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```Spinors - Part 2

Spinors - Part 1
----------------

"No one fully understands spinors. Their algebra is
formally understood but their general significance
is mysterious.  In some sense they describe the
'square root' of geometry and, just as understanding
the square root of -1 took centuries, the same might
be true of spinors." - Sir Michael Atiyah

Spinors naturally describe spin 1/2 objects in physics.
A 2π rotation does not result in the same quantum
state.  However, a 4π rotation does.  This is due to
a theorem in topology called ORIENTATION ENTANGLEMENT.
This is illustrated below.  After a 2π rotation, the
spiral flips between clockwise and counterclockwise
orientations.   It returns to its original configuration
after spinning a full 4π.

Animation courtesy of Wikipedia

Before we launch into the subject let us review some
of the important matrices and identities we will need.

The Dirac Matrices
------------------

The Dirac basis:

-            -
| 1  0 :  0  0 |
| 0  1 :  0  0 |    -    -
γ0 = | .....:...... | = | I  0 |
| 0  0 : -1  0 |   | 0 -I |
| 0  0 :  0 -1 |    -    -
-            -

The Chiral (Weyl) basis:

-           -
| 0  0 : 1  0 |
| 0  0 : 0  1 |    -   -
γ0 = | .....:..... | = | 0 I |
| 1  0 : 0  0 |   | I 0 |
| 0  1 : 0  0 |    -   -
-           -

The Dirac and Chiral bases:

-             -
|  0  0 :  0  1 |
|  0  0 :  1  0 |    -      -
γ1 = | ......:...... | = |  0  σx |
|  0 -1 :  0  0 |   | -σx  0 |
| -1  0 :  0  0 |    -      -
-             -

-             -
|  0  0 :  0 -i |
|  0  0 :  i  0 |    -      -
γ2 = | ......:...... | = |  0  σy |
|  0  i :  0  0 |   | -σy  0 |
| -i  0 :  0  0 |    -      -
-             -

-             -
|  0  0 :  1  0 |
|  0  0 :  0 -1 |    -      -
γ3 = | ......:...... | = |  0  σz |
| -1  0 :  0  0 |   | -σz  0 |
|  0  1 :  0  0 |    -      -
-             -

The σ matrices are the PAULI SPIN matrices.

Useful identities:

Both the Dirac and Chiral bases obey the following:

1.  (γ0)† = γ0

2.  (γ0)-1 = γ0

3.  γ0γμγ0 = (γμ)†

4.  (γi)† = -γi

5.  {γμ,γν} = 2ημνI - The CLIFFORD ALGEBRA.

6.  [γμ,γν] = 2γμγν for μ ≠ ν, 0 otherwise.

= 2γμγν - 2ημν

7.  All matrices are unitary (γμ(γμ)† = I)

8.  Only γ0 is symmetric (γ0 = (γ0)T).

9.  γμγν = -γνγμ

10. γ0 is hermitian but γi anti-hermitian (γi)† = -γi.
Multiplying γμ by i makes γi hermitian but then γ0 becomes
anti-hermitian.

11. γ0γμγν = γμγνγ0
Note:  Moving through an odd number of γ's changes the sign.
Moving through an even number of γ leaves the sign unchanged.

12. γ0γμγν = -γνγμγ0

SO(3) Review
------------

The following three basic rotation matrices rotate
vectors by an angle θ about the x, y, or z-axis,
in 3D.

-                 -
|  1    0       0   |
R1(θ)  = |  0 cos(θ) -sin(θ) |
|  0 sin(θ)  cos(θ) |
-                 -

-                -
|  cos(θ) 0 sin(θ) |
R2(θ)  = |   0     0  0     |
| -sin(θ) 0 cos(θ) |
-                -

-                -
| cos(θ) -sin(θ) 0 |
R3(θ)  = | sin(θ)  cos(θ) 0 |
|  0       0     1 |
-                -

With generators:

-      -
| 0  0 0 |
G1(θ)  = | 0  0 i |
| 0 -i 0 |
-      -

-      -
| 0 0 -i |
G2(θ)  = | 0 0  0 |
| i 0  0 |
-      -

-      -
|  0 i 0 |
G3(θ)  = | -i 0 0 |
|  0 0 0 |
-      -

And Lie algebra:

[Gi,Gj] = iεijkGk

SU(2) and SO(3)
---------------

Are there other G's that we can write down that satisfy
the Lie algebra?  The answer is yes.  These are the
Pauli matrices of SU(2).  The commutation relationship
(Lie algebra) is:

[σi,σj] = 2iεijkσk

We can write this equivalently as:

(1/4)[σi,σj] = iεijkσk/2

For SO(3) rotational transformation on 3D vectors are
represented by:

ΛR = exp(iθG)

If we write:

Gi = σi/2

We can perform a rotation as follows:

R = exp(iεijkσk/2)

If we let εijk = θ, this can be expanded as a Taylor
series as follows:

R(θ) = 1 + iσθ/2 - (σθ/2)2/2 - i(σθ/2)3/6 + (σθ/2)4/24 ...

= 1 - (σθ/2)2/2 + (σθ/2)4/24 + i{σθ/2 - (σθ/2)3/6}

Now σieven = 1 and σiodd = σi

Therefore,

R(θ) = 1 - (θ/2)2/2 + (θ/2)4/24 + iσ{θ/2 - (θ/2)3/6} ...

= Icos(θ/2) + iσsin(θ/2)

-   -
= Icos(θ/2) + i| 0 1 |sin(θ/2) for σ1
| 1 0 |
-   -
-    -
= Icos(θ/2) + i| 0 -i |sin(θ/2) for σ2
| i  0 |
-    -
-    -
= Icos(θ/2) + i| 1  0 |sin(θ/2) for σ3
| 0 -1 |
-    -

If we pick σ1 we get:
-                   -
R(θ) = exp(iσ1θ/2) = |  cos(θ/2) isin(θ/2) |
| isin(θ/2)  cos(θ/2) |
-                   -

with detR(θ) = +1 and R†R = I

Now, unlike SO(3) transformations, which are real 3 x 3
matrices acting on 3 vectors, we have 2 x 2 matrices
that are complex.  In this case it doesn't make sense
for such matrices to act on vectors with real coordinates
(x, y and z).  So what do they act on?  It is at this
point we introduce a new 2 component object called a
SPINOR, χ, that transforms as χ' = exp(θ.σ/2)χ.  Spinors
are abstract mathematical constructs that do not have
coordinates like a vector and should not be regarded
as being physically rotatable in spacetime like a vector.
However, as we will see in the Spinors - Part 2 notes,
spinors can be combined to form scalars and vectors that
do transform in the 'traditional' way.  In this sense,
the spinor representation is more fundamental than the
vector representation with coordinates x,y and z.

Compare to a SO(3) rotation about the x axis.

-                 -
|  1    0       0   |
= |  0 cos(θ) -sin(θ) |
|  0 sin(θ)  cos(θ) |
-                 -

Now when θ is small (i.e. near the identity) we get:

-   -
SU(2):  Λ = | 1 0 |
| 0 1 |
-   -
-     -
SO(3):  Λ = | 1 0 0 |
| 0 1 0 |
| 0 0 1 |
-     -

Therefore, SU(2) ~ SO(3) near the identity meaning that
they have the same Lie algebra.  Mathematically we say
that the groups are 'locally isomorphic', meaning that
as long as we consider only small rotations, we canâ€™t
detect any difference between the two.

However, when θ = 2π we get:

-     -
SU(2):  Λ = | -1  0 |
|  0 -1 |
-     -
-     -
SO(3):  Λ = | 1 0 0 |
| 0 1 0 |
| 0 0 1 |
-     -

Of course, at 4π both agree!

So this transformation has the effect of a 3D rotation
in space.  However, a rotation by 2π only rotates the
object by 180°.  This is clearly different from how
a vector in spacetime would behave.

SO(3,1) Review
--------------

The following three basic rotation matrices rotate
vectors by an angle θ about the t, x, y, or z-axis,
in 4D.

