Gamma matrices

In mathematical physics, the gamma matrices, $\{ \gamma^0, \gamma^1, \gamma^2, \gamma^3 \}$, also known as the Dirac matrices, are a set of conventional matrices with specific anticommutation relations that ensure they generate a matrix representation of the Clifford algebra Cℓ_1,3(R). It is also possible to define higher-dimensional gamma matrices. When interpreted as the matrices of the action of a set of orthogonal basis vectors for contravariant vectors in Minkowski space, the column vectors on which the matrices act become a space of spinors, on which the Clifford algebra of spacetime acts. This in turn makes it possible to represent infinitesimal spatial rotations and Lorentz boosts. Spinors facilitate spacetime computations in general, and in particular are fundamental to the Dirac equation for relativistic spin-½ particles.

In Dirac representation, the four contravariant gamma matrices are

$\gamma^0 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix},\quad \gamma^1 = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix}$

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

$\gamma^0$ is the time-like matrix and the other three are space-like matrices.

Analogous sets of gamma matrices can be defined in any dimension and signature of the metric. For example, the Pauli matrices are a set of "gamma" matrices in dimension 3 with metric of Euclidean signature (3,0). In five spacetime dimensions, the four gammas above together with the fifth gamma matrix to be presented below generate the Clifford algebra.

Mathematical structure

The defining property for the gamma matrices to generate a Clifford algebra is the anticommutation relation

$\displaystyle\{ \gamma^\mu, \gamma^\nu \} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu} I_4$

where $\{ , \}$ is the anticommutator, $\eta^{\mu \nu}$ is the Minkowski metric with signature (+ − − −) and $I_4$ is the 4 × 4 identity matrix.

This defining property is more fundamental than the numerical values used in the specific representation of the gamma matrices. Covariant gamma matrices are defined by

$\displaystyle \gamma_\mu = \eta_{\mu \nu} \gamma^\nu = \left\{\gamma^0, -\gamma^1, -\gamma^2, -\gamma^3 \right\},$

and Einstein notation is assumed.

Note that the other sign convention for the metric, (− + + +) necessitates either a change in the defining equation:

$\displaystyle\{ \gamma^\mu, \gamma^\nu \} = -2 \eta^{\mu \nu} I_4$

or a multiplication of all gamma matrices by $i$, which of course changes their hermiticity properties detailed below. Under the alternative sign convention for the metric the covariant gamma matrices are then defined by

$\displaystyle \gamma_\mu = \eta_{\mu \nu} \gamma^\nu = \left\{-\gamma^0, +\gamma^1, +\gamma^2, +\gamma^3 \right\}$.

Physical structure

The Clifford Algebra Cl_1,3(R) over spacetime V can be regarded as the set of real linear operators from V to itself, End(V), or more generally, when complexified to Cl_1,3(R)_C, as the set of linear operators from any 4-dimensional complex vector space to itself. More simply, given a basis for V, Cl_1,3(R)_C is just the set of all 4 × 4 complex matrices, but endowed with a Clifford algebra structure. Spacetime is assumed to be endowed with the Minkowski metric η_μν. A space of bispinors, U_x, is also assumed at every point in spacetime, endowed with the bispinor representation of the Lorentz group. The bispinor fields Ψ of the Dirac equations, evaluated at any point x in spacetime, are elements of U_x, see below. The Clifford algebra is assumed to act on U_x as well (by matrix multiplication with column vectors Ψ(x) in U_x for all x). This will be the primary view of elements of Cl_1,3(R)_C in this section.

For each linear transformation S of U_x, there is a transformation of End(U_x) given by SES⁻¹ for E in Cl_1,3(R)_C ≈ End(U_x). If S belongs to a representation of the Lorentz group, then the induced action E ↦ SES⁻¹ will also belong to a representation of the Lorentz group, see Representation theory of the Lorentz group.

If S(Λ) is the bispinor representation acting on U_x of an arbitrary Lorentz transformation Λ in the standard (4-vector) representation acting on V, then there is a corresponding operator on End(U_x) = Cl_1,3(R)_C given by

$\gamma^\mu \mapsto S(\Lambda)\gamma^\mu S(\Lambda)^{-1} = { {({\Lambda}^{-1})}^\mu}_\nu \gamma^\nu := {\Lambda_\nu}^\mu \gamma^\nu,$

showing that the γ^μ can be viewed as a basis of a representation space of the 4-vector representation of the Lorentz group sitting inside the Clifford algebra. This means that quantities of the form

$a\!\!\!/ := a_\mu\gamma^\mu$

should be treated as 4-vectors in manipulations. It also means that indices can be raised and lowered on the γ using the metric η_μν as with any 4-vector. The notation is called the Feynman slash notation. The slash operation maps the unit vectors e_μ of V, or any 4-dimensional vector space, to basis vectors γ_μ. The transformation rule for slashed quantities is simply

${a\!\!\!/}^\mu \mapsto {\Lambda^\mu}_\nu {a\!\!\!/}^\nu.$

One should note that this is different from the transformation rule for the γ^μ, which are now treated as (fixed) basis vectors. The designation of the 4-tuple (γ^μ) = (γ⁰, γ¹, γ², γ³) as a 4-vector sometimes found in the literature is thus a slight misnomer. The latter transformation corresponds to an active transformation of the components of a slashed quantity in terms of the basis γ^μ, and the former to a passive transformation of the basis γ^μ itself.

