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Subtraction point not physical mass

  1. Aug 5, 2013 #1
    In conventional renormalization, for the self-energy, is it possible to make a subtraction from a point not equal to the physical mass?

    [tex]\frac{1}{p^2-m_o^2-\Sigma(\mu^2)-\Sigma'(\mu^2)(p^2-\mu^2)-...} [/tex]

    Now define [itex]m_o^2+\Sigma(\mu^2)\equiv m(\mu^2) [/itex]

    Then:

    [tex]\frac{1}{p^2-m(\mu)^2-\Sigma'(\mu^2)(p^2-\mu^2)-...} [/tex]

    But you can't seem to write this in the form [itex]\frac{Z}{p^2-m(\mu)^2-\text{finite}}[/itex]

    unless you choose [itex]\mu^2=m(\mu^2) [/itex]. But this choice corresponds to the physical mass.

    But in BPZ renormalization, you have no problems working with a mass that depends on scale μ, and a scale is like a subtraction point is it not?
     
  2. jcsd
  3. Aug 5, 2013 #2
    It is possible to choose a subtraction point besides the physical mass. I'm not familiar with how you're writing the self-energy (what's the utility in expanding about [itex]p^2 = \mu^2[/itex]?), but writing the bare propagator as

    $$ \frac{1}{p^2-m_o^2-\Sigma_{0}(p^2)} $$

    we can choose any subtraction scheme to obtain a finite self energy [itex]\Sigma_R[/itex], and then simply write the propagator as

    $$ \frac{1}{p^2-m_R^2-\Sigma_{R}(p^2)} .$$

    Now [itex]m_R[/itex] is a parameter dependent on your subtraction scheme, and thus dependent on [itex]\mu[/itex], and it will take the value [itex]m_R^2 = m_P^2 - \Sigma_R(m_P^2)[/itex]. The dependence of renormalized mass on the energy scale [itex]\mu[/itex] is physically important, especially for studying critical exponents.

    EDIT: And of course, the [itex]\mu[/itex] dependence of [itex]Z = (1-\Sigma'_R(m_P^2))^{-1}[/itex] is also very important
     
    Last edited: Aug 5, 2013
  4. Aug 5, 2013 #3
    Textbooks start with the bare propagator as you have, but then make an expansion about the physical mass:

    $$ \frac{1}{p^2-m_o^2-\Sigma_{0}(p^2)}=\frac{1}{p^2-m_o^2-\Sigma_{0}(m_p^2)-\Sigma'_{0}(m_p^2)(p^2-m_p^2)-...} $$

    They then say the definition of the physical mass is that [itex]m_o^2+\Sigma_{0}(m_p^2)=m_p^2 [/itex], so that

    $$ \frac{1}{p^2-m_o^2-\Sigma_{0}(p^2)}=\frac{1}{p^2-m_o^2-\Sigma_{0}(m_p^2)-\Sigma'_{0}(m_p^2)(p^2-m_p^2)-...}=
    \frac{1}{(p^2-m_p^2)-\Sigma'_{0}(m_p^2)(p^2-m_p^2)-...}
    $$

    Then the (1-Ʃ')-1 can be factored out, which is Z:

    $$
    =\frac{(1-\Sigma'(m_p))^{-1}}{(p^2-m_p^2)-\Sigma_{R}(p^2)}
    $$

    That is how they got the relation: Z=(1-Ʃ')-1.

    But this doesn't seem to work if Taylor expand Ʃ about a different point than the physical mass.

    So textbooks basically manipulate Ʃ0 directly and define the physical mass from it, whereas you define the physical mass from the renormalized ƩR and the finite m2(μ).
     
    Last edited: Aug 5, 2013
  5. Aug 6, 2013 #4
    If you will calculate the self energy correction by taking into account the inifinities of diagram and use some identity like 1/A+B=1/A-(1/A)B(1/A)+... you will see the result getting around the physical mass.This idea is just translated into a much convenient way without doing it always .They make an expansion around physical mass because that is the way it comes around.
     
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