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In Peskin & Schroeder, it is written that the quantity $\sigma^2\psi_L^*$ transforms as a right-handed spinor. What confuses me is that I only get the correct result when considering the following: \begin{equation} \psi_L^*\to(\psi_L')^*=\exp\left((i\boldsymbol{\theta}-\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma^*}}{2}\right)\psi_L^*, \end{equation} from which the desired result follows by multiplying with $\sigma^2$ and using $\sigma^2\boldsymbol{\sigma}^*=-\boldsymbol{\sigma}\sigma^2$. However, to me, this looks like we considered the transformation of the spinor $\psi_L$ and not the transformation of the quantity $\sigma^2\psi_L^*$, where I understand the transformation of the latter as: \begin{equation} \sigma^2\psi_L^*\to(\sigma^2\psi_L^*)'=\exp\left((-i\boldsymbol{\theta}-\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2}\right)\sigma^2\psi_L^*=\sigma^2\exp\left((i\boldsymbol{\theta}+\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}^*}{2}\right)\psi_L^*=\sigma^2(\psi_L')^*, \end{equation} which does look odd to me. What I expect when reading "show that $\sigma^2\psi_L^*$ transforms as a right-handed spinor" is something along the line of: \begin{equation} \sigma^2\psi_L^*\to(\sigma^2\psi_L^*)'=S[\Lambda]\sigma^2\psi_L^*, \end{equation} which is not what I get when doing the calculations, as shown above.

I hope that I made it clear what confuses me.

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  • $\begingroup$ Your second equation is correct. It is just a definition of what $(\sigma^2 \psi_L^*)'$ is $\endgroup$
    – nwolijin
    Commented Apr 16, 2021 at 14:02
  • $\begingroup$ @nwolijin Yes, but it isn't the desired transformation as shown in my third equation, or am I missing something? $\endgroup$
    – pjHart1000
    Commented Apr 16, 2021 at 14:43

1 Answer 1

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Left handed spinors $\psi_{\text{L}}$ transform as $$ \psi_{\text{L}} \mapsto \exp\left((-\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2} \right)\psi_{\text{L}}\tag{1} $$ and right handed spinors $\psi_{\text{R}}$ transform as $$ \psi_{\text{R}} \mapsto \exp\left((-\mathrm{i}\boldsymbol{\theta} -\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2} \right)\psi_{\text{R}}. \tag{2} $$ Using equation (1), the object $\sigma_2\psi_{\text{L}}^*$ transforms as \begin{align} \sigma_2\psi_{\text{L}}^* \quad\mapsto & \quad\sigma_2\left[\exp\left((-\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2} \right)\psi_{\text{L}}\right]^* \\ &= \sigma_2\exp\left((\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}^*}{2} \right)\psi_{\text{L}}^* \\ &= \sigma_2\exp\left((\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{-\sigma_2\boldsymbol{\sigma}\sigma_2}{2} \right)\psi_{\text{L}}^*\\ &= \underbrace{\sigma_2^2}_{1}\exp\left(-(\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2} \right)\sigma_2\psi_{\text{L}}^*, \end{align} that is, in the same way as the right handed spinor in equation (2).

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  • $\begingroup$ Just to be sure: In the third line, you used $\sigma_2\boldsymbol{\sigma}^*=-\boldsymbol{\sigma}\sigma_2\Rightarrow\boldsymbol{\sigma}^*=-\sigma_2\boldsymbol{\sigma}\sigma_2$ since $\sigma_2$ is its own inverse, right? $\endgroup$
    – pjHart1000
    Commented Apr 16, 2021 at 16:27
  • $\begingroup$ Yep exactly! :) $\endgroup$
    – xzd209
    Commented Apr 16, 2021 at 16:33
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    $\begingroup$ That is the essence of the error that you made in your calculation: $\sigma_2$ does not transform, it is just a matrix! To find the transformation of the object $\sigma_2\psi_{\text{L}}^*$ you have to apply the known transformation law of $\psi_{\text{L}}$. You derive the transformation of $\sigma_2\psi_{\text{L}}^*$ from that of $\psi_{\text{L}}$; you cannot simply assume that $\sigma_2\psi_{\text{L}}^*$ will transform in the same way as $\psi_{\text{L}}$. $\endgroup$
    – xzd209
    Commented Apr 17, 2021 at 21:55
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    $\begingroup$ Suppose you have a contravariant four vector $A^\mu$, and that you multiply it by the Minkowski metric $\eta_{\mu\nu}$ to obtain $A_\nu = \eta_{\mu\nu}A^\mu$. The object $A_\nu$ of course transforms as a covariant four vector. The reason this is possible is that $\eta_{\mu\nu}$ is an invariant tensor of the vector representation of the Lorentz group: it obeys $\Lambda{^\mu}_\rho\Lambda{^\nu}_\sigma \eta_{\mu\nu} = \eta_{\rho\sigma}$, so is unaffected by a transformation. Note that it is incorrect to say that $\eta$ transforms under the Lorentz group - it is just a matrix. $\endgroup$
    – xzd209
    Commented Apr 18, 2021 at 9:28
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    $\begingroup$ The relation of $\sigma_2$ to $\psi_{\text{L}}$ is directly analogous: $\sigma_2$ is an invariant tensor of the left-handed spinor representation of the Lorentz group: $(S_{\text{L}}){_a}^c (S_{\text{L}}){_b}^d (\sigma_2)_{cd} = (\sigma_2)_{ab}$ ($a, b, c, d = 1,2$ are spinor indices in Van-der Waerden notation). This means that it is possible to multiply a left-handed spinor by $\sigma_2$ in a Lorentz covariant way, and importantly the form of $\sigma_2$ is the same independent of the reference frame used. $\endgroup$
    – xzd209
    Commented Apr 18, 2021 at 9:35

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