User:YohanN7/Bispinor

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 $$ 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 $C^{4}$, and its elements will be bispinors.

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

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

where $I_{4}$ 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 $V_{γ} = span{γμ }$ of the Clifford algebra are given by

Only six of the matrices $SO(3;1)^{+}$ are linearly independent. This follows directly from their definition since $σ^{μν}$. They act on the subspace $σ^{μν}$ the $σ^{μν} =−σ^{νμ}$ span in the passive sense, according to

In $$, the second equality follows from property $$ of the Clifford algebra.

Lie algebra embedding of so(3;1) in Cℓ4(C)
Now define an action of $V_{γ}$ on the $γ^{μ}$, and the linear subspace $π_{γ}(M^{μν})$ they span in $Σ_{γ}^{μν}$, given by

The last equality in $$, which follows from $$ and the property $$ of the gamma matrices, shows that the $V_{γ}$ constitute a representation of $σ^{μν}$ since the commutation relations in $$ are exactly those of $γ^{μ}$. The action of $π_{γ}:so(3;1)→gl(

V_{γ}

); M^{μν} → Σ^{μν}$ can be either be thought of as 6-dimensional matrices $π(M^{μν})$ multiplying the basis vectors $Σ^{μν}$, since the space in $so(3;1)$ spanned by the $π_{γ}$ is 6-dimensional, or it can be thought of as the action by commutation on the $γμ$. In the following, $V_{γ}$

The $γ^{μ}$ and the $so(3;1)$ are both (disjoint) subsets of the basis elements of Cℓ4(C), generated by the 4-dimensional Dirac matrices $σ^{μν}$ in 4 spacetime dimensions. The Lie algebra of $V_{σ} ⊂ Cℓ_{4}(C)$ is thus embedded in Cℓ4(C) by $Cℓ_{4}(C) ≈

M^{n}_{C}$ as the real subspace of Cℓ4(C) spanned by the $σ^{μν}$. For a full description of the remaining basis elements other than $so(3;1)$ and $so(3;1)$ of the Clifford algebra, please see the article Dirac algebra.

Bispinors introduced
Now introduce any 4-dimensional complex vector space U where the γμ act by matrix multiplication. Here $π(M^{μν})$ will do nicely. Let $Σ^{μν}$ be a Lorentz transformation and define the action of the Lorentz group on U to be


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

Since the $σ^{μν}$ according to $$ constitute a representation of $M_{n}(C)$, the induced map

according to general theory either is a representation or a projective representation of $σ^{μν}$. 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 $σ^{ρσ}$ in order to obtain the spin representation $$. One such choice, appropriate for the ultra-relativistic limit, is

where the $π(M^{μν}) =

σ^{μν}$ are the Pauli matrices. In this representation of the Clifford algebra generators, the $γ^{μ}$ become

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 $σ^{μν}$ representation, i.e. a bispinor representation. Now using the recipe of exponentiation of the Lie algebra representation to obtain a representation of $γ^{μ}$,

a projective 2-valued representation is obtained. Here $so(3;1)$ is a vector of rotation parameters with $π$, and $σ^{μν}$ is a vector of boost parameters. With the conventions used here one may write

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

as representative for $γ^{μ}$. It seen that the representation is irreducible when space parity inversion included.

An example
Let $σ^{μν}$ so that $$ generates a rotation around the z-axis by an angle of $U = C^{4}$. Then $Λ = e^{ω_{μν}M^{μν}}|undefined$ but $σ^{μν}$. Here, $$ denotes the identity element. If $so(3;1)$ is chosen instead, then still $SO(3;1)^{+}$, but now $γμ$.

This illustrates the double valued nature of a spin representation. The identity in $σ_{i}$ gets mapped into either $σ^{μν}$ or $(1⁄2,0)⊕(0,1⁄2)$ depending on the choice of Lie algebra element to represent it. In the first case, one can speculate that a rotation of an angle $SO(3;1)^{+}$ will turn a bispinor into minus itself, and that it requires a $φ$ rotation to rotate a bispinor back into itself. What really happens is that the identity in $0 ≤ φ^{i} ≤2π$ is mapped to $χ$ in $P = diag(1,−1,−1,−1)$ with an unfortunate choice of $$.

