Spinor Representations and Gamma Matrices

Scalar and vector fields are not the end of Lorentz representation theory. The proper orthochronous Lorentz group has the double cover $SL(2,\mathbb C)$, whose smallest nontrivial finite-dimensional representations are two-component spinors. These are not ordinary tensors: a $2\pi$ rotation acts on them by a minus sign. That minus sign is the representation-theoretic seed of spin one-half.

The previous page distinguished finite-dimensional field representations from the little-group representations carried by one-particle states. We now construct the finite-dimensional representations needed for spin one-half fields. The two elementary objects are the two inequivalent Weyl spinors. A massive relativistic fermion combines both, and a chosen chiral gamma-matrix basis then identifies them with the left- and right-chiral components of a Dirac spinor. The gamma matrices package this pair into a single Dirac spinor and turn the two first-order Weyl equations into the compact Dirac operator $\gamma^\mu p_\mu-m$.

The reader should watch two separate ideas. A Weyl spinor is a Lorentz-representation object; a gamma matrix is an intertwiner that maps between the two inequivalent Weyl representations. The Dirac equation works because momentum, written as $p\cdot\sigma$ or $p\cdot\bar\sigma$, is exactly such an intertwiner.

The Lorentz algebra and two spinor representations

Chiral-basis dictionary

This page uses the chiral, or Weyl, gamma-matrix basis. Define

$$\sigma^\mu=(\mathbf 1,\sigma^1,\sigma^2,\sigma^3), \qquad \bar\sigma^\mu=(\mathbf 1,-\sigma^1,-\sigma^2,-\sigma^3),$$

where

$$\sigma^1= \begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad \sigma^2= \begin{pmatrix}0&-i\\ i&0\end{pmatrix},\qquad \sigma^3= \begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$

Thus

$$p\cdot\sigma=p_\mu\sigma^\mu=p^0\mathbf 1-\mathbf p\cdot\boldsymbol\sigma, \qquad p\cdot\bar\sigma=p_\mu\bar\sigma^\mu=p^0\mathbf 1+\mathbf p\cdot\boldsymbol\sigma.$$

This differs by a spatial sign from the coordinate matrix $X=x^0\mathbf 1+\mathbf x\cdot\boldsymbol\sigma$ used on the previous page because $p\cdot\sigma$ uses the lowered index $p_\mu$. We suppress dotted and undotted spinor indices in the main text; the matrix products indicate which chirality is mapped to which. The transformation laws below fix the otherwise ambiguous choice of which Weyl matrix is called $A$.

The gamma matrices are

$$\gamma^\mu= \begin{pmatrix} 0&\sigma^\mu\\ \bar\sigma^\mu&0 \end{pmatrix}, \qquad \gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3= \begin{pmatrix} -\mathbf 1&0\\ 0&\mathbf 1 \end{pmatrix}.$$

In this chiral basis a Dirac spinor is ordered as

$$\Psi=\begin{pmatrix}\psi_L\\ \psi_R\end{pmatrix},$$

so the projectors $P_L=(1-\gamma^5)/2$ and $P_R=(1+\gamma^5)/2$ pick the upper and lower two components respectively.

Later in the page we write the spinor Lorentz generator as $\Sigma^{\mu\nu}=i[\gamma^\mu,\gamma^\nu]/4$. This is half of the $\sigma^{\mu\nu}=i[\gamma^\mu,\gamma^\nu]/2$ notation fixed in the convention page.

Let $M^{\mu\nu}=-M^{\nu\mu}$ be Lorentz generators. Separate rotations and boosts by

$$J_i={1\over2}\epsilon_{ijk}M^{jk}, \qquad K_i=M^{0i}.$$

Their commutation relations are

$$[J_i,J_j]=i\epsilon_{ijk}J_k, \qquad [J_i,K_j]=i\epsilon_{ijk}K_k, \qquad [K_i,K_j]=-i\epsilon_{ijk}J_k.$$

The minus sign in $[K_i,K_j]$ is the algebraic trace of the noncompactness of the Lorentz group. It is also why finite-dimensional Lorentz representations are generally not unitary.

