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Units, Constants and Useful Formulas
Regularization and Renormalization

Regularization and Renormalization are a collection of
enormously important and mathematically difficult techniques
that are employed in QFT to solve the problem of divergences
in Feynman loop integrals. Because the mathematics is lengthy
and difficult, we will skip a lot of the more complicated
calculations in favor of illustrating the concepts.
We start by saying that regularization and renormalization
are 2 different things. Simply put, Regularization makes a
divergent expression (quantity) finite. Renormalization
discards this expression (makes it zero). Regularization
essentially assumes the existence of 'new' unknown physics
at unobserved scales (small size or large energy levels) and
enables current theories to give accurate predictions as an
"effective theory" within its intended scale of use.
We will concentrating on 2 of the most popular techiques in
use today  DIMENSIONAL REGULARIZATION and the MINIMAL
SUBTRACTION RENORMALIZATION scheme.
Dimensional Regularization

In general, Feynman loop integrals have the form:
I = (1/2π)^{4}∫ dk^{4}p/(2π)^{4} 1/((k^{2}  m^{2}))^{n}
These integrals are either logarithmically or quadratically
divergent.
To evaluate these types of integrals and get them into
the form required for dimensional regularization requires
the use of 3 techniques
 FEYNMAN PARAMETERIZATION: Feynman noted that,
_{1} _{1} δ(x_{1} + ... + x_{n}  1)
1/(A_{1} ... A_{n}) = (n  1)!∫dx_{1} ... ∫dx_{n} 
^{0} ^{0} [x_{1}A_{1} + ... + x_{n}A_{n}]^{n}
For n = 2 we get:
_{1} _{1} δ(x + y  1)
1/(AB) = ∫dx ∫dy 
^{0} ^{0} [xA + yB]^{2}
If we let y = 1  x => δ(x + 1  x  1) = δ(0) = 1
And the integral becomes:
_{1} _{1} 1
1/(AB) = ∫dx ∫dy 
^{0} ^{0} [xA + (1  x)B]^{2}
 WICK ROTATION: After introducing Feynman parameters and
completing the square, one is often left with an integral
over a loop 4momentum in Minkowski space. For example:
I = ∫d^{4}k/(2π)^{4} i/(k^{2}  m^{2})^{n}
= ∫d^{3}k ∫dk_{0} 1/(k_{0}^{2}  k^{2}  m^{2})^{n}
However, this has poles at k^{0} = ±√(k^{2} + m^{2})
To rectify this the term iε is added to displace these
poles away from the integration path.
k_{0} plane:
Im(k^{0})


x 
iε 
+ Re(k^{0})
 iε
 x


x = poles of the Feynman propagator.
So the integral now becomes the integral of a contour
in the complex plane:
I = ∫d^{3}k∫dk_{0}1/(k^{0}^{2}  k^{2}  m^{2} + iε)^{n}
If we rotate Re(k_{0}) counter clockwise by 90° we do not
encounter any problem with crossing the poles. If we
do this we end up with:
(k^{E})^{2} = (k^{0})^{2} + (k^{1})^{2} + (k^{2})^{2} + (k^{3})^{2}.
This is the momentum in Euclidean space. Now the iε
term no longer plays a role and we can set ε = 0. This
rotation is refered to as a Wick rotation. Once Wick
rotated, the integrals can be evaluated in standard
ways.
∫d^{4}k/(2π)^{4} 1/(k^{2}  m^{2} + iε)^{2} => i∫d^{4}k/(2π)^{4} 1/(k_{E}^{2}  m^{2})^{2}
The Wick rotation is just a trick to do integrals. It
has no physical meaning.
 TENSOR INTEGRALS: Often integrals have terms such as
k^{μ}k^{ν} in the numerator:
I^{μν} = d^{4}k/(2π)^{4} k^{μ}k^{ν}/(k^{2}  m^{2})^{n}
^{ } = g^{μμ}∫d^{4}k/(2π)^{4} k^{2}/(k^{2}  m^{2})^{n}
If there is just one k in the numerator we get:
∫d^{4}k/(2π)^{4} kp/(k^{2}  m^{2})^{n} = 0
This is because we are integrating over all k and the
integrand is antisymmetric under k > k. The only
terms we need to keep are those with even powers of k
in the numerator.
