Using the anticommutators of the gamma matrices, one can show that for any ${\displaystyle a_{\mu }}$  and ${\displaystyle b_{\mu }}$ ,

${\displaystyle {\begin{aligned}{a\!\!\!/}{a\!\!\!/}=a^{\mu }a_{\mu }\cdot I_{4}=a^{2}\cdot I_{4}\\{a\!\!\!/}{b\!\!\!/}+{b\!\!\!/}{a\!\!\!/}=2a\cdot b\cdot I_{4}.\end{aligned}}}$ 

where ${\displaystyle I_{4}}$  is the identity matrix in four dimensions.

In particular,

${\displaystyle {\partial \!\!\!/}^{2}=\partial ^{2}\cdot I_{4}.}$ 

Further identities can be read off directly from the gamma matrix identities by replacing the metric tensor with inner products. For example,

${\displaystyle {\begin{aligned}\gamma _{\mu }{a\!\!\!/}\gamma ^{\mu }&=-2{a\!\!\!/}\\\gamma _{\mu }{a\!\!\!/}{b\!\!\!/}\gamma ^{\mu }&=4a\cdot b\cdot I_{4}\\\gamma _{\mu }{a\!\!\!/}{b\!\!\!/}{c\!\!\!/}\gamma ^{\mu }&=-2{c\!\!\!/}{b\!\!\!/}{a\!\!\!/}\\\gamma _{\mu }{a\!\!\!/}{b\!\!\!/}{c\!\!\!/}{d\!\!\!/}\gamma ^{\mu }&=2({d\!\!\!/}{a\!\!\!/}{b\!\!\!/}{c\!\!\!/}+{c\!\!\!/}{b\!\!\!/}{a\!\!\!/}{d\!\!\!/})\\\operatorname {tr} ({a\!\!\!/}{b\!\!\!/})&=4a\cdot b\\\operatorname {tr} ({a\!\!\!/}{b\!\!\!/}{c\!\!\!/}{d\!\!\!/})&=4\left[(a\cdot b)(c\cdot d)-(a\cdot c)(b\cdot d)+(a\cdot d)(b\cdot c)\right]\\\operatorname {tr} ({a\!\!\!/}{\gamma ^{\mu }}{b\!\!\!/}{\gamma ^{\nu }})&=4\left[a^{\mu }b^{\nu }+a^{\nu }b^{\mu }-\eta ^{\mu \nu }(a\cdot b)\right]\\\operatorname {tr} (\gamma _{5}{a\!\!\!/}{b\!\!\!/}{c\!\!\!/}{d\!\!\!/})&=4i\varepsilon _{\mu \nu \lambda \sigma }a^{\mu }b^{\nu }c^{\lambda }d^{\sigma }\\\operatorname {tr} ({\gamma ^{\mu }}{a\!\!\!/}{\gamma ^{\nu }})&=0\\\operatorname {tr} ({\gamma ^{5}}{a\!\!\!/}{b\!\!\!/})&=0\\\operatorname {tr} ({\gamma ^{0}}({a\!\!\!/}+m){\gamma ^{0}}({b\!\!\!/}+m))&=8a^{0}b^{0}-4(a.b)+4m^{2}\\\operatorname {tr} (({a\!\!\!/}+m){\gamma ^{\mu }}({b\!\!\!/}+m){\gamma ^{\nu }})&=4\left[a^{\mu }b^{\nu }+a^{\nu }b^{\mu }-\eta ^{\mu \nu }((a\cdot b)-m^{2})\right]\\\operatorname {tr} ({a\!\!\!/}_{1}...{a\!\!\!/}_{2n})&=\operatorname {tr} ({a\!\!\!/}_{2n}...{a\!\!\!/}_{1})\\\operatorname {tr} ({a\!\!\!/}_{1}...{a\!\!\!/}_{2n+1})&=0\end{aligned}}}$ 

where:

• ${\displaystyle \varepsilon _{\mu \nu \lambda \sigma }}$  is the Levi-Civita symbol
• ${\displaystyle \eta ^{\mu \nu }}$  is the Minkowski metric
• ${\displaystyle m}$  is a scalar.

This section uses the (+ − − −) metric signature. Often, when using the Dirac equation and solving for cross sections, one finds the slash notation used on four-momentum: using the Dirac basis for the gamma matrices,

${\displaystyle \gamma ^{0}={\begin{pmatrix}I&0\\0&-I\end{pmatrix}},\quad \gamma ^{i}={\begin{pmatrix}0&\sigma ^{i}\\-\sigma ^{i}&0\end{pmatrix}}\,}$ 

as well as the definition of contravariant four-momentum in natural units,

${\displaystyle p^{\mu }=\left(E,p_{x},p_{y},p_{z}\right)\,}$ 

we see explicitly that

${\displaystyle {\begin{aligned}{p\!\!/}&=\gamma ^{\mu }p_{\mu }=\gamma ^{0}p^{0}-\gamma ^{i}p^{i}\\&={\begin{bmatrix}p^{0}&0\\0&-p^{0}\end{bmatrix}}-{\begin{bmatrix}0&\sigma ^{i}p^{i}\\-\sigma ^{i}p^{i}&0\end{bmatrix}}\\&={\begin{bmatrix}E&-{\vec {\sigma }}\cdot {\vec {p}}\\{\vec {\sigma }}\cdot {\vec {p}}&-E\end{bmatrix}}.\end{aligned}}}$ 

Similar results hold in other bases, such as the Weyl basis.

• Weyl basis
• Gamma matrices
• Four-vector
• S-matrix