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Units, Constants and Useful Formulas
Spinors  Part 1
Spinors  Part 2

In Spinors  Part 1 we identified the following:
1. The generators, S^{>ρσ}, are defined as:
S^{ρσ} = (1/4)[γ^{ρ},γ^{σ}]
^{ } = (1/2)γ^{ρ}γ^{σ}  (1/2)η^{ρσ}
2. The Lorentz transformation for a spinor, S[Λ]
S[Λ] = exp((1/2)Ω_{ρσ}S^{ρσ})
3. The Lorentz transformation for a vector, Λ
Λ = exp((1/2)Ω_{ρσ}M^{ρσ})
4. The S^{μν} operates operate on a spinor field, ψ.
Under a Lorentz transformation these fields
transform as:
ψ^{α} = S[Λ]^{α}_{β}ψ^{β}
5. (M^{ρσ})^{μ}_{ν} = η^{ρμ}δ^{σ}_{ν}  η^{σμ}δ^{ρ}_{ν}
6. {γ^{μ},γ^{ν}} = 2η^{μν} ... CLIFFORD ALGEBRA.
7. (γ^{0})^{†} = γ^{0}
8. (γ^{0})^{1} = γ^{0}
9. γ^{0}γ^{μ}γ^{0} = (γ^{μ})^{†}
10. (γ^{i})^{†} = γ^{i}
In addition, we will need:
11. [S^{μν},γ^{μ}] = γ^{μ}η^{νρ}  γ^{ν}η^{ρμ}
Proof:
[S^{μν},γ^{μ}] = (1/2)[γ^{μ}γ^{ν},γ^{ρ}]
^{ } = (1/2)γ^{μ}γ^{ν}γ^{ρ}  (1/2)γ^{ρ}γ^{μ}γ^{ν}
^{ } = (1/2)γ^{μ}{γ^{ν},γ^{ρ}}  (1/2)γ^{μ}γ^{ρ}γ^{ν}
^{ }  (1/2){γ^{ρ},γ^{ν}}γ^{ν} + (1/2)γ^{μ}γ^{ρ}γ^{ν}
Use 6. to get:
^{ } = γ^{μ}η^{νρ}  γ^{ν}η^{ρμ}
12. [S^{μν},S^{ρσ}] = η^{νρ}S^{μσ}  η^{μρ}S^{νσ} + η^{μσ}S^{νρ}  η^{νσ}S^{μρ}
... LORENTZ ALGEBRA
Proof:
[S^{μν},S^{ρσ}] = (1/2)[S^{μν},γ^{ρ}γ^{σ}]
^{ } = (1/2)[S^{μν},γ^{ρ}]γ^{σ} + (1/2)γ^{ρ}[S^{μν},γ^{σ}]
Use 7. to get:
^{ } = γ^{μ}γ^{σ}η^{νρ}  γ^{ν}γ^{σ}η^{ρμ} + γ^{ρ}γ^{μ}η^{νσ}  γ^{ρ}γ^{ν}η^{σμ}
Rearrange 1. γ^{μ}γ^{σ} = 2S^{μσ} + η^{μσ} and substitute to
get:
[S^{μν},S^{ρσ}] = η^{νρ}S^{μσ}  η^{μρ}S^{νσ} + η^{μσ}S^{νρ}  η^{νσ}S^{μρ}
The Dirac Adjoint

