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). 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 space time acts. This in turn makes it possible to represent infinitesimal spatial rotations and Lorentz boosts. Spinors facilitate space-time 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

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

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

Analogue 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).

Mathematical structure

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

$\displaystyle\{ \gamma^\mu, \gamma^\nu \} = \gamma^\mu \gamma^\nu %2B \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu} I$

where $\eta^{\mu \nu} \,$ is the Minkowski metric with signature (+ − − −) and $\ I \,$ is the unit matrix.

This defining property is considered to be more fundamental than the numerical values used in 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:

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

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

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

Physical structure

The 4-tuple $\displaystyle\gamma^\mu=(\gamma^0,\gamma^1,\gamma^2,\gamma^3) = \gamma^0 e^0 %2B \gamma^1 e^1 %2B \gamma^2 e^2 %2B \gamma^3 e^3$ is often loosely described as a 4-vector (where e⁰ to e³ are the basis vectors of the 4-vector space). But this is misleading. Instead $\displaystyle\gamma^\mu$ is more appropriately seen as a mapping operator, taking in a 4-vector $\displaystyle a_\mu$ and mapping it to the corresponding matrix in the Clifford algebra representation.

This is symbolised by the useful Feynman slash notation,

$a\!\!\!/�:= \gamma^\mu a_\mu.$

Slashed quantities like $a\!\!\!/$ "live" in the multilinear Clifford algebra, with its own set of basis directions — they are immune to changes in the 4-vector basis.

On the other hand, one can define a transformation identity for the mapping operator $\displaystyle\gamma^\mu$ . If $\displaystyle\lambda$ is the spinor representation of an arbitrary Lorentz transformation $\displaystyle\Lambda$ , then we have the identity

$\displaystyle\gamma^\mu=\Lambda^\mu{}_\nu\lambda\gamma^\nu\lambda^{-1}.$

This says essentially that an operator mapping from the old 4-vector basis $\displaystyle\{e^0,e^1,e^2,e^3\}$ to the old Clifford algebra basis $\displaystyle\{\gamma^0,\gamma^1,\gamma^2,\gamma^3\}$ is equivalent to a mapping from the new 4-vector basis $\displaystyle\Lambda^\mu{}_\nu\{e^0,e^1,e^2,e^3\}$ to a correspondingly transformed new Clifford algebra basis $\displaystyle\lambda\{\gamma^0,\gamma^1,\gamma^2,\gamma^3\}\lambda^{-1}$ . Alternatively, in pure index terms, it shows that $\gamma^\mu$ transforms appropriately for an object with one contravariant 4-vector index and one covariant and one contravariant Dirac spinor index.

Given the above transformation properties of $\gamma^\mu$ , if $\psi$ is a Dirac spinor then the product $\gamma^\mu\psi$ transforms as if it were the product of a contravariant 4-vector with a Dirac spinor. In expressions involving spinors, then, it is often appropriate to treat $\gamma^\mu$ as if it were simply a vector.

There remains a final key difference between $\gamma^\mu$ and any nonzero 4-vector: $\gamma^\mu$ does not point in any direction. More precisely, the only way to make a true vector from $\gamma^\mu$ is to contract its spinor indices, leaving a vector of traces

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

This property of the gamma matrices is essential for them to serve as coefficients in the Dirac equation.

Expressing the Dirac equation

In natural units, the Dirac equation may be written as

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

where ψ is a Dirac spinor. Here, if $\gamma^\mu$ were an ordinary 4-vector, then it would pick out a preferred direction in spacetime, and the Dirac equation would not be Lorentz invariant.

Switching to Feynman notation, the Dirac equation is

$(i \not\!\partial - m) \psi = 0.$

Applying $-(i \not\!\partial %2B m)$ to both sides yields

$(\not\!\partial^2 %2B m^2) \psi = (\partial^2 %2B m^2) \psi = 0,$

which is the Klein-Gordon equation. Thus, as the notation suggests, the Dirac particle has mass m.

The Fifth Gamma Matrix, $\gamma^5$

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. 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

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

where $(\delta^{\alpha\beta\gamma\delta}_{\mu\nu\varrho\sigma})_{\alpha,...,\sigma \in \{0,...,3\}} =\delta(4)$ is the generalized Kronecker symbol (completely antisymmetric tensor w.r.t upper and lower indices separately) in $n=4$ dimensions, which is the unit operator on 4-forms. If $\varepsilon_{\alpha \dots \beta}$ denotes the Levi-Civita symbol in n dimensions, we can use the property

$\varepsilon_{\alpha \dots \beta}\varepsilon^{\alpha\dots\beta} = \sum_{\text{perm. of }\{0,\dots,n-1\} } 1 = n!$

to show the identity

$\delta^{\alpha\beta\gamma\delta}_{\mu\nu\varrho \sigma} = \frac{1}{4!} \varepsilon^{\alpha\beta\gamma\delta} \varepsilon_{\mu\nu\varrho\sigma}$ .

