- #1

- 7

- 2

- Homework Statement
- Why is the "physical" electron charge e_0 / (1-Pi(0)) and not e_0 / (1-Pi(q^2))

- Relevant Equations
- (physical charge) = e = sqrt(Z_3) e_0 = sqrt(Z_3) (bare charge)

On p. 246 in the Peskin QFT textbook, below is stated

where Z

$$Z_3=\frac{1}{1-\Pi(0)}$$

and e is the bare charge.

In advance, the exact photon two point function is calculated as

$$\frac{-ig_{\mu\nu}}{q^2(1-\Pi(q^2))}$$

Though the photon propagator should be divided by

$$1-\Pi(q^2)$$

, why the defined "physical" charge is divided by

$$1-\Pi(0)$$

?

Why only the low-q

the "physical" electron charge measured in experiments is $$\sqrt{Z_3} e$$

where Z

_{3}is defined as the residue of the q^{2}= 0 pole, explicitly as$$Z_3=\frac{1}{1-\Pi(0)}$$

and e is the bare charge.

In advance, the exact photon two point function is calculated as

$$\frac{-ig_{\mu\nu}}{q^2(1-\Pi(q^2))}$$

Though the photon propagator should be divided by

$$1-\Pi(q^2)$$

, why the defined "physical" charge is divided by

$$1-\Pi(0)$$

?

Why only the low-q

^{2}scattering is being considered here?