The one-loop correction in Lehmann-Kallen form

1. Mar 14, 2014

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?

2. Mar 14, 2014

Avodyne

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

3. Mar 15, 2014

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.

4. Mar 20, 2014

Avodyne

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

5. Mar 20, 2014

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 $s=m^2$.)

6. Mar 21, 2014

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 $4m^2$.

7. Mar 21, 2014

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.

8. Mar 21, 2014

JILIN

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