-        -
| 0 0 0  0 |
J1 = | 0 0 0  0 |
| 0 0 0 -i |
| 0 0 i  0 |
-        -

-       -
| 0  0 0 0 |
J2 = | 0  0 0 i |
| 0  0 0 0 |
| 0 -i 0 0 |
-       -

-       -
| 0 0  0 0 |
J3 = | 0 0 -i 0 |
| 0 i  0 0 |
| 0 0  0 0 |
-       -

And Lie algebra:

[Ji,Jj] = iεijkJk

[ji,Kj] = iεijkKk

[Ki,Kj] = -iεijkJk

The algebra can be recast more simply be redefining:

J+i = (1/2)(Ji + iKi)

J-i = (1/2)(Ji - iKi)

The commutation relations are:

[J+i,J+j] = iεijkJ+k

[J-i,J-j] = iεijkJ-k

[J+i,J-j] = 0

From these calculations it is clear that J+ and J-
form their own groups which follow the same commutation
rules as SO(3).  Thus, we can write:

SO(3,1) = SO(3) x SO(3)

But from before we found that SO(3) ~ SU(2).  Therefore,

SO(3,1) ~ SU(2) x SU(2)

Note:  Technically, we should write so(3,1) etc. instead
of SO(3,1) etc.  since we are talking about the Lie
algebra of the groups.

Since a rotation of 0 to 2π covers all of SO(3) but
only one half of SU(2), SU(2) is referred to as the
DOUBLE COVER of SO(3,1).  SU(2) has 2 dimensions and,
therefore, SO(3,1) has 2 x 2 = 4 dimensions as we would
expect.

So what this means is that a Hilbert space that is
Lorentz invariant can be expressed as a pair of SU(2)
representations as follows:

Spin 0: (0,0) = Scalar

Spin 1/2:  (1/2,0) = Left handed spinor

Spin 1/2:  (0,1/2) = Right handed spinor

So we can have 2 spin 1/2 representations. One transforms
under left but not right and vice versa. These are the
WEYL SPINORS that we discuss in the next section.

Spinors in 4 Dimensions
-----------------------

To get to get to 4D we replace the Pauli matrices with
the Gamma matrices.  The commutator becomes:

Sρσ = (1/4)[γρ,γσ]

Then it can be shown that the commutatator [Sμν,Sρσ]
satisfies the Lorentz algebra.

[Sρσ,Sτν] = ηστSρν - ηρτSσν + ηρνSστ - ησνSρτ

The Chiral Basis
----------------

There are many possible versions of the γ matrices
that satisfy the Clifford algebra.  However, it turns
out there is one unique irreducible representation
of the Clifford algebra known as the Chiral or Weyl
representation.

-   -
In the Chiral basis γ0 = | 0 I |
| I 0 |
-   -

Rotations are given by:

-      -  -      -     -      -  -      -
Sij = (1/4)|  0  σi ||  0  σj | - |  0  σj ||  0  σi |
| -σi  0 || -σj 0  |   | -σj 0  || -σi  0 |
-      -  -      -     -      -  -      -

-               -
= (1/4)| [σj,σi]    0    |
|    0    [σj,σi] |
-               -

-     -
= -(i/2)εijk| σk  0 |
^     | 0  σk |
|      -     -
|
The 1/2 will be a critical factor in the
behaviour of rotations.

And, for boosts:

S01 = (1/2)γ0γi

Thus,

-    -  -      -
S0i  = (1/2)| 0  1 ||  0  σj |
| 1  0 || -σj 0  |
-    -  -      -

-       -
= (1/2)| -σi   0 |
|  0   σi |
-       -

The chiral basis has the advantage that the generators
are in block diagonal form and hence the represenation
is irreducible (see the note on Basic Representation
Theory).  We can see that this would not be the case
in the Dirac representation where,

-    -
γ0 = | I  0 |
| 0 -I |
-    -

And boosts are be given by:

-    - -      -
S0i  = (1/2)| 1  0 ||  0  σj |
| 0 -1 || -σj 0  |
-    - -      -

-     -
= (1/2)| 0  σi |
| σi 0  |
-     -

Therefore, the reducibility is not manifest.  The fact
that the Weyl representation is irreducible is essentially
the motivation for using this representation for our
calculations.

Infinitesimal Generators for Rotations
--------------------------------------

We had from before (after renaming indeces):

-     -
Sij = -(i/2)εijk| σk  0 |
| 0  σk |
-     -

A Lorentz transformation for a VECTOR is given by:

Λ = exp((1/2)ΩijMij)  (Mij is complex)

Where Ωij consists of 6 numbers corresponding to the
6 generators.  They tell us what kind of Lorentz
transformation we are performing (i.e., rotate by
θ = π/7 about the z-direction and boost at speed
v = 0.2c in the x direction.

