User:YohanN7/Bispinor

A bispinor representation

Introduction

This outline describes one type of bispinors as elements of a particular representation space of the (½,0)⊕ (0,½) representation of the Lorentz group. This representation space is related to, but not identical to, the (½,0)⊕ (0,½) representation space contained in the Clifford algebra over Minkowski spacetime as described in the article Spinors. Language and terminology is used as in Representation theory of the Lorentz group. The only property of Clifford algebras that is essential for the presentation is the defining property given in D1 below. The basis elements of so(3;1) are labeled M^μν.

A representation of the Lie algebra so(3;1) of the Lorentz group O(3;1) will emerge among matrices that will be chosen as a basis (as a vector space) of the complex Clifford algebra over spacetime. These 4×4 matrices are then exponentiated yielding a representation of SO(3;1)⁺. This representation, that turns out to be a (⁠1/2⁠,0)⊕(0,⁠1/2⁠) representation, will act on an arbitrary 4-dimensional complex vector space, which will simply be taken as C⁴, and its elements will be bispinors.

For reference, the commutation relations of so(3;1) are

$[M^{\mu \nu },M^{\rho \sigma }]=i(\eta ^{\sigma \mu }M^{\rho \nu }+\eta ^{\nu \sigma }M^{\mu \rho }-\eta ^{\rho \mu }M^{\sigma \nu }-\eta ^{\nu \rho }M^{\mu \sigma })$ (M1)

with the spacetime metric η = diag(−1,1,1,1).

The Gamma Matrices

Let γ^μ denote a set of four 4-dimensional Gamma matrices, here called the Dirac matrices. The Dirac matrices satisfy

$\{\gamma ^{\mu },\gamma ^{\nu }\}=2\eta ^{\mu \nu }I_{4},$ (D1)

where I₄ is a 4×4 unit matrix, and η^μν is the spacetime metric with signature (-,+,+,+). This is the defining condition for a generating set of a Clifford algebra. Further basis elements σ^μν of the Clifford algebra are given by

$\sigma ^{\mu \nu }=-{\frac {i}{4}}[\gamma ^{\mu }\gamma ^{\nu }-\gamma ^{\nu }\gamma ^{\mu }].$ (C1)

Only six of the matrices σ^μν are linearly independent. This follows directly from their definition since σ^μν =−σ^νμ. They act on the subspace V_γ the γ^μ span in the passive sense, according to

$[\sigma ^{\mu \nu },\gamma ^{\rho }]=-i\gamma ^{\mu }\eta ^{\nu \rho }+i\gamma ^{\nu }\eta ^{\mu \rho }.$ (C2)

In (C2), the second equality follows from property (D1) of the Clifford algebra.

Lie algebra embedding of so(3;1) in Cℓ₄(C)

Now define an action of so(3;1) on the σ^μν, and the linear subspace V_σ ⊂ Cℓ₄(C) they span in Cℓ₄(C) ≈ Mⁿ_C, given by

$\pi (M^{\mu \nu })(\sigma ^{\rho \sigma })=[\sigma ^{\mu \nu },\sigma ^{\rho \sigma }]=i(\eta ^{\sigma \mu }\sigma ^{\rho \nu }+\eta ^{\sigma \nu }\sigma ^{\rho \mu }-\eta ^{\mu \rho }\sigma ^{\nu \sigma }-\eta ^{\nu \rho }\sigma ^{\mu \sigma }),$. (C4)

The last equality in (C4), which follows from (C2) and the property (D1) of the gamma matrices, shows that the σ^μν constitute a representation of so(3;1) since the commutation relations in (C4) are exactly those of so(3;1). The action of π(M^μν) can be either be thought of as 6-dimensional matrices Σ^μν multiplying the basis vectors σ^μν, since the space in M_n(C) spanned by the σ^μν is 6-dimensional, or it can be thought of as the action by commutation on the σ^ρσ. In the following, π(M^μν) = σ^μν

The γ^μ and the σ^μν are both (disjoint) subsets of the basis elements of Cℓ₄(C), generated by the 4-dimensional Dirac matrices γ^μ in 4 spacetime dimensions. The Lie algebra of so(3;1) is thus embedded in Cℓ₄(C) by π as the real subspace of Cℓ₄(C) spanned by the σ^μν. For a full description of the remaining basis elements other than γ^μ and σ^μν of the Clifford algebra, please see the article Dirac algebra.