The elements σ^μν = γ^μγ^ν − γ^νγ^μ form a representation of the Lie algebra of the Lorentz group. This is a spin representation. When these matrices, and linear combinations of them, are exponentiated, they are bispinor representations of the Lorentz group, e.g., the S(Λ) of above are of this form. The 6-dimensional space the σ^μν span is the representation space of a tensor representation of the Lorentz group. For the higher order elements of the Clifford algebra in general, and their transformation rules, see the article Dirac algebra. But it is noted here that the Clifford algebra has no subspace being the representation space of a spin representation of the Lorentz group in the context used here.

Expressing the Dirac equation

In natural units, the Dirac equation may be written as

$(i \gamma^\mu \partial_\mu - m) \psi = 0$

where $\psi$ is a Dirac spinor.

Switching to Feynman notation, the Dirac equation is

$(i \partial\!\!\!/ - m) \psi = 0.$

The fifth gamma matrix, γ⁵

It is useful to define the product of the four gamma matrices as follows:

$\gamma^5 := i\gamma^0\gamma^1\gamma^2\gamma^3 = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$ (in the Dirac basis).

Although $\gamma^5$ uses the letter gamma, it is not one of the gamma matrices of Cℓ_1,3(R). The number 5 is a relic of old notation in which $\gamma^0$ was called "$\gamma^4$".

$\gamma^5$ has also an alternative form:

$\gamma^5 = \frac{i}{4!} \varepsilon_{\mu \nu \alpha \beta} \gamma^{\mu} \gamma^{\nu} \gamma^{\alpha} \gamma^{\beta}$

Proof

This can be seen by exploiting the fact that all the four gamma matrices anticommute, so

$\gamma^0\gamma^1\gamma^2\gamma^3 = \gamma^{[0}\gamma^1\gamma^2\gamma^{3]} = \frac{1}{4!} \delta^{0123}_{\mu\nu\varrho\sigma}\gamma^\mu\gamma^\nu\gamma^\varrho\gamma^\sigma$,

where $\delta^{\alpha\beta\gamma\delta}_{\mu\nu\varrho\sigma}$ is the type (4,4) generalized Kronecker delta in 4 dimensions. If $\varepsilon_{\alpha \dots \beta}$ denotes the Levi-Civita symbol in n dimensions, we can use the identity $\delta^{\alpha\beta\gamma\delta}_{\mu\nu\varrho \sigma} = \varepsilon^{\alpha\beta\gamma\delta} \varepsilon_{\mu\nu\varrho\sigma}$. Then we get

$\gamma^5 = i\gamma^0\gamma^1\gamma^2\gamma^3 = \frac{i}{4!} \varepsilon^{0123}\varepsilon_{\mu\nu\varrho\sigma} \,\gamma^\mu\gamma^\nu\gamma^\varrho \gamma^\sigma = \frac{i}{4!} \varepsilon_{\mu\nu\varrho\sigma} \,\gamma^\mu\gamma^\nu\gamma^\varrho \gamma^\sigma$

This matrix is useful in discussions of quantum mechanical chirality. For example, a Dirac field can be projected onto its left-handed and right-handed components by:

$\psi_L= \frac{1-\gamma^5}{2}\psi, \qquad\psi_R= \frac{1+\gamma^5}{2}\psi$.

Some properties are:

• It is hermitian:

$(\gamma^5)^\dagger = \gamma^5. \,$

• Its eigenvalues are ±1, because:

$(\gamma^5)^2 = I_4. \,$

• It anticommutes with the four gamma matrices:

$\left\{ \gamma^5,\gamma^\mu \right\} =\gamma^5 \gamma^\mu + \gamma^\mu \gamma^5 = 0. \,$

The set {γ⁰, γ¹, γ², γ³, iγ⁵} therefore, by the last two properties (keeping in mind that i² = −1) and those of the old gammas, forms the basis of the Clifford algebra in 5 spacetime dimensions for the metric signature (1,4).^[1] In metric signature (4,1), the set {γ⁰, γ¹, γ², γ³, γ⁵} is used, where the γ^μ are the appropriate ones for the (3,1) signature.^[2] This pattern is repeated for spacetime dimension 2n even and the next odd dimension 2n + 1 for all n ≥ 1.^[3] For more detail, see Higher-dimensional gamma matrices.

Identities

The following identities follow from the fundamental anticommutation relation, so they hold in any basis (although the last one depends on the sign choice for $\gamma^5$).

