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Renormalization procedure (Peskin and Schroeder)

  1. Apr 17, 2013 #1
    In Peskin and Schroeder at page 323 second paragraph the author state

    'To obtain finite results for an amplitude involving divergent diagrams, we have so far used the following procedure: Compute the diagrams using a regulator to obtain an expression that depends on the bare mass (m0), the bare coupling constant (e0) and some UV cutoff ([itex]\Lambda[/itex]). Then compute the physical mass (m) and the physical coupling constant (e), to whatever order is consisten with the rest of the calculation; these quantities will also depend on m0, e0 and [itex]\Lambda[/itex]. To calculate an S-matrix element one must also compute the field-strength renormalization Z (according to the LSZ formula). Combining all of these expression, eliminate m0 and e0 in favour of m and e; this step is the 'renormalization'. The resulting expression for the amplitude should be finite in the limit [itex] \Lambda \to \infty[/itex].'

    Where does the author actually use this procedure to compute an S-matrix element?

    Have I understood it correctly if the point is that when one inserts the relations m(m0), e(e0) and Z into the LSZ formula the divergences in the respective relations cancel each other out to a given order?
  2. jcsd
  3. Apr 17, 2013 #2


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    In any of the proceeding one-loop calculations this is used. For instance all of Chapter 7.

    Yes, exactly. When you calculate a diagram, you have infinities. However you also have m0 and e0 and Z, which are not physical quantities. So you work out m0(m,e), e0(m,e) and Z(m,e) so that you have only physical quantities in the diagram.

    However it turns out that the functions m0(m,e), e0(m,e) and Z(m,e) contain divergences equal and opposite to the ones present in the diagram, hence they cancel.

    In other words the theory doesn't really contain any divergence, it just appears to if you think that m0, e0 and Z are finite constants, rather than what they really are, divergent functions of the physical parameters.
  4. Apr 18, 2013 #3
    The author does indeed calculate mass-shifts, charge correction at one-loop in chapter 7, but I do not find that he actually use these results to compute an S-matrix element (for instance to compute a higher order scattering amplitude).

    How would I go about to compute such an amplitude for example in phi-four theory? Would I first compute the amplitude in terms of m0 and [itex]\lambda_0[/itex] to one-loop, then compute m(m0, [itex]\lambda_0[/itex]) and [itex]\lambda[/itex](m0,[itex]\lambda_0[/itex] ), also to one-loop, and then insert these relations into the amplitude?

    Should I then end up with the same results as (eqn 10.25) at page 327 i Peskin and Schroeder?
  5. May 1, 2013 #4


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    Eqn 10.25 on p.327 is an example of the procedure they are discussing. They calculate one-loop scattering in [tex]\phi^4[/tex] with a cutoff, in this case dimensional regularisation. Then they calculate [tex]\lambda_0[/tex] the bare coupling, in terms of the physical parameters [tex]m, \lambda[/tex] Of course the calulcation is only to second order, so they only calculate [tex]\lambda_0(\lambda,m)[/tex]
    to second order.

    They've also already made the following split in the Lagrangian:
    [tex]Z^2 \lambda_0 = \lambda + \delta_{\lambda}[/tex]
    i.e. they've isolated the directly physical part of the product of the bare coupling and the field strength factor and isolated the unphysical part in [tex]\delta_{\lambda}[/tex]
    This is because it is much easier to calculate the dependence of these counterterms:
    [tex]\delta_{\lambda}, \delta_{Z}, \delta_{m}[/tex]
    on the physical paramters, than it is to do the same calculations for the actual bare parameters:
    [tex]m_0, Z, \lambda_0[/tex]
    This is because the counterterms each appear in front of their own term in the action and hence appear as a vertex perturbatively, which means that the naturally "match" the diagrammatic structure of perturbation theory and it's easy to isolate the terms that contribute to them.

    This is not the case with the original bare quantities.

    Finally, when you use the counterterms instead of the bare quantities, the physical quantities are there from the start, they're isolate in the Lagrangian. So you don't have to "find them" via calculations.
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