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Electroweak Unification (GlashowWeinbergSalam)

Glashow, Weinberg and Salam showed that at very high energies
the electromagnetic and weak interactions can be merged into
a single electroweak interaction between particles. At its
core is the spontaneous symmetry breaking of SU(2)_{L} x U(1)_{Y}
down to U(1)_{EM}. L denotes left handed and Y denotes weak
hypercharge. The weak force only interacts with left handed
particles.
High energy: SU(3) x SU(2)_{L} x U(1)_{Y}
 Spontaneous Symmetry Breaking
v
Low energy: SU(3) x U(1)_{EM}
Since the electroweak interaction involves fermions we start
with the Dirac Lagrangian.
_ _
L = ψγ^{μ}D_{μ}ψ + mψψ
We can write ψ in terms its of left and right handed components
in terms of the projection operators involving the γ^{5} matrix
as follows:
ψ_{R} = (1 + γ^{5})ψ/2
Therefore,
_ ___________
ψ_{R} = (1 + γ^{5})ψ/2
= ((1 + γ^{5})ψ)^{†}γ^{0}/2
= ψ^{†}(1 + γ^{5})γ^{0}/2
= ψ^{†}γ^{0}(1  γ^{5})/2
_
= ψ(1  γ^{5})/2
ψ_{L} = (1  γ^{5})ψ/2
_^{ } _
ψ_{L} = (1 + γ^{5})ψ/2
The Dirac spinor, ψ, can be written as linear combinations
of lefthanded and righthanded states. Therefore,
ψ_{L} + ψ_{R} = ψ
and,
_ __{ } _
ψ_{L} + ψ_{R} = ψ
Therefore, we can rewrite the Lagrangian as:
_ __{ } __{ } _
L = (ψ_{R} + ψ_{L})γ^{μ}∂_{μ}(ψ_{R} + ψ_{L}) + m(ψ_{R} + ψ_{L})(ψ_{R} + ψ_{L})
Now,
_ _
ψ_{R}γ^{μ}ψ_{L} = (1  γ^{5})ψγ^{μ}(1  γ^{5})ψ
_
_{ } = γ^{μ}(1 + γ^{5})(1  γ^{5})ψψ
_
_{ } = γ^{μ}(1  (γ^{5})^{2})ψ<ψ
_{ } = 0
_
Likewise, ψ_{L}γ^{μ}ψ_{R} = 0
Also,
_ _
ψ_{R}ψ_{R} = (1  γ^{5})ψ/2(1 + γ^{5})ψ/2
_{ } _
_{ } = ψψ(1  (γ^{5})^{2})
_{ } = 0
_
Likewise, ψ_{L}ψ_{L} = 0
So the Lagrangian reduces to:
_ _ _ _
L = ψ_{R}γ^{μ}∂_{μ}ψ_{R} + ψ_{L}γ^{μ}∂_{μ}ψ_{L} + mψ_{R}ψ_{L} + mψ_{L}ψ_{R}
_ _ _ _
= ψ_{R}γ^{μ}∂_{μ}ψ_{R} + ψ_{L}γ^{μ}∂_{μ}ψ_{L} + m(ψ_{R}ψ_{L} + ψ_{L}ψ_{R})
The left and right handed ψs can be written as doublets.
       
 u_{L}   ν_{L}   u_{R}   ν_{R} 
 d_{L}   e_{L}   d_{R}   e_{R} 
       
However, right handed neutrinos have not been observed in
nature. Therefore, e_{R} can be written as a singlet. We are
only interested in interactions so we can ignore the mass term.
Thus, the Lagrangian reduces to:
_ _
L = e_{R}γ^{μ}∂_{μ}e_{R} + ψ_{L}γ^{μ}∂_{μ}ψ_{L}
SU(2)_{L}: exp(iθσ_{i})ψ_{L}
U(1)_{Y}: exp(iφY_{W})ψ_{L} and exp(iφY_{W}')e_{R}
To promote to a local symmetry we use the covariant derivative.
For the 1st term we get:
∂_{μ}e_{R} + ig''Y_{W}'B_{μ}e_{R}
For the 2nd term we get:
∂_{μ}ψ_{L} + iσW_{μ}ψ_{L} + iY_{W}'B_{μ}ψ_{L}
^ ^
 
