Introduction

This outline describes one type of bispinors as elements of a particular representation space of the (1/2, 0) ⊕ (0, 1/2) representation of the Lorentz group. This representation space is related to, but not identical to, the (1/2, 0) ⊕ (0, 1/2) 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

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

(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

${\displaystyle \left\{\gamma ^{\mu },\gamma ^{\nu }\right\}=2\eta ^{\mu \nu }I_{4},}$[4]

(D1)

where { , } is the anticommutator, 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

${\displaystyle \sigma ^{\mu \nu }=-{\frac {i}{4}}\left[\gamma ^{\mu }\gamma ^{\nu }-\gamma ^{\nu }\gamma ^{\mu }\right].}$[5]

(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

${\displaystyle \left[\sigma ^{\mu \nu },\gamma ^{\rho }\right]=-i\gamma ^{\mu }\eta ^{\nu \rho }+i\gamma ^{\nu }\eta ^{\mu \rho }.}$[6]

(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

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

(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 either be thought of as six-dimensional matrices Σ^μν multiplying the basis vectors σ^μν, since the space in M_n(C) spanned by the σ^μν is six-dimensional, or 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 four-dimensional Dirac matrices γ^μ in four 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

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

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

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

(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 ultrarelativistic limit, is

${\displaystyle {\begin{aligned}\gamma ^{0}&=-i{\biggl (}{\begin{matrix}0&I\\I&0\\\end{matrix}}{\biggr )},\\\gamma ^{i}&=-i{\biggl (}{\begin{matrix}0&\sigma _{i}\\-\sigma _{i}&0\\\end{matrix}}{\biggr )},\quad i=1,2,3\\\end{aligned}}.}$[7]

(E1)

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

${\displaystyle {\begin{aligned}\sigma ^{i0}&={\frac {i}{2}}{\biggl (}{\begin{matrix}\sigma _{i}&0\\0&-\sigma _{i}\\\end{matrix}}{\biggr )},\\\sigma ^{ij}&={\frac {1}{2}}\epsilon _{ijk}{\biggl (}{\begin{matrix}\sigma _{k}&0\\0&\sigma _{k}\\\end{matrix}}{\biggr )}\\\end{aligned}}.}$[8]

(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)⁺,

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

(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

${\displaystyle \psi ={\begin{pmatrix}\psi _{R}\\\psi _{L}\end{pmatrix}},}$

(E4)

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

${\displaystyle \beta =i\gamma ^{0}={\biggl (}{\begin{matrix}0&I\\I&0\\\end{matrix}}{\biggr )},}$[9]

(E5)

as representative for P = diag(1, −1, −1, −1). It is seen that the representation is irreducible when space parity inversion is 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 negates a bispinor, 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 U × U. This decomposition hints how to couple any bispinor field with other fields in a Lagrangian to yield Lorentz scalars.

Conventions

There are several choices of signature and representation that are in common use in the physics literature. The Dirac matrices are typically written as ${\displaystyle \gamma ^{\mu }}$ where ${\displaystyle \mu }$ runs from 0 to 3. In this notation, 0 corresponds to time, and 1 through 3 correspond to x, y, and z.

The + − − − signature is sometimes called the west coast metric, while the − + + + is the east coast metric. At this time the + − − − signature is in more common use, and our example will use this signature. To switch from one example to the other, multiply all ${\displaystyle \gamma ^{\mu }}$ by ${\displaystyle i}$.

After choosing the signature, there are many ways of constructing a representation in the 4×4 matrices, and many are in common use. In order to make this example as general as possible we will not specify a representation until the final step. At that time we will substitute in the "chiral" or Weyl representation.

Construction of Dirac spinor with a given spin direction and charge

First we choose a spin direction for our electron or positron. As with the example of the Pauli algebra discussed above, the spin direction is defined by a unit vector in 3 dimensions, (a, b, c). Following the convention of Peskin & Schroeder, the spin operator for spin in the (a, b, c) direction is defined as the dot product of (a, b, c) with the vector

${\displaystyle {\begin{aligned}\left(i\gamma ^{2}\gamma ^{3},\;\;i\gamma ^{3}\gamma ^{1},\;\;i\gamma ^{1}\gamma ^{2}\right)&=-\left(\gamma ^{1},\;\gamma ^{2},\;\gamma ^{3}\right)i\gamma ^{1}\gamma ^{2}\gamma ^{3}\\\sigma _{(a,b,c)}&=ia\gamma ^{2}\gamma ^{3}+ib\gamma ^{3}\gamma ^{1}+ic\gamma ^{1}\gamma ^{2}\end{aligned}}}$

