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Last modified: July 29, 2022

The Gauge Hierarchy Problem
---------------------------

The gauge hierarchy problem is perhaps the main motivation 
for supersymmetry.  To understand what the problem is, it 
is necessary to look more closely at the mass of the Higgs 
boson.

Consider the following 1-loop Feynman diagrams:



(a) Represents the bare Higgs mass.

(b) Represents the Higgs SELF INTERACTION.   
                                            
(c) Represents the Yukawa interaction.

The corresponding Lagrangian is:
                               _
L = ∂μφ∂μφ - μ2φ2/2 + λφ4/4 - gYψψφ

Case (b).  Applying Feynman rules we get: 

I = λ∫d4p/(2π)4 1/(p2 - mH2)

Where i/(p2 - mH2) is the massive bosonic propagator.

From the discussion on Regularization and Renormalization 
we had (using dimensional regularization):

I = (iλΔ/16π2)[2/ε - ln(μ2/Δ) + ln(4π) - γE + 1]
                                    __
  = (iλmH2/16π2)(1 + ln(μ2/mH2) after MS

Which is quadratically divergent.  

Likewise, case (c) is also calculated in the notes on 
Regularization and Renormalization.  The 1 loop Yukawa
correction to the Higgs mass is given by: 
                      
Π(k) = -4g2(-i)[1/8π2(k2/6 + m2) - 3/16π2(ln(μ2/Δ))]
  
Which, again, is quadratically divergent.  

Aside:  We can schematically see this as follows:

Π = -gY2∫ d4k/(2π)4 (γμk + m)2/(k2 - m2)2

(γμk + m)2 = k2 + 2km + m2

Therefore,

I = -gY2∫ d4k/(2π)4 (k2 + 2km + m2)/(k2 - m2)2

  = -gY2∫ d4k/(2π)4 (k2 + m2)/(k2 - m2)2

Write the numerator as k2 - m2 + 2m2 and the integral 
becomes:

I = -gY2∫ d4k/(2π)4 1/(k2 - m2) + 2m/(k2 - m2)2

Note:  There is a -ve in front of the g2 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)

The Problem
-----------

We can write the physical Higgs mass with loop corrections
(neglecting all factors of π etc.) as:

mH2/2 = m02/2 
        + λmH2(1 + log terms)            
        - gY2(p2 + 2mH2 + log terms)

The log terms are of lesser importance because they are 
very nuch smaller than the squared terms.  Again, we can 
write the above schematically as:

mH2/2 ~ m02/2 + λmH2 - gY2mH2

Unfortunately, these quadratically divergent terms 
cannot be removed by renormalization.  They are 
considered to be unphysical or 'unnatural'.  As a 
result, one would expect that these large quantum 
corrections to mH2 would make the Higgs mass huge as 
mH approaches the scale at which new physics appears 
(the Planck mass).  In fact, unless there is an incredible 
amount of fine tuning cancellation between the quadratic 
terms and the bare mass, the 1-loop correction to mH2 
would be more than 30 orders of magnitude larger than 
the measured mH of a few hundred GeV (~(1019)2/(100)2).
 
To compound issues, there are many, many combinations 
of possible Feynman diagrams leading to additional 
corrections to the mass, some positive and some negative
with different powers of the coupling constant.  For
example, every fermion in the SM has a Yukawa coupling 
to the Higgs field and will quadratically diverge.  In 
addition there are contributions from the gauge bosons. 
All of these terms would need to collectively cancel 
each other to achieve the physical mass! 

The Mass of the Higgs
---------------------

The Higgs field is described by:

V(φ) = -μ2φ2/2 + λφ4/4

This has a minimum (VEV) at dV(φ)/dφ = 0.  Therefore,

-μ2φ = λφ3

or φMIN = f = μ/√λ

We know that f ~ 246 GeV and μH ~ 125 GeV.  Therefore,
λ < 1.

Supersymmetry to the Rescue
---------------------------

What if the Higgs had a fermionic superpartner, the Higsino, 
~
H, that couples to the field with a coupling constant, λS.

This would be represented by (b) in the following diagram:



The cancellations would be exact if the mass and charge 
of the partner/superpartner pair were the same and 
λS = λY2. 

This is exactly the idea behind supersymmetric theories.

However, because they have yet to be observed in the 
laboratory, superpartners must be heavier than their 
partners.  The fact that there is not a perfect symmetry 
implies that supersymmetry is not an exact, unbroken 
symmetry.  The implication is that the true symmetry is 
somehow spontaneously broken.