Bispinors introduced

Now introduce any 4-dimensional complex vector space U where the γ^μ act by matrix multiplication. Here U = C⁴ will do nicely. Let Λ = e^ω_μνMμν be a Lorentz transformation and define the action of the Lorentz group on U to be

$u\rightarrow S(\Lambda )u=e^{i\pi (\omega _{\mu \nu }M^{\mu \nu })}u;\quad u^{\alpha }\rightarrow [e^{\omega _{\mu \nu }\sigma ^{\mu \nu }}]^{\alpha }{}_{\beta }u^{\beta }.$

Since the σ^μν according to (C4) constitute a representation of so(3;1), the induced map

$S:SO(3;1)^{+}\rightarrow \mathrm {GL} (U);\quad \Lambda \rightarrow e^{i\pi (X)};\quad e^{iX}=\Lambda ,\quad X=\omega _{\mu \nu }M^{\mu \nu }\in {\mathfrak {so}}(3;1)$ (C5)

according to general theory either is a representation or a projective representation of SO(3;1)⁺. It will turn out to be a projective representation. The elements of U, when endowed with the transformation rule given by S, are called bispinors or simply spinors.

A choice of Dirac matrices

It remains to choose a set of Dirac matrices γ^μ in order to obtain the spin representation S. One such choice, appropriate for the ultra-relativistic limit, is

${\begin{aligned}\gamma ^{0}&=-i{\biggl (}{\begin{matrix}0&I\\I&0\\\end{matrix}}{\biggr )},\\\gamma ^{i}&=-i{\biggl (}{\begin{matrix}0&\sigma _{i}\\-\sigma _{i}&0\\\end{matrix}}{\biggr )},\quad i=1,2,3\\\end{aligned}}.$ (E1)

where the σ_i are the Pauli matrices. In this representation of the Clifford algebra generators, the σ^μν become

${\begin{aligned}\sigma ^{i0}&={\frac {i}{2}}{\biggl (}{\begin{matrix}\sigma _{i}&0\\0&-\sigma _{i}\\\end{matrix}}{\biggr )},\\\sigma ^{ij}&={\frac {i}{2}}\epsilon _{ijk}{\biggl (}{\begin{matrix}\sigma _{i}&0\\0&\sigma _{i}\\\end{matrix}}{\biggr )}\\\end{aligned}}.$ (E23)

This representation is manifestly not irreducible, since the matrices are all block diagonal. But by irreducibility of the Pauli matrices, the representation cannot be further reduced. Since it is a 4-dimensional, the only possibility is that it is a (⁠1/2⁠,0)⊕(0,⁠1/2⁠) representation, i.e. a bispinor representation. Now using the recipe of exponentiation of the Lie algebra representation to obtain a representation of SO(3;1)⁺,

${\begin{aligned}S(\Lambda _{B})&=e^{i\pi (\phi \cdot \mathbf {J} )}={\biggl (}{\begin{matrix}e^{-{\frac {1}{2}}\chi \cdot \sigma }&0\\0&e^{{\frac {1}{2}}\chi \cdot \sigma }\\\end{matrix}}{\biggr )},\\S(\Lambda _{R})&=e^{i\pi (\chi \cdot \mathbf {K} )}={\biggl (}{\begin{matrix}e^{{\frac {i}{2}}\phi \cdot \sigma }&0\\0&e^{{\frac {i}{2}}\phi \cdot \sigma }\\\end{matrix}}{\biggr )}\\\end{aligned}},$ (E3)

a projective 2-valued representation is obtained. Here φ is a vector of rotation parameters with 0 ≤ φⁱ ≤2π, and χ is a vector of boost parameters. With the conventions used here one may write

$\psi ={\begin{pmatrix}\psi _{R}\\\psi _{L}\end{pmatrix}},$ (E4)

for a bispinor field. Here, the upper component correspond to a right Weyl spinor. To include space parity inversion in this formalism, one sets

Failed to parse (syntax error): {\displaystyle \beta = i\gamma^0 = \biggl(\begin{matrix} 0 & I\\ I & 0\\ \end{matrix}\biggr),\\ } (E5)

as representative for P = diag(1,−1,−1,−1). It seen that the representation is irreducible when space parity inversion included.

