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) $$(\sigma^{\mu})_{\alpha\beta}(\sigma_\mu)_{\gamma\delta}=2\epsilon_{\alpha\gamma}\epsilon_{\beta\delta}.\tag{3.77}$$ The author explains that

One 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?


4 Answers 4


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:

$$ \bar{\psi} \gamma _\mu \psi = \psi^\dagger\gamma^0\gamma^\mu\psi=\left( \begin{array}{cc}\psi_R^* & \psi_L^* \end{array} \right) \left( \begin{array}{cc} 0 & \sigma ^\mu \\ \bar{\sigma} ^\mu & 0 \end{array} \right) \left( \begin{array}{c} \psi _L \\ \psi _R \end{array} \right) = \psi_R^*\sigma^\mu\psi_R +\psi_L^* \bar{\sigma} ^\mu \psi _L. $$ One can then show that $\psi_L^*$ ($\psi_R^*$) transforms as a right-handed (left-handed) spinor. Clearly a $ \sigma ^\mu $ field then 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: $$ \psi^*_{L \, \dot{\alpha}} \left(\sigma^\mu\right)^{\dot{\alpha}}_{\phantom {\alpha}\alpha} \psi_L ^\alpha $$

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.

  • $\begingroup$ 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
    Commented Jun 7, 2016 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
    Commented Jun 7, 2016 at 15:13
  • $\begingroup$ Is this first equation correct? I think it should be $\bar{\psi} _R \sigma ^\mu \psi _R + \bar{\psi} _L \bar{\sigma} ^\mu \psi _L$, so I can't find what does "Clearly a $\sigma^{\mu}$ field connects a $\bar{\psi} _R$ field with a $\bar{\psi} _L$ field". $\endgroup$
    – Keyflux
    Commented May 29, 2020 at 12:22

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)).


It is convenient to use dotted notation. The Lorentz algebra is $so(1,3)$ is isomorphic to $su(2)_L \times su(2)_R$. We denote fundamental $su(2)_L$ indices by $\alpha$, $\beta$, etc. and fundamental $su(2)_R$ indices by ${\dot \alpha}$, ${\dot \beta}$, etc. Note that for unitary representations of $so(1,3)$, complex conjugation exchanges L and R representations and therefore also exchanges undotted and dotted indices.

There are several Lorentz invariant tensors of interest, particularly, $\delta^\alpha_\beta$, $\delta^{\dot \alpha}_{\dot \beta}$, $\varepsilon_{\alpha\beta}$, $\varepsilon_{ {\dot \alpha} {\dot \beta}}$ and the Pauli matrices $(\sigma^\mu)_{\alpha{\dot \beta}}$ To be absolutely clear, what I mean is that when sandwiched between spinors, they transform in a way indicated by their indices (see Jeff's answer to this question). I leave it to you as a homework exercise to prove what I have just said.

With all the indices explicit, then any equation must preserve index structure. For example, the quantity of interest for you is $$ (\sigma^\mu)_{\alpha {\dot \beta}} (\sigma_\mu)_{\gamma {\dot \delta}} $$ This is Lorentz invariant and has free indices $\alpha$, ${\dot \beta}$, $\gamma$ $\dot \delta$. The only Lorentz invariant quantity that one can construct out of these indices is $\varepsilon_{\alpha\gamma} \varepsilon_{ {\dot \beta} {\dot \delta}}$. Thus, we must have $$ (\sigma^\mu)_{\alpha {\dot \beta}} (\sigma_\mu)_{\gamma {\dot \delta}} = \lambda \varepsilon_{\alpha\gamma} \varepsilon_{ {\dot \beta} {\dot \delta}} $$ What's left then is to fix the constant $\lambda$. To do this, can set ${\dot \beta}=\gamma$ and $\alpha={\dot \delta}$ and sum over repeated indices (doing this breaks Lorentz invariance, but we don't care at this point). We then have the matrix equation $$ \text{tr}(\sigma^\mu \sigma_\mu) = \lambda \text{tr}(\varepsilon^2) = - \text{tr}(I_2) + \text{tr} ( \vec{\sigma} \cdot \vec{\sigma} ) = - \lambda \text{tr}(I_2) \quad \implies \quad \lambda = - 2 . $$ Thus, $$ (\sigma^\mu)_{\alpha {\dot \beta}} (\sigma_\mu)_{\gamma {\dot \delta}} = - 2 \varepsilon_{\alpha\gamma} \varepsilon_{ {\dot \beta} {\dot \delta}} $$ This matches your equation up to a sign which is convention dependent. I have used the convention that $\varepsilon_{12} = \varepsilon_{ {\dot 1} {\dot 2} } = 1$.

