QED Renormalization Counterterm Confusion

In summary, when Peskin evaluates the renormalization condition for QED, he sets ##q^2=0## in the full propagator including counterterm, but the ##q^{\mu}q^{\nu}## term is still present.
  • #1
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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?
 
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  • #2
thatboi said:
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
 
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  • #3
thatboi said:
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}##
 
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  • #4
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?
 
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  • #5
What do you mean by "deal with"
 
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  • #6
malawi_glenn said:
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?
 
  • #7
thatboi said:
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.
 
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