφ − 1)p2 − iδm2 ,
(5)
= −iδλ.
The counter-term vertices are denoted using a little square. In practice, we perform a
systematic expansion of all Green’s functions in powers of λR , which means that only the
corresponding part of the counter-terms needs to be taken into account. I will indicate this
by attaching a label to the counter-term vertex.
Renormalization at one loop
The complete one-loop renormalization of the theory was carried out in detail in the lecture.
I summarize here the main results in order to pin down the values of the counter-terms
at this order, which will be needed in the two-loop computation below. The scalar field
self-energy at order λR is given by
+
z1 /c1
iλR m2R µ−2
=
2(4π)2
1
4πµ2
+ 1 − γE + log 2
mR
+ iλR (z1 p2 − c1 m2R ),
(6)
whence we can extract the counter-terms
µ−2
.
c1 =
2(4π)2 z1 = 0,
(7)
(Here and in the following, I use for simplicity the MS renormalization scheme.) Likewise,
the corrections to the interaction vertex start at order λ2R and are given by the diagrams
+
3iλ2R µ−2
=
2(4π)2
+
+
1
4πµ2
− γE + log 2
mR
(8)
− id2 λ2R + finite contributions.
This allows us to determine
d2 =
3µ−2
.
2(4π)2 (9)
Two-loop self-energy
We finally get to the main point of this text. At order λ2R , the scalar field self-energy acquires
the following contributions,
c1
+
+
+
+
d2
.
(10)
z2 /c2
I evaluate the diagrams one by one, leaving the second, most difficult one to the end. In order
to avoid confusion, I write the momentum integrals directly in Euclidean space, keeping in
mind that the integration measure is (i dd p) and the propagator is written as −i/(p2 + m2 ).
This way, I will be able to use the formulas for momentum integration summarized in the
appendix, in particular Eq. (25). Also, for better legibility, the symmetry factors are initially
indicated in red. Let us start with the double-bubble diagram,
1
= (−iλR )2
4
Z
i dd p
−i
d
2
(2π) p + m2R
d
Z
i dd k
(2π)d
−i
2
k + m2R
2
d
iλ2R (m2R ) 2 −1 Γ(1 − d2 ) (m2R ) 2 −2 Γ(2 − d2 )
=
4 (4π)d/2 Γ(1) (4π)d/2 Γ(2)
2
iλ2R m2R µ−4 4πµ2
Γ()Γ( − 1)
=
4(4π)4
m2R
4πµ2
iλ2R m2R µ−4 1
1
= −
1 − 2γE + 2 log 2
+ finite contributions.
+
4(4π)4
2 mR
(11)
The counter-term diagrams are much easier. First,
c1
2
Z
1
i dd p
−i
2
= (−iλR )(−ic1 λR mR )
2
(2π)d p2 + m2R
d
ic1 λ2R m2R µ−2 4πµ2
ic1 λ2R m2R (m2R ) 2 −2 Γ(2 − d2 )
=
=
Γ()
2
(4π)d/2 Γ(2)
2(4π)2
m2R
1
4πµ2
iλ2R m2R µ−4 1
+
−γE + log 2
+ finite contributions,
=
4(4π)4
2 mR
(12)
where I used the result (7) for c1 . We can see that the worst divergences proportional to
1/2 in the diagrams (11) and (12) cancel. This is of course expected: the counter-term c1
removes the subdivergence corresponding to the upper loop in Eq. (11). Finally, we find
d2
d
i dd p
−i
id2 λ2R (m2R ) 2 −1 Γ(1 − d2 )
=
−
(2π)d p2 + m2R
2 (4π)d/2 Γ(1)
−id2 λ2R m2R µ−2 4πµ2
=−
Γ( − 1)
2(4π)2
m2R
3iλ2R m2R µ−4 1
1
4πµ2
=
+
1 − γE + log 2
+ finite contributions,
4(4π)4
2 mR
1
= (−id2 λ2R )
2
Z
(13)
where I again used the already known result (9) for d2 .
The setting-sun diagram
The remaining two-loop diagram in Eq. (10) is the only one which requires a nontrivial effort.
