In ''An intro to QFT (2018)'' chapter 3, Peskin does the following:

Let me introduce some notation first, let $v^s_k=\begin{pmatrix}\;\;\,\sqrt{k\cdot\sigma}\,\xi^{-s}\\-\sqrt{k\cdot\bar{\sigma}}\,\xi^{-s}\end{pmatrix}$ be a bispinor for the negative-energy solution to the Dirac eq. with momentum $k$ and spin state $\xi^{-s}$.

From previous chapters we know $\,\xi^{-s}\equiv-i\sigma_2(\xi^s)^*$ and $\,\sigma_2^*=-\sigma_2\,$, then, $\,(\xi^{-s})^*=-i\sigma_2\,\xi^s$. We also know $\,(\sqrt{k\cdot\sigma^*}\sigma_2=\sigma_2\sqrt{k\cdot\bar{\sigma}}\,)\,$ and $\,(\sqrt{k\cdot\bar{\sigma}^*}\sigma_2=\sigma_2\sqrt{k\cdot\sigma}\,)$ so we can compute $(v^s_k)^*$ as

$(v^s_k)^*=\begin{pmatrix}-i\sqrt{k\cdot\sigma^*}\sigma_2\,\xi^s\\\;\;i\sqrt{k\cdot\bar{\sigma}^*}\sigma_2\,\xi^s\end{pmatrix}=\begin{pmatrix}-i\sigma_2\sqrt{k\cdot\bar{\sigma}}\,\xi^s\\\;\;i\sigma_2\sqrt{k\cdot\sigma}\,\xi^s\end{pmatrix}=\begin{pmatrix}0 & -i\sigma_2 \\ i\sigma_2 & 0\end{pmatrix} \begin{pmatrix}\;\;\,\sqrt{k\cdot\sigma}\,\xi^s\\-\sqrt{k\cdot\bar{\sigma}}\,\xi^s\end{pmatrix}=-i\gamma^2 u^s_k$.

Here, $u^s_k$ is the bispinor of the positive-energy solution with momentum $k$ and spin state $\xi^s$.

We see that $\:\boxed{\,(v^s_k)^*=-i\gamma^2 u^s_k\,}\;$ but my question comes now as Peskin proceeds saying that the following expressions follow immediately from it:

$u^s_k=-i\gamma^2 (v^s_k)^*\;$ and $\;v^s_k=-i\gamma^2 (u^s_k)^*$.

How is that possible? They don't even hold for $\,u^s_k\neq0$. Did I missed something?


1 Answer 1


Schwartz uses negative signature, so $(-i\gamma^2)^2=1$.

  • 1
    $\begingroup$ Hahaha, yup. This is shameful. thx $\endgroup$
    – JuanC97
    May 24, 2020 at 8:21

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