Miscellaneous identities

Num	Identity
1	$\displaystyle\gamma^\mu\gamma_\mu= 4 I_4$
2	$\displaystyle\gamma^\mu\gamma^\nu\gamma_\mu=-2\gamma^\nu$
3	$\displaystyle\gamma^\mu\gamma^\nu\gamma^\rho\gamma_\mu=4\eta^{\nu\rho} I_4$
4	$\displaystyle\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma_\mu=-2\gamma^\sigma\gamma^\rho\gamma^\nu$
5	$\displaystyle\gamma^\mu\gamma^\nu\gamma^\rho= \eta^{\mu\nu}\gamma^\rho + \eta^{\nu\rho}\gamma^\mu - \eta^{\mu\rho}\gamma^\nu - i\epsilon^{\sigma\mu\nu\rho}\gamma_\sigma\gamma^5$

Proofs of 1 & 2

To show

$\displaystyle\gamma^\mu\gamma_\mu= 4 I_4$

one begins with the standard anticommutation relation

$\displaystyle\{ \gamma^\mu, \gamma^\nu \} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu} I_4.$

One can make this situation look similar by using the metric $\eta$:

$\gamma^\mu \gamma_\mu \,= \gamma^\mu \eta_{\mu \nu} \gamma^\nu = \eta_{\mu \nu} \gamma^\mu \gamma^\nu$	$= \frac{1}{2} (\eta_{\mu \nu} + \eta_{\nu \mu}) \gamma^\mu \gamma^\nu$ ($\eta$ symmetric)
	$= \frac{1}{2} (\eta_{\mu \nu}\gamma^\mu \gamma^\nu + \eta_{\nu \mu}\gamma^\mu \gamma^\nu)$ (expanding)
	$= \frac{1}{2} (\eta_{\mu \nu}\gamma^\mu \gamma^\nu + \eta_{\mu \nu}\gamma^\nu \gamma^\mu)$ (relabeling term on right)
	$= \frac{1}{2} \eta_{\mu \nu} \{\gamma^\mu, \gamma^\nu \} \,$
	$= \frac{1}{2} \eta_{\mu \nu} \left(2 \eta^{\mu \nu} I_4 \right) = \eta_{\mu \nu} \eta^{\mu \nu} I_4 = 4 I_4. \,$

To show

$\gamma^\mu\gamma^\nu\gamma_\mu=-2\gamma^\nu . \,$

We again will use the standard commutation relation. So start:

$\gamma^\mu \gamma^\nu \gamma_\mu \,$	$= \gamma^\mu \left(2 \eta_\mu^\nu I_4 - \gamma_\mu \gamma^\nu \right) \,$
	$= 2 \gamma^\mu \eta_\mu^\nu - \gamma^\mu \gamma_\mu \gamma^\nu \,$
	$= 2 \gamma^\nu - 4 \gamma^\nu = -2 \gamma^\nu. \,$

Proof of 3

To show

$\gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu = 4 \eta^{\nu\rho} I_4. \,$

Use the anticommutator to shift $\gamma^\mu$ to the right

$\gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu \,$	$= \{\gamma^\mu,\gamma^\nu\} \gamma^\rho \gamma_\mu - \gamma^\nu \gamma^\mu \gamma^\rho \gamma_\mu \,$
	$= 2\ \eta^{\mu\nu} \gamma^\rho \gamma_\mu - \gamma^\nu \{\gamma^\mu,\gamma^\rho\} \gamma_\mu + \gamma^\nu \gamma^\rho \gamma^\mu \gamma_\mu . \,$

Using the relation $\gamma^\mu\gamma_\mu=4 I$ we can contract the last two gammas, and get

$\gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu \,$	$= 2\ \gamma^\rho \gamma^\nu - \gamma^\nu 2 \eta^{\mu\rho} \gamma_\mu + 4\ \gamma^\nu \gamma^\rho \,$
	$= 2\ \gamma^\rho \gamma^\nu - 2\ \gamma^\nu \gamma^\rho + 4\ \gamma^\nu \gamma^\rho \,$
	$= 2\ (\gamma^\rho \gamma^\nu + \gamma^\nu \gamma^\rho) \,$
	$= 2\ \{\gamma^\nu, \gamma^\rho\}. \,$

Finally using the anticommutator identity, we get

$\gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu = 4\ \eta^{\nu \rho} I_4. \,$

Proof of 4

$\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma \gamma_\mu \,$	$= (2\eta^{\mu\nu}-\gamma^\nu \gamma^\mu) \gamma^\rho \gamma^\sigma \gamma_\mu \, \quad$ (anticommutator identity)
	$= 2\eta^{\mu\nu}\gamma^\rho \gamma^\sigma \gamma_\mu - 4 \gamma^\nu \eta^{\rho \sigma} \, \quad$ (using identity 3)
	$= 2\gamma^\rho \gamma^\sigma \gamma^\nu - 4 \gamma^\nu \eta^{\rho \sigma} \,$ (raising an index)
	$= 2(2\eta^{\rho \sigma} - \gamma^\sigma \gamma^\rho) \gamma^\nu - 4 \gamma^\nu \eta^{\rho \sigma} \,$ (anticommutator identity)
	$=-2\gamma^\sigma\gamma^\rho\gamma^\nu \,$ (2 terms cancel)

Proof of 5

If $\mu = \nu = \rho$ then $\epsilon^{\sigma \mu \nu \rho}=0$ and it is easy to verify the identity. That is the case also when $\mu = \nu \neq \rho$, $\mu = \rho \neq \nu$ or $\nu = \rho \neq \mu$. On the other hand, if all three indices are different, $\eta^{\mu \nu}=0$, $\eta^{\mu \rho}=0$ and $\eta^{\nu \rho}=0$ and both sides are completely antisymmetric (the left hand side because of the anticommutativity of the $\gamma$ matrices, and on the right hand side because of the antisymmetry of $\epsilon_{\sigma \mu \nu \rho}$. It thus suffices verifying the identities for the cases of $\gamma^0 \gamma^1 \gamma^2$, $\gamma^0 \gamma^1 \gamma^3$, $\gamma^0 \gamma^2 \gamma^3$ and $\gamma^1 \gamma^2 \gamma^3$.