Its evaluation is not straightforward and requires additional tricks beyond the basic Feynman
parametrization and momentum integrals. It can be computed in different ways; one of them,
based on integration by parts, is nicely explained in Section 5.6 of the notes by Hendrik van
Hees (http://fias.uni-frankfurt.de/˜hees/publ/lect.pdf). I will follow a different approach
which is sufficient to extract the divergent part of the diagram and hence to renormalize it
properly. While it may not be the easiest path leading to the desired result, it illustrates
several other computation techniques that appear to be useful in other contexts. Let us start
by writing the expression for the diagram from the Feynman rules,
Z
i dd k i dd q
−i
−i
−i
1
2
= (−iλR )
2
2
6
(2π)d (2π)d k 2 + mR q 2 + mR (p + k + q)2 + m2R
Z
dd k dd q
1
1
iλ2R
1
,
=
2 2
2
d
d
2
6
(2π) (2π) k + mR q + mR (p + k + q)2 + m2R
(14)
where p is the external momentum. From dimensional analysis, we expect the divergences
of this diagram to appear at orders p0 and p2 in the expansion in powers in p. We will
evaluate their coefficients separately, and thus first simply set p = 0. Let us write the
iλ2
second line of Eq. (14) at p = 0 as 6R I(m2R ). This depends on mR as the sole dimensionful
parameter, and therefore has to be proportional to m2d−6
= m2−4
. Instead of calculating
R
R
I(m2R ) directly, it is easier to evaluate its derivative I 0 (m2R ), which can be expressed as (the
three propagators of the diagram differ just by relabelling of the momentum variables, which
is why the differentiation produces three times the same integral)
Z
1
dd k dd q
0
2
I (mR ) = − 3
2
(2π)d (2π)d (k 2 + mR )(q 2 + m2R )2 [(k + q)2 + m2R ]
Z
1
dd k dd q
(15)
=−3
2
d
d
2
2
(2π) (2π) (k + mR )(q + m2R )2 (k + q)2
Z
1
dd k dd q
2
+ 3mR
.
2
2 2
d
d
2
2
(2π) (2π) (k + mR )(q + mR ) (k + q)2 [(k + q)2 + m2R ]
Why have I split one integral into two, apparently more complicated ones? The point is that
the former integral here, with one massless propagator, can actually be evaluated analytically,
while the latter is finite at d = 4 (check that it does not have any subdivergences!). As long
as we are only interested in the divergent part of the integral, we can then use Eq. (37) with
α = 1, β = 2 and γ = 1 to write
(m2R )d−4 Γ( d2 − 1)Γ(2 − d2 )Γ(3 − d2 )Γ(4 − d)
(4π)d
Γ( d2 )Γ(1)Γ(2)Γ(5 − d)
(m2 )−2 Γ(1 − )Γ()Γ(1 + )Γ(2)
3(m2R )−2 [Γ()]2
= −3 R 4−2
=−
(4π)
Γ(2 − )Γ(1 + 2)
2(4π)4−2 1 − I 0 (m2R ) = −3
(16)
plus finite contributions. Note that I used extensively the rule Γ(z + 1) = zΓ(z) to simplify
the product of Γ-functions. Now we can integrate with respect to m2R to obtain
I(m2R )
3(m2R )1−2
[Γ()]2
=−
2(4π)4−2 (1 − )(1 − 2)
3m2R µ−4 1
1
4πµ2
=−
+
3 − 2γE + 2 log 2
+ finite contributions.
2(4π)4 2 mR
(17)
For the p2 divergence in the setting-sun diagram, we introduce three Feynman parameters
using Eq. (26). The momentum integral thereby acquires the general form (28), studied in
the appendix. We readily identify the bilinear form in the denominator as
x(k 2 + m2R ) + y(q 2 + m2R ) + z[(p + k + q)2 + m2R ] = P T AP + 2b · P + m2R + zp2 ,
(18)
x+z
z
T
where P = (k, q) , A =
, and b = (zp, zp)T . Introducing the abbreviations
z
y+z
Σ = xy + yz + zx and Π = xyz, we find that det A = Π and M 2 = m2R + zp2 − bT A−1 b =
m2R + p2 Π/Σ. With the help of Eq. (28), the double momentum integral in Eq. (14) is
therefore rewritten as
Z 1
(M 2 )d−3 Γ(3 − d) 1
Γ(3)
dx dy dz δ(x + y + z − 1)
(4π)d
Γ(3) Σd/2
0
1−2
p2 Π
2
Z ∞
mR + Σ
Γ(−1 + 2)
dx
dy
dz
δ(x
+
y
+
z
−
1)
(19)
=
(4π)4−2
Σ2−
0
1−2
"
#
p2 Π
2
Z ∞
− (m2R )1−2
mR + Σ
Γ(−1 + 2)
(m2R )1−2
dx dy dz δ(x + y + z − 1)
+
=
.
