The parity transformation $\mathcal{P}$ and the time-reversal transformations $\mathcal{T}$ are defined as follows : \begin{equation} \mathcal{P}= \begin{bmatrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{bmatrix},\qquad \mathcal{T}= \begin{bmatrix} -1 & & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{bmatrix}. \end{equation} Those transformations are clearly element of the Poincaré group, and in particular of the Lorentz group but they are no restricted Lorentz transformation (or proper orthochronous Lorentz transformation). In QFT, Lorentz symmetries $\Lambda$ (and in general any symmetries) are implemented through the use of unitary/antiunitary operators $U(\Lambda)$. Those act as \begin{equation} U(\Lambda)A_\mu(x)U(\Lambda) = \Lambda_\mu^{~~\nu}A_\nu(\Lambda x) \end{equation} on a quantum vector field $A$ for example. We usually use the notations $P\equiv U(\mathcal{P})$ and $T\equiv U(\mathcal{T})$. Since QFTs are relativistic theories, they must be invariant under the Poincaré group and so in particular under the Lorentz group (proper and orthochronous or not).

Therefore, I don't understand why we can consider Lagrangian densities $\mathcal{L}$ that are not invariant under $P$ and $T$ (more precisely, Lagrangians that leads to actions that are not invariant), why is that? For example, it's a well known fact that electroweak interactions \begin{equation} \mathcal{L}_{EW}=\bar{\psi}_LW_\mu\gamma^\mu\psi_L = \frac{1}{2}W_\mu(\bar{\psi}\gamma^\mu\psi-\bar{\psi}\gamma^\mu\gamma^5\psi) \end{equation} are not invariant under $P$, where $\gamma^\mu$ are Dirac's matrices, $\psi$ is a fermionic field and $W$ is the $SU(2)_L$ gauge boson.

  • 2
    $\begingroup$ This is the classic abuse of notation: people say $\text{SO(1, 3)}$, or even worse, $\text{O(1, 3)}$, when they really mean $\text{SO(1, 3)}^+$ $\endgroup$ Dec 24, 2020 at 12:06

1 Answer 1


Just as the proper orthochronous Lorentz transformations form a subgroup, the same is true for the Poincare transformations.

You can't reverse the time or reflect the space by any combination of the proper orthochronous Lorentz transformation and shift. So $\mathcal{P}$ and $\mathcal{T}$ don't belong to the connected component of the Poincare group that is generated by exponentiation of its Lie algebra. In fact the full Poincare group of your preference has similar structure to the Lorentz group - the semidirect product of a proper orthochronous Poincare group and a discrete group $\{1,\mathcal{P},\mathcal{T},\mathcal{PT}\}$.

There's no inconsistency in violating the discrete symmetries and everything about special relativity kinematics and dynamics remains to be true. Your statement that "relativistic theories ... must be invariant under the Poincaré group and so in particular under the Lorentz group (proper and orthochronous or not)" is simply too restrictive.

Your mention that the Lorentz symmetries are represented by unitary seems irrelevant but note that $U(\mathcal{T})$ is not unitary, rather it is antiunitary (i.e. it is an antilinear operator that conserves the inner products)

  • $\begingroup$ Indeed for the antiunitarity, its corrected. So QFTs are not invariant under Poincaré group necessarily ? I thought that the EOMs always had to be invariant under Poincaré and that it implied that it the action must be too. $\endgroup$
    – xpsf
    Dec 24, 2020 at 12:13
  • $\begingroup$ @xpsf Not necessarily under full Poincare group. Your misunderstanding likely comes from the fact that in the literature "Poincare group" often means only its connected (i.e. proper orthochrounous) component. And you may debate whether it is "correct" to call it this way but the connected component catches all the essential part about the special relativitiy. $\endgroup$
    – OON
    Dec 24, 2020 at 12:29
  • $\begingroup$ Ok thank you very much. One last thing, I really thought that the QFTs has to be invariant under the hole Poincaré group but apparently it's not the case. Doesn't it matter? Time-reversal, parity and translations seems very important. So Poincaré group isn't really a fundamental symmetry of Nature, only $SO(3,1)^+$ is really fundamental, is that what it teaches us? Since some really realistic theories (e.g. EW interaction) are not invariant under parity for example. Maybe this comment should be a post by itself. $\endgroup$
    – xpsf
    Dec 24, 2020 at 13:08
  • $\begingroup$ @xpsf I'll repeat again - everything is fine with translations. Note that I was talking about the connected component of the Poincare group, not the Lorentz group. $\endgroup$
    – OON
    Dec 24, 2020 at 13:53
  • $\begingroup$ @xpsf as for the parity violation it is indeed very interesting question but currently we can't say anything about what does it tell us. The nature may be just inherently P, T asymmetric or these symmetries may be restored at higher energies, we just don't know $\endgroup$
    – OON
    Dec 24, 2020 at 13:55

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