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- ```Spinors - Part 1

Spinors - Part 2
----------------

In Spinors - Part 1 we identified the following:

1.  The generators, S>ρσ, are defined as:

Sρσ = (1/4)[γρ,γσ]

= (1/2)γργσ - (1/2)ηρσ

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

S[Λ] = exp((1/2)ΩρσSρσ)

3.  The Lorentz transformation for a vector, Λ

Λ = exp((1/2)ΩρσMρσ)

4.  The Sμν operates operate on a spinor field, ψ.
Under a Lorentz transformation these fields
transform as:

ψα = S[Λ]αβψβ

5.  (Mρσ)μν = ηρμδσν - ησμδρν

6.  {γμ,γν} = 2ημν  ... CLIFFORD ALGEBRA.

7.  (γ0)† = γ0

8.  (γ0)-1 = γ0

9.  γ0γμγ0 = (γμ)†

10.  (γi)† = -γi

In addition, we will need:

11.  [Sμν,γμ] = γμηνρ - γνηρμ

Proof:

[Sμν,γμ] = (1/2)[γμγν,γρ]

= (1/2)γμγνγρ - (1/2)γργμγν

= (1/2)γμ{γν,γρ} - (1/2)γμγργν

- (1/2){γρ,γν}γν + (1/2)γμγργν

Use 6. to get:

= γμηνρ - γνηρμ

12.  [Sμν,Sρσ] = ηνρSμσ - ημρSνσ + ημσSνρ - ηνσSμρ

... LORENTZ ALGEBRA

Proof:

[Sμν,Sρσ] = (1/2)[Sμν,γργσ]

= (1/2)[Sμν,γρ]γσ + (1/2)γρ[Sμν,γσ]

Use 7. to get:

= γμγσηνρ - γνγσηρμ + γργμηνσ - γργνησμ

Rearrange 1.  γμγσ = 2Sμσ + ημσ and substitute to
get:

[Sμν,Sρσ] = ηνρSμσ - ημρSνσ + ημσSνρ - ηνσSμρ

-----------------

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)ΩρσSρσ)

S[Λ]† = exp((1/2)Ωρσ(Sρσ)†)

So to be unitary we require (Sρσ)† = -Sρσ

Unfortunately, this is not possible because γ0 is
hermitian whereas the γi's are anti-hermitian.
Therefore, there is no way to pick γμ such that
all are anti-hermitian.  However, there is a way
around this.  Consider:

γ0S[Λ]†γ0 = γ0[exp(iΩμνSμν)]†γ0

= γ0exp((-i/2)Ωμν(Sμν)†γ0

= γ0(1 - (i/2)Ωμν(Sμν)†)γ0

Multiplying from the left and the right gives:

γ0S[Λ]†γ0 = 1 - (i/2)Ωμνγ0(Sμν)†γ0

= exp(-(i/2)Ωμνγ0(Sμν)†γ0)

Now,

(Sμν)† = (1/4)[(γμ)†,(γν)†]

= (1/4)((γμ)†(γν)† - (γν)†(γμ)†)

= (1/4)((γ0γμγ0)(γ0γνγ0) - (γ0γνγ0)(γ0γμγ0))

= (1/4)((γ0γμγ0)(γ0γνγ0) + (γ0γμγ0)(γ0γνγ0))

= (1/2)(γ0γμγνγ0)

= γ0(1/2)(γμγν)γ0

Therefore,

(Sμν)† = γ0Sμνγ0

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

γ0S[Λ]†γ0 = exp(-(i/2)ΩμνSμν)

= 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, xμxμ.

-          -  -  -
|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:
_
ψγμψ

Under a Lorentz transformation,
_
ψγμψ -> ψS[Λ]-1γμS[Λ]ψ

For this to be true we need S[Λ]-1γμS[Λ] = Λμνγν  ... 13.

Where,

Λ = exp((1/2)ΩρσMρσ) ~ 1 + (1/2)ΩρσMρσ ...

and,

S[Λ] = exp((1/2)ΩρσSρσ) ~ 1 + (1/2)ΩρσSρσ ...

