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What functions of fields describe particles?

  1. Mar 3, 2015 #1
    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)?
  2. jcsd
  3. Mar 3, 2015 #2
    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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