Redshift Academy


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∫ dk⁴p/(2π)⁴ 1/((k² - m²))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(x₁ + ... + x_n - 1)
    1/(A₁ ... A_n) = (n - 1)!∫dx₁ ... ∫dx_n --------------------
                           ⁰        ⁰     [x₁A₁ + ... + x_nA_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 4-momentum in Minkowski space.  For example:

           I = ∫d⁴k/(2π)⁴ i/(k² - m²)ⁿ

             = ∫d³k ∫dk₀ 1/(k₀² - k² - m²)ⁿ

    However, this has poles at k⁰ = ±√(k² + m²)

    To rectify this the term iε is added to displace these 
    poles away from the integration path.

    k₀ plane:

                Im(k⁰)
                |
                |
       x        |
      iε        |
     -----------+---------- Re(k⁰)
                |       iε
                |       x
                |
                |

     x = poles of the Feynman propagator.

     So the integral now becomes the integral of a contour 
     in the complex plane:
 
       I = ∫d³k∫dk₀1/(k⁰² - k² - m² + iε)ⁿ

     If we rotate Re(k₀) 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 = (k⁰)² + (k¹)² + (k²)² + (k³)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.  

     ∫d⁴k/(2π)⁴ 1/(k² - m² + iε)²  => -i∫d⁴k/(2π)⁴ 1/(k_E² - m²)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^μk^ν in the numerator:

     I^μν = d⁴k/(2π)⁴ k^μk^ν/(k² - m²)ⁿ

       = g^μμ∫d⁴k/(2π)⁴ k²/(k² - m²)ⁿ

     If there is just one k in the numerator we get:

     ∫d⁴k/(2π)⁴ kp/(k² - m²)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/(k² - 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)²i Γ(2-d/2)   1
I = μ^ε --------------- -----
          (4π)^d/2Γ(2)   Δ^2-d/2

Where Γ() is the Gamma function and 1/Δ^2-d/2 can be expressed
using the series representation:

     _∞
x^-n = Σn^m(-ln(x))
     ^m=0

  = 1 - nlog(x) ...

Which leads to:

1/Δ^2-d/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)^2-ε/2Γ(2)  Δ^ε/2
                            
        iΓ(ε/2)   
  =  ---------- (μ²/Δ)^ε/2
     (4π)^2-ε/2Γ(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 Euler-Mascheroni constant (~ 0.5772).  
Noting that Γ(2) = 1 we get:

(i/16π²)(2/ε - γ_E)(1 - εlnΔ) 

           = (i/16π²)(2/ε - ln(μ²/Δ) - γ_E - εγ_Eln(μ²/Δ))

           = (i/16π²)(2/ε - ln(μ²/Δ) - γ_E - O(ε)) 

The final result becomes:

I = (i/16π²)[2/ε - γ_E + ln(4π) + ln(μ²/Δ)]

  = (i/16π²)[2/ε + ln{4πexp(-γ_E)μ²/Δ}]
                     _
  = (i/16π²)[2/ε + ln{μ²/Δ}] ... 1.
      _
Where μ² = 4πexp(-γ_E)μ² 


Other useful integrals are:
                        
                       (-1)¹i Γ(1-d/2)  1
∫d⁴k/(2π)⁴ 1/(k² - Δ) = -------------- -----
                         (4π)^d/2Γ(1)   Δ^1-d/2

Γ(ε/2-1) = Γ(ε/2)/(ε/2 - 1)

       = -2/ε + γ_E - 1 

1/Δ^1-d/2 = 1/Δ^ε/2-1 = 1/Δ^ε/21/Δ^-1 = (1 - (ε/2)ln(μ²/Δ))Δ

-(i/16π²)(-2/ε + γ_E - 1)(1 - (ε/2)ln(μ²/Δ))Δ 

              = (i/16π²)Δ(2/ε - ln(μ²/Δ) - γ_E + 1 + O(ε))
Therefore,

I = (iΔ/16π²)[2/ε - ln(μ²/Δ) + ln(4π) - γ_E + 1] ... 2.

