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The one-loop correction in Lehmann-Kallen form

  1. Mar 14, 2014 #1
    Hi.
    I would like to ask a question about Chapter 15 in Srednicki's QFT book.

    In chapter 15, after eq. (15.12), he compares eq. (15.12)
    ## \mathrm{Im}\bm{\Delta}(k^2)=\frac{\mathrm{Im}\Pi (k^2)}{(k^2+m^2-\mathrm{Re}\Pi (k^2))^2 + (\mathrm{Im}\Pi (k^2))^2}##
    with eq. (15.8)
    ##\mathrm{Im}\bm{\Delta}(k^2)=\pi \delta(k^2+m^2)+\pi\rho(-k^2).##
    Then he gets eq. (15.13)
    ##\pi \rho(s)=\frac{\mathrm{Im}\Pi (-s)}{(-s+m^2-\mathrm{Re}\Pi (-s))^2 + (\mathrm{Im}\Pi (-s))^2}.##
    Why does ##\pi \delta(k^2+m^2)## disappears?
     
  2. jcsd
  3. Mar 14, 2014 #2

    Avodyne

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    (15.12) only applies if ##{\rm Im}\Pi(k^2)\ne0##. If ##{\rm Im}\Pi(k^2)=0##, then (15.11) applies.
     
  4. Mar 15, 2014 #3
    Thanks, Avodyne.
    But my question is the following one.

    I think that eqs. (15.12) and (15.8) give
    ##\pi \rho (s) = \frac{\mathrm{Im}\Pi(-s)}{(k^2+m^2-\mathrm{Re}\Pi(-s))^2 + (\mathrm{Im}\Pi(-s))^2} - \pi \delta(-s+m^2),##
    instead of eq. (15.13) in his text.
     
  5. Mar 20, 2014 #4

    Avodyne

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    That makes no difference in 15.8, because the integral over ##s## starts at ##4m^2##.
     
  6. Mar 20, 2014 #5
    Integral over ##s## in eq. (15.8) does not necessarily need to start at ##4m^2##.
    It starts at ##4m^2## just because we know ##\rho(s)=0## for ##s<4m^2## (this is clearly seen around eq. (13.11) in chapter 13).
    Thus, eq. (15.8) and also eq. (15.13) is valid for any ##s## (or any ##-k^2##), although we already know ##\rho(s)=0## for ##s<4m^2##.

    In fact, he used eq. (15.13) for ##s<4m^2## in the discussion below eq. (15.13).
    Here, he combined eq. (15.13) for ##s<4m^2## with the fact ##\rho(s)=0## for ##s<4m^2##.

    My understanding is the following.
    Eq. (15.13) should be
    ##\pi \rho (s) = \frac{\mathrm{Im}\Pi(-s)}{(-s+m^2-\mathrm{Re}\Pi(-s))^2+(\mathrm{Im}\Pi(-s))^2} - \pi \delta(s-m^2)##
    (this is valid for any ##s##).
    Next, we consider the region ##s<4m^2##.
    Combining my modified eq. (15.13) with \rho(s)=0 for ##s<4m^2##, we get ##\mathrm{Im}\Pi(-s)=0## for ##s<4m^2##, which is the same conclusion in the book.
    (For ##s \ne m^2##, we can drop the delta-function and the modified eq. (15.13) becomes same as that in the book.
    However, we have to be slightly careful at ##s=m^2##.)
     
  7. Mar 21, 2014 #6

    Avodyne

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    If you agree that ##\rho(s)=0## and ##{\rm Im}\Pi(-s)=0## for ##s<4m^2##, then your modified eq.(15.13) is not consistent, because the delta function does not vanish at ##s=m^2##, which is less than ##4m^2##.
     
  8. Mar 21, 2014 #7
    I appreciate your kind discussion (at the same time I'm sorry that I'm slow to understand).

    I agree that ##\rho(s)=0## and ##\mathrm{Im}\Pi(-s)=0## for ##4m^2>s \ne m^2##.
    But at ##s=m^2##, I think that an additional condition ##\mathrm{Re}\Pi(-m^2)=0## is needed; this condition and ##\mathrm{Im}\Pi(-m^2)=0## can cancel the delta-function ##-\pi\delta(s-m^2)##.
    Actually, in Fig. 14.5, ##\mathrm{Re}\Pi(-m^2)## looks zero.
     
  9. Mar 21, 2014 #8
    Oh, finally I understand!
    Thank you very much, Avodyne!!!!!
     
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