# QED Renormalization Counterterm Confusion

• I
thatboi
Hey all,
When looking at the renormalization conditions for QED (see page 332, equation 10.40 from Peskin), there is a condition that requires the photon propagator at ##q^2 = 0## to evaluate to 0. But looking at the expression for the photon propagator counterterm: ##-i(g^{\mu\nu}q^2 - q^{\mu}q^{\nu})\delta_{3}##, can I not rewrite ##q^{\mu}q^{\nu} = g^{\mu\nu}q_{\nu}q^{\nu} = g^{\mu\nu}q^{2}## and then the entire counterterm just disappears?

Gold Member
MHB
Hey all,
When looking at the renormalization conditions for QED (see page 332, equation 10.40 from Peskin), there is a condition that requires the photon propagator at ##q^2 = 0## to evaluate to 0. But looking at the expression for the photon propagator counterterm: ##-i(g^{\mu\nu}q^2 - q^{\mu}q^{\nu})\delta_{3}##, can I not rewrite ##q^{\mu}q^{\nu} = g^{\mu\nu}q_{\nu}q^{\nu} = g^{\mu\nu}q^{2}## and then the entire counterterm just disappears?
Try writing your expression for the second term explicitly. What you wrote was
##\displaystyle q^{\mu} q^{\nu} = \left ( \sum_{\nu} g^{\mu \nu} q_{\nu} \right ) q^{\nu} = g^{\mu \nu} \left ( \sum_{\nu} q_{\nu} q^{\nu} \right ) = g^{\mu \nu} q^2##

Does this make sense?

-Dan

• • thatboi, vanhees71 and malawi_glenn
Homework Helper
Gold Member
2022 Award
can I not rewrite ##q^{\mu}q^{\nu} = g^{\mu\nu}q_{\nu}q^{\nu} = g^{\mu\nu}q^{2}##
You need another index
##q^{\mu}q^{\nu} = g^{\mu\tau}q_{\tau}q^{\nu}## which is not equal to ##g^{\mu\nu}q^{2}##

• • dextercioby, thatboi, vanhees71 and 1 other person
thatboi
Great, thanks a lot. As a followup question (let me know if I should make a new thread for this): For the renormalization condition Peskin evaluates the photon propagator at ##q^2 =0##, how do I deal with the ##q^{\mu}q^{\nu}## term?

• ohwilleke
Homework Helper
Gold Member
2022 Award
What do you mean by "deal with"

• topsquark
thatboi
What do you mean by "deal with"
As in equation (10.44) of Peskin's book. To get the counterterm ##\delta_{3}##, they set ##q^2=0## in the full propagator including counterterm, but then what happens to the ##q^{\mu}q^{\nu}## term?

Gold Member
2022 Award
Great, thanks a lot. As a followup question (let me know if I should make a new thread for this): For the renormalization condition Peskin evaluates the photon propagator at ##q^2 =0##, how do I deal with the ##q^{\mu}q^{\nu}## term?
The photon is a gauge boson. That implies Ward-Takahashi identities which tell you that there is no mass generated by loop corrections and there's also no mass counterterm necessary to renormalize the photon propagator, i.e., there's only a wave-function renormalizing counter term. This implies that the photon-polarization tensor (aka photon self-energy tensor) is of the form
$$\Pi^{\mu \nu}(k)=k^2 \Pi(k) \left (g^{\mu \nu}-\frac{k^{\mu} k^{\nu}}{k^2} \right).$$
The Dyson equation then tells you that the photon propagor reads
$$D_{\gamma \perp}^{\mu \nu} = -\frac{\Theta^{\mu \nu}(k)}{k^2 (1-\Pi(k))}+D_{\gamma 0 \parallel}^{\mu \nu},$$
i.e., the longitudinal part is non-interacting, and the longitudinal photons are unphysical gauge-dependend pieces, which don't participate in any physical quantities, which are gauge invariant.

For details, see Sect. 6.6 in

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

where I used the particularly elegant and simple background-field gauge description of QED.

• thatboi, topsquark, malawi_glenn and 1 other person