(4π)4−2
Σ2−
Σ2−
0
The last manipulation may seem trivial, yet it almost solves the problem. The first term
in the square brackets is p-independent and is therefore already contained in Eq. (17). The
second term gives rise, as we will see shortly, to a convergent integral in the limit → 0.
Since we are only interested in the divergent part of the setting-sun diagram, we can perform
this limit safely, leaving us with
Z ∞
Π
2 −4 Γ(−1 + 2)
dx
dy
dz
δ(x
+
y
+
z
−
1)
pµ
(4π)4−2
Σ3
0
(20)
2 −4 Z ∞
pµ
xyz
=−
dx dy dz δ(x + y + z − 1)
+ finite contributions.
2(4π)4 0
(zy + yz + zx)3
The power of µ in front is as usual added in order to keep the correct dimension of the whole
expression after expansion in . The remaining integral looks still formidable, but can in
fact be evaluated rather straightforwardly by a series of steps: (i) expressing the negative
√
power of Σ by means of an integral over an auxiliary variable t; (ii) rescaling x, y, z by t;
(iii) introducing new variables u = yz, v = zx, w = xy (check that the integration measure
transforms as du dv dw = 2xyz dx dy dz!),
Z ∞
xyz
dx dy dz δ(x + y + z − 1)
(zy + yz + zx)3
0
Z ∞
Z ∞
1
dx dy dz δ(x + y + z − 1)xyz
dt t2 e−t(xy+yz+zx)
=
2 0
0
Z
Z ∞
xyz 2 −(xy+yz+zx)
(21)
1 ∞
dx dy dz x+y+z
√
−
1
=
dt
δ
t
e
t
2 0
t3/2 |
{z
} t3/2
0
=2tδ[(x+y+z)2 −t]
Z
=
∞
dx dy dz xyz e
0
−(xy+yz+zx)
1
=
2
Z
0
∞
1
du dv dw e−(u+v+w) = .
2
We can now collect all the pieces from Eqs. (17), (20) and (21) and write the divergent part
of the setting-sun diagram as
iλ2R m2R µ−4
1
4πµ2
1
p2
− 2+
−3 + 2γE − 2 log 2 −
=
+ ··· .
4(4π)4
mR
6m2R
(22)
Renormalization at two loops
Having evaluated all the necessary loop integrals, let us now put together all the contributions
from Eqs. (11), (12), (13) and (22). A number of cancellations occurs in the sum,
"
2
HH 4πµ
1
iλ2R m2R µ−4
1
H
HH
− 2−
1 − 2γH
E + 2 log
H
2
4(4π)4
mH
RH
1
1
+ 2+
+
3
1
+
2
1
1
− 2+
2
HH 4πµ
H
−
−3 + H
2γH
E − 2 log
H
H
2
mH
H
R
d2
p2
6m2R
#
+ iz2 λ2R p2 − ic2 λ2R m2R
iλ2R m2R µ−4
=
4(4π)4
1
p2
2
− −
2
6m2R c1
!
2
Z4πµ
−H
γH
+
log
E
2
Z
mZ
RZ
2
H
HH4πµ
H
3 − 3γH
E + 3 log
H
2
H
mH
RH
Z
(23)
z2 /c2
+ iz2 λ2R p2 − ic2 λ2R m2R .
This is the final result. It is now obvious that all divergences in the two-loop self-energy can
be removed by adjusting the counter-terms properly, namely to
µ−4
1
µ−4
2
,
c2 =
−
.
(24)
z2 =
24(4π)4 4(4π)4 2
Appendix: some useful formulas
First recall the basic formula for momentum integration in Euclidean space:
Z
d
dd p
1
(m2 ) 2 −α Γ(α − d2 )
=
.
(2π)d (p2 + m2 )α
(4π)d/2 Γ(α)
Next remember the identity that allows one to introduce Feynman parameters:
Z
1
Γ(α1 + · · · + αn ) 1
xα1 1 −1 · · · xαnn −1 δ(x1 + · · · + xn − 1)
=
.