SU(2)_{L} U(1)_{Y}
So we have 3 gauge fields W_{i} corresponding to SU(2)_{L} and 1
gauge field corresponding to U(1)_{Y}.
The kinetic terms for the gauge fields are given by:
U(1)_{Y}: F_{μν} = (i/g)[D_{μ},D_{ν}] = ∂_{μ}B_{ν}  ∂_{ν}B_{μ}
SU(2)_{L}: G_{μν}^{i} = ∂_{μ}W_{ν}^{i}  ∂_{ν}W_{μ}^{i} + f^{ijk}W_{μ}^{j}W_{ν}^{k}
Complex Fields

Complex fields are necessary when one deals with particles that
carry electric charge. To see this consider the following:
L = ∂_{μ}φ*∂^{μ}φ  m^{2}φ*φ
Where,
φ = φ_{1} + iφ_{2}
and,
φ* = φ_{1}  iφ_{2}
and φ_{1} and φ_{2} are 2 real scalar fields.
It is easy to see that the Lagrangian is invariant under the
U(1) symmetry transformation (φ > φexp(iθ) and φ^{*} > φ^{*}exp(iθ)).
This implies there is a conserved quantity.
Now e^{iε} = 1 + iε ... Taylor series
Thus,
φ > φ + iεφ > φ + εf_{φ} where f_{φ} = iφ
φ^{*} > φ^{*}  iεφ^{*} > φ^{*}  εf_{φ*} where f_{φ*} = iφ^{*}
Look for a conserved quantity using Noether's Theorem.
. .
π_{φ} = ∂L/∂φ = φ^{*}
. .
π_{φ*} = ∂L/∂φ^{*} = φ
Conserved quantity = ∫[π_{φ}f_{φ} + π_{φ*}f_{φ*}]d^{3}x
. .
= i∫[φ^{*}φ  φφ^{*}]d^{3}x
= i∫[πφ  π^{*}φ^{*}]d^{3}x
The quantity in [] is the charge density of the field, ρ.
Knowing this we can construct charge carrying fields out of
combinations of the W gauge fields by creating raising and
lowering operators from the Pauli matrices that describe
SU(2):
 
W^{0} =  1 0 
_{ }  0 1 
 
 
W^{+} = σ_{1} + iσ_{2} =  0 1 
_{ }  0 0 
 
 
W^{} = σ_{1} 1 iσ_{2} =  0 0 
_{ }  1 0 
 
We then apply these operators to the doublets to see their
effect:
   
ν_{L} =  1  and e_{L} =  0 
_{ }  0 _{ }  1 
   
W^{0}:
     
 1 0  1  =  1 
 0 1  0   0 
     
     
 1 0  0  =  0 
 0 1  1   1 
     
/ e/ν_{e}
W^{0} /
σ_{3}: /\/\/\/\/
\
\ e/ν_{e}
W^{0} does not change anything and must have neutral charge.
This is referred to as neutral current interaction.
W^{+}:
   
 0 1  1  = 0
 0 0  0 
   
     
 0 1  0  =  1 
 0 0  1   0 
     
/ ν_{e}
W^{+} /
σ^{+}: /\/\/\/\/
\
\ e
W^{+} changes e > ν and must 'add' charge to do so.
W^{}:
     
 0 0  1  =  0 
 1 0  0   1 
     
   
 0 0  0  = 0
 1 0  1 
   
/ e
W^{} /
σ^{}: /\/\/\/\/
\
\ ν_{e}
W^{} changes ν > e and must 'remove' charge to do so.
The W^{+} and W^{} interactions are referred to as charged current
interactions.
Before symmetry breaking there are 4 generators. After symmetry
breaking there is 1 generator. One might naively expect that
U(1)_{Y} > U(1)_{EM} but this is not the case. In actuality W^{0} from
SU(2)_{L} mixes with B_{μ} from U(1)_{Y} to form 2 orthogonal states given
by the following:
     
 γ^{ }  =  cosθ_{W} sinθ_{W}  B_{μ}^{ } 
 Z^{0}   sinθ_{W} cosθ_{W}  W_{μ}^{0} 
     
θ_{W} is found experimentally and is the angle by which spontaneous
symmetry breakin rotates the original W_{μ}^{0} and B_{μ} vector boson
plane, producing as a result the Z^{0} boson, and the photon.
The Higgs Mechanism

