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In Peskin and Schroeder's Quantum Field Theory, there is an identity of Pauli matrices which is connected to the Fierz identity, (equation 3.77) $$(\sigma^{\mu})_{\alpha\beta}(\sigma_\mu)_{\gamma\delta}=2\epsilon_{\alpha\gamma}\epsilon_{\beta\delta}.$$ The author explains that we can understand the identity by noting that the indices $\alpha,\gamma$ transform in the Lorentz representation of $\Psi_{L}$, while $\beta,\delta$ transform in the separate representation of $\Psi_{R}$, and the whole quantity must be a Lorentz invariant.

How can one see $\alpha,\gamma$ and $\beta,\delta$ transform in different representation?

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The Pauli matrices are invariant tensors that couple left and right-handed spinors. These spinors transform in different representations of the Lorentz group (as you mentioned) and hence are usually denoted with different indices. This is trivial to see in two component notation, however if you are not familiar with this notation this can also be seen from a four-component Lagrangian:

$$ \bar{\psi} \gamma _\mu \psi = \left( \begin{array}{cc}\bar{ \psi} _R & \bar{\psi} _L \end{array} \right) \left( \begin{array}{cc} 0 & \sigma ^\mu \\ \bar{\sigma} ^\mu & 0 \end{array} \right) \left( \begin{array}{c} \psi _L \\ \psi _R \end{array} \right) = \bar{\psi} _R \sigma ^\mu \psi _L + \bar{\psi} _L \bar{\sigma} ^\mu \psi _R $$ Clearly a $ \sigma ^\mu $ field connects a $ \psi _R $ field with a $ \psi _L $ field. We can write these contractions more explicitly by denoting the left-handed representation indices by greek indices and right-handed reprentation indices with dotted greek indices: $$ \bar{\psi}_{R \, \dot{\alpha}} \left(\sigma^\mu\right)^{\dot{\alpha}}_{\phantom {\alpha}\alpha} \psi_L ^\alpha $$

Note: you might be tempted to think of $\psi _R $ and $\psi_L$ not as separate fields, but just fields with projectors acting on them. This makes this whole topic very confusing and I would urge to get comfortable thinking in terms of two component fields as the fundamental objects making up fermions.

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  • $\begingroup$ Thanks for the answer. I have the same question. However, I did not really understand your answer, would you please give some more specific explanations in terms of simple language? I understand that $\bar{\psi}\gamma_\mu\psi$ is a Lorentz vector, $\psi$ is a Lorentz spinner which transforms according to a reducible representation of the group (due to some requirements from physics). When you show that $\bar{\psi}\gamma_\mu\psi$ can be written as a sum of two terms, but it does not(?) necessarily imply that each term is a also Lorentz vector. $\endgroup$ – gamebm Jun 7 '16 at 15:09
  • $\begingroup$ So, what means "$\sigma^\mu$ field connects a right Weyl field with a left Weyl field"? In other words, by just looking at one term, as you did, what is the meaning of the statement that $\alpha$ and $\dot{\alpha}$ belongs to different representation? Lastly, why Fierz identity can be understood (derived up to a constant) in terms of these arguments? Apologies for my ignorance and many thanks in advance! $\endgroup$ – gamebm Jun 7 '16 at 15:13
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I had the same question, and the link provided by Qmechanic seems to be founded on a solid understanding of the group theory. I was wondering if one may simply understand the transformation of the indices for this specific question based on the textbook alone while using a minimal amount of knowledge/arguments from the group theory. After consulting my colleague Alberto, here is the answer I got following this criterion.

The identity (Eq.(3.77))) reads $$(\sigma^{\mu})_{\alpha\beta}(\sigma_\mu)_{\gamma\delta}=2\epsilon_{\alpha\gamma}\epsilon_{\beta\delta} \ .$$

We start by the right hand side of the equality. Firstly, we can show that anti-symmetric symbol $\epsilon_{\alpha\gamma}$ is Lorentz invariant if both indices transform as left-handed Weyl spinors

$$\Psi_L\rightarrow U_L\Psi_L=\exp(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2})\Psi_L \ \ .$$

The above statement is equivalent to the following identity $$\epsilon_{\alpha\gamma}\rightarrow \exp(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2}) _{\alpha\alpha'} \exp(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2})_{\gamma\gamma'}\epsilon_{\alpha'\gamma'}=\epsilon_{\alpha\gamma} $$ or $$\exp(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2})\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \exp(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2})^T=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$

The above identity can be shown without much difficulty by noting $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}=i\sigma^2$, $\sigma^\dagger=\sigma$ and the transverse of Eq.(3.38), so that

$$\exp(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2})\sigma^2 \exp(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2})^T\\ =\exp(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2})\sigma^2 [\sum_n\frac{1}{n!}(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2})^n]^T\\ =\exp(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2}) \exp(+i{\mathbf \theta}\cdot \frac{\sigma^*}{2}+\mathbf{\beta}\cdot\frac{\sigma^*}{2})^T \sigma^2\\ =\exp(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2}) \exp(+i{\mathbf \theta}\cdot \frac{\sigma^\dagger}{2}+\mathbf{\beta}\cdot\frac{\sigma^\dagger}{2}) \sigma^2\\ =\exp(-i{\mathbf \theta}\cdot \frac{\sigma}{2}-\mathbf{\beta}\cdot\frac{\sigma}{2}) \exp(+i{\mathbf \theta}\cdot \frac{\sigma}{2}+\mathbf{\beta}\cdot\frac{\sigma}{2}) \sigma^2=\sigma^2$$

A very similar argument shows that $\epsilon_{\alpha\gamma}$ is also invariant if both indices transform as right-handed Weyl spinors. So one may choose to look at the right hand side of the identity as an invariant tensor where $\alpha,\gamma$ transforms in the Lorentz representation of $\Psi_L$, while $\beta,\delta$ transforms in the separate Lorentz representation of $\Psi_R$, as pointed out in the textbook.

Now we move to the left hand side of the identity. It is more difficult mathematically (but still feasible) to show (while a smart guess also strongly points to) that $(\sigma^\mu)_{\alpha\beta}$ is also a Lorentz invariant tensor when $\mu$ transforms as a Lorentz vector defined by Eq.(3.19), $\alpha$ transforms as a left-handed spinor and $\beta$ transforms as a right-handed spinor. So that when $\mu$ is contracted out on the left hand side of the identity, the remaining free parameters transform exactly the same way as those on the right hand side. This immediately leads to the conclusion that the identity is correct up to a constant number, which can be then fixed by evaluating only one term (instead of all $2^4=16$ of them).

Usually, the above arguments are given in terms of the language of group theory in a more elegant way, and one good reference is Quantum Field Theory by Srednicki (see the text between Eq.(34.18) and Eq.(35.20)).

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