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A well-known fact about QFTs in curved spacetimes is that there is a phenomenon of particle production in expanding universes, these being described by the line element $$ds^2=-dt^2+b^2(t)d\vec x^2.$$

This is related to the fact that the action $S$ of a field in this background includes the factor $b(t)$ in some of its terms, and it therefore affects the equations of motion of the mode functions and the notion of particle.

However, for some special cases, e.g. massless scalar fields with trivial coupling to the curvature, the function $b(t)$ is no longer present in the action. As a consequence, no particle production occurs.

Intuitively, I would say that particle production is associated to the breaking of a symmetry under transformations of the scale factor, $b(t_1)\rightarrow b(t_2)$. When this symmetry is respected, no particle production occurs.

Is this intuition correct? If so, which is the associated symmetry (mathematically speaking)?

Edit: from the discussion in the comments, I see two possibilities: that particle production is related to the breaking of time-translation symmetry, or that particle production is related to the breaking of Weyl symmetry. Also, it could be the case that particle production has nothing to do with symmetry breaking. Is there any formal derivation relating particle production with the breaking of any symmetry?

TopoLynch
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1 Answers1

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Actually there is no particle production in QFTCS. Particle states in QFTCS are not well defined, since the definition of a particle relies on a choice of time coordinate to decompose the field into Fock representations* modes.

The phenomenon of particle production is related to diffeomorphisms invariance. The Fock space, and hence the vacuum state, is not diffeomorphism invariant, hence a state that in a reference frame is seen as a vacuum, in another reference frame (not obtained through Lorentz transf.) is seen as a state full of particles.

The concept of symmetry breaking is not very relevant in this sense, since what breaks the definiton of vacuum is not the lack of time translation invariance but the lack of invariance under time-diffeomorphisms. The breaking of time translations for an observer in an FLRW universe has the effect of changing the energy of particles, redshift of photons being a primary example, but does not affect the definition of a vacuum state, which remains a vacuum state also if you don't have time translation invariance.

*Fock spaces rely on a choice of time to define the plane waves, and all inertial frames possess the same Fock representations.

LolloBoldo
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