                        
                         (-1)^2-1i Γ(1-d/2) d     1
∫d⁴k/(2π)⁴ k²/(k² - Δ)² = ---------------- --- ------
                            (4π)^d/2Γ(2)    2   Δ^1-d/2

Γ(ε/2-1) = -2/ε + γ_E - 1 

1/Δ^1-d/2 = 1/Δ^ε/2-1 = 1/Δ^ε/21/Δ^-1 = (1 - (ε/2)ln(μ²/Δ))Δ

-(i/16π²)(-2/ε + γ_E - 1)(2 - ε/2)(1 - (ε/2)ln(μ²/Δ))Δ 

                = (i/16π²)2Δ(2/ε - ln(μ²/Δ) - γ_E + 1 + O(ε))

Therefore,

I = ((iΔ/8π²)[2/ε - ln(μ²/Δ) + ln(4π) - γ_E + 1] ... 3.
                                       
                          (-1)^2-1i Γ(1-d/2) g^μν   1
∫d⁴k/(2π)⁴ k^μk^ν/(k² - Δ)² = ---------------- -- ------
                            (4π)^d/2Γ(2)      2  Δ^1-d/2

Γ(ε/2-1) = -2/ε + γ_E - 1 

1/Δ^1-d/2 = 1/Δ^ε/2-1 = 1/Δ^ε/21/Δ^-1 = (1 - (ε/2)ln(μ²/Δ))Δ

-(ig^μν/16π²)(-2/ε + γ_E - 1)(2 - ε)(1 - εln(μ²/Δ))Δ 

                = (i/16π²)2g^μνΔ(2/ε - ln(μ²/Δ) - γ_E + 1 + O(ε))

Therefore,

I = ((ig^μνΔ/8π²)[2/ε - ln(μ²/Δ) + ln(4π) - γ_E + 1] ... 4.

The general forms of above solutions is:
           
I = (-1)ⁿi(1/4π)^d/2(1 or d/2 or g^μν)f(Γ,Δ)
          
The (-1)ⁿi 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)ⁿi term where n is 
the power of the denominator, (k² - 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^kΣa^k_i/εⁱ 
    ^{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 = (g²/16π²)[a¹/ε]

Therefore, in our simple case:

δ = g²/16π²ε for MS
                                   __
δ = (g²/16π²)(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 = ψ₀(iγ^μD_μ - m₀)ψ₀ - (1/4)(F_μνF^μν)₀
     _        _             _
  = iψ₀γ^μ∂ψ₀ + ψ₀γ^μe₀A₀ψ₀ - m₀ψ₀ψ₀ - (1/4)(F_μνF^μν)₀

The Feynman rules are:

Vertex (A):  -ieγ^μ

Fermionic propagator (B):  (iγ^μp + m)/(p² - m² + iε)

Photon propagator (C):   -ig^μν/p²

Electron Self Energy, Σ(p)
--------------------------



                          iγ^α(γ^μk + m) γ^β(-ig^μν)
Σ(p) = (-ie)²∫d⁴k/(2π)⁴ ----------------------------
                       (k² - m² + iε)((p - k)² + iε) 

Applying Feynman parameters to our integral leads to:

                    ₁             (2γ^μk - 4m)
Σ(p) = e²∫d⁴k/(2π)⁴ ∫dx ---------------------------------
                   ⁰   [(k² - m²)(1 - x) + (p - k)²x + iε]²

For the denominator, D, we get:

D = k² - m² - k²x + m²x + p²x + k²x - 2kpx

  = k² - m² + m²x + p²x - 2kpx

  = k² - m² + m²x + p²x² - p²x² + p²x - 2kpx

  = (k - px)² - m² + m²x - p²x² + p²x

  = (k - px)² - Δ

Where Δ = (1 - x)(m² - p²x)

Shifting k -> k + px results in:

         ₁             (2γ^μk + 2γ^μpx - 4m) 
Σ(p) = e²∫dx ∫d⁴k/(2π)⁴ -----------------
         ⁰               [k² - Δ + iε]²

Now, from before we said ∫dk k/(...) = 0.  Therefure,

         ₁                               1
Σ(p) = e²∫dx (2γ^μpx - 4m) ∫d⁴k/(2π)⁴ -------------
         ⁰                          [k² - Δ + iε]²