The difference in masses jeopardizes the solution to 
the hierarchy problem because exact cancellation is 
spoiled - but not totally.  The prediction is that 
the mass difference is small enough to still provide 
a high degree of cancellation.  

R-Parity
--------

R-parity is a set of rule that define possible partner/
superpartner interactions.  If we define 'normal' SM 
particles, N as + and supersymmetric particles, S as -1, 
the rules are:

N -> NN

N -> SS

S -> NS

For 2-body particle transitions the rules are:

NN -> NN

NN -> SS

SS -> SS

NS -> NS

Therefore, R parity is a multiplicative quantum number.

For example
     ~                                ~~              
H -> H -> H is not allowed while H -> FF -> H or
     ~~
H -> HH -> is allowed.

(a), (b), (c) and (d) are valid.  However, a diagram like 
the one below would not be.



Dark Matter
-----------

As we said before, superpartners can only be pair produced 
from SM particles and a superpartner can only produce a
another superpartner and a normal partner.  Thus,

          S            
         /            
        /            
N ------     
        \            
         \            
          S            
and,
          S            
         /            
        /            
S ------     
        \            
         \            
          N           


Because of their large mass, superpartners decay and are 
not freely floating around in the Universe.  Superpartners 
will decay in a chain until the lightest superpartner (LSP) 
is produced (cascade decay).  Thus,


                               LSP
                               /
                              /
                             ------ N
                            /
                           .
                          .
                       S /
                        /
                       ------ N
                    S /
                     /
             S ------
                     \
                      \
                       N


The resulting LSP) is thought to be stable and electrically 
neutral.  If it were charged the theory would be cosmologically 
disfavored.  In addition,  LSPs must interact weakly with 
ordinary matter and are not seen in detectors since they 
interact only by exchange of heavy virtual particles.  These 
are exactly the characteristics required for Dark Matter, 
thought to make up most of the matter in the universe and to 
hold galaxies together.

Fermionic and Gauge Boson Mass
------------------------------

The Hierachy problem does not impact fermions or the gauge
bosons.  In both cases the self energy varies logarithmically
and are not quadratically divergent.  (see the discussion on 
regularization and renormalizationfor the fermionic self 
energy calculation).  More fundamentally, for fermions it is
the chiral symmetry that protects the mass.  The Lagarangian 
has a symmetry as the fermionic masses goes to 0.  This 
guarantees that all corrections to the mass must also vanish 
as the mass goes to 0.  We can see this from the discussion 
on regularization and renormalization.  There we showed that
the fermionic self energy correction to the mass is given by
(schematically): 

δm ~ αmln(μ2/Δ) for 1 loop.

δm is very small.  For example, for μ2/m2 = 1030 the log = 69
Now, α = 1/137 ∴ δm ~ (1/137)* 69 ~ 0.5*m and as m -> 0 δm -> 0

For the gauge bosons, it is the gauge invariance that protects 
the mass.  A mass term for a gauge field is m2AμAμ.  However, 
such terms are not invariant and break the local symmetry.  
Therefore, m2 must be zero.  

The Vacuum Catastrphe
---------------------

While Supersymmetry offers a possible solution to the Gauge
Hierarchy problem.  It does not explain the vacuum catastrophe
problem that is discussed in another note.

The Fine Tuning Problem
----------------------- 

The problem with the Higgs mass being so finely tuned is 
part of a wider question as why the Universe in generai 
is so finely tuned.  The current standard model of particle 
physics has 25 freely adjustable parameters.  In addition 
gravity has the cosmological constant.  A small change in 
several of the dimensionless physical constants would make 
the Universe radically different from what we observe today.
The famous scientist Stephen Hawking has stated, "The laws of 
science, as we know them at present, contain many fundamental 
numbers, like the size of the electric charge of the electron 
and the ratio of the masses of the proton and the electron. 
The remarkable fact is that the values of these numbers seem 
to have been very finely adjusted to make possible the 
development of life."  If any of several fundamental constants 
were only slightly different, the Universe would be unlikely 
to be conducive to the establishment and development of matter, 
astronomical structures, elemental diversity, or life as we 
know it.  For example, if the coupling constant for the strong 
force was 2% larger, while the other constants were unchanged,
the physics of stars would be drastically different.  

Possible explanations for fine-tuning are the subject
of vigorous discussions among philosophers, scientists, 
theologians, and proponents of creationism. 

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