Pseudoscalar action in classical field theory

I was reading Landau and Lifschitz's "Classical Field Theory" and came across a comment that the action for electromagnetism must be a scalar, not a pseudoscalar (footnote in section 27). So I was wondering, is it possible/interesting to construct a classical field theory with a pseudoscalar action? If not, why not?

(Note: I was motivated to look at this because of a question from Sean Carroll's "Spacetime and Geometry", which asks us to show that adding a pseudoscalar term ($\vec{E} \cdot \vec{B}$) to the Lagrangian doesn't change Maxwell's equations.)

Answer

Electromagnetism is parity-symmetric. Because all other terms in the action - such as $mv^2-V(x)$ for particles - are parity-even, the electromagnetic contribution has to be parity-even, too. Otherwise the different terms would transform differently and the combined theory would violate parity. "Parity-even" simply means that the Lagrangian density is a scalar, not a pseudoscalar. It's the same thing.

The actions' being invariant under

$$(x,y,z)\to(-x,-y,-z),$$ including the sign ($S\to +S$), is simply what we mean by the parity symmetry, and scalars (as opposed to pseudoscalars) are the objects that preserve their signs under this operation, too.

$E\cdot B$ is a term that doesn't affect equations of motion because it is a total derivative (which gets integrated to a constant, unaffected by variations of the fields, as long as the variations of the fields at $t=\pm \infty$ vanish):

$$E\cdot B \sim \epsilon_{\alpha \beta\gamma\delta} F^{\alpha\beta} F^{\gamma\delta} \sim \partial^\alpha( \epsilon_{\alpha \beta\gamma\delta} A^\beta F^{\gamma\delta})$$ You see that it is a total derivative; the $\partial^\alpha F^{\gamma\delta}$ term contracted with the $\epsilon$ symbol vanishes identically because it is the Bianchi identity (in the language of forms, $d^2 A\equiv 0$).

In non-Abelian theories, however, terms of the type

$${\rm Tr }\, F_{\mu\nu} F^{*\mu \nu}$$ change the physics even though they are total derivatives. It's because they get integrated to a nontrivial integral in the Euclidean spacetime where the configuration of the gauge field is topologically nontrivial - an instanton.

In its Feynman's path-integral formulation, quantum mechanics calculates the transition amplitudes as the sum over the normal histories as well as the instantons, and the additive shifts in instantons matters. Because the instanton action above is integer - after a proper normalization - the coefficient $\theta$ in front of it is defined modulo $2\pi$ - as an angle - because a change of the action $S$ by $2\pi i$ doesn't matter since the path integral only depends on $\exp(iS)$. For example, in QCD, the term

$$\theta{\rm Tr }\, F_{\mu\nu} F^{*\mu \nu}$$ is known to affect the physics but experimentally, the coefficient $\theta$ is smaller than $10^{-9}$ which is surprising and unnatural: we would expect $\theta$ to be of order one. The $\theta$-term above, if nonzero, is also P- (pseudoscalar) and CP-odd, and it would lead to new sources of CP-violation which is not observed (the only CP-violation that has been observed comes from the phase of the CKM matrix mixing quark masses).

This smallness of the $\theta$-angle, which is apparently not explained and not needed, not even for life (so even the anthropic principle fails to help), is called the strong CP-problem. The main candidate explanation why the observed $\theta$ is small, even though it doesn't have to be, is the Peccei-Quinn mechanism using the axions. $\theta$ gets promoted to a light scalar field in a way...