We can now use equation 1. from above to get:
              ₁                         _
Σ(p) = (iα/4π)∫dx (2γ^μpx - 4m)(2/ε + ln(μ²/Δ))
              ⁰
Where α = e²/4π
               __
After applying MS we get:
               ₁
Σ_R(p) = (iα/4π)∫dx (2γ^μpx - 4m_R)(ln(μ²/Δ))
               ⁰
               ₁
      = (iα/4π)∫dx (2γ^μpx - 4m_R)(ln(μ²/(1 - x)(m_R² - p²x))
               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^μp - γ^μk + m)γ^β(γ^νk + m)}
Π(p) = -(-ie)²∫d⁴p/(2π)⁴ ---------------------------------
                        ((p - k)² - m² + iε)(p² - m² + iε)

Note:  There is a -ve in front of the e² 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 SPIN-STATISTICS 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(x₁)φ(x₂) = φ(x₂)φ(x₁)

While

ψ(x₁)ψ(x₂) = -ψ(x₂)ψ(x₁)

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) - m²)

  = 2p^μp^ν - p^μk^ν - p^νk^μ - g^μν(p.(p - k) - m²)

The integral becomes:
                   
                                    N
Π(p) = -4e²∫d⁴p/(2π)⁴ --------------------------------
                     ((p - k)² - m² + iε)(p² - m² + iε)
             
We can now apply Feynman parameters to get:
                   
                     ₁                   N
Π(p) = -4e²∫d⁴p/(2π)⁴∫dx ------------------------------------
                    ⁰   [x((p - k)² - m²) + (1 - x)(p² - m²)]²

For the denominator, D, we get: 

D = [p²x + k²x - 2pkx - m²x + p² - m² - p²x + m²x]²

  = [k²x - 2pkx + p² - m²]²

  = [k²x - 2pkx + k²x² - k²x² + p² - m²]²

  = [(p - kx)² - k²x² + k²x - m²]²

  = [(p - kx)² - k²x(x - 1) - m²]²

  = [(p - kx)² - Δ]²

Where Δ = m² - k²x(1 - x)

We can now make the shift p -> p + kx to get for the 
numerator, N:

For the 2p^μp^ν term we get:

2p^μp^ν = 2(p^μ + xk^μ)(p^ν + xk^ν) 

    = 2(p^μp^ν + p^μxk^ν + p^νxk^μ + x²k^μk^ν)

For the p^μk^ν and p^νk^μ terms we get:

p^μk^ν -> (p^μ + k^μx)k^ν = p^μk^ν + k^μk^νx

p^νk^μ -> (p^ν + k^νx)k^μ = p^νk^μ + k^νk^μx

For the g^μν(p.(p - k) - m²) term we get:

g^μν(p + kx)(p + kx - k) - m²)

= g^μν(p² + kpx - kp + pxk + k²x² - k²x - m²)

Putting it all together for the numerator we get:

N = 2(p^μp^ν + p^μxk^ν + p^νxk^μ + x²k^μk^ν) 

    - (p^μk^ν + k^μk^νx) - (p^νk^μ + k^νk^μx)

    - (g^μν(p² + kpx - kp + pxk + k²x² - k²x - m²))

Neglecting terms linear in p leads to:

N = 2p^μp^ν + 2x²k^μk^ν - k^μk^νx - k^νk^μx 

    - (g^μν(p² + k²x² - k²x - m²))

  = 2p^μp^ν + 2x²k^μk^ν - 2k^μk^νx - (g^μν(p² + k²x² - k²x - m²))

  = 2p^μp^ν + 2k^μk^νx(x - 1) - g^μν(p² + k²x(x - 1) - m²)

  = 2p^μp^ν - 2k^μk^νx(1 - x) - g^μν(p² - k²x(1 - x) - m²)

Now, in d dimensions g_μνg^μν = d.  Therefore,

dp^μk^ν = g_μνg^μνp^μk^ν 

      = p²g^μν

∴ p^μp^ν = (1/d)p²g^μν

So we can write:

N = (2/d)p²g^μν - k^μk^ν2x(1 - x) - g^μν(p² - k²x(1 - x) - m²)

  = (-1 + 2/d)p²g^μν - k^μk^ν2x(1 - x) - g^μν(-k²x(1 - x) - m²)