An example

Let X=2πM¹² so that X generates a rotation around the z-axis by an angle of 2π. Then Λ = e^iX = I ∈ SO(3;1)⁺ but e^iπ(X) = -I ∈ GL(U). Here, I denotes the identity element. If X = 0 is chosen instead, then still Λ = e^iX = I ∈ SO(3;1)⁺, but now e^iπ(X) = I ∈ GL(U).

This illustrates the double valued nature of a spin representation. The identity in SO(3;1)⁺ gets mapped into either -I ∈ GL(U) or I ∈ GL(U) depending on the choice of Lie algebra element to represent it. In the first case, one can speculate that a rotation of an angle 2π will turn a bispinor into minus itself, and that it requires a 4π rotation to rotate a bispinor back into itself. What really happens is that the identity in SO(3;1)⁺ is mapped to -I in GL(U) with an unfortunate choice of X.

It is impossible to continuously choose X for all g ∈ SO(3;1)⁺ so that S is a continuous representation. Suppose that one defines S along a loop in SO(3;1) such that X(t)=2πtM¹², 0 ≤ t ≤ 1. This is a closed loop in SO(3;1), i.e. rotations ranging from 0 to 2π around the z-axis under the exponential mapping, but it is only "half"" a loop in GL(U), ending at -I. In addition, the value of I ∈ SO(3;1) is ambiguous, since t = 0 and t = 2π gives different values for I ∈ SO(3;1).

The Dirac algebra

The representation S on bispinors will induce a representation of SO(3;1)⁺ on End(U), the set of linear operators on U. This space corresponds to the Clifford algebra itself so that all linear operators on U are elements of the latter. This representation, and how it decomposes as a direct sum of irreducible SO(3;1)⁺ representations, is described in the article on Dirac algebra. One of the consequences is the decomposition of the bilinear forms on {{math|U×U}. This decomposition hints how to couple any bispinor field with other fields in a Lagrangian to yield Lorentz scalar's.

The Dirac algebra

The 4-Vector representation of SO(3;1)⁺ in Cℓ_n(C)

General results in finite dimensional representation theory also show that the induced action on End(U) ≈ Cℓ_n(C), given explicitly by

$\Gamma \rightarrow S(\Lambda )\Gamma S(\Lambda ){-1}=e^{{\frac {i}{2}}\omega _{\mu \nu }\sigma ^{\mu \nu }}\Gamma e^{{\frac {-i}{2}}\omega _{\mu \nu }\sigma ^{\mu \nu }}=e^{ad_{{\frac {i}{2}}\sigma ^{mu\nu }}}(\Gamma ),\quad \Gamma \in End(U)$ (C6)

is a representation of SO(3;1)⁺. This is a bona fide representation of SO(3;1)⁺, i.e., it is not projective. This is a consequence of the Lorentz group being doubly connected. But the γ^μ form part of the basis for End(U). Therefore, the corresponding map for the γ^μ is

$\gamma ^{\rho }\rightarrow S(\Lambda )^{-1}\gamma ^{\rho }S(\Lambda )=e^{{\frac {-i}{2}}\omega _{\mu \nu }\sigma ^{\mu \nu }}\gamma ^{\rho }e^{{\frac {i}{2}}\omega _{\mu \nu }\sigma ^{\mu \nu }}{\overset {?}{=}}\Lambda _{\sigma }{}^{\rho }\gamma ^{\sigma }=[e^{{\frac {-i}{2}}\omega _{\mu \nu }M^{\mu \nu }}]^{\rho }{}_{\sigma }\gamma ^{\sigma }$ (C7)

Claim: The space V_γ endowed with the Lorentz group action defined above is a 4-vector representation of SO(3;1)⁺. This holds if the equality with a question mark in (C7) holds.

