# How to show that $\sigma^2\psi_L^*$ transforms as a right-handed spinor? (Peskin&Schroeder)

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: $$$$\psi_L^*\to(\psi_L')^*=\exp\left((i\boldsymbol{\theta}-\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma^*}}{2}\right)\psi_L^*,$$$$ from which the desired result follows by multiplying with $$\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^*$$, where I understand the transformation of the latter as: $$$$\sigma^2\psi_L^*\to(\sigma^2\psi_L^*)'=\exp\left((-i\boldsymbol{\theta}-\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2}\right)\sigma^2\psi_L^*=\sigma^2\exp\left((i\boldsymbol{\theta}+\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}^*}{2}\right)\psi_L^*=\sigma^2(\psi_L')^*,$$$$ which does look odd to me. What I expect when reading "show that $$\sigma^2\psi_L^*$$ transforms as a right-handed spinor" is something along the line of: $$$$\sigma^2\psi_L^*\to(\sigma^2\psi_L^*)'=S[\Lambda]\sigma^2\psi_L^*,$$$$ which is not what I get when doing the calculations, as shown above.

I hope that I made it clear what confuses me.

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

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).
• 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? Commented Apr 16, 2021 at 16:27
• 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 $\psi_{\text{L}}$. Commented Apr 17, 2021 at 21:55
• 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. Commented Apr 18, 2021 at 9:28
• 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. Commented Apr 18, 2021 at 9:35