Dimensional regularization assumes that the spacetime
dimension is not 4 but is analytically continued to 4  ε.
As an example, for the case where n = 2, we write:
I = lim μ^{ε}∫ d^{4  ε}k/(2π)^{4  ε} 1/(k^{2}  Δ)^{2}
^{ε > 0}
Where μ is an arbitrary parameter with the dimension of
energy that forces the result to be dimensionless. The
integral is now convergent and has the solution:
(1)^{2}i Γ(2d/2) 1
I = μ^{ε}  
(4π)^{d/2}Γ(2) Δ^{2d/2}
Where Γ() is the Gamma function and 1/Δ^{2d/2} can be expressed
using the series representation:
_{ ∞}
x^{n} = Σn^{m}(ln(x))
^{ m=0}
_{ } = 1  nlog(x) ...
Which leads to:
1/Δ^{2d/2} = 1  (2  d/2)lnΔ
If we now replace d with 4  ε we get:
iΓ(ε/2) 1
I = μ^{ε/2}  
(4π)^{2ε/2}Γ(2) Δ^{ε/2}
iΓ(ε/2)
=  (μ^{2}/Δ)^{ε/2}
(4π)^{2ε/2}Γ(2)
Expansion of Γ around ε = 0 (using Weierstrass's definition
of Γ) gives:
Γ(ε/2) = 2/ε  γ_{E} + O(ε) + ...
Where γ_{E} is the EulerMascheroni constant (~ 0.5772).
Noting that Γ(2) = 1 we get:
(i/16π^{2})(2/ε  γ_{E})(1  εlnΔ)
= (i/16π^{2})(2/ε  ln(μ^{2}/Δ)  γ_{E}  εγ_{E}ln(μ^{2}/Δ))
= (i/16π^{2})(2/ε  ln(μ^{2}/Δ)  γ_{E}  O(ε))
The final result becomes:
I = (i/16π^{2})[2/ε  γ_{E} + ln(4π) + ln(μ^{2}/Δ)]
= (i/16π^{2})[2/ε + ln{4πexp(γ_{E})μ^{2}/Δ}]
_{ } _
= (i/16π^{2})[2/ε + ln{μ^{2}/Δ}] ... 1.
_
Where μ^{2} = 4πexp(γ_{E})μ^{2}
Other useful integrals are:
(1)^{1}i Γ(1d/2) 1
∫d^{4}k/(2π)^{4} 1/(k^{2}  Δ) =  
(4π)^{d/2}Γ(1) Δ^{1d/2}
Γ(ε/21) = Γ(ε/2)/(ε/2  1)
= 2/ε + γ_{E}  1
1/Δ^{1d/2} = 1/Δ^{ε/21} = 1/Δ^{ε/2}1/Δ^{1} = (1  (ε/2)ln(μ^{2}/Δ))Δ
(i/16π^{2})(2/ε + γ_{E}  1)(1  (ε/2)ln(μ^{2}/Δ))Δ
= (i/16π^{2})Δ(2/ε  ln(μ^{2}/Δ)  γ_{E} + 1 + O(ε))
Therefore,
I = (iΔ/16π^{2})[2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E} + 1] ... 2.
(1)^{21}i Γ(1d/2) d 1
∫d^{4}k/(2π)^{4} k^{2}/(k^{2}  Δ)^{2} =   
(4π)^{d/2}Γ(2) 2 Δ^{1d/2}
Γ(ε/21) = 2/ε + γ_{E}  1
1/Δ^{1d/2} = 1/Δ^{ε/21} = 1/Δ^{ε/2}1/Δ^{1} = (1  (ε/2)ln(μ^{2}/Δ))Δ
(i/16π^{2})(2/ε + γ_{E}  1)(2  ε/2)(1  (ε/2)ln(μ^{2}/Δ))Δ
= (i/16π^{2})2Δ(2/ε  ln(μ^{2}/Δ)  γ_{E} + 1 + O(ε))
Therefore,
I = ((iΔ/8π^{2})[2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E} + 1] ... 3.