dx
·
·
·
dx
1
n
Aα1 1 · · · Aαnn
Γ(α1 ) · · · Γ(αn ) 0
(x1 A1 + · · · + xn An )α1 +···+αn
(25)
(26)
The above two formulas just repeat what has already been shown in the lecture and thus
do not require a proof. (If you think that you have only seen Feynman parametrization
with αi = 1 for all i, take derivatives of the formula you know with respect to Ai !) When
combining several propagators using the Feynman parameters, one often has to deal with
the following, more general class of integrals,
Z d
1
d p1
dd pN
·
·
·
,
(27)
(2π)d
(2π)d (P T AP + 2b · P + m2 )α
where A is a symmetric N ×N matrix acting in the momentum space P = (p1 , . . . , pN )T , and
b is a vector of length N . By the usual completing the square and shifting the integration
variables, the polynomial in the denominator can be rewritten as a strictly quadratic form
with a constant piece P T AP + M 2 , where M 2 = m2 − bT A−1 b. The matrix A can be
diagonalized by a rotation of the integration variables. Finally, the momenta need to be
rescaled in order to bring the denominator into the canonical form P 2 + M 2 . Altogether,
the identity (25) generalizes as
Z
dd p1
dd pN
1
···
d
d
T
(2π)
(2π) (P AP + 2b · P + m2 )α
(m2 − bT A−1 b)
=
(4π)N d/2
Nd
−α
2
Γ(α − N2d )
1
.
Γ(α) (det A)d/2
Next come two auxiliary identities that we will need below. First,
Z ∞
tα−1
aα−β Γ(α)Γ(β − α)
dt
=
.
(a + bt)β
bα
Γ(β)
0
(28)
(29)
α−β R 1
Upon the substitution z = a/(a+bt), the integral becomes a bα 0 dz (1−z)α−1 z β−α−1 , where
we recognize the familiar definition of the B-function, and the result follows by recalling the
identity B(α, β) = Γ(α)Γ(β)/Γ(α + β). Second,
Z
∞
dx dy
0
xα−1 y β−1 −(x+y) Γ(α)Γ(β)Γ(α + β − γ)
e
=
.
(x + y)γ
Γ(α + β)
This can be most easily obtained from the generating function
Z ∞
Γ(α)Γ(β)
f (t) =
dx dy xα−1 y β−1 e−t(x+y) =
.
tα+β
0
(30)
(31)
Fill in the missing details as a simple homework! Finally, I want to derive a class of momentum integrals which become very handy when evaluating the setting-sun diagram,
Z
dd p dd k
1
1
1
Iα,β,γ =
.
(32)
d
d
2
2
α
2
2
β
(2π) (2π) (p + m ) (k + m ) [(p + k)2 ]γ
Let us start by introducing three Feynman parameters x, y, z according to Eq. (26), upon
which the momentum denominator acquires the form
x+z
z
2
2
2
2
2
T
2
x(p +m )+y(k +m )+z(p+k) = P AP +(x+y)m , where A =
, (33)
z
y+z
and P = (p, k)T . Eq. (28) then allows us to integrate over momentum, which leads to the
first intermediate result,
Z
xα−1 y β−1 z γ−1 δ(x + y + z − 1)
Γ(α + β + γ − d) (m2 )d−α−β−γ 1
dx
dy
dz
Iα,β,γ =
. (34)
Γ(α)Γ(β)Γ(γ)
(4π)d
(x + y)α+β+γ−d (xy + yz + zx)d/2
0
The integration over x, y, z can obviously be extended to infinity since the δ-function forces
them to be at most one anyway. Next, use a trick, representing the inverse power of x + y as
Z ∞
Γ(α + β + γ − d)
dt tα+β+γ−d−1 e−t(x+y) .
(35)
=
(x + y)α+β+γ−d
0
Rescaling subsequently all x, y, z by t and using that δ( x+y+z
− 1) = tδ(x + y + z − t) allows
t
one to carry out the integration over t trivially. This brings the integral into a much more
human form
Z
(m2 )d−α−β−γ ∞
xα−1 y β−1 z γ−1 −(x+y)
1
dx
dy
dz
Iα,β,γ =
e
.
(36)
Γ(α)Γ(β)Γ(γ)
(4π)d
(xy + yz + zx)d/2
0
The calculation is completed by using Eq. (29) to integrate over z and then Eq. (30) to
integrate over x, y (fill in the details!). The desired final result reads
Z
dd p dd k
1
1
1
d
d
2
2
α
2
2
β
(2π) (2π) (p + m ) (k + m ) [(p + k)2 ]γ
(37)
(m2 )d−α−β−γ Γ( d2 − γ)Γ(α + γ − d2 )Γ(β + γ − d2 )Γ(α + β + γ − d)
.
=
(4π)d
Γ( d2 )Γ(α)Γ(β)Γ(α + β + 2γ − d)
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