We work infinitesimally with Ωρσ = ερσ.  Therefore,

(1 - (1/2)ερσSρσ)γμ(1 + (1/2)ερσSρσ) = (1 + (1/2)ερσMρσ)γν

(1 - (1/2)ερσSρσ)(γμ + (1/2)ερσSρσγμ) = (1 + (1/2)ερσMρσ)γν

γμ + (1/2)ερσSρσγμ - (1/2)ερσSρσγμ + O(ερσ2) = γμ + (1/2)ερσMρσγν

(Sρσγμ - Sρσγμ) = Mρσγν

[Sρσ,γμ] = (Mρσ)μνγν

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

γμηνρ - γνηρμ = (ηρμδσν - ησμδρν)γν

γμηνρ - γνηρμ = ηρμγσ - ησμγρ
_
So we have proven 13. and ψγμψ does indeed transform
as a vector.

Tensor
------

To form a tensor (spin 2) from two spinors one writes:
_
ψSμνψ

Under a Lorentz transformation,
_
ψSμνψ -> ψS[Λ]-1SμνS[Λ]ψ
_
= ψS[Λ]-1((1/2)[γμ,γν])S[Λ]ψ
_
= (1/2)ψS[Λ]-1(γμγν - γνγμ)S[Λ]ψ
_
= ψ(1/2)(S[Λ]-1γμS[Λ]S[Λ]-1γνS[Λ]

- S[Λ]-1γνS[Λ]S[Λ]-1γμS[Λ])ψ

Now from 13. S[Λ]-1γμS[Λ] = Λμνγν

ψSμνψ -> ψ(1/2)(ΛμαγαΛνβγβ - ΛνβγβΛμαγα)ψ
_
= ψ(1/2)ΛμαΛνβ[γα,γβ]ψ
_
= ΛμαΛνβψSαβψ
_
So ψSμνψ does indeed transform as a tensor.

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

Consider the expression:
_
ψΓAψ

We have already found expressions for scalars,
vectors and tensors.  Can we find ΓA's that
create other objects that transform correctly
under a Lorentz transformation.  Again, the
answer is yes.  We can write ΓA in terms of the
following combinations of 16 γ matrices.

ΓA =

1                    :  scalar (1 component)
γμ                   :  vector (4 components)
Sμν = (i/2)[γμ,γν]   :  Tensor (6 components)
γμγ5                 :  pseudo-vector (4 components)
γ5                   :  pseudo-scalar (4 component)

Spinor Indeces
--------------

Consider the Weyl spinors mentioned above and
define:

-  -
ψL = | ξα |
| 0  |
-  -
-  -
ψR = | 0  |
| _  |
| χα'|
-  -

-  -
ψ = ψL + ψR = | ξα |
| _  |
| χα'|
-  -

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:
_
ξα' = (ξα)†

and,
_
χα = (χα')†

-_  -
ψ† = | ξα' |
| χα  |
-  -
_
Now, ψ = γ0ψ†

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

_    -   -  - _ -
ψ = | 0 1 || ξα' |
| 1 0 || χα  |
-   -  -   -

-   _ -
= | χα ξα' |
-     -

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

εabεbc = δac = 1

εabεbc = δac = 1

εa'b'εb'c' = δa'c' = 1

εa'b'εb'c' = δa'c' = 1

In matrix form:

-    -
εαβ = εα'β' = |  0 1 | = iσ2
| -1 0 |
-    -

-    -
εαβ = εα'β' = | 0 -1 | = -iσ2
| 1  0 |
-    -

We can use these operators as follows:
_         _    _         _
ξα = εαβξβ  ξα = εαβξβ  χα' = εα'β'χβ'  χα' = εα'β'χβ'

Therefore,

εαβξβ = -ξα

-    -  -  -     -   -
|  0 1 || ξα | = |  0  |
| -1 0 || 0  |   | -ξα |
-    -  -  -     -   -

εαβξβ = ξα

-    -  -  -     -  -
| 0 -1 || ξα | = | 0  |
| 1  0 || 0  |   | ξα |
-    -  -  -     -  -

Note: εαβ = εα'β' etc. εα'β and εαβ' are not allowed.