(1)^{21}i Γ(1d/2) g^{μν} 1
∫d^{4}k/(2π)^{4} k^{μ}k^{ν}/(k^{2}  Δ)^{2} =   
(4π)^{d/2}Γ(2) 2 Δ^{1d/2}
Γ(ε/21) = 2/ε + γ_{E}  1
1/Δ^{1d/2} = 1/Δ^{ε/21} = 1/Δ^{ε/2}1/Δ^{1} = (1  (ε/2)ln(μ^{2}/Δ))Δ
(ig^{μν}/16π^{2})(2/ε + γ_{E}  1)(2  ε)(1  εln(μ^{2}/Δ))Δ
= (i/16π^{2})2g^{μν}Δ(2/ε  ln(μ^{2}/Δ)  γ_{E} + 1 + O(ε))
Therefore,
I = ((ig^{μν}Δ/8π^{2})[2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E} + 1] ... 4.
The general forms of above solutions is:
I = (1)^{n}i(1/4π)^{d/2}(1 or d/2 or g^{μν})f(Γ,Δ)
The (1)^{n}i term represents th
NOTE: The integrals are specified in Minkowski space. The
solutions involving the Γ functions have already taken into
account the Wick rotation via the (1)^{n}i term where n is
the power of the denominator, (k^{2}  Δ)^{n}.
The main advantage of dimensional regularization is that it
preserves gauge invariance. Now, there is still a divergence
in this expression as ε > 0. How do we get rid of it?.
There are 2 renormalization schemes that can be applied.
The Minimal Subtraction (MS) scheme or, more commonly, the
Modified Minimal Subtraction (MS) scheme. Both methods
introduce a COUNTERTERM, δ, into the Lagrangian that removes
the 2/ε or the 2/ε  γ_{E} + ln(4π) terms respectively. With
dimensional regularization the δ's take the form:
_{L k}
δ = Σc^{k}Σa^{k}_{i}/ε^{i}
^{k=1 i=1}
Where L is the number of loops, c is the coupling term and
a^{k} is a numerical coefficient.
For 1 loop:
δ = (g^{2}/16π^{2})[a^{1}/ε]
Therefore, in our simple case:
δ = g^{2}/16π^{2}ε for MS
_{ } __
δ = (g^{2}/16π^{2})(2/ε  γ_{E} + ln(4π)) for MS
Photon and Electron Self Energy

To illustrate these techniques, we will focus on the simpler
theory of QED. In particlar, will consider the electron and
photon self energy to 1 loop only in our calculations.
Consider the QED Lagrangian:
_
L = ψ_{0}(iγ^{μ}D_{μ}  m_{0})ψ_{0}  (1/4)(F_{μν}F^{μν})_{0}
_ _ _
= iψ_{0}γ^{μ}∂ψ_{0} + ψ_{0}γ^{μ}e_{0}A_{0}ψ_{0}  m_{0}ψ_{0}ψ_{0}  (1/4)(F_{μν}F^{μν})_{0}
The Feynman rules are:
Vertex (A): ieγ^{μ}
Fermionic propagator (B): (iγ^{μ}p + m)/(p^{2}  m^{2} + iε)
Photon propagator (C): ig^{μν}/p^{2}
Electron Self Energy, Σ(p)

iγ^{α}(γ^{μ}k + m) γ^{β}(ig^{μν})
Σ(p) = (ie)^{2}∫d^{4}k/(2π)^{4} 
(k^{2}  m^{2} + iε)((p  k)^{2} + iε)