Lets take this term by term.  For the 1st term we get:

∫d⁴p/(2π)⁴(-1 + 2/d)p²g^μν/(p² - Δ)² 

        = (-1 + 2/d)g^μν∫d⁴p/(2π)⁴ p²/(p² - Δ)² 

        = -ig^μν(d/2)(-1 + 2/d)(d/2)Γ(1-d/2)(1/Δ)^1-d/2/(4π)^d/2

        = -ig^μν(1 - d/2)Γ(1-d/2)(1/Δ)^1-d/2/(4π)^d/2  

Now Γ(2-d/2) = (1-d/2)Γ(1-d/2) and Δ^1-d/2 = Δ^2-d/2Δ. 
 
Therefore,

∫d⁴p/(2π)⁴(-1 + 2/d)p²g^μν/(p² - Δ)² 

        = -iΓ(2-d/2)(1/Δ)^2-d/2/(4π)^d/2g^μνΔ

For the 2nd term we get:

k^μk^ν∫d⁴p/(2π)⁴ 1/(p² - Δ)² = ik^μk^νΓ(2-d/2)(1/Δ)^2-d/2/(4π)^d/2

For the 3rd term we get:

g^μν(-k²x(1 - x) - m²)∫d⁴p/(2π)⁴ 1/(p² - Δ)² 

         = ig^μν(-k²x(1 - x) - m²)Γ(2-d/2)(1/Δ)^2-d/2/(4π)^d/2

Therefore,

Π(k) = iΓ(2-d/2)(1/Δ)^2-d/2/(4π)^d/2[-g^μνΔ - k^μk^ν - g^μν(-k²x(1 - x) - m²)

     = i[-g^μν(m² - k²x(1 - x)) - 2x(1 - x)k^μk^ν - g^μν(-k²x(1 - x) - m²)]

     = i[-g^μν(-2k²x(1 - x)) - 2x(1 - x)k^μk^ν]

     = i[g^μνk² - k^μk^ν]∫dx x(1 - x)(-8e²/(4π)^d/2)∫dx x(1 - x)Γ(2-d/2)(1/Δ)^2-d/2

We can now write this as:

Π(k) = (g^μνk² - k^μk^ν)iπ(k²)

Where, 

π(k²) = -8e²/(4π)^d/2∫dx x(1 - x)Γ(2-d/2)(1/Δ)^2-d/2

     = -8e²/(4π)^d/2∫dx x(1 - x)(2/ε - ln(μ²/Δ) + ln(4π) - γ_E - O(ε)) 

Π(k) is a rank 2 tensor formed from the 4-momentum 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^μk^ν.  It, 
therefore, has the form:

Π^μν(k) = Ag^μν + Bk^μk^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_μΠ^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_μΠ^μν = Ak_μg^μν + Bk²k^ν

   = Ak^ν + Bk²p^ν = 0

   = k^ν(A + Bk²) = 0

For k^ν ≠ 0 we get A = -Bk²

Substituting back into our original equation gives:

Π^μν(k) = -Bk²g^μν + Bk^μk^ν

  = (-k²g^μν + k^μk^ν)B

  = (-k²g^μν + k^μk^ν)π(k²)  

Where,
                   ₁                                  
π(k²) = -8e²/(4π)^d/2∫dx x(x - 1)(2/ε - ln(μ²/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^μp - γ^μk + m)γ^β(γ^νk + m)] = p.(p - k) + m²

                               = 4(p² - pk + m²)

The integral becomes:
                   
                               (p² - pk + m²)
Π(p) = -4g²∫d⁴p/(2π)⁴ --------------------------------
                     ((p - k)² - m² + iε)(p² - m² + iε)
             
We can now apply Feynman parameters to get:
                   
                     ₁             (p² - pk + m²)
Π(p) = -4g²∫d⁴p/(2π)⁴∫dx ------------------------------------
                    ⁰   [x((p - k)² - m²) + (1 - x)(p² - m²)]²

For the denominator calculation is identical to the photon
self energy case: 