Proof of claim

In (C7) it is asserted, as a guess, that the representation on End(U) of SO(3;1)⁺ reduces to the 4-vector representation on the space spanned by the γ^μ. Now use the relationship Ad_exp(X) = exp(ad(X)) between ad and Ad and assume that the ω_μν are small.

$e^{-i\omega _{\mu \nu }\sigma ^{\mu \nu }}\gamma ^{\rho }e^{i\omega _{\mu \nu }\sigma ^{\mu \nu }}{\overset {?}{=}}[e^{i\omega _{\mu \nu }M^{\mu \nu }}]_{\sigma }^{\rho }\gamma ^{\sigma }{\overset {Ad/ad}{\Leftrightarrow }}e^{ad_{-i\omega _{\mu \nu }\sigma ^{\mu \nu }}}(\gamma ^{\rho }){\overset {?}{=}}[e^{i\omega _{\mu \nu }M^{\mu \nu }}]_{\sigma }^{\rho }\gamma ^{\sigma }{\overset {\omega ^{\mu \nu }small}{\Leftrightarrow }}$

$ad_{-i\omega _{\mu \nu }\sigma ^{\mu \nu }}(\gamma ^{\rho }){\overset {?}{=}}[i\omega _{\mu \nu }M^{\mu \nu }]_{\sigma }^{\rho }\gamma ^{\sigma }{\overset {def}{\Leftrightarrow }}[-i\omega _{\mu \nu }\sigma ^{\mu \nu },\gamma ^{\rho }]{\overset {?}{=}}[i\omega _{\mu \nu }M^{\mu \nu }]_{\sigma }^{\rho }\gamma ^{\sigma }{\overset {linearity}{\Leftrightarrow }}$

$-[\sigma ^{\mu \nu },\gamma ^{\rho }]{\overset {?}{=}}[M^{\mu \nu }]_{\sigma }^{\rho }\gamma ^{\sigma }{\overset {(C2)}{\Leftrightarrow }}-i\gamma ^{\mu }\eta ^{\nu \rho }+i\gamma ^{\nu }\eta ^{\mu \rho }=[M^{\mu \nu }]_{\sigma }^{\rho }\gamma ^{\sigma }$

By inspection of ???,

$(M^{\mu \nu })_{\sigma }^{\rho }\gamma ^{\sigma }=i(\eta ^{\mu \rho }\delta _{\sigma }^{\nu }-\eta ^{\nu \rho }\delta _{\sigma }^{\mu })\gamma ^{\sigma }=i\eta ^{\mu \rho }\gamma ^{\nu }-i\eta ^{\nu \rho }\gamma ^{\mu },$

so the assertion is proved for ω^μν small.

This means that the subspace V_γ ⊂ Cℓ_n(C) is mapped into itself, and further that there is no proper subspace of V_γ that is mapped into itself under the action of SO(3;1).

The tensor representation of SO(3;1)+ in Cℓ_n(C)

Let Λ = e^{⁠i/2⁠ω_μνMμν} be a Lorentz transformation, and let S(Λ) denote the action of Λ on U and consider how σ^μν transform under the induced action on V_σ ⊂ End(U).

$S(\Lambda )^{-1}\sigma ^{\mu \nu }S(\Lambda )=S^{-1}(\Lambda )[\gamma ^{\mu }\gamma ^{\nu }-\gamma ^{\nu }\gamma ^{\mu }]S(\Lambda )={\frac {-i}{4}}S^{-1}[\gamma ^{\mu }S(\Lambda )S^{-1}(\Lambda )\gamma ^{\nu }-\gamma ^{\nu }S(\Lambda )S^{-1}(\Lambda )\gamma ^{\mu }]S(\Lambda )={\frac {-i}{4}}(\Lambda _{\mu }^{\rho }\gamma ^{\mu }\Lambda _{\nu }^{\sigma }\gamma ^{\nu }-\Lambda _{\nu }^{\sigma }\gamma ^{\nu }\Lambda _{\mu }^{\rho }\gamma ^{\mu })={\frac {-i}{4}}\Lambda _{\mu }^{\rho }\Lambda _{\nu }^{\sigma }(\gamma ^{\mu }\gamma ^{\nu }-\gamma ^{\nu }\gamma ^{\mu })=\Lambda _{\mu }^{\rho }\Lambda _{\nu }^{\sigma }\sigma ^{\mu \nu },$ (C8)

where the known transformation rule of the γ^μ given in (C7) has been used. Thus the 6-dimensinal space V_σ = span{σ^μν} is a representation space of a (1,0)⊕(0,1) tensor representation.

