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Application of LSZ reduction in NLO computations (external legs)

  1. Oct 25, 2013 #1
    Hello everyone,

    I have a question about virtual corrections of an external fermion line.

    According to the LSZ reduction formula and what some other people say, we neglect the self energies of external legs and multiply the wavefunction therefor just with a squareroot of the wave function renormalization constant.

    So considering the feynman rules in momentum space this would giive for an incoming electron:

    u → √z u

    So in the end I would get just the Born diagram multplied by a term which is proportioanl to 1/ε plus some finite parts ( depending on your renormalization scheme)

    So, my question is: How do I get rid of that divergence? Someone told me it is connected to the vertex correction. And in principal one does not compute that diagram at all. It will be considered in the vertex correction.
    But I do not understand that.

    I also found an somehow wrong and old fashion approach for external self energies. The problem there is the "on shell propagator". But there it is easier for me to see how I can cancel the divergence by adding a counterterm. But in the end that diagram should also still be divergent, right? And someone told me that I have to introduce a factor of 1/2 by hand to get the correct result like that one with the LSZ formula.

    So, my questions are:

    How to apply LSZ for external self energies? What do I have to compute for example for e+ + e- → μ+ + μ- ?
    How do I consider there the "external leg self energies"?

    What are the differences between LSZ and the old approach? Are the counterterms for the vertex correction the same?

    How do I get rid of the divergence introduced by squareroot of Z?

    In most of the books they introduce LSz before renormalization and never use it for an NLO computation. I know that it is good for the transition from greens functions to matrix elements.

    Is there any book, paper, reference where they really compute a full NLO process with those factors of Z^(1/2) ?

    Thanks a lot!
  2. jcsd
  3. Oct 25, 2013 #2
    You could look at Srednicki chapter 27 for an example of a loop calculation explicitly including the Z factor (Srednicki calls it R, since it is the residue of the pole of the propagator). See equation (27.19), where he explicitly inserts a factor of ##\sqrt{R}## for each external propagator.

    LSZ tells you that the (say) 2 to 2 scattering amplitude is the residue of the pole of the momentum-space four point function as the external particle momenta go on shell. But this requires that the field whose four-point function you are computing have a properly normalized amplitude to create a one-particle state. The amplitude to create a one particle state is ##Z^{1/2}## where ##Z## is the residue of the pole in the two-point function of the field.

    Now, if you compute the four-point function without worrying about this you get

    ##\langle 0 | \phi(p_1) \phi(p_2) \phi(p_3) \phi(p_4) | 0 \rangle = \frac{iZ}{p_1^2 - m^2}\frac{iZ}{p_2^2 - m^2}\frac{iZ}{p_3^2 - m^2}\frac{iZ}{p_4^2 - m^2} \times \mbox{amptutated 4-point function}##

    The first four terms represent the FULL external propagators including all self-energy corrections. In principle the full propagators are more complicated, but *near the on-shell pole* they have the above simple form. The "amputated 4-point function" is what you get by summing all the Feynman diagrams but neglecting the external legs and all their loop corrections. The expression above indeed has a pole as all four external momenta go on-shell; the residue of the pole is

    ##Z^4 \times \mbox{amputated 4-point function}##

    But recall that our field has the wrong amplitude to create a one-particle state; the amplitude is ##Z^{1/2}## when it should be ##1##. So we need to multiply by ##(Z^{-1/2})^4##. This serves to *cancel* a (divergent) error in our result. We end up with

    ##(Z^{1/2})^4 \times \mbox{amputated 4-point function}##

    as the scattering amplitude.
  4. Oct 27, 2013 #3
    It is not connected to the vertex correction,however if you use renormalized charge at the vertex then the contribution of self energy to an external fermion line can be omitted.
    In the usual approach,the correction factor to external line is indeterminate.A careful treatment done by Dyson based on adiabatic switching reveals the factor to be √z and not z,which is what you wrote as factor of 1/2(In old notation, correction factor is 1-(B/2) and not 1-B according to it).You may understand it on the ground that when fermion appears in an internal line,we get a factor of z for it.Now the fermion propagator is bilinear in fermion spinor, in contrast to an external fermion line in which the electron spinor appears linearly.The field theoretic way of describing it is discussed in Gasiorowicz 'elementary particle physics' chapter 13.
    it is not connected to it.LSZ formalism and old approach don't yield different result.
    with those factor of √z included.You can as well leave the correction of external self energy,when you use renormalized charge at vertex.
    By the usual regularization procedure.
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