Bispinor

In, a bispinor is an object with four complex components which transform in a specific way under : specifically, a bispinor is an element of a 4-dimensional  considered as a (½,0)⊕(0,½)  of the. Bispinors are, for example, used to describe relativistic spin-½ wave functions.

In the, a bispinor
 * $$\psi=\left(\begin{array}{c}\psi_L\\ \psi_R\end{array}\right)$$

consists of two (two-component) Weyl spinors $$\psi_L$$ and $$\psi_R$$ which transform, correspondingly, under (½,0) and (0,½) representations of the $$SO(1,3)$$ group (the Lorentz group without s). Under parity transformation the Weyl spinors transform into each other.

The Dirac bispinor is connected with the Weyl bispinor by a unitary transformation to the ,

\psi\rightarrow{1\over\sqrt2}\left[ \begin{array}{cc}1&1\\-1&1 \end{array} \right]\psi= {1\over\sqrt2}\left(\begin{array}{c}\psi_R+\psi_L\\ \psi_R-\psi_L \end{array}\right). $$ The Dirac basis is the one most widely used in the literature.

Expressions for Lorentz transformations of bispinors
A bispinor field $$\psi(x)$$ transforms according to the rule
 * $$\psi^a(x) \to {\psi^\prime}^a(x^\prime) = S[\Lambda]^a_b \psi^b(\Lambda^{-1}x^\prime) = S[\Lambda]^a_b \psi^b(x)$$

where $$\Lambda$$ is a. Here the coordinates of physical points are transformed according to $$x^\prime = \Lambda x$$, while $$S$$, a matrix, is an element of the spinor representation (for spin $1/2$) of the Lorentz group.

In the Weyl basis, explicit transformation matrices for a boost $$\Lambda_{boost}$$ and for a rotation $$\Lambda_{rot}$$ are the following:
 * $$S[\Lambda_{boost}] = \left(

\begin{array}{cc} e^{+\chi\cdot\sigma / 2}&0\\ 0&e^{-\chi\cdot\sigma / 2} \end{array} \right)$$
 * $$S[\Lambda_{rot}] = \left(

\begin{array}{cc} e^{+i\phi\cdot\sigma / 2}&0\\ 0&e^{+i\phi\cdot\sigma / 2} \end{array} \right)$$

Here $$\chi$$ is the boost parameter, and $$\phi^i$$ represents rotation around the $$x^i$$ axis. $$\sigma_i$$ are the. The exponential is the, in this case the defined by putting the matrix into the usual power series for the exponential function.

Properties
A of bispinors can be reduced to five irreducible (under the Lorentz group) objects: where $$\bar{\psi}\equiv\psi^\dagger\gamma^0$$ and $$\{\gamma^\mu,\gamma^5\}$$ are the.
 * , $$\bar{\psi}\psi$$ ;
 * , $$\bar{\psi}\gamma^5\psi$$ ;
 * , $$\bar{\psi}\gamma^\mu\psi$$ ;
 * , $$\bar{\psi}\gamma^\mu\gamma^5\psi$$ ;
 * , $$\bar{\psi}(\gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu)\psi$$ ,

A suitable Lagrangian for the relativistic spin-½ field can be built out of these, and is given as

\mathcal{L}={i\over2}\left( \bar{\psi}\gamma^\mu\partial_\mu\psi-\partial_\mu\bar{\psi}\gamma^\mu\psi\right)-m\bar{\psi}\psi\;. $$ The can be derived from this Lagrangian by using the.

Introduction
This outline describes one type of bispinors as elements of a particular 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 over  as described in the article. Language and terminology is used as in. The only property of Clifford algebras that is essential for the presentation is the defining property given in $$ below. The are labeled $so(3;1)$.

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

For reference, the commutation relations of $C^{4}$ are

with the spacetime metric $so(3;1)$.

The gamma matrices
Let γμ denote a set of four 4-dimensional gamma matrices, here called the Dirac matrices. The Dirac matrices satisfy

where $η = diag(−1,1,1,1)$ is the, ${, }$ is a $I_{4}$ unit matrix, and $4×4$ is the spacetime metric with signature (+,-,-,-). This is the defining condition for a generating set of a. Further basis elements $η^{μν}$ of the Clifford algebra are given by

Only six of the matrices $V_{γ} = span{γ^{μ}}|undefined$ are linearly independent. This follows directly from their definition since $SO(3;1)^{+}$. They act on the subspace $σ^{μν}$ the $σ^{μν}$ span in the, 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 $σ^{μν} =−σ^{νμ}$ on the $V_{γ}$, and the linear subspace $γ^{μ}$ they span in $π_{γ}(M^{μν})$, given by

The last equality in $$, which follows from $$ and the property $$ of the gamma matrices, shows that the $Σ_{γ}^{μν}$ constitute a representation of $V_{γ}$ since the s in $$ are exactly those of $σ^{μν}$. The action of $γ^{μ}$ can either be thought of as six-dimensional matrices $π_{γ}:so(3;1)→gl(

V_{γ}

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

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

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

Bispinors introduced
Now introduce any 4-dimensional complex vector space U where the γμ act by matrix multiplication. Here $so(3;1)$ will do nicely. Let $π(M^{μν})$ 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 $σ^{μν}$, the induced map

according to either is a representation or a  of $M_{n}(C)$. 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 $σ^{μν}$ in order to obtain the spin representation $$. One such choice, appropriate for the, is

where the $σ^{ρσ}$ are the. In this representation of the Clifford algebra generators, the $π(M^{μν}) =

σ^{μν}$ become

This representation is manifestly not irreducible, since the matrices are all. 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 $γ^{μ}$ representation, i.e. a bispinor representation. Now using the recipe of exponentiation of the Lie algebra representation to obtain a representation of $σ^{μν}$,

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

for a bispinor field. Here, the upper component corresponds to a right. To include in this formalism, one sets

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

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

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

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

The Dirac algebra
The representation $$ on bispinors will induce a representation of $2π$ on $4π$, 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. One of the consequences is the decomposition of the bilinear forms on $-I$. This decomposition hints how to couple any bispinor field with other fields in a Lagrangian to yield s.