Applying Feynman parameters to our integral leads to:
_{1} (2γ^{μ}k  4m)
Σ(p) = e^{2}∫d^{4}k/(2π)^{4} ∫dx 
^{0} [(k^{2}  m^{2})(1  x) + (p  k)^{2}x + iε]^{2}
For the denominator, D, we get:
D = k^{2}  m^{2}  k^{2}x + m^{2}x + p^{2}x + k^{2}x  2kpx
= k^{2}  m^{2} + m^{2}x + p^{2}x  2kpx
= k^{2}  m^{2} + m^{2}x + p^{2}x^{2}  p^{2}x^{2} + p^{2}x  2kpx
= (k  px)^{2}  m^{2} + m^{2}x  p^{2}x^{2} + p^{2}x
= (k  px)^{2}  Δ
Where Δ = (1  x)(m^{2}  p^{2}x)
Shifting k > k + px results in:
_{1} (2γ^{μ}k + 2γ^{μ}px  4m)
Σ(p) = e^{2}∫dx ∫d^{4}k/(2π)^{4} 
^{0} [k^{2}  Δ + iε]^{2}
Now, from before we said ∫dk k/(...) = 0. Therefure,
_{1} 1
Σ(p) = e^{2}∫dx (2γ^{μ}px  4m) ∫d^{4}k/(2π)^{4} 
^{0} [k^{2}  Δ + iε]^{2}
We can now use equation 1. from above to get:
_{1} _
Σ(p) = (iα/4π)∫dx (2γ^{μ}px  4m)(2/ε + ln(μ^{2}/Δ))
^{0}
Where α = e^{2}/4π
__
After applying MS we get:
_{1}
Σ_{R}(p) = (iα/4π)∫dx (2γ^{μ}px  4m_{R})(ln(μ^{2}/Δ))
^{0}
_{1}
= (iα/4π)∫dx (2γ^{μ}px  4m_{R})(ln(μ^{2}/(1  x)(m_{R}^{2}  p^{2}x))
^{0}
Photon Self Energy, Π(p)

For fermionic loops the rule is to multiply the propagators
and take the trace of the matrix product. The probability
amplitude for 1 loop is:
Tr{γ^{α}(γ^{μ}p  γ^{μ}k + m)γ^{β}(γ^{ν}k + m)}
Π(p) = (ie)^{2}∫d^{4}p/(2π)^{4} 
((p  k)^{2}  m^{2} + iε)(p^{2}  m^{2} + iε)
Note: There is a ve in front of the e^{2} term because it
represents a fermionic loop. The sign difference between
fermionic and bosonic loops can be seen in the following
diagram.
Bosons Fermions
 
A B AB A B AB
     
+ + + +  
  +  + 
Therefore, bosonic loops make a positive contribution
to the vacuum energy while fermionic loops make a
negative contribution. The diagram also illustrates
the SPINSTATISTICS THEOREM. If A and B swap positions
there is no change in sign for the bosons but there is a
change for the fermions. Thus,
φ(x_{1})φ(x_{2}) = φ(x_{2})φ(x_{1})
While
ψ(x_{1})ψ(x_{2}) = ψ(x_{2})ψ(x_{1})
For the numerator, N, using various γ matrix identites
yields:
N = p^{μ}(p^{ν}  k^{ν}) + p^{ν}(p^{μ}  k^{μ})  g^{μν}(p.(p  k)  m^{2})
= 2p^{μ}p^{ν}  p^{μ}k^{ν}  p^{ν}k^{μ}  g^{μν}(p.(p  k)  m^{2})
The integral becomes:
N
Π(p) = 4e^{2}∫d^{4}p/(2π)^{4} 
((p  k)^{2}  m^{2} + iε)(p^{2}  m^{2} + iε)