Now complexify the algebra and define

$$N_i^{(+)}={1\over2}(J_i+iK_i), \qquad N_i^{(-)}={1\over2}(J_i-iK_i).$$

A direct calculation gives

$$[N_i^{(+)},N_j^{(+)}]=i\epsilon_{ijk}N_k^{(+)}, \qquad [N_i^{(-)},N_j^{(-)}]=i\epsilon_{ijk}N_k^{(-)}, \qquad [N_i^{(+)},N_j^{(-)}]=0.$$

Thus the complexified Lorentz algebra is the direct sum of two commuting complex simple algebras,

$$\mathfrak{so}(1,3)_{\mathbb C}\simeq \mathfrak{sl}(2,\mathbb C)_L\oplus \mathfrak{sl}(2,\mathbb C)_R.$$

Physicists often call these the two $SU(2)$ factors because the displayed commutators have the $\mathfrak{su}(2)$ form and their finite-dimensional irreducible representations carry the usual spin labels. Strictly speaking, after complexification the algebra on each side is $\mathfrak{sl}(2,\mathbb C)$, not the real compact algebra $\mathfrak{su}(2)$.

The finite-dimensional irreducible representations are labeled by a pair

$$(j_L,j_R), \qquad j_L,j_R=0,{1\over2},1,{3\over2},\ldots,$$

where the two entries specify which of the two commuting complex factors is active. In the chiral convention used below, $({1\over2},0)$ is called left-handed and $(0,{1\over2})$ is called right-handed. The representations have dimension

$$\dim(j_L,j_R)=(2j_L+1)(2j_R+1).$$

The simplest representations are

$$\begin{array}{ccl} (0,0)&:&\text{scalar},\\ ({1\over2},0)&:&\text{left Weyl spinor},\\ (0,{1\over2})&:&\text{right Weyl spinor},\\ ({1\over2},{1\over2})&:&\text{four-vector}. \end{array}$$

Under ordinary spatial rotations, the physical rotation generator is

$$J_i=N_i^{(+)}+N_i^{(-)}.$$

Therefore the representation $(j_L,j_R)$ contains ordinary spins

$$j=|j_L-j_R|, |j_L-j_R|+1, \ldots, j_L+j_R.$$

Both $({1\over2},0)$ and $(0,{1\over2})$ contain spin $1/2$ under rotations. They differ not by their rotation spin, but by how boosts act.

The complexified Lorentz algebra splits into two commuting $\mathfrak{sl}(2,\mathbb C)$ factors, each with the familiar angular-momentum representation labels. The two Weyl representations are inequivalent under boosts. Their tensor product gives a vector, and their direct sum gives the four-component Dirac spinor used below.

A useful way to remember the difference between the two Weyl representations is this. The matrices

$$A, \qquad (A^\dagger)^{-1}$$

are equivalent for pure rotations, because then $A$ is unitary, but they are inequivalent for boosts. Fix $A$ by the Hermitian-vector transformation $X'=AXA^\dagger$ used below. With that choice, the chiral components transform as

$$\psi_R\mapsto A\psi_R, \qquad \psi_L\mapsto(A^\dagger)^{-1}\psi_L.$$

The upper component is still $\psi_L=P_L\Psi$ and the lower component is $\psi_R=P_R\Psi$; the subscripts refer to chirality, not to whether the corresponding matrix was named $A$. Some references instead attach the symbol $A$ to the other Weyl factor. Stating the Hermitian-vector map and the two transformation laws removes that naming ambiguity.

Four-vectors as two-by-two matrices

The bridge between spinors and vectors is the Pauli matrix identity

$$(\mathbf a\cdot\boldsymbol\sigma)(\mathbf b\cdot\boldsymbol\sigma) =(\mathbf a\cdot\mathbf b)\mathbf 1 +i(\mathbf a\times\mathbf b)\cdot\boldsymbol\sigma.$$

For a real four-vector $x^\mu$, define the Hermitian matrix

$$X=x^0\mathbf 1+\mathbf x\cdot\boldsymbol\sigma = \begin{pmatrix} x^0+x^3&x^1-ix^2\\ x^1+ix^2&x^0-x^3 \end{pmatrix}.$$

