
Supersymmetry. Theory, Experiment, and Cosmology
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420 Spinors
Then using η1 = η2 and η2 = −η1, one may write, using the Einstein convention
ξ2η1 − ξ1η2 = ξ1η1 + ξ2η2 ≡ ξαηα.
Let us note that the spinor ξα introduced in (B.2) transforms as:
ξ = εαβ ξβ |
= εαβ M β δ ξδ = εγδ (M −1)γ |
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(B.5) |
where we have used (B.3). |
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One may then check that ξαηα is invariant under a Lorentz transformation: |
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ξ αηα = M αβ ξβ M −1T α |
γ ηγ = M −1M γ β ξβ ηγ = ξβ ηβ . |
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We will use from now on the following notation: |
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ξαηα ≡ ξη, |
ξξ ≡ ξ2. |
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(B.6) |
Our convention is that spinors always anticommute. It follows that
ξη = ηξ. |
(B.7) |
The complex conjugate (ξα) transforms with the matrix M . To make the di erence explicit, one introduces a di erent notation
ξ¯α˙ |
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(B.8) |
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ξ¯β . |
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¯α˙ ¯α˙
The ξ form an inequivalent spinor representation: ξ , α˙ = 1, 2 = 2R = (0, 1/2).
˙
We use the antisymmetric tensors ε ˙ and εα˙ β to raise and lower dotted indices
α˙ β
˙ ˙
(ε˙ ˙ = 1 = ε12), and we note:
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(B.9) |
ξα˙ |
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≡ ξ |
The convention is the following: when indices are not written, undotted indices are descending and dotted indices are ascending.
We will also take as a convention to reverse the order of spinors when performing complex conjugation:
(ξη)† = (ξαηα)† = η¯α˙ ξ¯α˙ = η¯ξ¯ = ξη¯¯. |
(B.10) |
B.1.2 Relation of spinors with vectors
As we have just seen, one may form a spin-zero object ξη out of two spinors. One may also form a spin-one object. In order to do so, let us introduce the Pauli matrices σµ,
µ = 0, . . . , 3,
σ0 |
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0 1 |
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1 0 |
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σ2 = |
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σ3 = |
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(B.11) |
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and form the vector V |
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424 Spinors
B.2 Spinors in higher dimensions
We first review the representations of the group SO(2n) before defining spinors in D dimensions.
B.2.1 Representations of SO(2n)
SO(2n) is defined as the group of 2n-dimensional orthogonal matrices Λ of determinant unity, i.e.
ΛT Λ = 1, Det Λ = 1. |
(B.42) |
Turning to the algebra, Λ = eJ 1 + J + · · · , these two conditions become |
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JT = −J. |
(B.43) |
There are 2n(2n − 1)/2 independent 2n-dimensional matrices satisfying (B.43). Hence the dimension of the SO(2n) algebra is n(2n − 1). A basis of generators is
(1 ≤ K, L ≤ 2n) |
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JKL = |
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(B.44) |
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From this one obtains the commutation relations
[JIJ , JKL] = δJK JIL − δJLJIK + δILJJK − δIK JJL , |
(B.45) |
which define the SO(2n) algebra.
A maximal set of commuting generators (known as the Cartan subalgebra) is given by the n generators3 Nk ≡ J2k−1,2k, k = 1, . . . , n. The number of such generators is the rank of the algebra, hence the rank of SO(2n) is n.
The representation (B.44) is the vector representation of dimension 2n. The dimension of the algebra gives the dimension of the adjoint representation: n(2n − 1). In the following, we build the spinor representations of dimension 2n−1.
The standard procedure is to consider a set of n fermion creation and annihilation
operators ak† , ak, k = 1, . . . , n: |
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&ak, al†' = δkl, [ak, al] = 0 = &ak† , al†' . |
(B.46) |
One then constructs a Fock space by defining a vacuum |0 and acting on it with creation operators:
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form a set of 2n independent states. |
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3Each of them is a block diagonal matrix with only nonzero block |
0 1 |
in the (2k − 1, 2k) |
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−1 0 |
entry.
426 Spinors
In the case of Minkowski spacetime M4, with gµν = (+1, −1, −1, −1), this tangent space group is the proper Lorentz group SO(1, 3) (if we restrict to det Λ = +1). For a D-dimensional manifold, with metric gM N = (+1, −1, −1, . . . , −1), the tangent space group is SO(1, D − 1).
