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In physics, can we legally add two physical quantities that behave differently under time-reversal or parity? For example, let $\vec{a}$ and $\vec{b}$ are two observables. Let $\vec{a}$ flips sign under time-reversal (or parity) but $\vec{b}$ does not. For example, $\vec{a}$ is velocity and $\vec{b}$ is the angular momentum of a particle. Can we legally add them (of course, by suitable making their dimensions the same) to define a new observable $\vec{c}=\vec{a}+\alpha\vec{b}$ (where $\alpha$ is some appropriate dimensional constant) such that it has no definite behaviour under time-reversal (or parity)? If this is illegal or meaningless, I would like to see/know why. Thanks.

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    $\begingroup$ an interesting question. Well, certain behaviour under parity and time-reversal is not neccessary to be imposed from the philosphical point of view. However, i a not aware of any sensible quantity, composed as a sum of vector and pseudovector $\endgroup$ Jul 30, 2020 at 6:44
  • $\begingroup$ @ChiralAnomaly I think the ocalMathmatician's answer and the comments are good enough but I had another thing in mind: the 3-vector part of the Pauli-Lubanski vector $W_\mu$ is $\vec{W}=E\vec{J}-\vec{P}\times \vec{K}$ and it left me to wonder whether this quantity $\vec{W}$ must have to have definite parity/time-reversal behaviour. en.wikipedia.org/wiki/… $\endgroup$ Aug 1, 2020 at 17:36

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I don't know if this gives the answer you were expecting but I want to give you the following example where a suitable quantity is constructed by adding together a vector with a pseudovector.

What I'm talking about is the Fermi V-A theory of weak interactions. From the experimental results of the time, Fermi built up the first theory of weak decays, which we now know as to be mediated by the weak interaction, through the following hamiltonian $$\mathcal{H}_F = -\frac{G_F}{\sqrt{2}}\bar{\nu}_\mu\gamma^\sigma(1-\gamma_5)\mu\,\bar{e}^-\gamma_\sigma(1-\gamma_5)\nu_e\tag{1}$$ Although Fermi firstly hypothesized the hamiltonian for the decay of the neutron, I gave you the hamiltonian for the decay of the muon since it has a simpler form and it's easier to see that the hamiltonian $(1)$ is build up from the difference of a vector quantity and a pseudovector quantity. Given that $\nu_\mu$, $\mu$, $e^-$, $\nu_e$ are all spinors of the related particle, which I'll now call simply by a general $\psi$, we know that $$\bar\psi\gamma^\mu\psi\qquad\text{Vector}\\\bar{\psi}\gamma^\mu\gamma_5\psi\qquad\text{Pseudovector}$$ and it's easy to see that the Fermi hamiltonian contains two such quantities $$\bar{\psi}\gamma^\mu\psi-\bar{\psi}\gamma^\mu\gamma_5\psi$$ which is a difference of a vector and a pseudovector.

The hamiltonian is a well defined quantity with a not well defined parity. This result is very important for weak decays since we know that many of them are not parity invariant.

We know now, with the advent of the Standard Model, that the Fermi theory is nothing more than a low energy behaviour of the Standard Model theory. But the V-A structure is still present in the SM lagrangian since, again, weak decays are not always parity invariant.

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    $\begingroup$ Maybe the message here is: You can reasonably add vector+pseudovector exactly if parity is not a good symmetry. $\endgroup$
    – Toffomat
    Jul 30, 2020 at 9:50
  • $\begingroup$ @Toffomat Yeah, tldr is that! $\endgroup$ Jul 30, 2020 at 9:53

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