Linear Algebra – Do Gamma Matrices Form a Basis? How to Understand Dirac and Clifford Algebra

clifford-algebradirac-matriceslinear algebra

Do the four gamma matrices form a basis for the set of matrices $GL(4,\mathbb{C})$? I was actually trying to evaluate a term like $\gamma^0 M^\dagger \gamma^0$ in a representation independent way, where $M, M^\dagger$ are $4\times 4$ matrices.

Best Answer

As previous answers have correctly noted gamma matrices do not forma a basis of $M(4,\mathbb{C})$. Nevertheless you can construct one from them in the following way

  • 1 the identity matrix $\mathbb{1}$
  • 4 matrices $\gamma^\mu$
  • 6 matrices $\sigma^{\mu\nu}=\gamma^{[\mu}\gamma^{\nu]}$
  • 4 matrices $\sigma^{\mu\nu\rho}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho]}$
  • 1 matrix $\sigma^{\mu\nu\rho\delta}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\delta]}=i\epsilon^{\mu\nu\rho\delta}\gamma^5$

these 16 matrices form the basis that we were looking for.

Furthermore they are used to construct the spinor bilinears multiplying by $\bar{\psi}$ on the left and $\psi$ on the right, which transform in the Lorentz indices as follows

  • $\bar{\psi}\psi$ scalar
  • $\bar{\psi}\gamma^\mu\psi$ vector
  • $\bar{\psi}\sigma^{\mu\nu}\psi$ 2nd rank (antisymmetric) tensor
  • $\bar{\psi}\sigma^{\mu\nu\rho}\psi$ pseudovector
  • $\bar{\psi}\gamma^5\psi$ pseudoscalar

the fact that they form a basis of of $M(4,\mathbb{C})$ is very important because these are then the only independent spinor bilinears (i.e $\bar{\psi}M\psi$) that can be constructed, any other can be expressed a linear combination of these. A different issue is if it would make any sense to sum any of these since they are different types of tensors under Lorentz group transformations.

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