I'm looking to prove the following identity:

$$k_a \gamma^a \gamma^\nu K_b \gamma^b p_c \gamma^c \gamma_\nu P_d \gamma^d = 4(p\cdot K)(P\cdot k)$$ I tried this many times but always seem to be stuck in a loop with two $\gamma$-s multiplying my final expression. What I've used during my calculation are the following two things $$\gamma^\nu \gamma_\nu =4$$ as well as the anti-commutation relations the $\gamma$-s satisfy $$\{\gamma^\mu,\gamma^\nu\} = 2\eta^{\mu\nu}$$ The final expression I reach is the following $$4(p\cdot K)k_aP_d\gamma^a \gamma^d$$

I'd be happy to get some help with this.

  • 1
    $\begingroup$ Is there supposed to be a trace on the left hand side of the identity? $\endgroup$ – G. Smith Feb 18 at 1:10

Looks like the identity is not always correct. If $p$ and $K$ have only the zero-th components $p_0\neq 0$ and $K_0\neq 0$, we get your "final expression" in the left-hand side. If then $k$ has only the zero-th component $k_0\neq 0$ and $P$ has only the first component $P_1\neq 0$, we obtain an incorrect equality $4p_0 K_0 k_0 P_1 \gamma^0 \gamma^1=0$.


Your identity appears to equate a 4x4 matrix to a number. If one inserts an identity matrix on the right-hand-side to fix this, the equality does not hold. Therefore I think you have omitted a trace on the left-hand-side. However, then the right-hand-side is off by a factor of 4.

I think that what you are trying to show is actually

$$\text{Tr}(k_a \gamma^a \gamma^\nu K_b \gamma^b p_c \gamma^c \gamma_\nu P_d \gamma^d) = 16(p\cdot K)(P\cdot k).$$

This follows from

$$\gamma^\nu \gamma^b \gamma^c \gamma_\nu = 4 \eta^{bc} I$$


$$\text{Tr}(\gamma^a \gamma^d) = 4 \eta^{ad},$$

both of which are standard identities listed here:


  • $\begingroup$ Question for MathJax experts: Is there an easy way to make a Dirac slash? $\endgroup$ – G. Smith Feb 18 at 6:13
  • $\begingroup$ \not x becomes $\not x$. $\endgroup$ – J.G. Feb 18 at 6:38

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