What functions of fields describe particles?

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SUMMARY

The discussion centers on the relationship between fields and particles, specifically examining the scalar field Φ(x) and its transformations, such as Φ(x) + a and bΦ(x) + a. It highlights that the action ∫∂Φ∂Φdx^4 remains unchanged under these transformations, indicating that the propagator <0|[Φ(x)+a,Φ(y)+a]|0> is invariant. The conversation also explores the implications of redefining the field as χ(x) = f(Φ(x)), questioning the criteria for identifying a field representing a particle. Additionally, it addresses the linearity of metric expansions and the potential for non-linear representations in quantum gravity.

PREREQUISITES
  • Understanding of scalar fields in quantum field theory
  • Familiarity with action principles and propagators
  • Knowledge of metric tensor and its inverse in general relativity
  • Basic concepts of renormalization theory
NEXT STEPS
  • Study the implications of field transformations in quantum field theory
  • Research the role of the metric tensor in quantum gravity
  • Learn about Taylor series expansions in the context of field theory
  • Investigate the principles of renormalization and its applications in particle physics
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The discussion is beneficial for theoretical physicists, particularly those specializing in quantum field theory and general relativity, as well as researchers exploring the foundations of quantum gravity.

nuclearhead
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I was thinking about the connection between fields and particles. For instance the scalar field Φ(x) and the field Φ(x)+a both represent the same scalar particle. Because the action ∫∂Φ∂Φdx^4 is unaltered and the propagator <0|[Φ(x)+a,Φ(y)+a]|0> is presumably the same. What about if we replace Φ(x) with bΦ(x)+a?

But what if we rewrote the action in terms of a field χ(x) = f( Φ(x) ) where χ is some function, f, of Φ. Then we had an action in terms of χ. How do we know how to get the field representing the particle? Does it matter?

For example if χ = Φ^2 then S = ∫χ^(-1)∂χ∂χ dx^4. Why do we not say that χ is a field of a scalar particle? Are there any rules to this?

What got me thinking about this is that the metric is split into a constant background part and a graviton part: g(x)=η+h(x) but why not split the inverse-metric that way instead? e.g. g(x)^(-1) =η+h(x). Or why can't we have g(x) = ηexp( h(x) )=η+h(x)+h(x)^2+... for example? Does it always have to be linear? But then if the metric is split linearly into a graviton part the inverse graviton part is non-linear.

Related is that in quantum gravity we would take the path integral over the metric g, but why not over its inverse g^(-1)?
 
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You say that the field ##\Phi## represents a scalar particle because the Hamiltonian/action/equations of motion for ##\Phi## resembles the one of a scalar particle. You can do any transformation on the field which keeps the Hamiltonian/action/eom/propagators/... the same up to a scalar for example.

In renormalization theory, for example, what you do is exactly to find a proper ##b##, such that in the calculations ##b\Phi## "cancels some infinities".

Regarding the metric. When you do the taylor expansion you stop to the linear part because it's easy. You can also go further, and find first order corrections to the linear expansion, and so on. This also answer why you do a expansion as a taylor series and not as a more difficult one.

You can also expand the inverse metric, in fact you know that you can go from the metric to it's inverse just by multiplying it by itself two times.
 

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