Quick question...For some reason I'm having trouble finding an identity or discussion for the commutator of the gamma matrices at the moment...i.e $$\gamma^u\gamma^v-\gamma^v \gamma^u$$ but I am not finding this anywhere. I have an idea of what it may be, but then again I'm not always right. Can anyone fill me in here fill me in? (I already know the anticommutator, i.e $$\gamma^u\gamma^v+\gamma^v\gamma^u=2g^{uv}I.)$$

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    $\begingroup$ There is no identity for $[\gamma^\mu , \gamma^\nu]$. $\endgroup$ – Prahar Nov 3 '14 at 1:46
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    $\begingroup$ Ok then, but to clarify a discussion of what it represents would be useful. I read its related to the Lie Algebra somewhere but as to further details (as in details beyond being a commutator)... not finding them. $\endgroup$ – IntuitivePhysics Nov 3 '14 at 2:52
  • $\begingroup$ @Prahar The commutators definitely form a representation of the Lorentz algebra as JamalS says below. These can in turn be used to represent finite Lorentz transformations when exponentiated. $\endgroup$ – R. Rankin Nov 15 '18 at 1:02
  • $\begingroup$ @R.Rankin - The commutators of the Lorentz generators, (namely $S_{\mu\nu}$ in JamalS's answer) form the Lorentz algebra. The commutators of the Dirac matrices satisfy no such algebra. $\endgroup$ – Prahar Nov 15 '18 at 1:59
  • $\begingroup$ @Prahar I meant 1/4 the commutator. JamalS clearly defines: $$S^{\mu\nu}=\frac{1}{4}[\gamma^{\mu},\gamma^{\nu}]$$ where of course the Dirac matrices are but one possible example of the $\gamma$s (we could have chosen Weyl, Majorana or whatever matrices satisfy the anticommutation relations for the Minkowski space). $\endgroup$ – R. Rankin Nov 15 '18 at 2:07

Although the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}$ is the most famous, there is an expression for the commutator:

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

The matrix defined by $[\gamma^\mu,\gamma^\nu]$ actually has a purpose: it forms a representation of the Lorentz algebra. If we define $S^{\mu\nu}$ as $1/4$ the commutator, then we have,

$$[S^{\mu\nu},S^{\rho\sigma}] = \eta^{\nu\rho}S^{\mu\sigma} - \eta^{\mu\rho}S^{\nu\sigma} + \eta^{\mu\sigma}S^{\nu\rho} - \eta^{\nu\sigma}S^{\mu\rho}= \eta^{\rho[\nu}S^{\mu]\sigma} + \eta^{\sigma [\mu}S^{\nu]\rho}$$

which is the Lorentz algebra. One can verify this by simply using the first commutator, and the rule for the commutator involving a product.

There is a particularly important use for the commutator, namely defining $\sigma^{\mu\nu} = \frac{i}{2} [\gamma^\mu,\gamma^\nu]$, the action of a spin-$\frac32$ particle is given by,

$$\mathcal L = -\frac{1}{2}\bar{\psi}_\mu \left( \varepsilon^{\mu\lambda \sigma \nu} \gamma_5 \gamma_\lambda \partial_\sigma -im\sigma^{\mu\nu}\right)\psi_v,$$

which can be used to describe the superpartner to the graviton, namely the gravitino, thus making it necessary for supergravity theories.

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  • $\begingroup$ Right, but is it useful anywhere? (considering that it doesn't look independent of the Clifford one, I mean from the point of view of the other answer). $\endgroup$ – 299792458 Nov 3 '14 at 7:30
  • $\begingroup$ @New_new_newbie: Yes, a spinor transforms as $\psi^a \to S[\Lambda]^a_b \psi^b(\Lambda^{-1}x)$, where $S[\Lambda] = \exp(\frac{1}{2}M_{ab}S^{ab})$, i.e. it is used as a basis for the generators. (Well, in my convention, $1/4$ the commutator is used.) $\endgroup$ – JamalS Nov 3 '14 at 7:37
  • $\begingroup$ Thanks for the great reply. I'm not entirely sure what $M_{ab}$ is though. Again, I have a guess, but could you clarify. Thanks. $\endgroup$ – IntuitivePhysics Nov 9 '14 at 3:36
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    $\begingroup$ @TheDarkSide Added another use I just thought of, after two years :) $\endgroup$ – JamalS Nov 20 '16 at 15:20

The OP wrote in a comment on the commutators of gamma-matrices: "to clarify a discussion of what it represents would be useful. I read its related to the Lie Algebra somewhere but as to further details (as in details beyond being a commutator)... not finding them."

In some sense, the Dirac gamma matrices can be identified with mutually orthogonal unit vectors (orts) of the Cartesian basis in 3+1 spacetime, with their anticommutators corresponding to scalar products of the orts (this approach is used, e.g., in the book "The Theory of Spinors" by Cartan, the discoverer of spinors). Then the non-vanishing commutators of gamma-matrices (say, in the form $\sigma^{\mu\nu}=\frac{1}{2}[\gamma^\mu,\gamma^\nu]$) can be identified with the so called bivectors (2-dimensional planes in the spacetime spanned by two orts).

@TheDarkSide asked if the commutator is useful anywhere. Some uses are mentioned in other answers, but let me tell you how it was useful for me. Some time ago, I showed that the Dirac equation (which is a system of four first-order equations for four components of the Dirac spinor) is generally equivalent to just one fourth-order equation for one component of the Dirac spinor (http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf, published in the Journal of Mathematical Physics). Recently, I derived a relativistically covariant form of the equation ((https://arxiv.org/abs/1502.02351 , eq. (27)), where the following linear combination of the commutators of gamma-matrices is heavily used: $F=\frac{1}{2}F_{\mu\nu}\sigma^{\mu\nu}$, where $F_{\mu\nu}$ is the electromagnetic field.

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The above answers are good, yet I'm suprised that no one has mentioned that the commutator of the Dirac matrices is required in the description of ANY fermion in a general non-flat space. In such a space the Dirac equation reads (utilizing the tetrad formulation):




where $\omega$ is the so-called spin connection and $\sigma$ is defined as in JamalS's answer above. If you want to do quantum mechanics with fermions in curved spacetime, you will certainly need to utilize the commutator of Dirac matrices.

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$\gamma^u\gamma^v=-\gamma^v\gamma^u$ for $u\neq v$

Therfore $\gamma^u\gamma^v-\gamma^v\gamma^u=2\gamma^u\gamma ^v$ for $u \neq v$

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  • $\begingroup$ This is not true since the anticommutator is not zero (which your first line suggests). $\endgroup$ – Eva Mar 4 '17 at 17:48
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    $\begingroup$ @Eva The first line clearly states $\mu\ne\nu$. The anticommutator is $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g_{\mu\nu}I$, which is zero, except on the diagonals when $\mu=\nu$. $\endgroup$ – Geoff Pointer Apr 7 '17 at 17:27

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