  • $\begingroup$ @Pahar The one thing I don't understand is, why the only Lorentz invariant quantity possible is $\varepsilon_{\alpha \gamma} \varepsilon_{\dot{\beta} \dot{\delta}}$? $\endgroup$ Commented Oct 15, 2020 at 18:50
  • $\begingroup$ There is a difference of sign with your proof and the one of Peskin. Is it due to different metric choices? $\endgroup$ Commented Oct 15, 2020 at 18:52
  • $\begingroup$ @SamapanBhadury - (1) what else could it be? I've already written down all the Lorentz invariant quantities for you. What remains is to construct something from those but with the right index structure. (2) ya, that sounds right. my metric convention is (-+++) $\endgroup$
    – Prahar
    Commented Oct 15, 2020 at 18:54
  • $\begingroup$ why not $\delta_{\alpha\beta} \delta{\gamma\delta}$? Why can't we construct with such $\delta$-functions? $\endgroup$ Commented Oct 15, 2020 at 19:09
  • $\begingroup$ @SamapanBhadury - $\delta$ has one upper index and one lower index. The LHS has all four indices lowered. $\endgroup$
    – Prahar
    Commented Oct 15, 2020 at 19:10

Another way to see this is to use the transformation laws for the $\gamma^\mu$ matrices. Note that we can write $$\Lambda_{1/2}=\left(\begin{matrix} \Lambda_{1/2L} & 0 \\ 0 & \Lambda_{1/2R} \end{matrix}\right),$$ where $\Lambda_{1/2}$ is the matrix that performs a Lorentz transformation on a Dirac spinor. It is easy to confirm that this is true because $\Lambda_{1/2}$ is made from combinations of the identity and the block diagonal matrices $S^{\mu\nu}$ (assuming we are using the chiral representation, as in Peskin and Schroesder). It is easy to see that $\Lambda_{1/2L}$ is the operator that performs a Lorentz transformation on a left-handed spinor, and $\Lambda_{1/2R}$ does the same for a right-handed spinor. To give a concrete example, for infinitesimal transformations, equation 3.37 says that $$\Lambda_{1/2L}=1-i\theta\cdot\sigma/2-\beta\cdot\sigma/2\text{, and }\Lambda_{1/2R}=1-i\theta\cdot\sigma/2+\beta\cdot\sigma/2.$$

Now, we can use the transformation laws of the $\gamma^\mu$ matrices: $$\left(\begin{matrix} \Lambda^{-1}_{1/2L} & 0 \\ 0 & \Lambda^{-1}_{1/2R} \end{matrix}\right)\left(\begin{matrix} 0 & \sigma^\mu \\ \bar{\sigma}^\mu & 0 \end{matrix}\right)\left(\begin{matrix} \Lambda_{1/2L} & 0 \\ 0 & \Lambda_{1/2R} \end{matrix}\right)$$ $$=\Lambda^{-1}_{1/2}\gamma^\mu\Lambda_{1/2}={\Lambda^\mu}_\nu\gamma^\nu=\left(\begin{matrix} 0 & {\Lambda^\mu}_\nu\sigma^\nu \\ {\Lambda^\mu}_\nu\bar\sigma^\nu & 0 \end{matrix}\right).$$ Performing the matrix multiplication, and focusing on the top right block, we get $$\Lambda^{-1}_{1/2L}\sigma^\mu\Lambda_{1/2R}={\Lambda^\mu}_\nu\sigma^\nu.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.