Its determinant is the Minkowski norm:

$$\det X=(x^0)^2-\mathbf x^2=x^2.$$

If $A\in SL(2,\mathbb C)$, then

$$X\mapsto X'=AXA^\dagger$$

preserves the determinant because $\det A=1$. Thus every $A\in SL(2,\mathbb C)$ determines a Lorentz transformation $\Lambda(A)$ by

$$AX(x)A^\dagger=X(\Lambda x).$$

The two matrices $A$ and $-A$ give the same Lorentz transformation, so $SL(2,\mathbb C)$ is a double cover of the proper orthochronous Lorentz group.

The same construction works for momentum, but here the lowered index matters. With the convention stated above,

$$p\cdot\sigma=p_\mu\sigma^\mu=p^0\mathbf 1-\mathbf p\cdot\boldsymbol\sigma, \qquad p\cdot\bar\sigma=p_\mu\bar\sigma^\mu=p^0\mathbf 1+\mathbf p\cdot\boldsymbol\sigma.$$

The plus-sign matrix is $p\cdot\bar\sigma=p^0\mathbf1+\mathbf p\cdot\boldsymbol\sigma$. With the same $A$ that defines $X'=AXA^\dagger$, the two momentum matrices transform as

$$p\cdot\bar\sigma\mapsto A(p\cdot\bar\sigma)A^\dagger,$$ $$p\cdot\sigma\mapsto (A^\dagger)^{-1}(p\cdot\sigma)A^{-1}.$$

Consequently $p\cdot\bar\sigma$ maps $\psi_L$ into the $\psi_R$ representation, while $p\cdot\sigma$ maps $\psi_R$ into the $\psi_L$ representation. These transformation laws are the precise content behind the arrows in the coupled Weyl equations.

They obey

$$(p\cdot\sigma)(p\cdot\bar\sigma) =(p\cdot\bar\sigma)(p\cdot\sigma)=p^2\mathbf 1.$$

This single identity is the two-component version of “squaring the Dirac operator.” It is the reason first-order spinor equations can imply the second-order relativistic mass shell.

In index notation, $p\cdot\sigma$ maps a right-handed spinor into a left-handed one, while $p\cdot\bar\sigma$ maps a left-handed spinor into a right-handed one. The compact matrix notation suppresses dotted and undotted spinor indices, but the direction of this map is the reason the massive equations below couple the two chiralities in the order they do.

The sign pattern is worth pausing over. The Hermitian coordinate matrix on the previous page used $X=x^0\mathbf 1+\mathbf x\cdot\boldsymbol\sigma$. Here $p\cdot\sigma=p_\mu\sigma^\mu=p^0\mathbf 1-\mathbf p\cdot\boldsymbol\sigma$ because the lower-index momentum is $p_\mu=(p^0,-\mathbf p)$ in the mostly-minus convention. Many spinor sign errors are just this index-lowering step in disguise.

A four-vector can be represented by a $2\times2$ Hermitian matrix. The determinant gives the invariant $p^2$, while multiplication by $p\cdot\sigma$ and $p\cdot\bar\sigma$ maps between the two Weyl spinor representations.

Coupling left and right spinors

A massive Dirac spin one-half particle is described most economically by combining the two inequivalent Weyl spinors. In the chiral basis these become the left- and right-chiral components of a Dirac spinor. Momentum naturally maps one chirality to the other, and the Dirac mass term ties the two chiralities together.

Let $\psi_L$ transform as $({1\over2},0)$ and $\psi_R$ as $(0,{1\over2})$. The first-order massive equations are

$$(p\cdot\sigma)\psi_R=m\psi_L, \qquad (p\cdot\bar\sigma)\psi_L=m\psi_R.$$

Written out,

$$(p^0-\boldsymbol\sigma\cdot\mathbf p)\psi_R=m\psi_L, \qquad (p^0+\boldsymbol\sigma\cdot\mathbf p)\psi_L=m\psi_R.$$