The GlashowWeinbergSalam theory does not explain why the
gauge bosons are massive. Simply adding a term mW_{μ}W^{μ} to the
Lagrangian doesn't work because this term is not gauge invariant
since (A_{μ} + ∂_{μ}θ)(A^{μ} + ∂^{μ}θ) is clearly not equal to m'A_{μ}A^{μ}.
Furthermore, the left and right spinors transform differently
because their Y_{W} are different (e: L = 1, R = 2). Therefore,
__{ } _
there is no way that m(ψ_{L}ψ_{R} + ψ_{L}ψ_{R}) is gauge invariant. In
order for these particles to become massive we need to introduce
the Higgs mechanism. This is discussed in more detail elsewhere
on this site but briefly there is a Higgs field described by an
weak isospin doublet whose symmetry is spontaneously broken. In
the process the field aquires a vacuum expectation energy, f,
consisting of a condensate of Higgs boson, H. The breaking gives
the gauge bosons and the fermions mass. Schematically, the
symmetry breaking looks like:
       
 φ^{+}  =  H^{+} + iH^{}  >  0  =  0^{ } 
 φ^{0}   H^{ } + iH^{0}   f + H   φ^{0} 
       
Under infinitesinal rotations, the vacuum, φ_{0}, should be symmetric.
Therefore,
φ_{0} > exp(iθZ)φ_{0} where Z is the rotation.
This means that,
φ_{0} > (1 + iθZ)φ_{0}
So Zφ_{0} must equal 0.
SU(2)_{L}:
     
σ_{3}φ_{0} =  1 0  0  =  0  ≠ 0 > broken
_{ }  0 1  f + H   f + H 
     
U(1)_{Y}:
     
Y_{W}φ_{0} =  1 0  0  =  0  ≠ 0 > broken
_{ }  0 1  f + H   f + H 
     
U(1)_{EM}:
   
Qφ_{0} = (1/2)(σ_{3} + Y_{W}) =  1 0  0  = 0 > unbroken
_{ }  0 0  f + H 
   
Weak Isospin and Weak Hypercharge

SU(2)_{L} is the generator of I_{3} and U(1)_{Y} is the generator of Y_{W}.
Before symmetry breaking, I_{3} and Y_{W} are individually conserved
as is their combination Q (= I_{3} + Y_{W}/2). After the symmetry
is broken neither is individually conserved. However, a
specific combination of them is conserved because Q must be
conserved thereby ensuring that U(1)_{EM} remains unbroken. After
the symmetry breaking the Higgs field aquires a vacuum expectation
value or condensate of Higgs bosons. For reasons that will become
apparent, the Y_{W} for the Higgs bosons is chosen to be +1. This
implies that I_{3} must equal 1/2 to maintain the charge neutrality
of the vacuum.
The Higgs condensate becomes a source and sink of I_{3} and Y_{W}.
Particles constantly interact with this condensate in the
following manner.
W^{+} W^{}
 ^
v 
e_{R} e_{L} e_{R} e_{L} ν_{L} ν_{L} e_{L}
o > o > o > o > o > o > o
^  ^
H  H H
  
 v 

///////////////////////////////////////////////////////
Higgs Condensate
H

I_{3}: 1/2 1/2 1/2
Y_{W}: 1 1 1
Leptons

I_{3}: 0 1/2 0 1/2 1/2 1/2 1/2
Y_{W}: 2 1 2 1 1 1 1
Q:_{ } 1 1 1 1 0 0 1
Bosons

I_{3}: 1 1
Y_{W}: 0 0
Q:_{ } 1 1
In

I_{3}: 1/2 1/2
Y_{W}: 1 1
Q:_{ } 0 1
Out

I_{3}: 1/2 1/2
Y_{W}: 1 1
Q:_{ } 0 1
Likewise for the fermions:
W^{} W^{+}
 ^
v 
u_{L} u_{R} u_{L} d_{L} d_{R} d_{L} u_{L}
o > o > o > o > o > o > o
^   ^
H H H H
   
 v v 

///////////////////////////////////////////////////////
Higgs Condensate
H

I_{3}: 1/2 1/2 1/2 1/2
Y_{W}: 1 1 1 1
Quarks

I_{3}: 1/2 0 1/2 1/2 0 1/2 1/2
Y_{W}: 1/3 4/3 1/3 1/3 2/3 1/3 1/3
Q:_{ } 2/3 2/3 2/3 1/3 1/3 1/3 2/3
Bosons

I_{3}: 1 1
Y_{W}: 0 0
Q:_{ } 1 1
In

I_{3}: 1/2 1/2
Y_{W}: 1/3 1/3
Q:_{ } 1/3 2/3
Out

I_{3}: 1/2 1/2
Y_{W}: 1/3 1/3
Q:_{ } 1/3 2/3
Therefore, I_{3} and Y_{W} are conserved at vertices but not in
changing the handedness of the particles.