We can now apply Feynman parameters to get:
_{1} N
Π(p) = 4e^{2}∫d^{4}p/(2π)^{4}∫dx 
^{0} [x((p  k)^{2}  m^{2}) + (1  x)(p^{2}  m^{2})]^{2}
For the denominator, D, we get:
D = [p^{2}x + k^{2}x  2pkx  m^{2}x + p^{2}  m^{2}  p^{2}x + m^{2}x]^{2}
= [k^{2}x  2pkx + p^{2}  m^{2}]^{2}
= [k^{2}x  2pkx + k^{2}x^{2}  k^{2}x^{2} + p^{2}  m^{2}]^{2}
= [(p  kx)^{2}  k^{2}x^{2} + k^{2}x  m^{2}]^{2}
= [(p  kx)^{2}  k^{2}x(x  1)  m^{2}]^{2}
= [(p  kx)^{2}  Δ]^{2}
Where Δ = m^{2}  k^{2}x(1  x)
We can now make the shift p > p + kx to get for the
numerator, N:
For the 2p^{μ}p^{ν} term we get:
2p^{μ}p^{ν} = 2(p^{μ} + xk^{μ})(p^{ν} + xk^{ν})
^{ } = 2(p^{μ}p^{ν} + p^{μ}xk^{ν} + p^{ν}xk^{μ} + x^{2}k^{μ}k^{ν})
For the p^{μ}k^{ν} and p^{ν}k^{μ} terms we get:
p^{μ}k^{ν} > (p^{μ} + k^{μ}x)k^{ν} = p^{μ}k^{ν} + k^{μ}k^{ν}x
p^{ν}k^{μ} > (p^{ν} + k^{ν}x)k^{μ} = p^{ν}k^{μ} + k^{ν}k^{μ}x
For the g^{μν}(p.(p  k)  m^{2}) term we get:
g^{μν}(p + kx)(p + kx  k)  m^{2})
= g^{μν}(p^{2} + kpx  kp + pxk + k^{2}x^{2}  k^{2}x  m^{2})
Putting it all together for the numerator we get:
N = 2(p^{μ}p^{ν} + p^{μ}xk^{ν} + p^{ν}xk^{μ} + x^{2}k^{μ}k^{ν})
 (p^{μ}k^{ν} + k^{μ}k^{ν}x)  (p^{ν}k^{μ} + k^{ν}k^{μ}x)
 (g^{μν}(p^{2} + kpx  kp + pxk + k^{2}x^{2}  k^{2}x  m^{2}))
Neglecting terms linear in p leads to:
N = 2p^{μ}p^{ν} + 2x^{2}k^{μ}k^{ν}  k^{μ}k^{ν}x  k^{ν}k^{μ}x
 (g^{μν}(p^{2} + k^{2}x^{2}  k^{2}x  m^{2}))
= 2p^{μ}p^{ν} + 2x^{2}k^{μ}k^{ν}  2k^{μ}k^{ν}x  (g^{μν}(p^{2} + k^{2}x^{2}  k^{2}x  m^{2}))
= 2p^{μ}p^{ν} + 2k^{μ}k^{ν}x(x  1)  g^{μν}(p^{2} + k^{2}x(x  1)  m^{2})
= 2p^{μ}p^{ν}  2k^{μ}k^{ν}x(1  x)  g^{μν}(p^{2}  k^{2}x(1  x)  m^{2})
Now, in d dimensions g_{μν}g^{μν} = d. Therefore,
dp^{μ}k^{ν} = g_{μν}g^{μν}p^{μ}k^{ν}
= p^{2}g^{μν}
∴ p^{μ}p^{ν} = (1/d)p^{2}g^{μν}
So we can write:
N = (2/d)p^{2}g^{μν}  k^{μ}k^{ν}2x(1  x)  g^{μν}(p^{2}  k^{2}x(1  x)  m^{2})
= (1 + 2/d)p^{2}g^{μν}  k^{μ}k^{ν}2x(1  x)  g^{μν}(k^{2}x(1  x)  m^{2})
Lets take this term by term. For the 1st term we get:
∫d^{4}p/(2π)^{4}(1 + 2/d)p^{2}g^{μν}/(p^{2}  Δ)^{2}
= (1 + 2/d)g^{μν}∫d^{4}p/(2π)^{4} p^{2}/(p^{2}  Δ)^{2}
= ig^{μν}(d/2)(1 + 2/d)(d/2)Γ(1d/2)(1/Δ)^{1d/2}/(4π)^{d/2}
= ig^{μν}(1  d/2)Γ(1d/2)(1/Δ)^{1d/2}/(4π)^{d/2}
Now Γ(2d/2) = (1d/2)Γ(1d/2) and Δ^{1d/2} = Δ^{2d/2}Δ.