$-i \epsilon^{\sigma 0 1 2} \gamma_\sigma \gamma^5 = -i\epsilon^{3 0 1 2} (-\gamma^3) (i\gamma^0 \gamma^1 \gamma^2 \gamma^3) = -\epsilon^{3 0 1 2} \gamma^0 \gamma^1 \gamma^2 = \epsilon^{0 1 2 3} \gamma^0 \gamma^1 \gamma^2$

$-i \epsilon^{\sigma 0 1 3} \gamma_\sigma \gamma^5 = -i\epsilon^{2 0 1 3} (-\gamma^2) (i\gamma^0 \gamma^1 \gamma^2 \gamma^3) = \epsilon^{2 0 1 3} \gamma^0 \gamma^1 \gamma^3 = \epsilon^{0 1 2 3} \gamma^0 \gamma^1 \gamma^3$

$-i \epsilon^{\sigma 0 2 3} \gamma_\sigma \gamma^5 = -i\epsilon^{1 0 2 3} (-\gamma^1) (i\gamma^0 \gamma^1 \gamma^2 \gamma^3) = -\epsilon^{1 0 2 3} \gamma^0 \gamma^2 \gamma^3 = \epsilon^{0 1 2 3} \gamma^0 \gamma^2 \gamma^3$

$-i \epsilon^{\sigma 1 2 3} \gamma_\sigma \gamma^5 = -i\epsilon^{0 1 2 3} (\gamma^0) (i\gamma^0 \gamma^1 \gamma^2 \gamma^3) = \epsilon^{0 1 2 3} \gamma^1 \gamma^2 \gamma^3$

Trace identities

The gamma matrices obey the following trace identities:

Num	Identity
0	$\operatorname{tr} (\gamma^\mu) = 0$
1	trace of any product of an odd number of $\gamma^\mu$ is zero
2	trace of $\gamma^5$ times a product of an odd number of $\gamma^\mu$ is still zero
3	$\operatorname{tr} (\gamma^\mu\gamma^\nu) = 4\eta^{\mu\nu}$
4	$\operatorname{tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=4(\eta^{\mu\nu}\eta^{\rho\sigma}-\eta^{\mu\rho}\eta^{\nu\sigma}+\eta^{\mu\sigma}\eta^{\nu\rho})$
5	$\operatorname{tr}(\gamma^5)=\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^5) = 0$
6	$\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^5) = 4i\epsilon^{\mu\nu\rho\sigma}$
7	$\operatorname{tr} (\gamma^{\mu 1}\dots\gamma^{\mu n}) = \operatorname{tr} (\gamma^{\mu n}\dots\gamma^{\mu 1})$

Proving the above involves the use of three main properties of the Trace operator:

• tr(A + B) = tr(A) + tr(B)
• tr(rA) = r tr(A)
• tr(ABC) = tr(CAB) = tr(BCA)

Proof of 0

From the definition of the gamma matrices,

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

We get

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

or equivalently,

$\frac{\gamma^\mu\gamma^\mu}{\eta^{\mu\mu}}=I\,$

where $\eta^{\mu\mu}$ is a number, and $\gamma^\mu\gamma^\mu$ is a matrix.

$\operatorname{tr}(\gamma^\nu)=\frac{1}{\eta^{\mu\mu}}\operatorname{tr}(\gamma^\nu\gamma^\mu\gamma^\mu)$ (inserting the identity and using tr(rA) = r tr(A))

$=-\frac{1}{\eta^{\mu\mu}}\operatorname{tr}(\gamma^\mu\gamma^\nu\gamma^\mu)$ (from anti-commutation relation, and given that we are free to select $\mu\ne\nu$)

$=-\frac{1}{\eta^{\mu\mu}}\operatorname{tr}(\gamma^\nu\gamma^\mu\gamma^\mu)$ (using tr(ABC) = tr(BCA))

$=-\operatorname{tr}(\gamma^\nu)$ (removing the identity)

This implies $\operatorname{tr}(\gamma^\nu) = 0$

Proof of 1

To show

$\operatorname{tr} ( \mathrm{odd \ num \ of \ } \gamma) = 0 \,$

First note that

$\operatorname{tr} (\gamma^\mu) = 0. \,$

We'll also use two facts about the fifth gamma matrix $\gamma^5 \,$ that says:

$\left(\gamma^5 \right)^2 = I_4, \quad \mathrm{and} \quad \gamma^\mu \gamma^5 = - \gamma^5 \gamma^\mu \,$

So lets use these two facts to prove this identity for the first non-trivial case: the trace of three gamma matrices. Step one is to put in one pair of $\gamma^5 \,$'s in front of the three original $\gamma \,$'s, and step two is to swap the $\gamma^5 \,$ matrix back to the original position, after making use of the cyclicity of the trace.