D = [(p - kx)² - Δ]²

Where Δ = m² - k²x(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) + m²

  = p² + 2kpx + k²x² - pk - k²x + m²

  = p² - k²x(1 - x) + 2kpx - pk + m²

And for the denominator, D:

D = [p² - Δ]²

Therefore, the integral becomes:
                   
           ₁              p² - k²x(1 - x) + 2kpx - pk + m²
Π(k) = -4g²∫dx ∫d⁴p/(2π)⁴ --------------------------------
           ⁰                        [p² - Δ]²

As before we can omit the terms linear in p and rearrange 
to get:
                   
           ₁              p² + m² - k²x(1 - x)
Π(k) = -4g²∫dx ∫d⁴p/(2π)⁴ --------------------
           ⁰                   [p² - Δ]²
                   
            ₁                p²                                     
Π(k) = -4g²[∫dx ∫d⁴p/(2π)⁴ -------- 
            ⁰             [p² - Δ]²                               
                ₁                                   1  
              - ∫dx (m² - k²x(1 - x)) ∫d⁴p/(2π)⁴ ---------]
                ⁰                                [p² - Δ]²

Π(k) = (-4g²)(-i/(4π)^d/2[(d/2)Γ(1-d/2)(1/Δ)^1-d/2 

        + ΔΓ(2-d/2)(1/Δ)^2-d/2]

Now Γ(2-d/2)(1/Δ)^(2-d/2) = (1 - d/2)Γ(1-d/2)(1/Δ)^(1-d/2)

Therefore,

Π(k) = -4g²/(4π)^d/2[(-i(d/2)Γ(1-d/2)(1/Δ)^(1-d/2) 

              - (-i)(1 - d/2)(d/2)Γ(1-d/2)(1/Δ)^(1-d/2)]

     = -4g²/(4π)^d/2[-iΓ(1-d/2)(d/2) - (-i)(1 - d/2)(Γ(1-d/2))(1/Δ)^(1-d/2)]

     = -4g²/(4π)^d/2[-iΓ(1-d/2)((d/2) - (1 - d/2))(1/Δ)^(1-d/2)]

     = -4g²/(4π)^d/2[-i[(d - 1)Γ(1-d/2)(1/Δ)^1-d/2]

     = -4g²/(4π)^d/2[-i(3 - ε)Γ(1-d/2)(1/Δ)^1-d/2]

     = -4g²(-i/(4π)^d/2)[(3 - ε)Δ(2/ε - ln(μ²/Δ) + ln(4π) - γ_E + 1)

     = -4g²(-i)(1/(4π)^d/2)[(3 - ε)Δ  

        + (3 - ε)Δ(2/ε - ln(μ²/Δ) + ln(4π) - γ_E)]

     = -4g²(-i)(1/(4π)^d/2)[3Δ + 3Δ(2/ε - ln(μ²/Δ) + ln(4π) - γ_E) - Δ]

     = -4g²(-i)(1/(4π)^d/2)[2Δ + 3Δ(2/ε - ln(μ²/Δ) + ln(4π) - γ_E)]

     = -4g²(-i)[(1/8π²)(k²/6 + m²) 

        + (3/16π²)∫dx (m² - k²x(1 - x)(2/ε - ln(μ²/Δ) + ln(4π) - γ_E)] 

Check: Using equations 2. and 3. from above we get:

Π(k) = -4g²[-i/8π²Δ(2/ε - ln(μ²/Δ) + ln(4π) - γ_E + 1) 

          - i/16π²(2/ε - ln(μ²/Δ) + ln(4π) - γ_E)]

     = -4g²(-i)[1/8π²Δ + 1/8π²Δ(2/ε - ln(μ²/Δ) - ln(4π) - γ_E) 

          + 1/16π²(2/ε - ln(μ²/Δ) + ln(4π) - γ_E)]

     =  -4g²(-i)[1/8π²Δ + 3/16π²(2/ε - ln(μ²/Δ) + ln(4π) - γ_E)]

After MS we get:                  
                                                                   
Π(k) = -4g²(-i)[1/8π²(k²/6 + m²) - 3/16π²(ln(μ²/Δ))]

This is quadratically divergent.  This is significant in 
the context of the GAUGE HIERARCHY PROBLEM disussed in the
notes on SUPERSYMMETRY.