QED - running of coupling (beta function)

[tomdodd4598]

TL;DR

In φ^4 theory, the beta function is determined by analysing the amplitude of the vertex. Can we do the same in QED?

Hey there,

I am a little confused about the way most textbooks and notes I've read find the beta function for QED. They find it by looking at how the photon propagator varies with momentum $q$, in particular in the context of a $2\rightarrow2$ scattering process which is proportional to $e^2$, and arguing that the moving in the amplitude of the propagator can be interpreted in the movement of the value of the square of the coupling (e.g. Peskin and Schroeder pages 245-247).

In $\varphi^4$ theory, the story is rather different. Instead, we look at how the vertex amplitude varies with momentum, and the renormalisation of the field and mass is unrelated.

Can we not take a similar approach to determining the beta function for QED, i.e. look at how the vertex amplitude varies with momentum?

 

[HomogenousCow]

They're not actually looking at how amplitudes vary with asymptotic momenta, but rather the "momenta" at which renormalization conditions are set. With that said, you absolutely can do this for QED by looking at the 3 point function.

 

[tomdodd4598]

They're not actually looking at how amplitudes vary with asymptotic momenta, but rather the "momenta" at which renormalization conditions are set. With that said, you absolutely can do this for QED by looking at the 3 point function.

In that case, I must be making some sort of mistake. I have successfully found the Dirac form factor as given on page 196 of Peskin and Schroeder:

$m$ is the electron mass and $\mu$ is a fictitious photon mass.

However, focussing on the logarithmic piece, it is easy to see that the amplitude actually goes down as the magnitude of $q^2$ increases (it's a space-like momentum), and furthermore I end up getting a $\beta$-function which is wrong by a factor of $-3/2$.

 

[tomdodd4598]

It's been a fair few months now, and I am still struggling to find an example of how this calculation of the QED vertex can reveal the positive beta function of QED. Does anyone know what I'm doing wrong? I'm assuming P&S's result is correct, and I have reproduced it.

 

[vanhees71]

One calculation is here (Sect. 7.3.2):

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

The calculation is in background field gauge and for QCD. For QED just forget the three diagrams with the gluon and ghost loops, because in QED you only have the diagram with the charged-particle loop (usually electrons/positrons instead of quarks of course).

The amazing feature of the background field gauge is that you have simple Ward-Takahashi identities in both QCD and QED and thus you don't need to caluate loop corrections for the vertex function to get the $\beta$ function but the calculation of the gluon/photon self energy is sufficient.

 

[tomdodd4598]

One calculation is here (Sect. 7.3.2):

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

The calculation is in background field gauge and for QCD. For QED just forget the three diagrams with the gluon and ghost loops, because in QED you only have the diagram with the charged-particle loop (usually electrons/positrons instead of quarks of course).

The amazing feature of the background field gauge is that you have simple Ward-Takahashi identities in both QCD and QED and thus you don't need to caluate loop corrections for the vertex function to get the $\beta$ function but the calculation of the gluon/photon self energy is sufficient.

I'm familiar with the self-energy calculation yielding the $\beta$-function, and the argument for why that happens (Peskin & Schroeder), but what I'm looking for is getting the $\beta$-function from the vertex, as we would with something such as $\phi^4$ theory where we don't have the Ward-Takahashi identities, if possible as HomogenousCow suggested, or if not possible, why it isn't. I understand you don't need to do it this way, but I can't get a three-point function which gives rise to the correct result (agrees with P&S, but the resulting $\beta$-function is out by a factor of -2), which is what is baffling me.