Note that the above is a root of unity, that is, it squares to 1. Consequently, we can make a projection operator from it that projects out the sub-algebra of the Dirac algebra that has spin oriented in the (a, b, c) direction:

${\displaystyle P_{(a,b,c)}={\frac {1}{2}}\left(1+\sigma _{(a,b,c)}\right)}$

Now we must choose a charge, +1 (positron) or −1 (electron). Following the conventions of Peskin & Schroeder, the operator for charge is ${\displaystyle Q=-\gamma ^{0}}$, that is, electron states will take an eigenvalue of −1 with respect to this operator while positron states will take an eigenvalue of +1.

Note that ${\displaystyle Q}$ is also a square root of unity. Furthermore, ${\displaystyle Q}$ commutes with ${\displaystyle \sigma _{(a,b,c)}}$. They form a complete set of commuting operators for the Dirac algebra. Continuing with our example, we look for a representation of an electron with spin in the (a, b, c) direction. Turning ${\displaystyle Q}$ into a projection operator for charge = −1, we have

${\displaystyle P_{-Q}={\frac {1}{2}}\left(1-Q\right)={\frac {1}{2}}\left(1+\gamma ^{0}\right)}$

The projection operator for the spinor we seek is therefore the product of the two projection operators we've found:

${\displaystyle P_{(a,b,c)}\;P_{-Q}}$

The above projection operator, when applied to any spinor, will give that part of the spinor that corresponds to the electron state we seek. So we can apply it to a spinor with the value 1 in one of its components, and 0 in the others, which gives a column of the matrix. Continuing the example, we put (a, b, c) = (0, 0, 1) and have

${\displaystyle P_{(0,0,1)}={\frac {1}{2}}\left(1+i\gamma _{1}\gamma _{2}\right)}$

and so our desired projection operator is

${\displaystyle P={\frac {1}{2}}\left(1+i\gamma ^{1}\gamma ^{2}\right)\cdot {\frac {1}{2}}\left(1+\gamma ^{0}\right)={\frac {1}{4}}\left(1+\gamma ^{0}+i\gamma ^{1}\gamma ^{2}+i\gamma ^{0}\gamma ^{1}\gamma ^{2}\right)}$

The 4×4 gamma matrices used in the Weyl representation are

${\displaystyle {\begin{aligned}\gamma _{0}&={\begin{bmatrix}0&1\\1&0\end{bmatrix}}\\\gamma _{k}&={\begin{bmatrix}0&\sigma ^{k}\\-\sigma ^{k}&0\end{bmatrix}}\end{aligned}}}$

for k = 1, 2, 3 and where ${\displaystyle \sigma ^{i}}$ are the usual 2×2 Pauli matrices. Substituting these in for P gives

${\displaystyle P={\frac {1}{4}}{\begin{bmatrix}1+\sigma ^{3}&1+\sigma ^{3}\\1+\sigma ^{3}&1+\sigma ^{3}\end{bmatrix}}={\frac {1}{2}}{\begin{bmatrix}1&0&1&0\\0&0&0&0\\1&0&1&0\\0&0&0&0\end{bmatrix}}}$

Our answer is any non-zero column of the above matrix. The division by two is just a normalization. The first and third columns give the same result:

${\displaystyle \left|e^{-},\,+{\frac {1}{2}}\right\rangle ={\begin{bmatrix}1\\0\\1\\0\end{bmatrix}}}$

More generally, for electrons and positrons with spin oriented in the (a, b, c) direction, the projection operator is

${\displaystyle {\frac {1}{4}}{\begin{bmatrix}1+c&a-ib&\pm (1+c)&\pm (a-ib)\\a+ib&1-c&\pm (a+ib)&\pm (1-c)\\\pm (1+c)&\pm (a-ib)&1+c&a-ib\\\pm (a+ib)&\pm (1-c)&a+ib&1-c\end{bmatrix}}}$

where the upper signs are for the electron and the lower signs are for the positron. The corresponding spinor can be taken as any non zero column. Since ${\displaystyle a^{2}+b^{2}+c^{2}=1}$ the different columns are multiples of the same spinor. The representation of the resulting spinor in the Dirac basis can be obtained using the rule given in the bispinor article.