Apply $p\cdot\bar\sigma$ to the first equation:

$$(p\cdot\bar\sigma)(p\cdot\sigma)\psi_R =m(p\cdot\bar\sigma)\psi_L.$$

Use the second equation on the right and the sigma identity on the left:

$$p^2\psi_R=m^2\psi_R.$$

Similarly,

$$p^2\psi_L=m^2\psi_L.$$

Thus both Weyl components satisfy the Klein–Gordon mass shell,

$$p^2=m^2.$$

The massless limit is especially transparent. If $m=0$, the equations decouple:

$$(p\cdot\bar\sigma)\psi_L=0, \qquad (p\cdot\sigma)\psi_R=0.$$

This is not just a simplification of algebra. It means that chirality becomes a separately conserved Lorentz label for a free massless fermion. Interactions may still distinguish or mix chiralities depending on their form, but the free kinetic equation itself no longer needs both pieces.

A massless left-chiral field and a massless right-chiral field are independent Lorentz representations. A massive spinor ties them together.

This is the conceptual reason the Dirac spinor is a direct sum,

$$\Psi= \begin{pmatrix} \psi_L\\ \psi_R \end{pmatrix} \in ({1\over2},0)\oplus(0,{1\over2}).$$

It is not irreducible under the proper Lorentz group, but it is the natural object for a massive parity-symmetric spin one-half theory.

Gamma matrices and the Clifford algebra

The gamma matrices package the two equations above into one matrix equation. In the chiral basis,

$$\gamma^\mu= \begin{pmatrix} 0&\sigma^\mu\\ \bar\sigma^\mu&0 \end{pmatrix}.$$

Then

$$\gamma^\mu p_\mu= \begin{pmatrix} 0&p\cdot\sigma\\ p\cdot\bar\sigma&0 \end{pmatrix}.$$

The equation

$$(\gamma^\mu p_\mu-m)\Psi=0$$

is exactly the pair

$$(p\cdot\sigma)\psi_R=m\psi_L, \qquad (p\cdot\bar\sigma)\psi_L=m\psi_R.$$

The gamma matrices satisfy the Clifford algebra

$$\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\mathbf 1_4.$$

To verify it, multiply the block matrices:

$$\gamma^\mu\gamma^\nu= \begin{pmatrix} \sigma^\mu\bar\sigma^\nu&0\\ 0&\bar\sigma^\mu\sigma^\nu \end{pmatrix}.$$

Therefore

$$\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu = \begin{pmatrix} \sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu&0\\ 0&\bar\sigma^\mu\sigma^\nu+\bar\sigma^\nu\sigma^\mu \end{pmatrix}.$$

The Pauli matrices give

$$\sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu =2\eta^{\mu\nu}\mathbf 1_2,$$

and similarly for the lower block. Hence the Clifford algebra follows.

Now square the Dirac operator:

$$(\gamma^\mu p_\mu)^2 ={1\over2}\{\gamma^\mu,\gamma^\nu\}p_\mu p_\nu =p^2\mathbf 1_4.$$

The antisymmetric commutator part drops out because $p_\mu p_\nu$ is symmetric. Therefore

$$(\gamma^\mu p_\mu-m)(\gamma^\nu p_\nu+m) =(p^2-m^2)\mathbf 1_4.$$

This is the precise sense in which the Dirac operator is a Lorentz-covariant square root of the Klein–Gordon operator.

In the chiral basis, $\gamma^\mu$ is off-diagonal and maps left-chiral spinors to right-chiral spinors and back. The Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}$ is equivalent to the basic identities of the Pauli matrices.

Lorentz generators on Dirac spinors

Gamma matrices also give a compact representation of the Lorentz algebra. Define

$$\Sigma^{\mu\nu}={i\over4}[\gamma^\mu,\gamma^\nu].$$

The conventions page denotes $\sigma^{\mu\nu}=\frac{i}{2}[\gamma^\mu,\gamma^\nu]$; thus the generator used here is $\Sigma^{\mu\nu}=\sigma^{\mu\nu}/2$. This is the matrix that appears in the finite spinor transformation $S(\Lambda)=\exp[-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu}]$.