The full Clifford algebra

For elements of the Dirac algebra, define the antisymmetrization of products of three and four gamma matrices by

$A^{\rho \sigma \tau }=\gamma ^{[\mu }\gamma ^{\sigma }\gamma ^{\tau ]}=\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }-\gamma ^{\rho }\gamma ^{\tau }\gamma ^{\sigma }-\gamma ^{\sigma }\gamma ^{\rho }\gamma ^{\tau }+\gamma ^{\tau }\gamma ^{\rho }\gamma ^{\sigma }+\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{\rho }-\gamma ^{\tau }\gamma ^{\sigma }\gamma ^{\rho },$ (C10)

$P^{\rho \sigma \tau \eta }=\gamma ^{[\mu }\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{\eta ]}=\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{\eta }-\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\eta }\gamma ^{\tau }+\ldots$ (C11)

respectively. In the latter equation there are {{{1}}} terms with a plus or minus sign according to the parity of the permutation taking the indices from the order in the left hand sign to the order appearing in the term. For the σ^μν, one may in this formalism write

$S^{\mu \nu }=\gamma ^{[\mu }\gamma ^{\nu ]}=\gamma ^{\mu }\gamma ^{\nu }-\gamma ^{\nu }\gamma ^{\mu }\ =\sigma ^{\mu \nu }.$ (C12)

In four spacetime dimensions, there are no totally antisymmetric tensors of higher order than four.

A change of basis in the Clifford algebra

Now by observing that γ^τ and γ^η anticommute since τ and η are different in C10 (or the terms would cancel), it is found that all terms can be brought into a particular order of choice with respect to the indices. This order is chosen to be (0,1,2,3). The sign of each term depends on the number n of transpositions of the indices required to obtain the order (0,1,2,3). For n odd the sign is − and for n even the sign is +. This is precisely captured by the totally antisymmetric quantity

$\epsilon ^{\rho \sigma \tau \eta }=Sgn(P):P\in S_{4},P(0,1,2,3)=(\rho ,\sigma ,\tau ,\eta ).$ (C25)

Using this, and defining

$\gamma _{5}=-i\gamma ^{0}\gamma ^{1}\gamma ^{2}\gamma ^{3}$ (C26)

then corresponding to the chosen order, C10 becomes

$P^{\rho \sigma \tau \eta }=4!i\epsilon ^{\rho \sigma \tau \eta }\gamma _{5}.$ (C27)

Using the same technique for the rank 3 objects one obtains

$A^{\rho \sigma \tau }=3!i\epsilon ^{\rho \sigma \tau \eta }\gamma _{5}\gamma _{\eta }.$ (C28)

Proof

By considering a term in C21 one observes that for i ∈ {1,2,3}

$\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }=\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }(\gamma ^{i})^{2}=(\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{i})g^{\mu i}\gamma _{\mu }=(\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{i})\delta ^{\mu i}\gamma _{\mu }=(\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{i})\gamma _{i}$

while

$\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }=-\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }(\gamma ^{0})^{2}=-(\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{0})g^{\mu 0}\gamma _{\mu }=-(\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{0})(-\delta _{0}^{\mu })\gamma _{\mu }=(\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{0})\gamma _{0}$

so

$\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\eta }=(\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{\eta })\gamma _{\eta }$

holds for all η ∈ {0,1,2,3}. But

$\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{\eta }=\epsilon ^{\rho \sigma \tau \eta }(\gamma ^{0}\gamma ^{1}\gamma ^{2}\gamma ^{3})=i\epsilon ^{\rho \sigma \tau \eta }\gamma _{5},$

