The one-loop correction in Lehmann-Kallen form

[JILIN]

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?

 

[Avodyne]

(15.12) only applies if ##{\rm Im}\Pi(k^2)\ne0##. If ##{\rm Im}\Pi(k^2)=0##, then (15.11) applies.

 

[JILIN]

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.

 

[Avodyne]

That makes no difference in 15.8, because the integral over ##s## starts at ##4m^2##.

 

[JILIN]

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##.)

 

[Avodyne]

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##.

 

[JILIN]

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.

 

[JILIN]

Oh, finally I understand!
Thank you very much, Avodyne!