Therefore, the Dirac Lagrangion excluding the mass term
is:

-   _  -  -    -  -    -           _     _  _
L = i| χα ξα' || 0 σμ || ∂μξα  | = iχασμ∂μχα' + ξα'σμ∂μξα
-      - | _    ||   _   |
| σμ 0 || ∂μχα' |
-   -  -     -
^
|
γμ in the Weyl representation.

Spinor Transformations
----------------------

Let us now reconstruct the spinor transformation laws
that incorporate indeces.

-      -      -    -
In the Weyl basis γμ = |  0  σk | -> | 0 σμ |
| -σk 0  |    | _    |
-      -     | σμ 0 |
-    -
Where,

σμ = (1,σi) is substituted for σk
_
σμ = (1,-σi) is substituted for -σk

Which for rotations leads to:

-      -  -      -
Sμν = (1/4)| 0   σμ || 0   σν |
| _      || _      |
| σμ  0  || σν  0  |
-      -  -      -

-  _      _            -
= (1/4)| σμσν - σνσμ      0      |
|              _     _    |
|      0      σμσν - σνσμ |
-                      -

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

_      _
(Sμν)αβ = (1/4)(σμσν - σνσμ)αβ
_               _      _
(Sμν)α'β' = (1/4)(σμσν - σνσμ)α'β'

So these correspond to:

ξα -> (exp((1/2)ΩμνSμν))αβξβ

and,
_                   _
χα' -> (exp((1/2)ΩμνSμν))α'β'χβ'

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,

PψL,R(t,x) = ψR,L(t,-x)

= γ0ψR,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:

-   -  -  -     - _ -
ψP = | 0 1 || ξα  | = | χα' |
| 1 0 || _   |   | ξα  |
-   - | χα' |    -   -
-  -

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:
_
ψC = CψT

Where C is defined as the matrix:

-       -     -       -
C = | iσ2  0  | = | εαβ   0  |
|  0 -iσ2 |   | 0   εα'β'|
-       -     -       -

C satisfies CT = C† = C-1 = -C and has the property

C-1ψC = ψC

So for our spinor, ψ, we get:

_     -  -
ψT = | χα  |
| _   |
| ξα' |
-  -

-        - -   -     -       -
| εαβ   0  || χα  | = | εαβχα   |
| 0   εα'β'|| _   |   |       _ |
-        - | ξα' |   | εα'β'ξα'|
-   -     -       -

-  -
= | χβ |
| _  |
| ξβ'|
-  -

-  -               -  -
| ξα  |            | χβ |
| _   | <- h.c. -> | _  |
| χα' |            | ξβ'|
-  -               -  -

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 ≡ ξα'χα'
_
= ξα'εα'β'χβ'
_
= -χβ'εα'β'ξα' (- because χ and ξ anticommute)
_
= χβ'εβ'α'ξα'
_
= χβ'ξβ'
_
= χξ

Similarly,

ξχ ≡ ξαχα

= ξαεαβχβ

= -χβεαβξα (- because χ and ξ anticommute)

= χβεβαξα

= χβξβ

= χξ

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

Vectors
-------

Rewrite:

σμ = (σμ)αα'
_     _
σμ = (σμ)αα'

Such that,
_
σμαα' = εαβεα'β'σμββ'

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

ξσμχ ≡ ξασμαα'χα'
_
= -χα'σμαα'ξα
_
= -εα'β'χβ'σμαα'εαβξβ
_
= -εα'β'χβ'σμαα'εαβξβ
_
= -χβ'(εβ'α'εβα)σμαα'ξβ
_  _
= -χβ'σμβ'βξβ  (σμαα' = (1,σi) and σμαα' = (1,-σi))
__
= -χσμξ

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