so that

$A^{\rho \sigma \tau }=\gamma ^{\rho }\gamma ^{\sigma }\gamma ^{\tau }-\gamma ^{\rho }\gamma ^{\tau }\gamma ^{\sigma }-\gamma ^{\sigma }\gamma ^{\rho }\gamma ^{\tau }+\gamma ^{\tau }\gamma ^{\rho }\gamma ^{\sigma }+\gamma ^{\sigma }\gamma ^{\tau }\gamma ^{\rho }-\gamma ^{\tau }\gamma ^{\sigma }\gamma ^{\rho }=$

$i\epsilon ^{\rho \sigma \tau \eta }\gamma _{5}\gamma _{\eta }-i\epsilon ^{\rho \tau \sigma \eta }\gamma _{5}\gamma _{\eta }-i\epsilon ^{\sigma \rho \tau \eta }\gamma _{5}\gamma _{\eta }$

$+i\epsilon ^{\tau \rho \sigma \eta }\gamma _{5}\gamma _{\eta }+i\epsilon ^{\sigma \tau \rho \eta }\gamma _{5}\gamma _{\eta }-i\epsilon ^{\tau \sigma \rho \eta }\gamma _{5}\gamma _{\eta },$

which, by permuting indices in the Levi-Civita symbol (and counting transpositions), finally becomes

$A^{\rho \sigma \tau }=3!i\epsilon ^{\rho \sigma \tau \eta }\gamma _{5}\gamma _{\eta }$

Space Inversion

Space inversion (or parity) can be included in this formalism by setting

$\pi (P)=i\gamma ^{0}\equiv \beta .$ (C30)

One finds, using (D1)

$\{\beta ,\gamma ^{i}\}=i\{\gamma ^{0},\gamma ^{i}\}=i(\gamma ^{0}\gamma ^{i}+\gamma ^{i}\gamma ^{0})=0\Rightarrow \beta \gamma ^{i}\beta ^{-1}=-\gamma ^{i},$ (C31)

$[\beta ,\gamma ^{0}]=i[\gamma ^{0},\gamma ^{0}]=i\gamma ^{0}\gamma ^{i}-\gamma ^{i}i\gamma ^{0}=0\Rightarrow \beta \gamma ^{0}\beta ^{-1}=\gamma ^{0}$ (C32)

These properties of β are just the right ones for it to be a representative of space inversion as is seen by comparison with an ordinary 4-vector x^μ that under parity transforms as x⁰→x⁰, xⁱ→xⁱ. In general, the transformation of a product of gamma matrcies is even or odd depending on how many indices are space indices.

Details

The σ^μν transform according to

$\beta ^{-1}(\gamma ^{\mu }\gamma ^{\nu }-\gamma ^{\mu }\gamma ^{\nu })\beta =\beta ^{-1}\gamma ^{\mu }\gamma ^{\nu }\beta -\beta ^{-1}\gamma ^{\nu }\gamma ^{\mu }\beta =[\beta ^{-1}\gamma ^{\mu }\beta ][\beta ^{-1}\gamma ^{\nu }\beta ]-[\beta ^{-1}\gamma ^{\nu }\beta ][\beta ^{-1}\gamma ^{\mu }\beta ]=\pm (\mu )\gamma ^{\mu }\pm (\nu )\gamma ^{\nu }-\pm (\nu )\gamma ^{\nu }\pm (\mu )\gamma ^{\mu }$

For μ and ν spacelike, this becomes

$(-\gamma ^{\mu })(-\gamma ^{\nu })-(-\gamma ^{\nu })(-\gamma ^{\mu })=\sigma ^{\mu \nu }.$

For μ = 0 and ν spacelike, this becomes

$(\gamma ^{\mu })(-\gamma ^{\nu })-(-\gamma ^{\nu })(\gamma ^{\mu })=-\sigma ^{\mu \nu }.$

For μ and ν both zero, one obtains

$(\gamma ^{\mu })(\gamma ^{\nu })-(\gamma ^{\nu })(\gamma ^{\mu })=\sigma ^{\mu \nu }.$