Using only the Clifford algebra, one finds

$$[\Sigma^{\mu\nu},\gamma^\rho] =i\left(\eta^{\nu\rho}\gamma^\mu- \eta^{\mu\rho}\gamma^\nu\right).$$

This is the defining statement that the $\gamma^\rho$ transform as the components of a Lorentz vector under similarity transformations in spinor space. The phrase “gamma matrices transform as a vector” is a shorthand for this commutator identity. The matrices themselves are fixed matrices; what transforms is the spinor basis and therefore the bilinear object in which the gamma matrix appears.

For a finite Lorentz transformation,

$$S(\Lambda)=\exp\left(-{i\over2}\omega_{\mu\nu}\Sigma^{\mu\nu}\right),$$

with $\Lambda=\exp(\omega)$ in the vector representation. Then

$$S(\Lambda)^{-1}\gamma^\mu S(\Lambda) =\Lambda^\mu{}_{\nu}\gamma^\nu.$$

This identity makes the Dirac equation Lorentz covariant. If

$$(\gamma^\mu p_\mu-m)\Psi(p)=0,$$

then the transformed spinor

$$\Psi'(p')=S(\Lambda)\Psi(p), \qquad p'=\Lambda p,$$

obeys

$$(\gamma^\mu p'_\mu-m)\Psi'(p')=0.$$

In block form, the generators split into the two Weyl pieces:

$$\Sigma^{\mu\nu}=\begin{pmatrix} \Sigma_L^{\mu\nu}&0\\ 0&\Sigma_R^{\mu\nu} \end{pmatrix},$$

where

$$\Sigma_L^{\mu\nu}={i\over4}(\sigma^\mu\bar\sigma^\nu-\sigma^\nu\bar\sigma^\mu), \qquad \Sigma_R^{\mu\nu}={i\over4}(\bar\sigma^\mu\sigma^\nu-\bar\sigma^\nu\sigma^\mu).$$

With $A$ fixed by $X'=AXA^\dagger$, the finite matrix has the corresponding block form

$$S(\Lambda)= \begin{pmatrix} (A^\dagger)^{-1}&0\\ 0&A \end{pmatrix}.$$

Rotations act with the usual Pauli spin matrices on both chiralities, while boosts act with opposite signs. This is the matrix version of the representation-theoretic statement

$$\Psi\in({1\over2},0)\oplus(0,{1\over2}).$$

The commutators $\Sigma^{\mu\nu}={i\over4}[\gamma^\mu,\gamma^\nu]$ generate Lorentz transformations on Dirac spinors. Their defining property is that conjugating $\gamma^\rho$ by $S(\Lambda)$ reproduces the vector transformation of the index $\rho$.

Chirality, gamma five, and parity

The matrix

$$\gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3$$

has three central properties:

$$(\gamma^5)^2=1, \qquad \{\gamma^5,\gamma^\mu\}=0, \qquad [\gamma^5,\Sigma^{\mu\nu}]=0.$$

The first property says that $\gamma^5$ has eigenvalues $\pm1$. The second says that $\gamma^\mu$ flips chirality. The third says that chirality is preserved by proper Lorentz transformations.

The projectors

$$P_L={1-\gamma^5\over2}, \qquad P_R={1+\gamma^5\over2}$$

satisfy

$$P_L^2=P_L, \qquad P_R^2=P_R, \qquad P_LP_R=0, \qquad P_L+P_R=1.$$

In our chiral basis,

$$P_L\Psi= \begin{pmatrix} \psi_L\\ 0 \end{pmatrix}, \qquad P_R\Psi= \begin{pmatrix} 0\\ \psi_R \end{pmatrix}.$$

Parity is different from a proper Lorentz transformation. It sends $\mathbf x\to-\mathbf x$, hence it interchanges the two inequivalent Weyl representations:

$$({1\over2},0)\longleftrightarrow(0,{1\over2}).$$

This is why a single Weyl spinor cannot by itself furnish a parity-invariant Dirac theory. A Dirac spinor contains both chiralities, so parity can act within the same field multiplet.

The Dirac adjoint will become important in the next pages. It is defined by

$$\bar\Psi=\Psi^\dagger\gamma^0.$$

With this definition, $\bar\Psi\Psi$ is a Lorentz scalar and $\bar\Psi\gamma^\mu\Psi$ is a Lorentz vector. These bilinears are the building blocks of spinor Lagrangians and QED interactions.

