I am trying to prove the Lorentz invariance of the (left-handed) Weyl Lagrangian: $$\mathcal L=i\psi^\dagger\bar\sigma^\mu\partial_\mu\psi$$

A Lorentz transformation is realized as $\psi\to M\psi$, where $M=\exp[-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}]$, and $S$ are the generators of the representation, i.e., $S^{0i}=-\frac{i}{2}\sigma^i$ and $S^{ij}=\frac{1}{2}\epsilon^{ijk}\sigma^k$. Therefore, the Lagrangian transforms as $$\mathcal L=i\psi^\dagger\bar\sigma^\mu\partial_\mu\psi\to i\psi^\dagger M^\dagger\bar\sigma^\mu M\partial_\mu \psi$$ so that if it actually is invariant, we must have $M^\dagger\bar\sigma^\mu M=\bar\sigma^\mu$ (and, in particular, $M^\dagger M=\mathbb I$). I think this is a nice relation, which expresses the invariance of the symbol $\bar\sigma^\mu$.

Now the thing is, I think this relation is not true. Take, for example, a boost in the $z$-direction with rapidity $\omega$. This means $$M=\exp[-i\omega S^{03}]=\exp\left[\frac{\omega}{2}\begin{pmatrix}-1& 0\\0&1\end{pmatrix}\right]=\begin{pmatrix}e^{-\omega/2}&0\\0&e^{\omega/2}\end{pmatrix}$$

In this case, it is easly seen that $M^\dagger M\neq\mathbb I$. My question is: what went wrong in my argument?

  • $\begingroup$ Possible duplicate by OP: physics.stackexchange.com/q/195460/2451 $\endgroup$
    – Qmechanic
    Commented Jul 23, 2015 at 20:17
  • $\begingroup$ Certaily a duplicate. The question is more clearly stated in this post, which also has the solution. Should I delete the older one? $\endgroup$ Commented Jul 23, 2015 at 21:03
  • $\begingroup$ Yes. In the future, please edit rather than repost. $\endgroup$
    – Qmechanic
    Commented Jul 23, 2015 at 21:09

2 Answers 2


Lorentz invariance refers to the action $S=\int\mathcal{L}(x)\,\mathrm{d}x$, not to the Lagrangian. To determine the condition on the Lagrangian which we must have, we make the coordinate change $x\to \Lambda x=:x'$ (a Lorentz transformation) and use the general fact that the Jacobian of a Lorentz transformation is unity, so $$\int\mathcal{L}(x)\,\mathrm{d}x\to\int\mathcal{L}(x')\,\mathrm{d}x'=\int\mathcal{L}(\Lambda x)\,\mathrm{d}x$$ We must thus show $\mathcal{L}(x)\to\mathcal{L}(\Lambda x)$, where $\mathcal{L}(x)=\mathrm{i}\xi^\dagger(x)\bar\sigma^\mu\partial_\mu\xi(x)$.

The Weyl spinors transform as $\xi(x)\to D(\Lambda^{-1})\xi(\Lambda x)$, so the first step is$$\mathrm{i}\xi^\dagger(x)\bar\sigma^\mu\partial_\mu\xi(x)\to\mathrm{i}\xi^\dagger(\Lambda x)D^\dagger(\Lambda^{-1})\bar\sigma^\mu\partial_\mu D(\Lambda^{-1})\xi(\Lambda x)$$ The Pauli 4-vector $\bar\sigma^\mu =(1,\vec\sigma)$ transforms as $\bar\sigma^\mu\to D^\dagger(\Lambda)\bar\sigma^\mu D(\Lambda)=\Lambda^\mu{}_\nu\bar\sigma^\nu$, so the transformed Lagrangian is $$\mathrm{i}\xi^\dagger(\Lambda x)\bar\sigma^\nu(\Lambda^{-1})^\mu{}_\nu\partial_\mu \xi(\Lambda x)$$ Finally, the partial derivative transformation rule is $\partial_\mu=\Lambda^\nu{}_\mu\partial_\nu'$, and the transformed Lagrangian is $$\mathrm{i}\xi^\dagger(\Lambda x)\bar\sigma^\nu(\Lambda^{-1})^\mu{}_\nu\Lambda^\nu{}_\mu\partial_\nu' \xi(\Lambda x)=\mathrm{i}\xi^\dagger(\Lambda x)\bar\sigma^\nu\partial_\nu' \xi(\Lambda x)=\mathcal{L}(\Lambda x)$$ This shows the Lagrangian $\mathrm{i}\xi^\dagger\bar\sigma^\mu\partial_\mu\xi$ leads to a Lorentz-invariant action, as desired.

