I am following Peskin. During the derivation of the Dirac field (boosting the solution from the rest frame), we use that $(\vec{p}\cdot \vec{\sigma})^2 = |p|^2$. Where $\vec{p}$ is momentum vector, and $\vec{\sigma}$ is the Pauli matrices-vector. I dont see how this is correct. I believe I saw a short proof using the anti-commutator of the sigma matrices. It went something like: $$ (\vec{p}\cdot \vec{\sigma})^2 = p^i\sigma^i p^j \sigma^j = p^i p^j \frac{1}{2}\{\sigma^i, \sigma^j\} = |p|^2, $$

I understand the first equality, but not the rest. Can you present a short explaination/proof?

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    $\begingroup$ Tell us why it is wrong. $\endgroup$ – AMS Sep 21 '16 at 18:27
  • $\begingroup$ I hope it's true, just can't see why it's true. Its probably obvious $\endgroup$ – Mikkel Rev Sep 21 '16 at 18:29
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    $\begingroup$ Simplest way is to write the matrix $\mathbf{p}\cdot\mathbf{\sigma}$ explicitly. $\endgroup$ – Javier Sep 21 '16 at 18:33

As far as I know, what is true is that $$(\vec{p}\cdot \vec{\sigma})^2=\vec{p} \cdot \vec{p}=|\vec{p}|^2$$ and so what you want to prove is not true unless your $\vec{p}$ is a unit vector. Is this the reason why you are confused? Otherwise for the last step you just need to use the fact that $$\{\sigma_i,\sigma_j\}=2\delta_{ij}I$$ which you can easily prove yourself by checking the actual matrix multiplications. On the other hand the step $$p^i\sigma^i p^j \sigma^j = p^i p^j \frac{1}{2}\{\sigma^i, \sigma^j\}$$ is done by substituting the product $\sigma_i\sigma_j$ with its symmetric part. This is possible since any antisymmetric part would disappear after contraction with the symmetric tensor $p^ip^j$.

  • $\begingroup$ What do you mean by: "substituting the product σiσj with its symmetric part"? Thanks $\endgroup$ – Mikkel Rev Sep 22 '16 at 11:43
  • $\begingroup$ Each tensor $T_{ij}$ can be written as the sum of a symmetric part $1/2 (T_{ij}+T_{ji})$ and an antisymmetric part $1/2 (T_{ij}-T_{ji})$. If the tensor is $T_{ij}=\sigma_i\sigma_j$ its symmetric part is exactly the anticommutator of the sigmas (and their commutator is the antisymmetric part) $\endgroup$ – DelCrosB Sep 22 '16 at 11:49
  • $\begingroup$ Thanks. $\sigma^i \sigma^j = \frac{1}{2}(\sigma^i \sigma^j + \sigma^j \sigma^i) + \frac{1}{2}(\sigma^i \sigma^j - \sigma^j \sigma^i) = \frac{1}{2}\{\sigma^j ,\sigma^i\} + \frac{1}{2}(\sigma^i \sigma^j - \sigma^j \sigma^i)$. And how does the last part vanish when I contract it with $p_i p _j$? $\endgroup$ – Mikkel Rev Sep 22 '16 at 11:59
  • $\begingroup$ Yes, exactly. And the last two terms give $1/2[\sigma_i, \sigma_j]$ $\endgroup$ – DelCrosB Sep 22 '16 at 12:02
  • $\begingroup$ If $S^{ij}$ is symmetric in $i$ and $j$ and $A_{ij}$ is antisymmetric, then $S^{ij}A_{ij}=0$ This is a nice simple exercise and a very handy trick to know if you work with tensors. $\endgroup$ – DelCrosB Sep 22 '16 at 12:05

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