It is impossible to continuously choose $$ for all $X=2πM^{12}$ so that $$ is a continuous representation. Suppose that one defines $$ along a loop in $2π$ such that $Λ = e^{iX} = I ∈ SO(3;1)^{+}$. This is a closed loop in $e^{iπ(X)} = -I ∈ GL(U)$, i.e. rotations ranging from 0 to $X = 0$ around the z-axis under the exponential mapping, but it is only "half"" a loop in $Λ = e^{iX} = I ∈ SO(3;1)^{+}$, ending at $e^{iπ(X)} = I ∈ GL(U)$. In addition, the value of $SO(3;1)^{+}$ is ambiguous, since $-I ∈ GL(U)$ and $I ∈ GL(U)$ gives different values for $2π$.

The Dirac algebra
The representation $$ on bispinors will induce a representation of $4π$ on $SO(3;1)^{+}$, 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 $-I$ representations, is described in the article on Dirac algebra. One of the consequences is the decomposition of the bilinear forms on U×U}. This decomposition hints how to couple any bispinor field with other fields in a Lagrangian to yield Lorentz scalar's.

The 4-Vector representation of SO(3;1)+ in Cℓn(C)
General results in finite dimensional representation theory also show that the induced action on $GL(U)$, given explicitly by

is a representation of $g ∈ SO(3;1)^{+}$. 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 γμ form part of the basis for End(U). Therefore, the corresponding map for the γμ is

Claim: The space $X(t)=2πtM^{12}, 0 ≤ t ≤ 1$ 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 $$ holds. Proof of claim

In $S$ it is asserted, as a guess, that the representation on End(U) of $2π$ reduces to the 4-vector representation on the space spanned by the $GL(U)$. Now use the relationship Adexp(X) = exp(ad(X)) between ad and Ad and assume that the $-I$ 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}}]^\rho_\sigma \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}}]^\rho_\sigma\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}]^\rho_\sigma\gamma^\sigma \overset{def}{\Leftrightarrow}

[-i\omega_{\mu\nu}\sigma^{\mu\nu},\gamma^\rho] \overset{?}{=} [i\omega_{\mu\nu}M^{\mu\nu}]^\rho_\sigma\gamma^\sigma\overset{linearity}{\Leftrightarrow}$$


 * $$ -[\sigma^{\mu\nu}, \gamma^\rho] \overset{?}{=} [M^{\mu\nu}]^\rho_\sigma\gamma^\sigma

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

By inspection of $$,
 * $$ (M^{\mu\nu})^{\rho}_{\sigma}\gamma^\sigma = i(\eta^{\mu\rho}\delta^\nu_\sigma - \eta^{\nu\rho}\delta^\mu_\sigma)\gamma^\sigma = i\eta^{\mu\rho}\gamma^\nu - i\eta^{\nu\rho}\gamma^\mu,$$

so the assertion is proved for $I ∈ SO(3;1)$ small.

This means that the subspace $t = 0$ is mapped into itself, and further that there is no proper subspace of $t = 2π$ that is mapped into itself under the action of $I ∈ SO(3;1)$.

The tensor representation of SO(3;1)+ in Cℓn(C)
Let $SO(3;1)^{+}$ be a Lorentz transformation, and let $End(U)$ denote the action of Λ on U and consider how $SO(3;1)^{+}$ transform under the induced action on $End(U) ≈ Cℓ_{n}(C)$.

where the known transformation rule of the $SO(3;1)^{+}$ given in $$ has been used. Thus the 6-dimensinal space $SO(3;1)^{+}$ is a representation space of a $V_{γ}$ tensor representation.

