# 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.)

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## 2 Answers

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...

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In recent years, it has become apparent that a class of materials called topological insulators can be described by an action where the term $E\cdot B$ is added.

The action is

$$S_{top} = S_{em} + \frac{\theta}{2\pi}\frac{e^2}{\hbar c·2\pi} \int d^3xdt\, E·B.$$

For ordinary insulators, we have $\theta=0$ while for topological insulators, we have $\theta=\pi$.

Since the $E·B$ term is a total derivative, Maxwells equations are unchanged inside and outside the insulator. But the point is that $\theta$ has a gradient at the surface, and that's where something interesting happens. Namely, an outside electric field can induce surface currents and vice versa.

One might think that the action is not invariant under time-reversal, because then we would have to map $\theta\to-\theta$. But with periodic boundary conditions, it turns out that the value $\theta$ is only well-defined modulo $2\pi$. Thus, both $\theta=0$ and $\theta=\pi=2\pi-\pi$ are possible. Note that you need quantum mechanics to understand that $\theta$ can only be defined up to $2\pi$, it is not possible to see that classically.

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Nice. I didn't realize that topological insulators had such a simple explanation - at least compared to the ones I've found in the literature. +1. –  user346 Apr 2 '11 at 8:05
@Deepak Vaid, its funny to see what kind of explanation people think of as "simple". I have noticed that people with strong field theory background like to think about topological insulators in this way. People with more solid state physics background like to think in terms of all the nasty concrete details (band-structure, spin-orbit and so on). More mathematically inclined people find it simpler to think in terms of topology of space of certain Hamiltonians (using KR-theory, C*-algebras, characteristic classes). @Greg +1! –  Heidar Apr 2 '11 at 15:31
For a nice talk about this topic, take a look at this pirsa.org/10050088 . –  Heidar Apr 2 '11 at 15:36
@4tnemele - different people, different strokes. I do have more of a field theory background. Thanks for that talk reference. I'll definitely check it out. –  user346 Apr 2 '11 at 18:09
Well, the complications are of course all hidden in the fact there are materials with precisely this action and that $\theta$ is only defined up to $2\pi$. :-) –  Greg Graviton Apr 2 '11 at 18:27