Redshift Academy


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^ργ^σ - (1/2)η^ρσ

2.  The Lorentz transformation for a spinor, S[Λ]

    S[Λ] = exp((1/2)Ω_ρσS^ρσ)

3.  The Lorentz transformation for a vector, Λ

    Λ = exp((1/2)Ω_ρσM^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_βψ^β

5.  (M^ρσ)^μ_ν = η^ρμδ^σ_ν - η^σμδ^ρ_ν

6.  {γ^μ,γ^ν} = 2η^μν  ... CLIFFORD ALGEBRA.

7.  (γ⁰)^† = γ⁰

8.  (γ⁰)^-1 = γ⁰

9.  γ⁰γ^μγ⁰ = (γ^μ)^†

10.  (γⁱ)^† = -γⁱ

In addition, we will need:

11.  [S^μν,γ^μ] = γ^μη^νρ - γ^νη^ρμ

Proof:

[S^μν,γ^μ] = (1/2)[γ^μγ^ν,γ^ρ]

      = (1/2)γ^μγ^νγ^ρ - (1/2)γ^ργ^μγ^ν

      = (1/2)γ^μ{γ^ν,γ^ρ} - (1/2)γ^μγ^ργ^ν 
           
        - (1/2){γ^ρ,γ^ν}γ^ν + (1/2)γ^μγ^ργ^ν

Use 6. to get:

      = γ^μη^νρ - γ^νη^ρμ  

12.  [S^μν,S^ρσ] = η^νρS^μσ - η^μρS^νσ + η^μσS^νρ - η^νσS^μρ  

                          ... LORENTZ ALGEBRA

Proof:

[S^μν,S^ρσ] = (1/2)[S^μν,γ^ργ^σ]

      = (1/2)[S^μν,γ^ρ]γ^σ + (1/2)γ^ρ[S^μν,γ^σ]

Use 7. to get:

      = γ^μγ^ση^νρ - γ^νγ^ση^ρμ + γ^ργ^μη^νσ - γ^ργ^νη^σμ  

Rearrange 1.  γ^μγ^σ = 2S^μσ + η^μσ and substitute to 
get:

[S^μν,S^ρσ] = η^νρS^μσ - η^μρS^νσ + η^μσS^νρ - η^νσS^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_ρσS^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 γ⁰ is 
hermitian whereas the γⁱ's are anti-hermitian.  
Therefore, there is no way to pick γ^μ such that
all are anti-hermitian.  However, there is a way 
around this.  Consider:

γ⁰S[Λ]^†γ⁰ = γ⁰[exp(iΩ_μνS^μν)]^†γ⁰

       = γ⁰exp((-i/2)Ω_μν(S^μν)^†γ⁰

       = γ⁰(1 - (i/2)Ω_μν(S^μν)^†)γ⁰

Multiplying from the left and the right gives:

γ⁰S[Λ]^†γ⁰ = 1 - (i/2)Ω_μνγ⁰(S^μν)^†γ⁰

       = exp(-(i/2)Ω_μνγ⁰(S^μν)^†γ⁰)

Now,

(S^μν)^† = (1/4)[(γ^μ)^†,(γ^ν)^†] 

    = (1/4)((γ^μ)^†(γ^ν)^† - (γ^ν)^†(γ^μ)^†)
     
    = (1/4)((γ⁰γ^μγ⁰)(γ⁰γ^νγ⁰) - (γ⁰γ^νγ⁰)(γ⁰γ^μγ⁰))

    = (1/4)((γ⁰γ^μγ⁰)(γ⁰γ^νγ⁰) + (γ⁰γ^μγ⁰)(γ⁰γ^νγ⁰))

    = (1/2)(γ⁰γ^μγ^νγ⁰)

    = γ⁰(1/2)(γ^μγ^ν)γ⁰

Therefore,

(S^μν)^† = γ⁰S^μνγ⁰ 

Returning to where we left off and using this result we 
get:

γ⁰S[Λ]^†γ⁰ = exp(-(i/2)Ω_μνS^μν)

       = S[Λ]^-1

Multiplying both sides from the left and the right by γ⁰ 
gives:

S[Λ]^† = γ⁰S[Λ]^-1γ⁰

With this in mind, we now define the DIRAC ADJOINT:
_
ψ = ψ^†γ⁰

Under a boost we get:
_
ψψ -> ψ^†S[Λ]^†γ⁰S[Λ]ψ

   =  ψ^†γ⁰S[Λ]^-1γ⁰γ⁰S[Λ]ψ

   =  ψ^†γ⁰S[Λ]^-1S[Λ]ψ

   =  ψ^†γ⁰ψ

This looks like:

     -  -
    | ψ₁ |
ψ = | ψ₂ |
    | ψ₃ |
    | ψ₄ |
     -  -

              -              -
ψ^† = (ψ^*)^T = | ψ₁^* ψ₂^* ψ₃^* ψ₄^* |
              -              -

_           -                -
ψ = ψ^†γ⁰ = | ψ₁^* ψ₂^* -ψ₃^* -ψ₄^* |
            -                -
 _              
ψψ = ψ₁ψ₁^* + ψ₂ψ₂^* - ψ₃ψ₃^* - ψ₄ψ₄^*

Again, this is completely different to the vector case, x_μx^μ.

 -      -  -  -
|x⁰ x¹ x² x³ || x⁰ | = t² - x² - y² - z²
 -      - | x¹ |
          | x² |
          | x³ |
           -  -

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^†γ⁰ψ

Under a Lorentz transformation,
_
ψψ -> ψ^†(Λ^-1x)S[Λ]^†γ⁰S[Λ]ψ(Λ^-1x)

   = ψ^†(Λ^-1x)γ⁰ψ(Λ^-1x)
     _
   = ψ(Λ^-1x)ψ(Λ^-1x)
   _
So ψψ does indeed transform as a scalar.

Vector
------

To form a vector (spin 1) from 2 spinors one writes:
_
ψγ^μψ

Under a Lorentz transformation,
_
ψγ^μψ -> ψS[Λ]^-1γ^μS[Λ]ψ

For this to be true we need S[Λ]^-1γ^μS[Λ] = Λ^μ_νγ^ν  ... 13.

Where,

Λ = exp((1/2)Ω_ρσM^ρσ) ~ 1 + (1/2)Ω_ρσM^ρσ ...

and,

S[Λ] = exp((1/2)Ω_ρσS^ρσ) ~ 1 + (1/2)Ω_ρσS^ρσ ...

We work infinitesimally with Ω_ρσ = ε_ρσ.  Therefore,

(1 - (1/2)ε_ρσS^ρσ)γ^μ(1 + (1/2)ε_ρσS^ρσ) = (1 + (1/2)ε_ρσM^ρσ)γ^ν

(1 - (1/2)ε_ρσS^ρσ)(γ^μ + (1/2)ε_ρσS^ρσγ^μ) = (1 + (1/2)ε_ρσM^ρσ)γ^ν

γ^μ + (1/2)ε_ρσS^ρσγ^μ - (1/2)ε_ρσS^ρσγ^μ + O(ε_ρσ²) = γ^μ + (1/2)ε_ρσM^ρσγ^ν

(S^ρσγ^μ - S^ρσγ^μ) = M^ρσγ^ν

[S^ρσ,γ^μ] = (M^ρσ)^μ_νγ^ν

Using 11. on the LHS and 5. on the RHS gives:

γ^μη^νρ - γ^νη^ρμ = (η^ρμδ^σ_ν - η^σμδ^ρ_ν)γ^ν

γ^μη^νρ - γ^νη^ρμ = η^ρμγ^σ - η^σμγ^ρ
                          _
So we have proven 13. and ψγ^μψ does indeed transform 
as a vector.

Tensor
------

To form a tensor (spin 2) from two spinors one writes:
_
ψS^μνψ

Under a Lorentz transformation,
_
ψS^μνψ -> ψS[Λ]^-1S^μνS[Λ]ψ
        _
      = ψS[Λ]^-1((1/2)[γ^μ,γ^ν])S[Λ]ψ
             _
      = (1/2)ψS[Λ]^-1(γ^μγ^ν - γ^νγ^μ)S[Λ]ψ
        _
      = ψ(1/2)(S[Λ]^-1γ^μS[Λ]S[Λ]^-1γ^νS[Λ] 

        - S[Λ]^-1γ^νS[Λ]S[Λ]^-1γ^μS[Λ])ψ

Now from 13. S[Λ]^-1γ^μS[Λ] = Λ^μ_νγ^ν

ψS^μνψ -> ψ(1/2)(Λ^μ_αγ^αΛ^ν_βγ^β - Λ^ν_βγ^βΛ^μ_αγ^α)ψ
        _
      = ψ(1/2)Λ^μ_αΛ^ν_β[γ^α,γ^β]ψ
              _
      = Λ^μ_αΛ^ν_βψS^αβψ
   _
So ψS^μνψ does indeed transform as a tensor.