Now that we have a field, ψ, we need to construct
a Lorentz scalar and a Lorentz invariant equation
of motion.
If we try and construct a scalar ψψ^{†} we get:.
ψψ^{†} > S[Λ]ψψ^{†}S[Λ]^{†}
The problem with this is that S[Λ]S[Λ]^{†} ≠ 1 and is
certainly not real. To see this consider:
S[Λ] = exp((1/2)Ω_{ρσ}S^{ρσ})
S[Λ]^{†} = exp((1/2)Ω_{ρσ}(S^{ρσ})^{†})
So to be unitary we require (S^{ρσ})^{†} = S^{ρσ}
Unfortunately, this is not possible because γ^{0} is
hermitian whereas the γ^{i}'s are antihermitian.
Therefore, there is no way to pick γ^{μ} such that
all are antihermitian. However, there is a way
around this. Consider:
γ^{0}S[Λ]^{†}γ^{0} = γ^{0}[exp(iΩ_{μν}S^{μν})]^{†}γ^{0}
^{ } = γ^{0}exp((i/2)Ω_{μν}(S^{μν})^{†}γ^{0}
^{ } = γ^{0}(1  (i/2)Ω_{μν}(S^{μν})^{†})γ^{0}
Multiplying from the left and the right gives:
γ^{0}S[Λ]^{†}γ^{0} = 1  (i/2)Ω_{μν}γ^{0}(S^{μν})^{†}γ^{0}
^{ } = exp((i/2)Ω_{μν}γ^{0}(S^{μν})^{†}γ^{0})
Now,
(S^{μν})^{†} = (1/4)[(γ^{μ})^{†},(γ^{ν})^{†}]
^{ } = (1/4)((γ^{μ})^{†}(γ^{ν})^{†}  (γ^{ν})^{†}(γ^{μ})^{†})
^{ } = (1/4)((γ^{0}γ^{μ}γ^{0})(γ^{0}γ^{ν}γ^{0})  (γ^{0}γ^{ν}γ^{0})(γ^{0}γ^{μ}γ^{0}))
^{ } = (1/4)((γ^{0}γ^{μ}γ^{0})(γ^{0}γ^{ν}γ^{0}) + (γ^{0}γ^{μ}γ^{0})(γ^{0}γ^{ν}γ^{0}))
^{ } = (1/2)(γ^{0}γ^{μ}γ^{ν}γ^{0})
^{ } = γ^{0}(1/2)(γ^{μ}γ^{ν})γ^{0}
Therefore,
(S^{μν})^{†} = γ^{0}S^{μν}γ^{0}
Returning to where we left off and using this result we
get:
γ^{0}S[Λ]^{†}γ^{0} = exp((i/2)Ω_{μν}S^{μν})
^{ } = S[Λ]^{1}
Multiplying both sides from the left and the right by γ^{0}
gives:
S[Λ]^{†} = γ^{0}S[Λ]^{1}γ^{0}
With this in mind, we now define the DIRAC ADJOINT:
_
ψ = ψ^{†}γ^{0}
Under a boost we get:
_
ψψ > ψ^{†}S[Λ]^{†}γ^{0}S[Λ]ψ
= ψ^{†}γ^{0}S[Λ]^{1}γ^{0}γ^{0}S[Λ]ψ
= ψ^{†}γ^{0}S[Λ]^{1}S[Λ]ψ
= ψ^{†}γ^{0}ψ
This looks like:
 
 ψ_{1} 
ψ =  ψ_{2} 
 ψ_{3} 
 ψ_{4} 
 
 
ψ^{†} = (ψ^{*})^{T} =  ψ_{1}^{*} ψ_{2}^{*} ψ_{3}^{*} ψ_{4}^{*} 
 
_  
ψ = ψ^{†}γ^{0} =  ψ_{1}^{*} ψ_{2}^{*} ψ_{3}^{*} ψ_{4}^{*} 
 
_
ψψ = ψ_{1}ψ_{1}^{*} + ψ_{2}ψ_{2}^{*}  ψ_{3}ψ_{3}^{*}  ψ_{4}ψ_{4}^{*}
Again, this is completely different to the vector case, x_{μ}x^{μ}.
 ^{ }  
x^{0} x^{1} x^{2} x^{3}  x^{0}  = t^{2}  x^{2}  y^{2}  z^{2}
^{ }   x^{1} 
^{ }  x^{2} 
^{ }  x^{3} 
^{ }  
Bilinears

In the previous section we formed a scalar from 2
spinors. Are they other objects we can form by
combining 2 spinors? The answer is yes. Spinors
can be combined to form scalars, vectors, tensors
and more. The products of two spinors are called
BILINEARS.
Scalar

To form a scalar (spin 0) from 2 spinors one writes:
_
ψψ = ψ^{†}γ^{0}ψ
Under a Lorentz transformation,
_
ψψ > ψ^{†}(Λ^{1}x)S[Λ]^{†}γ^{0}S[Λ]ψ(Λ^{1}x)
= ψ^{†}(Λ^{1}x)γ^{0}ψ(Λ^{1}x)
_
= ψ(Λ^{1}x)ψ(Λ^{1}x)
_
So ψψ does indeed transform as a scalar.
Vector