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:

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

Some properties are:

It is hermitian:

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

Its eigenvalues are ±1, because:

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

It anticommutes with the four gamma matrices:

$\left\{ \gamma^5,\gamma^\mu \right\} =\gamma^5 \gamma^\mu %2B \gamma^\mu \gamma^5 = 0. \,$

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	$\displaystyle\gamma^\mu\gamma_\mu=4 I$
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	$\displaystyle\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma_\mu=-2\gamma^\sigma\gamma^\rho\gamma^\nu$
5	$\displaystyle\gamma^\mu\gamma^\nu\gamma^\lambda = \eta^{\mu\nu}\gamma^\lambda %2B \eta^{\nu\lambda}\gamma^\mu - \eta^{\mu\lambda}\gamma^\nu - i\epsilon^{\sigma\mu\nu\lambda}\gamma_\sigma\gamma^5$

Proofs of 1 & 2

To show

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

one begins with the standard anticommutation relation

$\displaystyle\{ \gamma^\mu, \gamma^\nu \} = \gamma^\mu \gamma^\nu %2B \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu} I.$

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

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

To show

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

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

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

Proof of 3

To show

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

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

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

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

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

Finally using the anticommutator identity, we get

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

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)

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	$\operatorname{tr} (\gamma^\mu\gamma^\nu) = 4\eta^{\mu\nu}$
3	$\operatorname{tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=4(\eta^{\mu\nu}\eta^{\rho\sigma}-\eta^{\mu\rho}\eta^{\nu\sigma}%2B\eta^{\mu\sigma}\eta^{\nu\rho})$
4	$\operatorname{tr}(\gamma^5)=\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^5) = 0$
5	$\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^5) = 4i\epsilon^{\mu\nu\rho\sigma}$

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,

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

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 $\mu!=\nu$ )

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

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

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

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

Proof of 2

To show

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

Begin with,

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

Proof of 3

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

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 2 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

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

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

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

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

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

Proof of 4

To show

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

begin with

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

For a proof of identity 5, 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.

Normalization

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

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

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

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

One checks immediately that these hermicity 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 hermicity conditions are not invariant under the action $\gamma^\mu \to \lambda \gamma^\mu \lambda^{-1}$ of a Lorentz transformation because $\lambda$ is not a unitary transformation. This is intuitively clear because time and space are treated on unequal footing.

Feynman slash notation

The contraction of the mapping operator $\gamma^\mu$ with a vector $a_\mu$ maps the vector out of the 4-vector representation. So, it is common to write identities using the Feynman slash notation, defined by

$a\!\!\!/�:= \gamma^\mu a_\mu.$

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

$a\!\!\!/b\!\!\!/ = a \cdot b - 2i a_\mu S^{\mu\nu} b_\nu$

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

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

$\operatorname{tr}(a\!\!\!/b\!\!\!/c\!\!\!/d\!\!\!/) = 4 \left[(a\cdot b)(c \cdot d) - (a \cdot c)(b \cdot d) %2B (a \cdot d)(b \cdot c) \right]$

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

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

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

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

where

$\epsilon_{\mu \nu \rho \sigma} \,$ is the Levi-Civita symbol and $S^{\mu\nu} = \frac{i}{4} [\gamma^\mu, \gamma^\nu].$

Other representations

The matrices are also sometimes written using the 2x2 identity matrix, I, 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:

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

Weyl 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:

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

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

$\psi_L=\frac12(1-\gamma^5)\psi=\begin{pmatrix} I & 0 \\0 & 0 \end{pmatrix}\psi,\quad \psi_R=\frac12(1%2B\gamma^5)\psi=\begin{pmatrix} 0 & 0 \\0 & I \end{pmatrix}\psi.$

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^[1] of the Weyl basis has:

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

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

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

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's also a 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.

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

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

Different from Minkowski space, in Euclidean space,

$\gamma^5 = \gamma^1 \gamma^2 \gamma^3 \gamma^4 = \gamma^{5%2B}.$

So in Chiral basis,

$\gamma^5=\gamma^1 \gamma^2 \gamma^3 \gamma^4 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}.$

Non-relativistic representation

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

References

^ 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". Ann. Inst. Henri Poincaré 6: 109. http://www.numdam.org/item?id=AIHP_1936__6_2_109_0.

External links

Dirac matrices on mathworld including their group properties

Gamma matrices

Contents

Mathematical structure

Physical structure

Expressing the Dirac equation

The Fifth Gamma Matrix, $\gamma^5$

Identities

Miscellaneous identities

Trace identities

Normalization

Feynman slash notation

Other representations

Dirac basis

Weyl basis

Majorana basis

Cℓ_1,3(C) and Cℓ_1,3(R)

Euclidean Dirac matrices

Chiral representation

Non-relativistic representation

See also

References

External links

Gamma matrices

Contents

Mathematical structure

Physical structure

Expressing the Dirac equation

The Fifth Gamma Matrix,

Identities

Miscellaneous identities

Trace identities

Normalization

Feynman slash notation

Other representations

Dirac basis

Weyl basis

Majorana basis

Cℓ1,3(C) and Cℓ1,3(R)

Euclidean Dirac matrices

Chiral representation

Non-relativistic representation

See also

References

External links

The Fifth Gamma Matrix, $\gamma^5$

Cℓ_1,3(C) and Cℓ_1,3(R)