Closed formula for product of gamma matrices

I was asking myself if there is a closed formula for the following product of gamma matrices:

$$\gamma_\mu\gamma_\nu \gamma_5.$$

I would like to express this matrix in terms of the basis

$${\mathbb{1}, \gamma_{5}, \gamma_{\mu}, \gamma_{\mu}\gamma_{5},[\gamma_{\nu},\gamma_{\mu}]} .$$

Thank you very much for any help.

• Hint: Use the Levi-Civita tensor $\epsilon^{\mu\nu\rho\sigma}$. – suresh Feb 2 '14 at 0:40
• Thank you, Suresh. I found $\gamma_\mu\gamma_\nu\gamma_5=i \varepsilon_{\mu\nu\alpha\beta} \gamma^\alpha \gamma^\beta$ when I take $\varepsilon^{0123}=1$. – Melquíades Feb 2 '14 at 0:53
• It is encouraged to add your own answer to your question if you have found the solution yourself. – kleingordon Feb 2 '14 at 0:59

Given a vector space with a basis, one way to find the coordinates of a given vector with respect to that basis would be to define an inner product which makes the basis orthogonal. Then if the vector in question is $v$ and the basis is $e_i$, then the $i$th coordinate of $v$ is $\frac{v \cdot e_i }{|e_i|^2}$.

This is the strategy I will use here, but the question is to find the appropriate inner product. Since all I seem to remember from QFT is doing traces of these matrices, the natural inner product seems to be multiplying the two matrices together and taking the trace.

I should say that to stick with this strategy I will view the indices as enumeration indices instead of lorentz indices. Thus I will not raise or lower them although summation will be implied over repeated indices. At the very end I will go back to usual notation.

The first step then is to verify that this basis is indeed orthogonal with respect to this inner product. This is easily checked once a few identies are recalled. First, the trace of $\gamma^5$ with fewer than four $\gamma$s (of the non-five variety) is zero; $\gamma^5$ anticommutes with $\gamma^\mu$; $\mathrm{Tr} \gamma^5 = 0$; the trace of an odd number of $\gamma$'s (of the non-five variety) is zero; $(\gamma^5)^2=1$; and $\mathrm{Tr}(\gamma^\mu \gamma^\nu) = 4 g^{\mu \nu}$.

Now we need to compute two things: the square norm of each basis element, and compute the inner product of our vector with each basis element.

Let's start with the first. We have $\mathrm{Tr}(1*1)=4$,

$\mathrm{Tr}(\gamma^5*\gamma^5)=\mathrm{Tr}(1)=4$,

$\mathrm{Tr}(\gamma^\mu *\gamma^\nu )=\mathrm{Tr}(\gamma^\mu \gamma^\nu) = 4g^{\mu\nu}$. This is zero for $\mu \ne \nu$ and it is $4(-1)^\mu$ for $\mu = \nu$, where $(-1)^\mu$ is just the $\mu$th diagonal entry of $g^{\mu\nu}$.

$\mathrm{Tr}(\gamma^\mu \gamma^5*\gamma^\nu \gamma^5)=\mathrm{Tr}(-\gamma^\mu \gamma^\nu \gamma^5\gamma^5) = -4g^{\mu\nu} = -4(-1)^\mu$ for $\mu = \nu$ and zero otherwise.

$\mathrm{Tr}([\gamma^\mu, \gamma^\nu]*[\gamma^\lambda \gamma^\eta])$ is just $4 \mathrm{Tr}(\gamma^\mu \gamma^\nu \gamma^\lambda \gamma^\eta)$ antisymmetrized over $(\mu \nu)$ and $(\lambda \eta)$, so it is the same as $16(g^{\mu \nu} g^{\lambda \eta} - g^{\mu \lambda} g^{\mu \eta} + g^{\mu \eta} g^{\lambda \nu})$ antisymmetrized over $(\mu \nu)$ and $(\lambda \eta)$. Thus it is the same as $-16(g^{\mu \eta} g^{\lambda \nu} -g^{\mu \lambda} g^{\mu \eta} )$ antisymmetrized over $(\mu \nu)$ and $(\lambda \eta)$. Since we don't consider $\mu = \nu$ or $\lambda = \eta$ here, this is only non-zero if $\{\mu, \nu \} = \{\lambda, \eta\}$ as a set. Without loss of generality we may assume $\mu = \lambda$ and $\nu = \eta$. Then the expression becomes $-16((-1)^\mu(-1)^\nu)$