Therefore,
∫d^{4}p/(2π)^{4}(1 + 2/d)p^{2}g^{μν}/(p^{2}  Δ)^{2}
= iΓ(2d/2)(1/Δ)^{2d/2}/(4π)^{d/2}g^{μν}Δ
For the 2nd term we get:
k^{μ}k^{ν}∫d^{4}p/(2π)^{4} 1/(p^{2}  Δ)^{2} = ik^{μ}k^{ν}Γ(2d/2)(1/Δ)^{2d/2}/(4π)^{d/2}
For the 3rd term we get:
g^{μν}(k^{2}x(1  x)  m^{2})∫d^{4}p/(2π)^{4} 1/(p^{2}  Δ)^{2}
= ig^{μν}(k^{2}x(1  x)  m^{2})Γ(2d/2)(1/Δ)^{2d/2}/(4π)^{d/2}
Therefore,
Π(k) = iΓ(2d/2)(1/Δ)^{2d/2}/(4π)^{d/2}[g^{μν}Δ  k^{μ}k^{ν}  g^{μν}(k^{2}x(1  x)  m^{2})
= i[g^{μν}(m^{2}  k^{2}x(1  x))  2x(1  x)k^{μ}k^{ν}  g^{μν}(k^{2}x(1  x)  m^{2})]
= i[g^{μν}(2k^{2}x(1  x))  2x(1  x)k^{μ}k^{ν}]
= i[g^{μν}k^{2}  k^{μ}k^{ν}]∫dx x(1  x)(8e^{2}/(4π)^{d/2})∫dx x(1  x)Γ(2d/2)(1/Δ)^{2d/2}
We can now write this as:
Π(k) = (g^{μν}k^{2}  k^{μ}k^{ν})iπ(k^{2})
Where,
π(k^{2}) = 8e^{2}/(4π)^{d/2}∫dx x(1  x)Γ(2d/2)(1/Δ)^{2d/2}
^{ } = 8e^{2}/(4π)^{d/2}∫dx x(1  x)(2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E}  O(ε))
Π(k) is a rank 2 tensor formed from the 4momentum of the
photon and g^{μν}. The only way to build such a tensor is to
make a linear combination of g^{μν} and k^{μ}k^{ν}. It,
therefore, has the form:
Π^{μν}(k) = Ag^{μν} + Bk^{μ}k^{ν}
We now need to introduce the WARD IDENTITY into the
discussion.
Ward Identity

Let Π(k) = ε_{μ}(k)Π^{μ}(k) be the amplitude for some QED process
involving an external photon with momentum, k, and ε_{μ}(k) is
the polarization vector of the photon. The Ward identity
is:
k_{μ}Π^{μ}(k) = 0
It tells us that the photon's polarization parallel to its
direction of propagation doesn't contribute to the scattering
amplitude.
If we now apply the Ward identity to Π^{μν} we get:
k_{μ}Π^{μν} = Ak_{μ}g^{μν} + Bk^{2}k^{ν}
_{ } = Ak^{ν} + Bk^{2}p^{ν} = 0
_{ } = k^{ν}(A + Bk^{2}) = 0
For k^{ν} ≠ 0 we get A = Bk^{2}
Substituting back into our original equation gives:
Π^{μν}(k) = Bk^{2}g^{μν} + Bk^{μ}k^{ν}
^{ } = (k^{2}g^{μν} + k^{μ}k^{ν})B
^{ } = (k^{2}g^{μν} + k^{μ}k^{ν})π(k^{2})
Where,
_{1}
π(k^{2}) = 8e^{2}/(4π)^{d/2}∫dx x(x  1)(2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E}  O(ε))
^{0}
So the contribution to the self energy by the photon itself
is 0. Zero energy implies 0 mass which is required by gauge
invariance. Therefore, the Ward identity is closely related
to gauge invariance.
With the Ward identity, the self energy of the photon is
logarithmically divergent. Without it, the self energy
would be quadratically divergent. We will see this in
the next section on Yukawa theory where we calculate the
1 loop fermionic correction to the massive Higgs boson.
Yukawa Theory

For the Yukawa 1 loop correction our integral is similar
to the photon self energy integral except that the coupling
constant becomes g, the photon is replaced by the Higgs
boson and trace of the numerator becomes simpler.