The corresponding Cli ord algebra reads |
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{ΓM , ΓN } = 2gM N , |
(B.54) |
where the gamma matrices have dimension 2[D/2], with [D/2] the integral part of D/2. As in the previous section, the generators of SO(1, 9) are the ΣM N = 1/4[ΓM , ΓN ].
Just as before, one may introduce for even spacetime dimension D, the matrix Γ(D+1) which anticommutes with all gamma matrices5:
Γ(D+1) ≡ ı(D−2)/2Γ0Γ1 · · · ΓD−1 . |
(B.55) |
We note that γ5 defined in (B.17) is precisely Γ(5) for D = 4. One then defines the chirality eigenstates by the condition:
Γ(D+1)ΨL,R = ΨL,R . |
(B.56) |
Thus, a Dirac spinor has 2D/2 degrees of freedom whereas a Weyl6 |
fermion has |
2(D−2)/2. |
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Can we always define a Majorana spinor? In order to answer this question, let us define the matrix B which intertwines the representation ΓM with the equivalent
representation −ΓM : |
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BΓM B−1 = −ΓM . |
(B.57) |
We have BB = ηI with η = ±1. Clearly, B = CΓ0 where C is defined as in (B.31) by
C−1ΓM C = −ΓM T . |
(B.58) |
If a spinor field Ψ of charge q satisfies the Dirac equation coupled with the electromagnetic field, one easily checks that B−1Ψ satisfies it with charge −q: it is associated with the antiparticle. One defines a Majorana spinor by the condition that particle and antiparticle are identical:
Ψ = BΨ . |
(B.59) |
This Majorana condition is easily shown (see Exercise 6) to be equivalent to the condition (B.32) given above. It obviously implies η = 1.
It turns out that η can be computed as a function of the dimension D (see [331]
for a proof): |
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(D + 1)' . |
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η = −√2 cos & |
(B.60) |
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5The di erence in the overall factor between (B.51) and (B.55) is due to the change of metric between Euclidean and Minkowski. In both cases, it is chosen in such a way that Γ(D+1) 2 = 1.
6In the case of odd dimension D , there are no Weyl spinors. If we write D = D + 1 where D is even, then Γ0, Γ1, . . . , ΓD−1 and iΓ(D+1) given in (B.55) yield a set of gamma matrices. In what follows, we only consider the case of even D.

428 Spinors
(σµν σρ)αβ˙ |
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gνρgµλ − gµρgνλ − i µνρλ σλαβ˙ |
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(σµσ¯νρ)αβ˙ |
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gµν gρλ − gµρgνλ − i µνρλ σλαβ˙ |
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(¯σµν σ¯ρ)αβ˙ |
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gνρgµλ − gµρgνλ + i µνρλ σ¯λαβ˙ |
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(¯σµσνρ)αβ˙ |
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gµν gρλ − gµρgνλ + i µνρλ σ¯λαβ˙ |
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(B.68) |
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Hints: To derive (B.67), use (B.25).
Exercise 5 We construct the generators of the Lorentz group in the spinor representation.
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Let Pµ be a covariant vector. Define Pαα˙ ≡ Pµσαµα˙ or alternatively |
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σ¯µαα˙ Pαα˙ . |
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(B.69) |
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Using (B.69), express the Lorentz transformation Lµ ν in the vector representation |
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in terms of M −1T |
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and σ matrices. |
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Making these transformations infinitesimal : |
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Lµ ν = δµν + ωµ ν |
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M −1T |
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α˙ |
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(M ) |
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show that |
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M −1T α |
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β = δαβ + ωµν (σνµ)α |
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(B.70) |
Exercise 6 Show that the Majorana condition (B.59) is equivalent to the condition (cf. (B.32) and (B.37) in four dimensions)
Ψ = Ψ |
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(B.71) |
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CΨ , |
¯ † 0 where Ψ = Ψ Γ .
Hints: Using (B.59) and C2 = −1, we have
CΨ¯ T = CΓ0T Ψ = CΓ0T BΨ = −Γ0CBΨ = −Γ0C2Γ0Ψ = Ψ .