Note:  Both Ωij and Mij are antisymmetric.  Therefore,
in index notation:

ΩijMij = ΩijMij + ΩjiMji

Since Ωij = -Ωji and Mij = -Mji the second term is
additive.  Hence the need for the factor of (1/2).

We can create the equivalent for our SPINOR as:

S[Λ] = exp((1/2)ΩijSij)

If we let,

Ωij = -Ωji = -εijkθk (accounting for the factor of 2)

-     -
and Sij = (i/2)| σk  0 |
| 0  σk |
-     -

The rotation transformation for the spinor, S[Λ]
becomes:

S[Λ] = exp(-iθSij)

Now, Sij is a diagonal matrix.  Therefore,

-                            -
S[Λ] = | exp(-(i/2)θσk)       0       |
|       0       exp(-(i/2)θσk) |
-                            -

-                                         -
= | cos(θ/2) - isin(θ/2)          0           |
|          0           cos(θ/2) - isin(θ/2) |
-                                         -

We can also derive this in a different way:

exp(-iθS/2) = 1 - iθS/2 - (θS)2/2.2! + i(θS)3/2.3! + (θS)4/2.4!... etc.

= [1 + (θS)2/2.2! + (θS)4/2.4! ...] - i[θS/2 + (θS)3/2.3! ...]

Now (S)even = I and (S)odd = S.  Thus,

= [1 + (θ)2/4 + (θ)4/48 ...] - iS[θ/2 + (θ)3/12 ...]

= Icos(θ/2) - iSijsin(θ/2)

-                                             -
= | [cos(θ/2) - isin(θ/2)]          0             |
|         0              [cos(θ/2) + isin(θ/2)] |
-                                             -

Spinor Fields
-------------

Just as the Lorentz transform operates on 4-vectors,
we need an object for these matrices to act on.  This
object is the DIRAC SPINOR field, ψα(x). It has 4
complex components labelled by α = 1, 2, 3, 4.  Under
a Lorentz transformation this field transforms as:

ψα(x) -> S[Λ]αβψβ(x)(Λ-1x)

Where

S[Λ] = exp((1/2)ΩijSij)

And,

Λ = exp((1/2)ΩijMij)

We can form two 2D representations by writing:

- -
ψDIRAC = m| ψL |
| ψR |
- -

ψL and ψR each have 2 components and are called the
left handed and right handed WEYL SPINORS.  In the
Weyl basis, the particle is regarded as being massless
with the spin quantized parallel or anti-parallel to
the direction of motion (see the note on Helicity and
Chirality).

So the transformation is:

-                                             -  - -
| [cos(θ/2) - isin(θ/2)]          0             || ψL |
|         0              [cos(θ/2) + isin(θ/2)] || ψR |
-                                             -  - -

Therefore, the left and right components behave the same
under rotations.

For a rotation of 2π about the z-axis this becomes:

-      -  -  -     -   -
| -1   0 || ψL | = | -ψL |
|  0  -1 || ψR |   | -ψR |
-      -  -  -     -   -

Which means:

S(θ + 2π) = -S(θ) as we found before.

Infinitesimal Generators for Boosts
-----------------------------------

S01 = (1/2)γ0γi

-   -
In the Chiral basis γ0 = | 0 I |
| I 0 |
-   -

Thus,

-    -  -      -
S0i  = (1/2)| 0  1 ||  0  σj |
| 1  0 || -σj 0  |
-    -  -      -

-       -
= (1/2)| -σi   0 |
|  0   σi |
-       -

We follow the same procedure from before except
that Ω0i = -Ωi0 = ξ.

We get:

-                    -
S[Λ] = exp(ξS0i) = | exp(ξσ/2)    0       |
|    0      exp(-ξσ/2) |
-                    -

Note:  We don't have to worry about the factor of
(1/2) here because we are only summing over one
index.

So the transformation is:

-                    -  - -
| exp(ξσ/2)      0     || ψL |
|    0      exp(-ξσ/2) || ψR |
-                    -  - -

Therefore, the left and right components behave
oppositely under boosts.

We can summarize these transformations in exponential
form in the following way.

ψL -> exp(-iθ.σ/2 + ξ.σ/2)ψL

ψR -> exp(-iθ.σ/2 - ξ.σ/2)ψR
```