The one-loop correction in Lehmann-Kallen form

In summary, the discussion in Chapter 15 of Srednicki's QFT book compares equations (15.12) and (15.8) and obtains equation (15.13). The disappearance of ##\pi\delta(k^2+m^2)## in equation (15.12) is explained by the fact that (15.12) only applies when ##{\rm Im}\Pi(k^2)\ne0##, and if ##{\rm Im}\Pi(k^2)=0##, then (15.11) applies. However, the question remains regarding the validity of equation (15.13) for any ##s##, as it does not match the modified equation presented in the conversation. It is
  • #1
JILIN
10
0
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?
 
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  • #2
(15.12) only applies if ##{\rm Im}\Pi(k^2)\ne0##. If ##{\rm Im}\Pi(k^2)=0##, then (15.11) applies.
 
  • #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.
 
  • #4
That makes no difference in 15.8, because the integral over ##s## starts at ##4m^2##.
 
  • #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##.)
 
  • #6
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##.
 
  • #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.
 
  • #8
Oh, finally I understand!
Thank you very much, Avodyne!
 

FAQ: The one-loop correction in Lehmann-Kallen form

1. What is the one-loop correction in Lehmann-Kallen form?

The one-loop correction in Lehmann-Kallen form is a mathematical expression used to calculate quantum corrections in field theory. It is a perturbative expansion of the scattering matrix that takes into account the effects of virtual particles.

2. Why is the one-loop correction important in field theory?

The one-loop correction is important because it accounts for the quantum effects of virtual particles, which are necessary for accurate predictions in field theory. Without this correction, calculations would not accurately reflect the behavior of particles at the quantum level.

3. How is the one-loop correction calculated?

The one-loop correction is calculated through a mathematical process called perturbation theory. This involves expanding the scattering matrix in terms of a small parameter and then solving for the corrections at each order.

4. Can the one-loop correction be applied to other theories besides field theory?

Yes, the one-loop correction can be applied to other theories, such as quantum electrodynamics and quantum chromodynamics. It is a general method for calculating quantum corrections in various physical theories.

5. What are the limitations of the one-loop correction in Lehmann-Kallen form?

The one-loop correction is a perturbative expansion and therefore, it is only accurate for small corrections. It also does not take into account higher-order corrections and can become unreliable at high energies. In some cases, other methods such as resummation may be necessary to accurately calculate quantum corrections.

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