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

respectively. In the latter equation there are $SO(3;1)^{+}$ 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 $SO(3;1)^{+}$, one may in this formalism write

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 $γ^{μ}$ and $ω_{μν}$ anticommute since τ and η are different in $$ (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 $ω^{μν}$. The sign of each term depends on the number n of transpositions of the indices required to obtain the order $V_{γ} ⊂ Cℓ_{n}(C)$. For n odd the sign is − and for n even the sign is +. This is precisely captured by the totally antisymmetric quantity

Using this, and defining

then corresponding to the chosen order, $$ becomes

Using the same technique for the rank 3 objects one obtains

Proof By considering a term in $$ one observes that for $V_{γ}$



\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^\mu_0)\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 $SO(3;1)$. 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

One finds, using $X$

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 $Λ = e^{{{sfrac|i|2}}ω_{μν}M^{μν}}|undefined$ that under parity transforms as $S(Λ)$, $σ^{μν}$. In general, the transformation of a product of gamma matrcies is even or odd depending on how many indices are space indices. Details The $V_{σ} ⊂ End(U)$ 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

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

Now consider the effect of β on the space $γ^{μ}$:


 * $$\beta^{-1}\gamma_5\gamma^{\nu}\beta = \beta\gamma_5\beta\beta\gamma^{\nu}\beta = \gamma_5\gamma^{\mu\dagger} = -\gamma_5\gamma^0, \mu = 0, +\gamma_5\gamma^i, \mu = i.$$

This behavior warrants the terminology pseudovector for $V_{σ} = span{σ^{μν}}|undefined$.

The pseudoscalar representation of O(3;1)+ in Cℓn(C)
For $(1,0)⊕(0,1)$ one obtains

for the action of the space inversion matrix β. The behavior of $4! = 24$ under proper orthocronous Lorentz transformations is simple. One has

Proof This can be seen by first writing out explicitly $$\{\gamma^0,\gamma_5\} = \gamma_0i\gamma_0\gamma_1\gamma_2\gamma_3 + i\gamma_0\gamma_1\gamma_2\gamma_3\gamma_0 = 0,$$ $$\{\gamma^1,\gamma_5\} = \gamma_1i\gamma_0\gamma_1\gamma_2\gamma_3 + i\gamma_0\gamma_1\gamma_2\gamma_3\gamma_1 = 0,$$ $$\{\gamma^2,\gamma_5\} = \gamma_2i\gamma_0\gamma_1\gamma_2\gamma_3 + i\gamma_0\gamma_1\gamma_2\gamma_3\gamma_2 = 0,$$ $$\{\gamma^3,\gamma_5\} = \gamma_3i\gamma_0\gamma_1\gamma_2\gamma_3 + i\gamma_0\gamma_1\gamma_2\gamma_3\gamma_3 = 0,$$ and then counting transpositions in each term required to make the two identical factors adjacent and then using the defining property $I$. 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 $σ^{μν}$ to the spot to the immediate left of the rightmost $γτ$. This introduces a minus sign, so the second term cancels the first. The rest of the cases are handled analogously.

For $γη$ the commutator with $(0,1,2,3)$ becomes, using $X$

Using the exponential expansion of S in powers of $(0,1,2,3)$ and $X$ the $i ∈ {1,2,3}$ transforms according to

and the space $η ∈ {0,1,2,3}$ is thus a representation space of the 1-dimensional pseudoscalar representation.

The pseudo-vector representation of O(3;1)+ in Cℓn(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 $x^{μ}$ are then found to be either

for proper transformations or

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

Summary
The space $x^{0}→x^{0}$ is together with the transformations $x^{i}→x^{i}$ 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 $S$,


 * $$S(P) = \beta.$$

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


 * $$C\ell(4,\mathbf{C}) = 1 \oplus V_\gamma \oplus V_\sigma \oplus \gamma_5V_\gamma \oplus \gamma_51$$

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 $σ^{μν}$, but instead transforming under the ⊕ representation, just like U in this article.