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I am self studying QFT from An Introduction to QFT and currently, I am completing problem 3.3(a). Here are some sample solutions that I am using to understand this problem: (https://theoreticalmaximum.files.wordpress.com/2017/07/intro-to-qft-solutions2.pdf). In the middle of page 9, the author writes:

$\frac{1}{\sqrt{2p\cdot k_0}} p_\mu p_\nu \gamma^\mu \gamma^\nu u_{R/L}(0) = \frac{1}{\sqrt{2p \cdot k_0}} p_\mu p_\nu\eta^{\mu \nu} u_{R/L}(0)$.

Isn't $\gamma^\mu \gamma^\nu = \eta^{\mu \nu} - i\sigma^{\mu \nu}$ (from the identity $\sigma^{\mu \nu} = \frac{i}{2}[\gamma^\mu, \gamma^\nu]$)? This seems to suggest that $p_\mu p_\nu\sigma^{\mu\nu} = 0$, but I do not completely understand why this is true. Can anyone explain?

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    $\begingroup$ A symmetric tensor saturated on an antisymmetric tensor always vanishes. Do you wish to withdraw your question? $\endgroup$ May 26 at 19:01
  • $\begingroup$ @CosmasZachos Why is that true? Sorry, I am new to tensor math. $\endgroup$
    – user758469
    May 26 at 19:02
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    $\begingroup$ Transpose the two indices to see that the object is minus itself! $\endgroup$ May 26 at 19:04
  • $\begingroup$ @CosmasZachos Thank you!! $\endgroup$
    – user758469
    May 26 at 19:04
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As @Cosmos Zachos comments, the contraction of an antisymmetric tensor with a symmetric one vanishes. By antisymmetry of $\sigma$,

$$p_\mu p_\nu \sigma^{\mu \nu} = -p_\nu p_\mu \sigma^{\nu \mu},$$ But as the indices are summed over we can relabel $\mu \to \nu$ on the RHS and we have

$$p_\mu p_\nu \sigma^{\mu \nu} = -p_\mu p_\nu \sigma^{\mu \nu},$$ which can only be true if both sides are equal to zero.

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