Building Clifford algebras by tensor products

The four-dimensional gamma matrices are not mysterious isolated objects. They are one instance of a general Clifford-algebra construction.

In two Euclidean dimensions, the Pauli matrices themselves may be used as gamma matrices:

$$\Gamma_1=\sigma_x, \qquad \Gamma_2=\sigma_y, \qquad \{\Gamma_a,\Gamma_b\}=2\delta_{ab}.$$

In four Euclidean dimensions, a convenient construction is

$$\Gamma_1=\sigma_x\otimes I, \qquad \Gamma_2=\sigma_y\otimes I,$$ $$\Gamma_3=\sigma_z\otimes\sigma_x, \qquad \Gamma_4=\sigma_z\otimes\sigma_y.$$

The factors of $\sigma_z$ make the old generators anticommute with the new ones. For example,

$$\Gamma_1\Gamma_3+ \Gamma_3\Gamma_1 =(\sigma_x\sigma_z+\sigma_z\sigma_x)\otimes\sigma_x=0.$$

The general even-dimensional pattern is

$$\Gamma_{2k-1} =\sigma_z^{\otimes(k-1)}\otimes\sigma_x\otimes I^{\otimes(n-k)},$$ $$\Gamma_{2k} =\sigma_z^{\otimes(k-1)}\otimes\sigma_y\otimes I^{\otimes(n-k)}, \qquad k=1,\ldots,n.$$

These $2n$ matrices act on a space of dimension $2^n$. Thus a spinor in even Euclidean dimension $D=2n$ has $2^n$ complex components before any further reality or chirality conditions are imposed.

In Lorentzian signature, one obtains gamma matrices satisfying

$$\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}$$

by assigning the appropriate signs, often by inserting factors of $i$ into the Euclidean matrices. The representation is not unique. In the even dimensions constructed here, all irreducible complex representations are equivalent up to similarity transformations; odd-dimensional complex Clifford algebras have two irreducible choices distinguished by the sign of the highest-grade gamma product.

The Clifford algebra in even dimension can be built recursively from Pauli matrices. Each new tensor slot adds two anticommuting generators and doubles the dimension of the spinor space.

Summary

The Lorentz algebra becomes transparent after complexification: it splits into two commuting $\mathfrak{sl}(2,\mathbb C)$ factors whose finite-dimensional representations carry the familiar angular-momentum labels. This produces two inequivalent two-component spinor representations, $({1\over2},0)$ and $(0,{1\over2})$. They look the same under rotations but transform differently under boosts.

The Pauli matrices connect spinors to vectors. A four-vector can be encoded as a $2\times2$ Hermitian matrix, and its determinant is the Minkowski norm. This is the practical meaning of the relation between $SL(2,\mathbb C)$ and the Lorentz group.

A massive spin one-half particle uses both chiralities. The equations

$$(p\cdot\sigma)\psi_R=m\psi_L, \qquad (p\cdot\bar\sigma)\psi_L=m\psi_R$$

square to the mass shell $p^2=m^2$. The gamma matrices combine these equations into

$$(\gamma^\mu p_\mu-m)\Psi=0,$$

and the Clifford algebra guarantees that this first-order operator is a Lorentz-covariant square root of the Klein–Gordon operator.

Common pitfalls

Chirality is not spin projection. A left-handed spinor is not “spin down,” and a right-handed spinor is not “spin up.” Both transform as spin $1/2$ under spatial rotations; chirality becomes helicity only in the massless positive-energy particle limit.

The symbol $A$ is not universal. Do not memorize $A$ versus $(A^\dagger)^{-1}$ without checking the Hermitian-vector map and the placement of dotted and undotted indices. Here $X'=AXA^\dagger$ fixes the choice, and the coupled equations with $p\cdot\sigma$ and $p\cdot\bar\sigma$ provide the consistency check.

Finite-dimensional boost matrices are not unitary. The Hilbert-space operator $U(\Lambda)$ is unitary, but the finite-dimensional component matrix $S(\Lambda)$ acting on spinor indices need not be.

