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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:
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
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)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.