Now that that is done we can compute our inner products with the matrix in question. The inner products of our vector with $1$, $\gamma^\lambda$, and $\gamma^\lambda \gamma^5$ are zero. However, the inner product with $\gamma^5$ is $4 g^{\mu \nu}$, and the inner product with $[\gamma^\lambda, \gamma^\eta]$ is $-8i \epsilon^{\lambda \eta \mu \nu}$.

Putting the two pieces from the last two paragraphs together, we get that our matrix is $g^{\mu \nu} \gamma^5 + \frac{i \epsilon ^{\mu \nu \lambda \eta}}{2(-1)^\lambda(-1)^\eta}\ \ [\gamma^\lambda, \gamma^\eta] = g^{\mu \nu} \gamma^5 + \frac{i}{2} \epsilon ^{\mu \nu} {}_{\lambda \eta}[\gamma^\lambda, \gamma^\eta]$. Note this expression differs from one given in the comments to the question.

This formula can be checked with mathematica code:

g = DiagonalMatrix[{1, -1, -1, -1}];

unit2d = IdentityMatrix[2];

(sigma1 = {{0, 1}, {1, 0}}) // MatrixForm

(sigma2 = {{0, -I}, {I, 0}}) // MatrixForm

(sigma3 = {{1, 0}, {0, -1}}) // MatrixForm

makeGamma[sigma_] :=
ArrayFlatten[{{0*unit2d, sigma}, {-sigma, 0*unit2d}}]

(gamma1 = makeGamma[sigma1]) // MatrixForm

(gamma2 = makeGamma[sigma2]) // MatrixForm

(gamma3 = makeGamma[sigma3]) // MatrixForm

(gamma0 =
ArrayFlatten[{{0*unit2d, unit2d}, {unit2d,
0*unit2d}}]) // MatrixForm

(gamma5 = DiagonalMatrix[{-1, -1, 1, 1}]) // MatrixForm

gammas = {gamma0, gamma1, gamma2, gamma3};

calculateDirect[i_, j_] := gammas[[i + 1]].gammas[[j + 1]].gamma5

calculateWithCoordinates[i_, j_] := Module[
{
otherIndices = Complement[{0, 1, 2, 3}, {i, j}],
lambda,
eta
},
lambda = otherIndices[[1]];
eta = otherIndices[[2]];
g[[i + 1, j + 1]]*gamma5 +
I*g[[lambda + 1, lambda + 1]]*g[[eta + 1, eta + 1]]*
Signature[{i, j, lambda,
eta}]*(gammas[[lambda + 1]].gammas[[eta + 1]] -
gammas[[eta + 1]].gammas[[lambda + 1]])/2
]

errorSoFar = 0;
For[i = 0, i < 4, i++,
For[j = 0, j < 4, j++,
errorSoFar +=
Total[Total[
Abs[calculateDirect[i, j] - calculateWithCoordinates[i, j]]
]
]
]
]
errorSoFar

• I just disagree about a small point: there is a factor two missing in your final equation: $\eta_{\mu\nu}\gamma^5+i \varepsilon_{\mu\nu\lambda\eta}[\gamma^\lambda,\gamma^\eta]/2$. – Melquíades Feb 2 '14 at 20:46
• Is $[\gamma^\lambda, \gamma ^\eta] = \frac{1}{2}(\gamma^\lambda \gamma ^\eta - \gamma ^\eta \gamma^\lambda)$? That's what I thought, and my mathematica code is consistent with this. In other words, if you interpret my final formula that way, then it is correct. – Brian Moths Feb 2 '14 at 21:17
• oh ya sorry I was confusing $[\gamma^\lambda,\gamma^\eta]$ with $\gamma^{[\lambda} \gamma^{\eta]}$. – Brian Moths Feb 3 '14 at 18:48