To form a vector (spin 1) from 2 spinors one writes:
_
ψγ^{μ}ψ
Under a Lorentz transformation,
_
ψγ^{μ}ψ > ψS[Λ]^{1}γ^{μ}S[Λ]ψ
For this to be true we need S[Λ]^{1}γ^{μ}S[Λ] = Λ^{μ}_{ν}γ^{ν} ... 13.
Where,
Λ = exp((1/2)Ω_{ρσ}M^{ρσ}) ~ 1 + (1/2)Ω_{ρσ}M^{ρσ} ...
and,
S[Λ] = exp((1/2)Ω_{ρσ}S^{ρσ}) ~ 1 + (1/2)Ω_{ρσ}S^{ρσ} ...
We work infinitesimally with Ω_{ρσ} = ε_{ρσ}. Therefore,
(1  (1/2)ε_{ρσ}S^{ρσ})γ^{μ}(1 + (1/2)ε_{ρσ}S^{ρσ}) = (1 + (1/2)ε_{ρσ}M^{ρσ})γ^{ν}
(1  (1/2)ε_{ρσ}S^{ρσ})(γ^{μ} + (1/2)ε_{ρσ}S^{ρσ}γ^{μ}) = (1 + (1/2)ε_{ρσ}M^{ρσ})γ^{ν}
γ^{μ} + (1/2)ε_{ρσ}S^{ρσ}γ^{μ}  (1/2)ε_{ρσ}S^{ρσ}γ^{μ} + O(ε_{ρσ}^{2}) = γ^{μ} + (1/2)ε_{ρσ}M^{ρσ}γ^{ν}
(S^{ρσ}γ^{μ}  S^{ρσ}γ^{μ}) = M^{ρσ}γ^{ν}
[S^{ρσ},γ^{μ}] = (M^{ρσ})^{μ}_{ν}γ^{ν}
Using 11. on the LHS and 5. on the RHS gives:
γ^{μ}η^{νρ}  γ^{ν}η^{ρμ} = (η^{ρμ}δ^{σ}_{ν}  η^{σμ}δ^{ρ}_{ν})γ^{ν}
γ^{μ}η^{νρ}  γ^{ν}η^{ρμ} = η^{ρμ}γ^{σ}  η^{σμ}γ^{ρ}
_
So we have proven 13. and ψγ^{μ}ψ does indeed transform
as a vector.
Tensor

To form a tensor (spin 2) from two spinors one writes:
_
ψS^{μν}ψ
Under a Lorentz transformation,
_
ψS^{μν}ψ > ψS[Λ]^{1}S^{μν}S[Λ]ψ
_
= ψS[Λ]^{1}((1/2)[γ^{μ},γ^{ν}])S[Λ]ψ
_
= (1/2)ψS[Λ]^{1}(γ^{μ}γ^{ν}  γ^{ν}γ^{μ})S[Λ]ψ
_
= ψ(1/2)(S[Λ]^{1}γ^{μ}S[Λ]S[Λ]^{1}γ^{ν}S[Λ]
 S[Λ]^{1}γ^{ν}S[Λ]S[Λ]^{1}γ^{μ}S[Λ])ψ
Now from 13. S[Λ]^{1}γ^{μ}S[Λ] = Λ^{μ}_{ν}γ^{ν}
ψS^{μν}ψ > ψ(1/2)(Λ^{μ}_{α}γ^{α}Λ^{ν}_{β}γ^{β}  Λ^{ν}_{β}γ^{β}Λ^{μ}_{α}γ^{α})ψ
_
= ψ(1/2)Λ^{μ}_{α}Λ^{ν}_{β}[γ^{α},γ^{β}]ψ
_
= Λ^{μ}_{α}Λ^{ν}_{β}ψS^{αβ}ψ
_
So ψS^{μν}ψ does indeed transform as a tensor.
Generalization

Consider the expression:
_
ψΓ^{A}ψ
We have already found expressions for scalars,
vectors and tensors. Can we find Γ^{A}'s that
create other objects that transform correctly
under a Lorentz transformation. Again, the
answer is yes. We can write Γ^{A} in terms of the
following combinations of 16 γ matrices.
Γ^{A} =
1 ^{ } : scalar (1 component)
γ^{μ} ^{ } : vector (4 components)
S^{μν} = (i/2)[γ^{μ},γ^{ν}] : Tensor (6 components)
γ^{μ}γ^{5} ^{ } : pseudovector (4 components)
γ^{5} ^{ } : pseudoscalar (4 component)
Spinor Indeces

Consider the Weyl spinors mentioned above and
define:
 
ψ_{L} =  ξ_{α} 
_{ }  0_{ } 
 
 
ψ_{R} =  0^{ }
_{ }  _^{ }
_{ }  χ^{α'}
 
 
ψ = ψ_{L} + ψ_{R} =  ξ_{α}_{ }
_{ }  __{ }
_{ }  χ^{α'}
 
This is the VAN DER WAERDEN NOTATION.
α = lefthanded spinor index.
α' = right handed spinor index.
_ does not mean conjugate  it is purely notation.
We can further define:
_
ξ_{α'} = (ξ_{α})^{†}
and,
_
χ^{α} = (χ^{α'})^{†}
_ 
ψ^{†} =  ξ_{α'} 
^{ }  χ^{α}^{ } 
 
