I am confused about the slash notation and especially taking the square of a slashed operator.

Defining $\displaystyle{\not} a \, = \, \gamma^\mu a_\mu$ we have $\,\,$ $\displaystyle{\not} a \displaystyle{\not} a = a^2 $

I tried to prove that, but I can't really doing it without assumption I didn't prove. That's my (I think wrong) procedure:

$\displaystyle{\not} a \displaystyle{\not} a = \gamma^\mu a_\mu \gamma^\nu a_\nu = \gamma^\mu \gamma^\nu a_\mu a_\nu $ assuming in the last equality that the they commute, now using the anticommutation relation of the gamma matrices :

$$\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu =2\eta^{\mu\nu}\tag{1}$$ I say that probably $\gamma^\mu \gamma^\nu= \eta^{\mu\nu}$ and substituing it in one i obtain

$$\eta^{\mu\nu}a_\mu a_\nu\,=\, a_\mu a^\nu \, = \, a^2$$

I don't think that's really a proof, can someone provide a right proof without the assumptions I've made?

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    $\begingroup$ Write $\gamma^\mu\gamma^\nu = (\gamma^\mu\gamma^\nu+\gamma^\mu\gamma^\nu)/2$ and use the symmetry of $a_\mu a_\nu$ w.r.t. index switching in one of the terms to change the gammas' order. $\endgroup$
    – Photon
    Jan 9, 2017 at 15:48
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    $\begingroup$ @DavidZ oh, I see. To OP: Well, in general $a_\mu\in\mathbb C$ is just a complex number, and therefore it commutes with anything. In any case, you can find a more or less detailed proof in the wikipedia entry (or just google "Feynman slash notation", the proof is rather standard) $\endgroup$ Jan 9, 2017 at 15:49
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    $\begingroup$ The basic assumption that you have made is that the $\gamma$ matrices and the $a$ operator (whatever it is) commute. You don't have to make the other assumption since you can add another expression with the $\mu$ and $\nu$ interchanged and cancel the extra factor of 2 at the end. Now your first assumption is generally valid as long as you are working in Cartesian coordinates (in curvilinear coordinates the $\gamma$ matrices can become coordinate dependent and would not commute with $a$ if it contains derivatives). Since you are free to choose Cartesian coordinates your proof is valid. $\endgroup$ Jan 9, 2017 at 15:55
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    $\begingroup$ Thank you all for clarification. @AccidentalFourierTransform you said that $a_\mu$ is usually just a complex number, but for example in the Dirac equation with this slashed notation we mean the 4-vector energy momentum contracted with the gamma matrices, When we quantize write as an operator: $-i \hbar \nabla$ could you clarify that? Does it commute with the gamma matrices even if it is an operator of that kind? $\endgroup$ Jan 9, 2017 at 16:37
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    $\begingroup$ @Runlikehell yes: even if it is a differential operator it commutes. Everything commutes with $\gamma^\mu$, except for other matrices (e.g., other gamma matrices, or products of the same). $\endgroup$ Jan 9, 2017 at 16:41

1 Answer 1


This: $$\gamma ^\mu \gamma ^\nu =\eta ^{\mu \nu}$$is wrong. It would give, for example, $$\gamma ^0 \gamma ^0=1=-\gamma ^1 \gamma ^1\\ \gamma ^0 \gamma ^1=0,$$ and you can see that these relations are inconsistent.

The relation comes from the following identities: $${\not} a \, {\not} a = a_\mu a_\nu \gamma ^\mu \gamma ^\nu=a_\mu a_\nu\left(\frac{\lbrace \gamma ^\mu ,\gamma ^\nu \rbrace }{2}+\frac{[ \gamma ^\mu ,\gamma ^\nu ]}{2}\right)=a_\mu a_\nu \frac{\lbrace \gamma ^\mu ,\gamma ^\nu \rbrace }{2}.$$ Notice: it is assumed that $a_\mu$ and $a_\nu$, whatever they are, commute.

The third equality is proved here.

The last equality is valid because contracting the symmetric tensor $a_\mu a_\nu$ with the antisymmetric indexed matrices $[\gamma ^\mu ,\gamma ^\nu]$ gives zero, that is, the contraction of a symmetric tensor ($a_\mu a_\nu \equiv A_{\mu\nu}$) and an antisymmetric tensor ($[\gamma ^\mu ,\gamma ^\nu] \equiv \Gamma^{\mu\nu}$) is zero. The proof is as follows.

\begin{eqnarray} A_{\mu\nu} \Gamma^{\mu\nu} &=& A_{\nu\mu} (-\Gamma^{\nu\mu}) \\ &=& -A_{\nu\mu}\Gamma^{\nu\mu} \\ &=& - A_{\mu\nu} \Gamma^{\mu\nu}, \quad [\because \mu, \nu \, \text{are dummy indices.}] \end{eqnarray} which implies that $A_{\mu\nu} \Gamma^{\mu\nu}$ is zero.

  • $\begingroup$ Thank you for your clear answer. You implicitly assumed right my assumption that $a_\mu$ and $\gamma^\nu$ commutes. Can you tell me why we're allowed to do that? $\endgroup$ Jan 9, 2017 at 16:29
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    $\begingroup$ $a_\mu$ (if it is an operator) and $\gamma^\mu$ are in different spaces, and therefore they commute. $\endgroup$ Jan 9, 2017 at 17:23
  • $\begingroup$ @Run like hell, you are welcome. Expanding a bit on Mikael's hint, the $\gamma ^\mu $'s which occur in all equations of quantum field theory commute with everything else which doesn't carry any Dirac index. This is totally clear if you write, for example ${\not} a$, element by element: $${\not}a _{\alpha \beta} = a_\mu \gamma ^{\mu }_{\alpha \beta}= \gamma ^{\mu }_{\alpha \beta} a_\mu,$$ since each $\gamma ^\mu _{\alpha \beta}$ is just a complex number. $\endgroup$
    – pppqqq
    Jan 9, 2017 at 18:42
  • $\begingroup$ How do we come to know that $a_{\mu} a_{\nu}$ is symmetric? $\endgroup$
    – rainman
    Dec 27, 2018 at 9:58
  • $\begingroup$ @omehoque right, thanks, it is symmetric if the $a_\mu$'s commute. $\endgroup$
    – pppqqq
    Dec 27, 2018 at 11:43

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