Graduate Observables when the symmetry is not broken?

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The discussion revolves around the observability of quantum excitations of a scalar field when symmetry is not broken, particularly in the context of inflationary cosmology. When the field is at its potential minimum with φ=0, as in chaotic inflation, the question arises whether the inflaton boson can be observed. The conversation highlights the distinction between global and local gauge symmetries, with examples from quantum chromodynamics (QCD) illustrating how symmetry breaking leads to observable particles like pions. It is noted that while excitations of fields typically correspond to observable particles, the situation with inflatons remains less clear. The importance of correctly phrasing the concept of "exciting the field" rather than "exciting space" is also emphasized.
mbond
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Hi,

Let be a scalar field φ that permeates all space. The quantum of the field has a mass m. The field is at the minimum of its potential. When this minimum is for φ≠0 (a broken symmetry), the quantum may be observed by exciting the field, as with the Higgs boson.

But if the symmetry is not broken, the field is at the minimum of its potential for φ=0. Is it then possible to observe its quantum?

I am asking this thinking of inflationary cosmology: in a scenario such as the "new inflation" the inflaton field potential has its minimum for φ≠0. The inflaton boson may then, at least in a thought experiment, be observable with an accelerator. But with the "chaotic inflation" the minimum is at φ=0. Is the inflaton boson observable then?

I would be grateful if you could enlighten me.
 
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You are a bit vague. What symmetry are you talking about? Is the symmetry a global one or a local gauge symmetry.

If you have a global symmetry (like the light-quark sector of QCD in the chiral limit, which has a ##\mathrm{SU}(2)_{\text{L}} \times \mathrm{SU}(2)_{\text{R}}## chiral symmetry), then you can have symmetry breaking. Then some scalar or pseudoscalar field has a non-vanishing vacuum expectation value. In the case of QCD that's the non-vanishing scalar quark condenstae, ##\langle \Omega|\bar{\psi} \psi|\Omega \rangle \neq 0##. This breaks the chiral symmetry to ##\mathrm{SU}(2)_{\text{V}}## (that it's the iso-vector subgroup that's the unbroken part is due to the small different quark up- and down-quark masses).

In hadron phenomenology that implies that there are 3 massless pseudoscalar particles, which are the pions. In reality they have a (small) mass, because the chiral symmetry is only approximate and broken by the light-quark masses. The order parameter of the chiral symmetry, the quark condensate, maps to a corresponding scalar-boson state, the ##f_0(500)## aka ##\sigma##-meson.

Usually for each quantum field you also have the corresponding excitations which are observed as particles. There's no doubt about the pions. It's already somewhat difficult for the ##\sigma##-meson, which is a very broad resonance rather then anything resembling a particle.

I can't say anything about inflatons since I'm not familiar enough with the quantum aspects of inflation.
 
mbond said:
the quantum may be observed by exiting the space

What do you mean by this?
 
PeterDonis said:
What do you mean by this?
exciting, sorry.
 
mbond said:
exciting, sorry

Ok, but then "exciting the space" isn't correct. You don't excite the "space", you excite the field--you add energy to it, in order to observe quanta of the field.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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