Generalization
--------------

Consider the expression:
_
ψΓ^Aψ

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^μγ⁵               :  pseudo-vector (4 components)
γ⁵                :  pseudo-scalar (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 = left-handed 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 = γ⁰ψ^†

            -    -     -   -
Where γ⁰ = | 0 I₂ | = | 0 1 | in the Weyl representation.
          | I₂ 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:

ε¹² = -ε²¹ = ε₂₁ = -ε₁₂ = 1

ε¹¹ = ε²² = ε₁₁ = ε₂₂ = 0

ε^abε_bc = δ^a_c = 1

ε_abε^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 | = iσ₂ 
        | -1 0 |
              -    -

              -    -
ε^αβ = ε^α'β' = | 0 -1 | = -iσ₂ 
        | 1  0 |
              -    -

We can use these operators as follows:
                       _         _    _         _
ξ_α = ε_αβξ^β  ξ^α = ε^αβξ_β  χ_α' = ε_α'β'χ^β'  χ^α' = ε^α'β'χ_β'

Therefore,

ε_αβξ^β = -ξ_α

 -    -  -  -     -   -
|  0 1 || ξ^α | = |  0 |
| -1 0 || 0 |   | -ξ_α |
 -    -  -  -     -   -

ε^αβξ_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 || ∂_μξ_α | = iχ^ασ^μ∂_μχ^α' + ξ_α'σ_μ∂_μξ_α
      -   - | _   ||   _|
            | σ_μ 0 || ∂_μχ^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,σⁱ) is substituted for σ_k
_
σ^μ = (1,-σⁱ) 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^μσ^ν - σ^νσ^μ      0      |
          |          _     _    |
          |      0      σ^μσ^ν - σ^νσ^μ |
           -                      -

The objects that obey the Lorentz algebra and generate 
the desired rotations are given by:

                  _      _      
(S^μν)_α^β = (1/4)(σ^μσ^ν - σ^νσ^μ)_α^β
 _               _      _      
(S^μν)^α'_β' = (1/4)(σ^μσ^ν - σ^νσ^μ)^α'_β'

So these correspond to:

ξ_α -> (exp((1/2)Ω_μνS^μν))_α^βξ_β

and,
_                   _
χ^α' -> (exp((1/2)Ω_μνS^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: 

x⁰ -> x⁰  xⁱ -> -xⁱ

Under parity, the left and right-handed spinors are 
exchanged.  Under parity, rotations don’t change sign
but boosts flip sign.  Therefore,

Pψ_L,R(t,x) = ψ_R,L(t,-x)

        = γ⁰ψ_R,L(t,-x)

if ψ(t,x) satisfies the Dirac equation, then the 
parity transformed spinor γ⁰ψ(t,-x) also satisfies 
the Dirac equation, meaning:

(iγ⁰∂_t + iγⁱ∂_i - m)γ⁰ψ(t,-x) = γ⁰(iγ⁰∂_t - iγⁱ∂_i - m)γ⁰ψ(t,-x)

Where the extra minus sign from passing γ⁰ through
γⁱ 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 = | iσ₂  0  | = | ε_αβ   0|
    |  0 -iσ₂ |   | 0   ε^α'β'|
     -       -     -       -

C satisfies C^T = C^† = C^-1 = -C and has the property

C^-1ψC = ψ^C

So for our spinor, ψ, we get:

_     -  -
ψ^T = | χ^α |
    | _ |
    | ξ_α' |
      -  -

 -        - -   -     -       -
| ε_αβ   0|| χ^α | = | ε_αβχ^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^αχ_α

   = ξ^αε_αβχ^β

   = -χ^βε_αβξ^α (- because χ and ξ anticommute)

   = χ^βε_βαξ^α

   = χ^βξ_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^αβε^α'β'σ^μ_ββ'

We can now form a vector out of 2 Weyl spinors as 
follows:

ξσ^μχ ≡ ξ^ασ^μ_αα'χ^α'
        _
     = -χ^α'σ^μ_αα'ξ^α
                _
     = -ε^α'β'χ_β'σ^μ_αα'ε^αβξ_β
            _
     = -ε^α'β'χ_β'σ^μ_αα'ε^αβξ_β
        _
     = -χ_β'(ε^β'α'ε^βα)σ^μ_αα'ξ_β
        _  _
     = -χ_β'σ^μβ'βξ_β  (σ^μ_αα' = (1,σⁱ) and σ^μαα' = (1,-σⁱ))
        __
     = -χσ^μξ

Note the negative sign in accordance with the spin 
statistics theorem.