I QED Renormalization Counterterm Confusion

[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?

 

[topsquark]

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

 

[malawi_glenn]

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 $g^{\mu\nu}q^{2}$

 

[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?

 

[malawi_glenn]

What do you mean by "deal with"

 

[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?

 

[vanhees71]

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.