$\operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho) \,$	$= \operatorname{tr} \left( \gamma^5 \gamma^5 \gamma^\mu \gamma^\nu \gamma^\rho \right) \,$
	$= -\operatorname{tr} \left(\gamma^5 \gamma^\mu \gamma^\nu \gamma^\rho \gamma^5 \right) \,$
	$= -\operatorname{tr} \left( \gamma^5 \gamma^5 \gamma^\mu \gamma^\nu \gamma^\rho \right) \,$
	$= -\operatorname{tr} \left(\gamma^\mu \gamma^\nu \gamma^\rho \right) \,$

This can only be fulfilled if

$\operatorname{tr} \left(\gamma^\mu \gamma^\nu \gamma^\rho \right) = 0 \,$

The extension to 2n+1 (n integer) gamma matrices, is found by placing two gamma-5s after (say) the 2n-th gamma matrix in the trace, commuting one out to the right (giving a minus sign) and commuting the other gamma-5 2n steps out to the left [with sign change (-1)^2n =1 ]. Then we use cyclic identity to get the two gamma-5s together and hence they square to identity, leaving us with the trace equalling minus itself, i.e. 0.

Proof of 2

If an odd number of gamma matrices appear in a trace followed by $\gamma^5$, our goal is to move $\gamma^5$ from the right side to the left. This will leave the trace invariant by the cyclic property. In order to do this move, we must anticommute it with all of the other gamma matrices. This means that we anticommute it an odd number of times and pick up a minus sign. A trace equal to the negative of itself must be zero.

Proof of 3

To show

$\operatorname{tr} (\gamma^\mu\gamma^\nu) = 4\eta^{\mu\nu}$

Begin with,

$\operatorname{tr} (\gamma^\mu\gamma^\nu) \,$	$= \frac{1}{2} \left(\operatorname{tr} (\gamma^\mu\gamma^\nu) + \operatorname{tr} (\gamma^\nu\gamma^\mu) \right) \,$
	$= \frac{1}{2} \operatorname{tr} (\gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu) = \frac{1}{2} \operatorname{tr} \left( \{\gamma^\mu, \gamma^\nu\} \right) \,$
	$= \frac{1}{2} 2 \eta^{\mu \nu} \operatorname{tr} (I_4) = 4 \eta^{\mu \nu} \,$

Proof of 4

$\operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma) \,$	$= \operatorname{tr} \left(\gamma^\mu \gamma^\nu (2\eta^{\rho \sigma} - \gamma^\sigma \gamma^\rho ) \right) \,$
	$= 2 \eta^{\rho \sigma} \operatorname{tr} \left(\gamma^\mu \gamma^\nu \right) - \operatorname{tr} \left( \gamma^\mu \gamma^\nu \gamma^\sigma \gamma^\rho \right) \quad \quad (1) \,$

For the term on the right, we'll continue the pattern of swapping $\gamma^\sigma \,$ with its neighbor to the left,

$\operatorname{tr} \left( \gamma^\mu \gamma^\nu \gamma^\sigma \gamma^\rho \right) \,$	$= \operatorname{tr} \left(\gamma^\mu (2 \eta^{\nu \sigma} - \gamma^\sigma \gamma^\nu ) \gamma^\rho \right) \,$
	$= 2 \eta^{\nu \sigma} \operatorname{tr} \left(\gamma^\mu \gamma^\rho \right) - \operatorname{tr} \left(\gamma^\mu \gamma^\sigma \gamma^\nu \gamma^\rho \right) \quad \quad (2) \,$

Again, for the term on the right swap $\gamma^\sigma \,$ with its neighbor to the left,

$\operatorname{tr} \left( \gamma^\mu \gamma^\sigma \gamma^\nu \gamma^\rho \right) \,$	$= \operatorname{tr} \left( (2 \eta^{\mu \sigma} - \gamma^\sigma \gamma^\mu ) \gamma^\nu \gamma^\rho \right) \,$
	$= 2 \eta^{\mu \sigma} \operatorname{tr} \left( \gamma^\nu \gamma^\rho \right) - \operatorname{tr} \left( \gamma^\sigma \gamma^\mu \gamma^\nu \gamma^\rho \right)\quad \quad (3) \,$

Eq (3) is the term on the right of eq (2), and eq (2) is the term on the right of eq (1). We'll also use identity number 3 to simplify terms like so:

$2 \eta^{\rho \sigma} \operatorname{tr} \left(\gamma^\mu \gamma^\nu \right) = 2 \eta^{\rho \sigma} (4 \eta^{\mu \nu}) = 8 \eta^{\rho \sigma} \eta^{\mu \nu} .\,$

So finally Eq (1), when you plug all this information in gives

$\operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma) = 8 \eta^{\rho \sigma} \eta^{\mu \nu} - 8 \eta^{\nu \sigma} \eta^{\mu \rho} + 8 \eta^{\mu \sigma} \eta^{\nu \rho} \,$

$- \ \operatorname{tr} \left( \gamma^\sigma \gamma^\mu \gamma^\nu \gamma^\rho \right) \quad \quad \quad \quad \quad \quad (4) \,$

The terms inside the trace can be cycled, so

$\operatorname{tr} \left( \gamma^\sigma \gamma^\mu \gamma^\nu \gamma^\rho \right) = \operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma). \,$

