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Spectral weight function and the mass shift of a scalar field

  1. Aug 13, 2011 #1
    In the Kallen-Lehmann spectral representation (http://en.wikipedia.org/wiki/Källén–Lehmann_spectral_representation) the interacting propagator is given as a weighted sum over free propagators. The pole of the integracting propagator is, of course, given by [itex]p^2=m^2[/itex], m being the physical mass of the particle.

    Could this spectral representation be used to find the mass shift [itex]\delta m^2 = m^2 - m_0^2[/itex] of a scalar field (m_0 is the bare mass of the noninteracting particle)?

  2. jcsd
  3. Aug 13, 2011 #2
    I think not but I'm not sure. In order to derive the spectral density function formula you are required to insert a basis of energy eigenstates which forces you to use the actual mass [itex]m[/itex] and not the bare mass [itex]m_0[/itex]. So that the free single particle propagator has the same pole location as the full propagator.

    You might think that since the spectral density is a sum over poles you would find that taking this sum would yield a "total pole" and the location of the pole would be different from the single particle propagator. However the location of the pole of the full propagator is completely determined by the location of the pole of the free single-particle propagator (expressed in terms of the physical mass). Usually the full propagator is written in terms of the spectral density the following way to make this idea more explicit.

    \Delta(k^2) = \frac{1}{k^2+m^2-i\epsilon}+\int^\infty_{(2m)^2}d\mu^2\rho(\mu^2) \frac{1}{k^2+\mu^2-i\epsilon}

    The integral of the poles gives something like a branch cut rather than contributing to the pole of the free propagator.
  4. Aug 13, 2011 #3
  5. Aug 13, 2011 #4
    Interesting. Sorry I led you astray!
  6. Aug 14, 2011 #5
    You didn't lead me astray, I was thinking the same thing as you and it got me nowhere so I tried PF :)
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