  • $\begingroup$ I've been thinking about this for some time, and I think this is the right answer. Nevertheless, I still think that there should be a way of proving the relation $D^\dagger \bar \sigma^\mu D=\Lambda^\mu_\nu \hat \sigma^\nu$, which expresses the transformation rule of the Pauli vector (instead of just stating it). Do you know how can this be proven, or any source where I can find the proof? $\endgroup$ Commented Jul 23, 2015 at 19:15
  • $\begingroup$ @qftishard I imagine you'd be able to find it in S. Weinberg, The Quantum Theory of Fields, Vol. I (1995). Chapter 5 would be a likely location for it. $\endgroup$
    – Ryan Unger
    Commented Jul 23, 2015 at 19:27
  • $\begingroup$ @qftishard K. Cahill, Physical Mathematics (2013) has a nearly complete partial proof on page 390, with some minor details left to the reader (a Taylor expansion and arithmetic). $\endgroup$
    – Ryan Unger
    Commented Jul 23, 2015 at 19:47
  • $\begingroup$ @qftishard The second reference answers your other question, but since the answer requires multiple pages of matrix algebra and group theory, I am hesitant to post an answer on that question. $\endgroup$
    – Ryan Unger
    Commented Jul 23, 2015 at 19:48
  • $\begingroup$ Well that's really helpful. Thank you very much. I'll check them out. $\endgroup$ Commented Jul 23, 2015 at 21:04

Regarding the concerns raised in the comments of the other answer: The 'transformation' law for the Pauli vector is always stated without proof because it is not a transformation law. It is a true fact about the Pauli vector but not how one would actually transform the object, which should be obvious since it is nonsense to selectively ignore indices of a tensor while transforming it. The Pauli vector is a tensor which resides in $\bar{S}\otimes S^*\otimes V^*$, where S is the relevant space of spinors (the overbar implies conjugate space, star is dual space) and $V$ the space of Minkowski vectors $\mathbb{R}^{1,3}$. This determines the object's transformation law under Lorentz transforms.

Via a contraction (what mathematicians would call the interior product, and what computer scientists call currying) we may treat the Pauli vector as a map between equivalent vector representations of the Lorentz group. In Particular we have the vector sub-representation of the spinor's tensor-product space: \begin{equation} V^* \in S\otimes \bar{S}^*\quad \mathrm{and \> via \> dualizing} \quad V \in S^*\otimes \bar{S}. \end{equation} That is to say, rank 2-tensors in the double cover of the Lorentz group, have a sub-representation which transform like vectors. Thus we may construct a mapping from Minkowski space to this subspace of rank-2 spin tensors given by the following: \begin{equation} \mathbb{R}^{1,3} = V \ni a \to a^\mu \sigma_\mu = {\not}a \in S\otimes \bar{S}^*. \end{equation} We can see this map is the source of the convenient Feynman slash notation. Importantly take note that despite the Pauli vector having a co or contra-variant index, it is serving the role in either case of a basis vector. I.e. we may consider the above map as a replacement of the standard basis in $\mathbb{R}^{1,3}$ with the Pauli vectors. This means the Pauli vectors spacetime index transforms inversely to what we are used to with vector components.

Of course, we may go the other way (being explicit with the spinor-indices): \begin{equation} SL(2,\mathbb{C}) = S\otimes\bar{S}^* \ni \psi \otimes \phi^\dagger \to \psi^a \phi^{\dot b} \sigma^{\>\>\mu}_{\dot b a} = \phi^\dagger \sigma^\mu \psi = \operatorname{Tr}(\psi \phi^\dagger\sigma^\mu) \in V = \mathbb{R}^{1,3}. \end{equation} The map in this direction implies the usual "rule" found in too many references. If we Lorentz transform $(D,\Lambda)$ the spinors and the vector, we anticipate our result should be consistent. With: $$ v^\mu = \phi^\dagger\sigma^\mu\psi, \> w = \Lambda v, \> \Psi = D\psi, \> \Phi = D\phi, $$ we have $$ \Phi^\dagger\sigma^\mu \Psi = w^\mu= \Lambda^\mu_{\>\nu}v^\nu=\phi^\dagger \left(\Lambda^\mu_{\>\nu}\sigma^\nu\right) \psi=\phi^\dagger \left(D^\dagger \sigma^\mu D\right) \psi. $$ This is not a transformation law, it is just a statement about the necessary compatibility between Lorentz transforms on spinors and vectors.

The point of all this is that there must be a natural map from $\bar{S}^*\otimes S \otimes V \to \mathbb{R}$ because the space of rank-2 spin-tensors $S\otimes \bar{S}^*$ contains $V^*$ as a sub-representation, and this gives us access to $V^* \otimes V$, which always has a natural map to scalars via the usual trace operation. Thus there is a natural map to scalars when combining at once a vector, spinor, and conjugate-dual spinor, and this map in components looks like the Pauli vector, just as the map in general from $V\otimes V^* \to \mathbb{R}$ looks like the Kronecker delta.

Finally, how does the Pauli vector transform? As stated, the contra-variant index on the Pauli vector is misleading, because it needs to transform as a basis-vector would, not as a vector component would. This implies the Pauli vector transforms via: $$ \sigma^\mu \to \left(\Lambda^{-1}\right)^\nu_{\>\mu}D^\dagger \sigma^\mu D = \left(\Lambda^{-1}\right)^\nu_{\>\mu}\Lambda^{\mu}_{\>\gamma} \sigma^\gamma = \sigma^\nu $$ Where we have used what I will call the compatibility rule of spinor and vector Lorentz transforms to derive the true transformation law for the Pauli vector, namely that is does not transform. Again, this is in perfect analogy with the Kronecker delta.

  • $\begingroup$ It's worth mentioning I had left handed Weyl spinors in mind while writing this, and some dual symbols change when you swap to right handed (to see which, transform a fermion current by a boost, which is sensitive to parity swapping, and see if the corresponding vector current transforms co or contra-variantly). Additionally, the duals on the $V$'s are not super important anyway, because the metric induces a map between $V$ and $V^*$. $\endgroup$
    – Craig
    Commented Apr 28, 2023 at 23:14

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