So really (4) is

$2 \ \operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma) = 8 \eta^{\rho \sigma} \eta^{\mu \nu} - 8 \eta^{\nu \sigma} \eta^{\mu \rho} + 8 \eta^{\mu \sigma} \eta^{\nu \rho} \,$

or

$\operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma) = 4 \left( \eta^{\rho \sigma} \eta^{\mu \nu} - \eta^{\nu \sigma} \eta^{\mu \rho} + \eta^{\mu \sigma} \eta^{\nu \rho} \right) \,$

Proof of 5

To show

$\operatorname{tr}(\gamma^5)= 0$,

begin with

$\operatorname{tr}(\gamma^5)$	$= \operatorname{tr} (\gamma^0 \gamma^0 \gamma^5)$	(because $\gamma^0 \gamma^0 = I_4 \,$)
	$=-\operatorname{tr} (\gamma^0 \gamma^5 \gamma^0)$	(anti-commute the $\gamma^5 \,$ with $\gamma^0 \,$)
	$=-\operatorname{tr} (\gamma^0 \gamma^0 \gamma^5)$	(rotate terms within trace)
	$=-\operatorname{tr}(\gamma^5) \,$	(remove $\gamma^0 \,$'s)

Add $\operatorname{tr}(\gamma^5)$ to both sides of the above to see

$2\operatorname{tr}(\gamma^5)= 0 \,$.

Now, this pattern can also be used to show

$\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^5) = 0 \,$.

Simply add two factors of $\gamma^\alpha \,$, with $\alpha \,$ different from $\mu \,$ and $\nu \,$. Anticommute three times instead of once, picking up three minus signs, and cycle using the cyclic property of the trace.

So,

$\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^5) = 0 \,$ .

Proof of 6

For a proof of identity 6, the same trick still works unless $(\mu \nu \rho \sigma) \,$ is some permutation of (0123), so that all 4 gammas appear. The anticommutation rules imply that interchanging two of the indices changes the sign of the trace, so $\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^5) \,$ must be proportional to $\epsilon^{\mu\nu\rho\sigma} \,$ $(\epsilon^{0123} = \eta^{0\mu}\eta^{1\nu}\eta^{2\rho}\eta^{3\sigma}\epsilon_{\mu \nu \rho \sigma}= \eta^{00}\eta^{11}\eta^{22}\eta^{33}\epsilon_{0123}=-1)\,$. The proportionality constant is $4i \,$, as can be checked by plugging in $(\mu \nu \rho \sigma)=(0123) \,$, writing out $\gamma^5 \,$, and remembering that the trace of the identity is 4.

Proof of 7

Denote the product of $n$ gamma matrices by $\Gamma = \gamma^{\mu 1} \gamma^{\mu 2} \dots \gamma^{\mu n}.$ Consider the Hermitian conjugate of $\Gamma$:

$\Gamma^\dagger$	$= \gamma^{\mu n \dagger} \dots \gamma^{\mu 2 \dagger} \gamma^{\mu 1 \dagger}$
	$= \gamma^0 \gamma^{\mu n} \gamma^0 \dots \gamma^0 \gamma^{\mu 2} \gamma^0 \gamma^0 \gamma^{\mu 1} \gamma^0$	(since conjugating a gamma matrix with $\gamma^0$ produces its Hermitian conjugate as described below)
	$= \gamma^0 \gamma^{\mu n} \dots \gamma^{\mu 2} \gamma^{\mu 1} \gamma^0$	(all $\gamma^0$s except the first and the last drop out)

Conjugating with $\gamma^0$ one more time to get rid of the two $\gamma^0$s that are there, we see that $\gamma^0 \Gamma^\dagger \gamma^0$ is the reverse of $\Gamma$. Now,

$\operatorname{tr} (\gamma^0 \Gamma^\dagger \gamma^0)$	$= \operatorname{tr} (\Gamma^\dagger)$	(since trace is invariant under similarity transformations)
	$= \operatorname{tr} (\Gamma^*)$	(since trace is invariant under transposition)
	$= \operatorname{tr} (\Gamma)$	(since the trace of a product of gamma matrices is real)

Normalization

The gamma matrices can be chosen with extra hermiticity conditions which are restricted by the above anticommutation relations however. We can impose

$\left( \gamma^0 \right)^\dagger = \gamma^0 \,$, compatible with $\left( \gamma^0 \right)^2 = I_4 \,$

and for the other gamma matrices (for k = 1, 2, 3)

$\left( \gamma^k \right)^\dagger = -\gamma^k \,$, compatible with $\left( \gamma^k \right)^2 = -I_4. \,$

One checks immediately that these hermiticity relations hold for the Dirac representation.

The above conditions can be combined in the relation

$\left( \gamma^\mu \right)^\dagger = \gamma^0 \gamma^\mu \gamma^0. \,$

The hermiticity conditions are not invariant under the action $\gamma^\mu \to S(\Lambda) \gamma^\mu {S(\Lambda)}^{-1}$ of a Lorentz transformation $\Lambda$ because $S(\Lambda)$ is not necessarily a unitary transformation due to the noncompactness of the Lorentz group.

Feynman slash notation

The Feynman slash notation is defined by

$a\!\!\!/ := \gamma^\mu a_\mu$

for any 4-vector a.