_
Now, ψ = γ^{0}ψ^{†}
   
Where γ^{0} =  0 I_{2}  =  0 1  in the Weyl representation.
^{ }  I_{2} 0   1 0 
   
_    _ 
ψ =  0 1  ξ_{α'} 
 1 0  χ^{α}^{ } 
   
 _ 
=  χ^{α} ξ_{α'} 
 
Spinor indices are raised and lowered using the
antisymmetric symbol, ε, that has the properties:
ε^{12} = ε^{21} = ε_{21} = ε_{12} = 1
ε^{11} = ε^{22} = ε_{11} = ε_{22} = 0
ε^{ab}ε_{bc} = δ^{a}_{c} = 1
ε_{ab}ε^{bc} = δ_{a}^{c} = 1
ε^{a'b'}ε_{b'c'} = δ^{a'}_{c'} = 1
ε_{a'b'}ε^{b'c'} = δ_{a'}^{c'} = 1
In matrix form:
 
ε_{αβ} = ε_{α'β'} =  0 1  = iσ_{2}
_{ }  1 0 
 
 
ε^{αβ} = ε^{α'β'} =  0 1  = iσ_{2}
_{ }  1 0 
 
We can use these operators as follows:
_ _ _ _
ξ_{α} = ε_{αβ}ξ^{β} ξ^{α} = ε^{αβ}ξ_{β} χ_{α'} = ε_{α'β'}χ^{β'} χ^{α'} = ε^{α'β'}χ_{β'}
Therefore,
ε_{αβ}ξ^{β} = ξ_{α}
     
 0 1  ξ^{α}  =  0_{ } 
 1 0  0_{ }   ξ_{α} 
     
ε^{αβ}ξ_{β} = ξ^{α}
     
 0 1  ξ_{α}  =  0_{ } 
 1 0  0_{ }   ξ^{α} 
     
Note: ε^{αβ} = ε^{α'β'} etc. ε^{α'β} and ε^{αβ'} are not allowed.
Therefore, the Dirac Lagrangion excluding the mass term
is:
 _      _ _ _
L = i χ^{α} ξ_{α'}  0 σ_{μ}  ∂_{μ}ξ_{α}^{ }  = iχ^{α}σ^{μ}∂_{μ}χ^{α'} + ξ_{α'}σ_{μ}∂_{μ}ξ_{α}
 ^{ }   _^{ }  _^{ }
^{ }  σ_{μ} 0  ∂_{μ}χ^{α'} 
   
^

γ^{μ} in the Weyl representation.
Spinor Transformations

Let us now reconstruct the spinor transformation laws
that incorporate indeces.
   
In the Weyl basis γ^{μ} =  0 σ^{k}  >  0 σ^{μ} 
^{ }  σ^{k} 0   _^{ } 
^{ }    σ^{μ} 0 
 
Where,
σ^{μ} = (1,σ^{i}) is substituted for σ_{k}
_
σ^{μ} = (1,σ^{i}) is substituted for σ_{k}
Which for rotations leads to:
   
S^{μν} = (1/4) 0 σ^{μ}  0 σ^{ν} 
^{ }  _ ^{ }  _ ^{ } 
^{ }  σ^{μ} 0  σ^{ν} 0 
   
 _ _ 
= (1/4) σ^{μ}σ^{ν}  σ^{ν}σ^{μ} 0 
 _{ } _ _ 
 0 σ^{μ}σ^{ν}  σ^{ν}σ^{μ} 
 
The objects that obey the Lorentz algebra and generate
the desired rotations are given by:
_ _
(S^{μν})_{α}^{β} = (1/4)(σ^{μ}σ^{ν}  σ^{ν}σ^{μ})_{α}^{β}
_ _ _
(S^{μν})^{α'}_{β'} = (1/4)(σ^{μ}σ^{ν}  σ^{ν}σ^{μ})^{α'}_{β'}
So these correspond to:
ξ_{α} > (exp((1/2)Ω_{μν}S^{μν}))_{α}^{β}ξ_{β}
and,
_ _
χ^{α'} > (exp((1/2)Ω_{μν}S^{μν}))^{α'}_{β'}χ^{β'}
Which is the same as the rotational transformation
that we derived in Spinors  Part 1.
Parity