Gamma matrices are not dynamical vectors. They are fixed matrices. Their vector character means the similarity identity $S^{-1}\gamma^\mu S=\Lambda^\mu{}_{\nu}\gamma^\nu$.

Matrix formulas depend on the gamma basis. Do not mix chiral-basis and Dirac-basis formulas without translating $\gamma^5$, $\gamma^0$, and the spinor components. The physics is basis-independent, but the block matrices are not.

Spinor indices do not use the spacetime metric. Dotted and undotted spinor indices have their own antisymmetric tensors; the spacetime metric acts on vector indices.

Two common Lorentz-generator symbols differ by a factor of two. The site convention is $\sigma^{\mu\nu}=i[\gamma^\mu,\gamma^\nu]/2$, while the generator used in $S(\Lambda)$ is $\Sigma^{\mu\nu}=i[\gamma^\mu,\gamma^\nu]/4$.

A massive solution cannot have only one chirality. The massless Weyl equations decouple, but the massive Dirac equation couples $\psi_L$ and $\psi_R$. Setting one chirality to zero is then inconsistent unless $m=0$.

Exercises

Exercise 1: sigma matrices and the mass shell

Using

$$\sigma^i\sigma^j=\delta^{ij}\mathbf 1+i\epsilon^{ijk}\sigma^k,$$

show that

$$(p\cdot\sigma)(p\cdot\bar\sigma)=p^2\mathbf 1.$$

Solution

By convention,

$$p\cdot\sigma=p^0\mathbf 1-\mathbf p\cdot\boldsymbol\sigma, \qquad p\cdot\bar\sigma=p^0\mathbf 1+\mathbf p\cdot\boldsymbol\sigma.$$

Therefore

$$(p\cdot\sigma)(p\cdot\bar\sigma) =(p^0\mathbf 1-\mathbf p\cdot\boldsymbol\sigma) (p^0\mathbf 1+\mathbf p\cdot\boldsymbol\sigma).$$

The cross terms cancel because $p^0$ is a scalar:

$$=(p^0)^2\mathbf 1-(\mathbf p\cdot\boldsymbol\sigma)^2.$$

Now

$$(\mathbf p\cdot\boldsymbol\sigma)^2 =p_i p_j\sigma^i\sigma^j =p_i p_j(\delta^{ij}\mathbf 1+i\epsilon^{ijk}\sigma^k).$$

The antisymmetric term vanishes because $p_i p_j$ is symmetric while $\epsilon^{ijk}$ is antisymmetric. Hence

$$(\mathbf p\cdot\boldsymbol\sigma)^2=\mathbf p^2\mathbf 1.$$

Thus

$$(p\cdot\sigma)(p\cdot\bar\sigma) =((p^0)^2-\mathbf p^2)\mathbf 1=p^2\mathbf 1.$$

Exercise 2: elementary Clifford-algebra checks

With

$$\gamma^\mu= \begin{pmatrix} 0&\sigma^\mu\\ \bar\sigma^\mu&0 \end{pmatrix},$$

verify explicitly that $(\gamma^0)^2=1$, $(\gamma^i)^2=-1$, and $\{\gamma^0,\gamma^i\}=0$.

Solution

Since $\sigma^0=\bar\sigma^0=\mathbf 1$,

$$\gamma^0= \begin{pmatrix} 0&\mathbf 1\\ \mathbf 1&0 \end{pmatrix}.$$

Therefore

$$(\gamma^0)^2= \begin{pmatrix} \mathbf 1&0\\ 0&\mathbf 1 \end{pmatrix}=\mathbf 1_4.$$

For a spatial index $i$,

$$\gamma^i= \begin{pmatrix} 0&\sigma^i\\ -\sigma^i&0 \end{pmatrix}.$$

Then

$$(\gamma^i)^2= \begin{pmatrix} -\sigma^i\sigma^i&0\\ 0&-\sigma^i\sigma^i \end{pmatrix} =-\mathbf 1_4,$$

because $(\sigma^i)^2=\mathbf 1$.