Here are some similar identities to the ones above, but involving slash notation:

$a\!\!\!/b\!\!\!/ = a \cdot b - i a_\mu \sigma^{\mu\nu} b_\nu$

$a\!\!\!/a\!\!\!/ =a^{\mu}a^{\nu}\gamma_{\mu}\gamma_{\nu}=\frac{1}{2}a^{\mu}a^{\nu}(\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu})=\eta_{\mu\nu}a^{\mu}a^{\nu}= a^2$

$\operatorname{tr}(a\!\!\!/b\!\!\!/) = 4 (a \cdot b)$

$\operatorname{tr}(a\!\!\!/b\!\!\!/c\!\!\!/d\!\!\!/) = 4 \left[(a\cdot b)(c \cdot d) - (a \cdot c)(b \cdot d) + (a \cdot d)(b \cdot c) \right]$

$\operatorname{tr}(\gamma_5 a\!\!\!/b\!\!\!/c\!\!\!/d\!\!\!/) = -4 i \epsilon_{\mu \nu \rho \sigma} a^\mu b^\nu c^\rho d^\sigma$

$\gamma_\mu a\!\!\!/ \gamma^\mu = -2 a\!\!\!/$

$\gamma_\mu a\!\!\!/ b\!\!\!/ \gamma^\mu = 4 a \cdot b \,$

$\gamma_\mu a\!\!\!/ b\!\!\!/ c\!\!\!/ \gamma^\mu = -2 c\!\!\!/ b\!\!\!/ a\!\!\!/ \,$

where

$\epsilon_{\mu \nu \rho \sigma} \,$ is the Levi-Civita symbol and $\sigma^{\mu\nu} = \frac{i}{2} [\gamma^\mu, \gamma^\nu].$

Other representations

The matrices are also sometimes written using the 2×2 identity matrix, $I_2$, and

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

where k runs from 1 to 3 and the σ^k are Pauli matrices.

Dirac basis

The gamma matrices we have written so far are appropriate for acting on Dirac spinors written in the Dirac basis; in fact, the Dirac basis is defined by these matrices. To summarize, in the Dirac basis:

$\gamma^0 = \begin{pmatrix} I_2 & 0 \\ 0 & -I_2 \end{pmatrix},\quad \gamma^k = \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix},\quad \gamma^5 = \begin{pmatrix} 0 & I_2 \\ I_2 & 0 \end{pmatrix}.$

Weyl (chiral) basis

Another common choice is the Weyl or chiral basis, in which $\gamma^k$ remains the same but $\gamma^0$ is different, and so $\gamma^5$ is also different, and diagonal,

$\gamma^0 = \begin{pmatrix} 0 & I_2 \\ I_2 & 0 \end{pmatrix},\quad \gamma^k = \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix},\quad \gamma^5 = \begin{pmatrix} -I_2 & 0 \\ 0 & I_2 \end{pmatrix},$

or in more compact notation:

$\gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu \\ \bar \sigma^\mu & 0 \end{pmatrix}, \quad \sigma^\mu \equiv (1,\sigma^i), \quad \bar\sigma^\mu \equiv (1,-\sigma^i).$

The Weyl basis has the advantage that its chiral projections take a simple form,

$\psi_L=\frac12(1-\gamma^5)\psi=\begin{pmatrix} I_2 & 0 \\0 & 0 \end{pmatrix}\psi,\quad \psi_R=\frac12(1+\gamma^5)\psi=\begin{pmatrix} 0 & 0 \\0 & I_2 \end{pmatrix}\psi.$

The idempotence of the chiral projections is manifest. By slightly abusing the notation and reusing the symbols $\psi_{L/R}$ we can then identify

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

where now $\psi_L$ and $\psi_R$ are left-handed and right-handed two-component Weyl spinors.

Another possible choice^[4] of the Weyl basis has

$\gamma^0 = \begin{pmatrix} 0 & -I_2 \\ -I_2 & 0 \end{pmatrix},\quad \gamma^k = \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix},\quad \gamma^5 = \begin{pmatrix} I_2 & 0 \\ 0 & -I_2 \end{pmatrix}.$

The chiral projections take a slightly different form from the other Weyl choice,

$\psi_R=\begin{pmatrix} I_2 & 0 \\0 & 0 \end{pmatrix}\psi,\quad \psi_L=\begin{pmatrix} 0 & 0 \\0 & I_2 \end{pmatrix}\psi.$

In other words,

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

where $\psi_L$ and $\psi_R$ are the left-handed and right-handed two-component Weyl spinors, as before.

Majorana basis

There is also the Majorana basis, in which all of the Dirac matrices are imaginary and spinors are real. In terms of the Pauli matrices, it can be written as

$\gamma^0 = \begin{pmatrix} 0 & \sigma^2 \\ \sigma^2 & 0 \end{pmatrix}, \quad \gamma^1 = \begin{pmatrix} i\sigma^3 & 0 \\ 0 & i\sigma^3 \end{pmatrix}$

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

The reason for making the gamma matrices imaginary is solely to obtain the particle physics metric (+,−,−,−) in which squared masses are positive. The Majorana representation however is real. One can factor out the $i$ to obtain a different representation with four component real spinors and real gamma matrices. The consequence of removing the $i$ is that the only possible metric with real gamma matrices is (−,+,+,+).