Space inversion commutes with the generators of rotation, but the boost operators, anticommute

$[\beta ,\sigma ^{ij}]=i\gamma ^{0}\gamma ^{i}\gamma ^{j}-i\gamma ^{0}\gamma ^{j}\gamma ^{i}=-\gamma ^{i}i\gamma ^{0}\gamma ^{j}+\gamma ^{j}i\gamma ^{0}\gamma ^{i}=\gamma ^{i}\gamma ^{j}i\gamma ^{0}-\gamma ^{j}\gamma ^{i}i\gamma ^{0}=[\sigma ^{ij},i\gamma ^{0}]=-[\beta ,\sigma ^{ij}]\Rightarrow [\beta ,\sigma ^{ij}]=0\Rightarrow \beta \sigma ^{ij}\beta ^{-1}=\sigma ^{ij},$ (C33)

$\{\beta ,\sigma ^{0j}\}=i\gamma ^{0}\gamma ^{0}\gamma ^{j}+i\gamma ^{0}\gamma ^{j}\gamma ^{0}=i(-\gamma ^{j}-\gamma ^{j}\gamma ^{0}\gamma ^{0})=i(-\gamma ^{j}+\gamma ^{j})=0\Rightarrow \beta \sigma ^{0j}\beta ^{-1}=-\sigma ^{0j}.$ (C34)

This is the correct behavior, since in the tandard representation with three-dimensional notation,

$\mathrm {P} \mathbf {J} \mathrm {P} ^{-1}=\mathbf {J} ,\quad \mathrm {P} \mathbf {K} \mathrm {P} ^{-1}=-\mathbf {K} .$ (C35)

Now consider the effect of β on the space V_γ5:

$\beta ^{-1}\gamma _{5}\gamma ^{\nu }\beta =\beta \gamma _{5}\beta \beta \gamma ^{\nu }\beta =\gamma _{5}\gamma ^{\mu \dagger }=-\gamma _{5}\gamma ^{0},\mu =0,+\gamma _{5}\gamma ^{i},\mu =i.$

This behavior warrants the terminology pseudovector for γ₅γ^μ.

The pseudoscalar representation of O(3;1)⁺ in Cℓ_n(C)

For γ₅ one obtains

$\beta \gamma _{5}\beta ^{-1}=-i\beta \gamma ^{0}\gamma ^{1}\gamma ^{2}\gamma ^{3}\beta ^{-1}=-i\beta \gamma ^{0}\beta ^{-1}\beta \gamma ^{1}\beta ^{-1}\beta \gamma ^{2}\beta ^{-1}\beta \gamma ^{3}\beta ^{-1}=+i\gamma ^{0}\gamma ^{1}\gamma ^{2}\gamma ^{3}=-\gamma _{5}$ (C33)

for the action of the space inversion matrix β. The behavior of γ₅ under proper orthocronous Lorentz transformations is simple. One has

$\{\gamma ^{\mu },\gamma _{5}\}=0.$ (C34)

Proof

This can be seen by first writing out explicitly $\{\gamma ^{0},\gamma _{5}\}=\gamma _{0}i\gamma _{0}\gamma _{1}\gamma _{2}\gamma _{3}+i\gamma _{0}\gamma _{1}\gamma _{2}\gamma _{3}\gamma _{0}=0,$ $\{\gamma ^{1},\gamma _{5}\}=\gamma _{1}i\gamma _{0}\gamma _{1}\gamma _{2}\gamma _{3}+i\gamma _{0}\gamma _{1}\gamma _{2}\gamma _{3}\gamma _{1}=0,$ $\{\gamma ^{2},\gamma _{5}\}=\gamma _{2}i\gamma _{0}\gamma _{1}\gamma _{2}\gamma _{3}+i\gamma _{0}\gamma _{1}\gamma _{2}\gamma _{3}\gamma _{2}=0,$ $\{\gamma ^{3},\gamma _{5}\}=\gamma _{3}i\gamma _{0}\gamma _{1}\gamma _{2}\gamma _{3}+i\gamma _{0}\gamma _{1}\gamma _{2}\gamma _{3}\gamma _{3}=0,$ and then counting transpositions in each term required to make the two identical factors adjacent and then using the defining property D1. In each case the two terms will cancel. For example, three transpositions are required in the second term in the first line to bring the leftmost γ⁰ to the spot to the immediate left of the rightmost γ⁰. This introduces a minus sign, so the second term cancels the first. The rest of the cases are handled analogously.