Finally,

$$\gamma^0\gamma^i= \begin{pmatrix} -\sigma^i&0\\ 0&\sigma^i \end{pmatrix}, \qquad \gamma^i\gamma^0= \begin{pmatrix} \sigma^i&0\\ 0&-\sigma^i \end{pmatrix}.$$

Adding gives zero:

$$\{\gamma^0,\gamma^i\}=0.$$

Together with the Pauli algebra, these identities imply the full Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}$.

Exercise 3: projectors from gamma five

Show that the projectors

$$P_L={1-\gamma^5\over2}, \qquad P_R={1+\gamma^5\over2}$$

are genuine projection operators and that $\gamma^\mu P_L=P_R\gamma^\mu$.

Solution

Since $(\gamma^5)^2=1$,

$$P_L^2={1\over4}(1-2\gamma^5+(\gamma^5)^2) ={1\over2}(1-\gamma^5)=P_L.$$

Similarly,

$$P_R^2=P_R.$$

Also,

$$P_LP_R={1\over4}(1-\gamma^5)(1+\gamma^5) ={1\over4}(1-(\gamma^5)^2)=0,$$

and

$$P_L+P_R=1.$$

Using $\{\gamma^5,\gamma^\mu\}=0$,

$$\gamma^\mu P_L ={1\over2}\gamma^\mu(1-\gamma^5) ={1\over2}(\gamma^\mu-\gamma^\mu\gamma^5).$$

But $\gamma^\mu\gamma^5=-\gamma^5\gamma^\mu$, so

$$\gamma^\mu P_L ={1\over2}(\gamma^\mu+\gamma^5\gamma^\mu) ={1+\gamma^5\over2}\gamma^\mu =P_R\gamma^\mu.$$

Thus $\gamma^\mu$ flips chirality.

Exercise 4: boost weights in a first-order equation

Using the chiral-basis dictionary above, let a boost along the $z$-axis act on the two Weyl components as

$$\psi_L\mapsto e^{-\eta\sigma^3/2}\psi_L, \qquad \psi_R\mapsto e^{+\eta\sigma^3/2}\psi_R,$$

and on light-cone momenta as

$$p_-=p^0-p^3\mapsto e^{-\eta}p_-, \qquad p_+=p^0+p^3\mapsto e^{+\eta}p_+.$$

For momentum along the $z$-axis, show that the equations

$$p_-\psi_{R,+}=m\psi_{L,+}, \qquad p_+\psi_{L,+}=m\psi_{R,+}$$

are covariant if $\sigma^3\psi_{L,+}=+\psi_{L,+}$ and $\sigma^3\psi_{R,+}=+\psi_{R,+}$.

Solution

For the $+$ component of a spinor, $\sigma^3$ has eigenvalue $+1$. Therefore

$$\psi_{L,+}\mapsto e^{-\eta/2}\psi_{L,+}, \qquad \psi_{R,+}\mapsto e^{+\eta/2}\psi_{R,+}.$$

The left-hand side of the first equation transforms as

$$p_-\psi_{R,+} \mapsto (e^{-\eta}p_-)(e^{+\eta/2}\psi_{R,+}) =e^{-\eta/2}p_-\psi_{R,+}.$$

The right-hand side transforms as

$$m\psi_{L,+}\mapsto e^{-\eta/2}m\psi_{L,+}.$$

So the first equation is covariant. The second equation transforms as

$$p_+\psi_{L,+} \mapsto (e^{+\eta}p_+)(e^{-\eta/2}\psi_{L,+}) =e^{+\eta/2}p_+\psi_{L,+},$$

and

$$m\psi_{R,+}\mapsto e^{+\eta/2}m\psi_{R,+}.$$

So the second equation is covariant as well. The opposite boost weights of left and right spinors are exactly what make the first-order equations Lorentz covariant.

Further reading

• Coleman, Sidney. Lectures on Quantum Field Theory. World Scientific, 2019, chapters 18–20.
• Srednicki, Mark. Quantum Field Theory. Cambridge University Press, 2007, sections 33–36 and 47.
• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995, sections 5.4–5.6.
• Zee, A. Quantum Field Theory in a Nutshell. 2nd ed., Princeton University Press, 2010, chapter II.3.