Tr[γ^{α}(γ^{μ}p  γ^{μ}k + m)γ^{β}(γ^{ν}k + m)] = p.(p  k) + m^{2}
_{ } = 4(p^{2}  pk + m^{2})
The integral becomes:
(p^{2}  pk + m^{2})
Π(p) = 4g^{2}∫d^{4}p/(2π)^{4} 
((p  k)^{2}  m^{2} + iε)(p^{2}  m^{2} + iε)
We can now apply Feynman parameters to get:
_{1} (p^{2}  pk + m^{2})
Π(p) = 4g^{2}∫d^{4}p/(2π)^{4}∫dx 
^{0} [x((p  k)^{2}  m^{2}) + (1  x)(p^{2}  m^{2})]^{2}
For the denominator calculation is identical to the photon
self energy case:
D = [(p  kx)^{2}  Δ]^{2}
Where Δ = m^{2}  k^{2}x(1  x)
Again, we can now make the shift p > p + kx to get for
the numerator, N:
N = (p + kx)^{2}  k(p + kx) + m^{2}
= p^{2} + 2kpx + k^{2}x^{2}  pk  k^{2}x + m^{2}
= p^{2}  k^{2}x(1  x) + 2kpx  pk + m^{2}
And for the denominator, D:
D = [p^{2}  Δ]^{2}
Therefore, the integral becomes:
_{1} p^{2}  k^{2}x(1  x) + 2kpx  pk + m^{2}
Π(k) = 4g^{2}∫dx ∫d^{4}p/(2π)^{4} 
^{0} [p^{2}  Δ]^{2}
As before we can omit the terms linear in p and rearrange
to get:
_{1} p^{2} + m^{2}  k^{2}x(1  x)
Π(k) = 4g^{2}∫dx ∫d^{4}p/(2π)^{4} 
^{0} [p^{2}  Δ]^{2}
_{1} p^{2}
Π(k) = 4g^{2}[∫dx ∫d^{4}p/(2π)^{4} 
^{0} [p^{2}  Δ]^{2}
_{1} 1
 ∫dx (m^{2}  k^{2}x(1  x)) ∫d^{4}p/(2π)^{4} ]
^{0} [p^{2}  Δ]^{2}
Π(k) = (4g^{2})(i/(4π)^{d/2}[(d/2)Γ(1d/2)(1/Δ)^{1d/2}
+ ΔΓ(2d/2)(1/Δ)^{2d/2}]
Now Γ(2d/2)(1/Δ)^{(2d/2)} = (1  d/2)Γ(1d/2)(1/Δ)^{(1d/2)}
Therefore,
Π(k) = 4g^{2}/(4π)^{d/2}[(i(d/2)Γ(1d/2)(1/Δ)^{(1d/2)}
 (i)(1  d/2)(d/2)Γ(1d/2)(1/Δ)^{(1d/2)}]
= 4g^{2}/(4π)^{d/2}[iΓ(1d/2)(d/2)  (i)(1  d/2)(Γ(1d/2))(1/Δ)^{(1d/2)}]
= 4g^{2}/(4π)^{d/2}[iΓ(1d/2)((d/2)  (1  d/2))(1/Δ)^{(1d/2)}]
= 4g^{2}/(4π)^{d/2}[i[(d  1)Γ(1d/2)(1/Δ)^{1d/2}]
= 4g^{2}/(4π)^{d/2}[i(3  ε)Γ(1d/2)(1/Δ)^{1d/2}]
= 4g^{2}(i/(4π)^{d/2})[(3  ε)Δ(2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E} + 1)
= 4g^{2}(i)(1/(4π)^{d/2})[(3  ε)Δ
+ (3  ε)Δ(2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E})]
= 4g^{2}(i)(1/(4π)^{d/2})[3Δ + 3Δ(2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E})  Δ]
= 4g^{2}(i)(1/(4π)^{d/2})[2Δ + 3Δ(2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E})]
= 4g^{2}(i)[(1/8π^{2})(k^{2}/6 + m^{2})
+ (3/16π^{2})∫dx (m^{2}  k^{2}x(1  x)(2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E})]
Check: Using equations 2. and 3. from above we get:
Π(k) = 4g^{2}[i/8π^{2}Δ(2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E} + 1)
 i/16π^{2}(2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E})]
= 4g^{2}(i)[1/8π^{2}Δ + 1/8π^{2}Δ(2/ε  ln(μ^{2}/Δ)  ln(4π)  γ_{E})
+ 1/16π^{2}(2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E})]
= 4g^{2}(i)[1/8π^{2}Δ + 3/16π^{2}(2/ε  ln(μ^{2}/Δ) + ln(4π)  γ_{E})]
After MS we get:
Π(k) = 4g^{2}(i)[1/8π^{2}(k^{2}/6 + m^{2})  3/16π^{2}(ln(μ^{2}/Δ))]
This is quadratically divergent. This is significant in
the context of the GAUGE HIERARCHY PROBLEM disussed in the
notes on SUPERSYMMETRY.