For σ^μν the commutator with γ₅ becomes, using (C34)

$[\sigma ^{\mu \nu },\gamma _{5}]=[[\gamma ^{\mu },\gamma ^{\nu }],\gamma _{5}]]=[\gamma ^{\mu }\gamma ^{\nu }-\gamma ^{\nu }\gamma ^{\mu },\gamma _{5}]=(\gamma ^{\mu }\gamma ^{\nu }\gamma _{5}-\gamma ^{\nu }\gamma ^{\mu }\gamma _{5}-\gamma _{5}\gamma ^{\mu }\gamma ^{\nu }+\gamma _{5}\gamma ^{\nu }\gamma ^{\mu })=(\gamma _{5}\gamma ^{\mu }\gamma ^{\nu }-\gamma _{5}\gamma ^{\nu }\gamma ^{\mu }-\gamma _{5}\gamma ^{\mu }\gamma ^{\nu }+\gamma _{5}\gamma ^{\nu }\gamma ^{\mu })=0.$ (C35)

Using the exponential expansion of S in powers of σ^μν and (C35) the γ₅ transforms according to

$S(\Lambda )^{-1}\gamma _{5}S(\Lambda )=\mathrm {det} (\Lambda )\gamma _{5}$ (C36)

and the space 1_γ5 = span{γ₅} is thus a representation space of the 1-dimensional pseudoscalar representation.

The pseudo-vector representation of O(3;1)⁺ in Cℓ_n(C)

Every orthocronous Lorentz transformation can be written wither as Λ or PΛ, where P is space inversion and Λ is orthocronous and proper. The Lorentz transformation properties of γ₅γ^μ are then found to be either

$S(\Lambda )^{-1}\gamma _{5}\gamma ^{\mu }S(\Lambda )=S(\Lambda )^{-1}\gamma _{5}S(\Lambda )S^{-1}(\Lambda )\gamma ^{\mu }S(\Lambda )=\gamma _{5}S^{-1}(\Lambda )\gamma ^{\mu }S(\Lambda )=\gamma _{5}\Lambda _{\mu }{}^{\rho }\gamma ^{\mu }$ (A1)

for proper transformations or

$\beta ^{-1}\gamma _{5}\gamma ^{\mu }\beta ={\begin{cases}-\gamma _{5}\gamma ^{0}&\mu =0\\+\gamma _{5}\gamma ^{i}&\mu =i\end{cases}}$ (A2)

for space inversion. These may be put together in a single equation equation in the context of bilinear covariants, see below.

Summary

The space U≈C⁴ is together with the transformations S(Λ) given by

$S(\Lambda )=e^{{\frac {i}{2}}\omega _{\mu \nu }\sigma ^{\mu \nu }}:\Lambda =e^{{\frac {i}{2}}\omega _{\mu \nu }M^{\mu \nu }}:M\in so(3,1)$,

or, if Λ is space inversion P,

$S(P)=\beta .$

a space of bispinors. Moreover, the Clifford algebra Cℓ_n(C) decomposes as a vector space according to

$C\ell (4,\mathbf {C} )=1\oplus V_{\gamma }\oplus V_{\sigma }\oplus \gamma _{5}V_{\gamma }\oplus \gamma _{5}1$

where the elements transform as:

• 1-dimensional scalars
• 4-dimensional vectors
• 6-dimensional tensors
• 4-dimensional pseudovectors
• 1-dimensional pseudoscalars

For a description of another type of bispinors, please see the Spinor article. The representation space corresponding to that description sits inside the Clifford algebra and is thus a linear space of matrices, much like the space V_γ, but instead transforming under the ()⊕() representation, just like U in this article.