Cℓ_1,3(C) and Cℓ_1,3(R)

The Dirac algebra can be regarded as a complexification of the real algebra Cℓ_1,3(R), called the space time algebra:

$Cl_{1,3}(\mathbb{C}) = Cl_{1,3}(\mathbb{R}) \otimes \mathbb{C}$

Cℓ_1,3(R) differs from Cℓ_1,3(C): in Cℓ_1,3(R) only real linear combinations of the gamma matrices and their products are allowed.

Two things deserve to be pointed out. As Clifford algebras, Cℓ_1,3(C) and Cℓ₄(C) are isomorphic, see classification of Clifford algebras. The reason is that the underlying signature of the spacetime metric loses its signature (3,1) upon passing to the complexification. However, the transformation required to bring the bilinear form to the complex canonical form is not a Lorentz transformation and hence not "permissible" (at the very least impractical) since all physics is tightly knit to the Lorentz symmetry and it is preferable to keep it manifest.

Proponents of geometric algebra strive to work with real algebras wherever that is possible. They argue that it is generally possible (and usually enlightening) to identify the presence of an imaginary unit in a physical equation. Such units arise from one of the many quantities in a real Clifford algebra that square to −1, and these have geometric significance because of the properties of the algebra and the interaction of its various subspaces. Some of these proponents also question whether it is necessary or even useful to introduce an additional imaginary unit in the context of the Dirac equation.

However, in contemporary practice, the Dirac algebra rather than the space time algebra continues to be the standard environment the spinors of the Dirac equation "live" in.

Euclidean Dirac matrices

In quantum field theory one can Wick rotate the time axis to transit from Minkowski space to Euclidean space. This is particularly useful in some renormalization procedures as well as lattice gauge theory. In Euclidean space, there are two commonly used representations of Dirac Matrices:

Chiral representation

$\gamma^{1,2,3} = \begin{pmatrix} 0 & i\sigma^{1,2,3} \\ -i\sigma^{1,2,3} & 0 \end{pmatrix}, \quad \gamma^4=\begin{pmatrix} 0 & I_2 \\ I_2 & 0 \end{pmatrix}$

Notice that the factors of $i$ have been inserted in the spatial gamma matrices so that the Euclidean Clifford algebra

$\{ \gamma^\mu , \gamma^\nu \} = 2 \delta^{\mu\nu}I_4$

will emerge. It is also worth noting that there are variants of this which insert instead $-i$ on one of the matrices, such as in lattice QCD codes which use the chiral basis.

In Euclidean space,

$\gamma_M^5 = i (\gamma^0\gamma^1\gamma^2\gamma^3 )_M = \frac{1}{i^2} (\gamma^4 \gamma^1\gamma^2\gamma^3 )_E = (\gamma^1 \gamma^2 \gamma^3 \gamma^4)_E = \gamma^5_E.$

Using the anti-commutator and noting that in Euclidean space $(\gamma^{\mu})^\dagger = \gamma^\mu$, one shows that

$(\gamma^5)^{\dagger} = \gamma^5$

In Chiral basis in Euclidean space,

$\gamma^5=\begin{pmatrix} -I_2 & 0\\ 0 & I_2 \end{pmatrix}$

which is unchanged from its Minkowski version.

Non-relativistic representation

$\gamma^{1,2,3} = \begin{pmatrix} 0 & -i \sigma^{1,2,3} \\ i \sigma^{1,2,3} & 0 \end{pmatrix}, \quad \gamma^4=\begin{pmatrix} I_2 & 0 \\ 0 & -I_2 \end{pmatrix}, \quad \gamma^5=\begin{pmatrix} 0 & -I_2 \\ -I_2 & 0 \end{pmatrix}$

See also

• Pauli matrices
• Gell-Mann matrices
• Higher-dimensional gamma matrices

References

1. The reason for the notation γ⁵ is because that set of matrices (Γ^A) = (γ^μ, iγ⁵) with A = (0, 1, 2, 3, 4) satisfy the five-dimensional Clifford algebra {Γ^A, Γ^B} = 2η^AB. Tong 2007, p. 93.
2. Weinberg 2002 Section 5.5.
3. de Wit & Smith 1996, p. 679.
4. Michio Kaku, Quantum Field Theory, ISBN 0-19-509158-2, appendix A

• Halzen, Francis; Martin, Alan (1984). Quarks & Leptons: An Introductory Course in Modern Particle Physics. John Wiley & Sons. ISBN 0-471-88741-2.
• A. Zee, Quantum Field Theory in a Nutshell (2003), Princeton University Press: Princeton, New Jersey. ISBN 0-691-01019-6. See chapter II.1.
• M. Peskin, D. Schroeder, An Introduction to Quantum Field Theory (Westview Press, 1995) [ISBN 0-201-50397-2] See chapter 3.2.
• W. Pauli (1936). "Contributions mathématiques à la théorie des matrices de Dirac". Annales de l'Institut Henri Poincaré 6: 109.
• Weinberg, S. (2002), The Quantum Theory of Fields 1, Cambridge University Press, ISBN 0-521-55001-7
• Tong, David (2007). "Quantum Field Theory". David Tong at University of Cambridge. p. 93. Retrieved 2015-03-07.
• de Wit, B.; Smith, J. (1986). Field Theory in Particle Physics. North-Holland Personal Library 1. North-Holland. ISBN 978-0444869999.Appendix E

External links

• Dirac matrices on mathworld including their group properties
• Hazewinkel, Michiel, ed. (2001), "Dirac matrices", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4