Parity is the operation of changing coordinates as:
x^{0} > x^{0} x^{i} > x^{i}
Under parity, the left and righthanded spinors are
exchanged. Under parity, rotations donâ€™t change sign
but boosts flip sign. Therefore,
Pψ_{L,R}(t,x) = ψ_{R,L}(t,x)
_{ } = γ^{0}ψ_{R,L}(t,x)
if ψ(t,x) satisfies the Dirac equation, then the
parity transformed spinor γ^{0}ψ(t,x) also satisfies
the Dirac equation, meaning:
(iγ^{0}∂_{t} + iγ^{i}∂_{i}  m)γ^{0}ψ(t,x) = γ^{0}(iγ^{0}∂_{t}  iγ^{i}∂_{i}  m)γ^{0}ψ(t,x)
Where the extra minus sign from passing γ^{0} through
γ^{i} is compensated by the derivative
acting on x instead of +x.
Thus, in the Weyl basis we have:
     _ 
ψ^{P} =  0 1  ξ_{α}^{ }  =  χ^{α'} 
^{ }  1 0  _^{ }   ξ_{α}^{ } 
^{ }    χ^{α'}   
 
Charge Conjugation

Charge conjugation is the operation of changing every
particle into its antiparticle. It is a discrete
symmetry and represents the 'C' in the CPT (Charge,
Parity, Time Reversal) symmetry. Charge conjugation is
accomplished using the charge conjugation operator, C:
_
ψ^{C} = Cψ^{T}
Where C is defined as the matrix:
   
C =  iσ_{2} 0  =  ε_{αβ} 0^{ }
 0 iσ_{2}   0 ε^{α'β'}
   
C satisfies C^{T} = C^{†} = C^{1} = C and has the property
C^{1}ψC = ψ^{C}
So for our spinor, ψ, we get:
_  
ψ^{T} =  χ^{α}^{ } 
^{ }  _^{ } 
^{ }  ξ_{α'} 
 
     
 ε_{αβ} 0^{ } χ^{α}^{ }  =  ε_{αβ}χ^{α}_{ }
 0 ε^{α'β'} _^{ }  _{ } _ 
_{ }   ξ_{α'}   ε^{α'β'}ξ_{α'}
   
 
=  χ_{β}^{ }
 _^{ }
 ξ^{β'}
 
   
 ξ_{α}_{ }   χ_{β}^{ }
 __{ }  < h.c. >  _^{ }
 χ^{α'}   ξ^{β'}
   
Therefore, the Hermitian conjugate of any left handed
Weyl spinor is a right handed Weyl spinor and vice
versa.
Scalars, Vectors and Tensors

Scalars

We can form a scalar out of 2 Weyl spinors as follows:
_ _
ξχ ≡ ξ^{α'}χ_{α'}
_
= ξ^{α'}ε_{α'β'}χ^{β'}
_
= χ^{β'}ε_{α'β'}ξ^{α'} ( because χ and ξ anticommute)
_
= χ^{β'}ε_{β'α'}ξ^{α'}
_
= χ^{β'}ξ_{β'}
_
= χξ
Similarly,
ξχ ≡ ξ^{α}χ_{α}
= ξ^{α}ε_{αβ}χ^{β}
= χ^{β}ε_{αβ}ξ^{α} ( because χ and ξ anticommute)
= χ^{β}ε_{βα}ξ^{α}
= χ^{β}ξ_{β}
= χξ
Note the positive sign in accordance with the spin
statistics theorem.
Vectors

Rewrite:
σ^{μ} = (σ^{μ})_{αα'}
_ _
σ^{μ} = (σ^{μ})^{αα'}
Such that,
_
σ^{μαα'} = ε^{αβ}ε^{α'β'}σ^{μ}_{ββ'}
We can now form a vector out of 2 Weyl spinors as
follows:
ξσ^{μ}χ ≡ ξ^{α}σ^{μ}_{αα'}χ^{α'}
_
= χ^{α'}σ^{μ}_{αα'}ξ^{α}
_
= ε^{α'β'}χ_{β'}σ^{μ}_{αα'}ε^{αβ}ξ_{β}
_
= ε^{α'β'}χ_{β'}σ^{μ}_{αα'}ε^{αβ}ξ_{β}
_
= χ_{β'}(ε^{β'α'}ε^{βα})σ^{μ}_{αα'}ξ_{β}
_ _
= χ_{β'}σ^{μβ'β}ξ_{β} (σ^{μ}_{αα'} = (1,σ^{i}) and σ^{μαα'} = (1,σ^{i}))
__
= χσ^{μ}ξ
Note the negative sign in accordance with the spin
statistics theorem.