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I am trying to understand the photon propagator renormalization procedure, followed in M. Srednicki's book Quantum Field Theory. Specifically, I am reading Chapter 62, titled "Loop Corrections in Spinor Electrodynamics" and I am focused on the renormalization of the photon propagator at the loop level.

The procedure is as follows:

  1. Write down the full QED Lagrangian comprised by the free Lagrangian, the interaction term and the counter-terms. The interaction term contains the renormalization constant $Z_1$ and the bare charge (I think despite that the author does not discuss anything about bare fields or bare masses/charges). The counter-term Lagrangian is given by $$\mathcal{L_{\text{c.t}}}=i(Z_2-1)\bar{\Psi}{\partial\!\!\!/}\Psi -(Z_m-1)m\bar{\Psi}\Psi-\frac{1}{4}(Z_3-1)F_{\mu\nu}F^{\mu\nu}$$

  2. Label the contributions to the photon propagator from the counter terms and all the loop level diagrams as $i\Pi^{\mu\nu}(k)=i\Pi(k^2)(g^{\mu\nu}k^2-k^{\mu}k^{\nu})$ and calculate them by including the diagram with the closed fermion loop and the counter term diagram.

  3. Perform the calculations to the very end and isolate the divergent behavior by analytically continuing the number of spacetime dimensions to $d=4-\varepsilon$. The divergences should be $\mathcal{O}(\varepsilon^{-1})$. For that step, the electron charge is redefined in a way such that the dimensionality of the latter is absorbed into a factor $\mu$, i.e. $e\rightarrow e\tilde{\mu}^{\epsilon/2}$.

  4. Cancel those divergences by choosing the counter-term $Z_3$ appropriately. Finiteness of the loop-corrected photon propagator implies that there will be a contribution in the counter term $\mathcal{O}(\varepsilon^{-1})$ to cancel the divergent part of the (associated with the number of dimensions of the spacetime), and upon imposing $\Pi(0)=0$ yields the exact form of the counter-term $Z_3$ $$Z_3=1-\frac{e^2}{6\pi^2}\bigg[\frac{1}{\varepsilon}-\ln(m/\mu)\bigg]+\mathcal{O}(e^4)$$ where $\mu^2=4\pi e^{-\gamma}\tilde{\mu}^2$.

I have three questions about the above steps:

  1. Why do we impose $\Pi(0)=0$? And how is this associated with the fact that we are choosing the "on-shell renormalization scheme"? Could this choice be exchanged with some other choice capable of fixing the counter term? I mean I realize that there are three renormalization schemes called "on-shell", "MS" and "MS bar" and that each of them removes the UV divergences of the theory with a certain way (depending each time on some specific conditions I guess), and that some conditions are needed in order to match the experimental results with the loop corrections, but I fail to see how this is applied here...

  2. Is the substitution $e\rightarrow e\tilde{\mu}^{\epsilon/2}$ what we call charge renormalization, i.e. the same as starting with a bare charge $e_0$ in the Lagrangian and simply substituting the bare charge $e_0$ with $e\tilde{\mu}$? This is what I read in other QFT books (i.e. Mandl's book) and I would like to make the correspondence with those books that I have read...

  3. Why is there not an anti-fermion diagram? A diagram in which the closed-fermion loop is comprised by anti-fermionic propagators instead of fermionic ones? Shouldn't be such a contribution as well?

I also have a bonus question: what if my Lagrangian contained another set of spinor fields, that correspond to another species of spin 1/2 particles. Then, we would add to the previous Lagrangian the following terms $$i\bar{\psi}{\partial\!\!\!/}\psi-\tilde{m}\bar{\psi}\psi+ i(\tilde{Z_2}-1)\bar{\psi}{\partial\!\!\!/}\psi-(\tilde{Z}_m-1)\tilde{m}\bar{\psi}\psi+ \tilde{Z_1}\tilde{e}\bar{\psi}{A\!\!\!/}\psi$$ where $\psi$ represents the new fermionic field (as opposed to $\Psi$), and $\tilde{e}$, $\tilde{m}$ represents its charge and its mass (respectively). Then, I find that $Z_3$ would be fixed in the following way $$Z_3=1-\frac{e^2}{6\pi^2}\bigg[\frac{1}{\varepsilon}-\ln(m/\mu)\bigg] -\frac{\tilde{e}^2}{6\pi^2}\bigg[\frac{1}{\varepsilon}-\ln(\tilde{m}/\mu)\bigg] +\mathcal{O}(e^4)$$ Is my "extension" of the photon propagator renormalization to the case in which there are two fermionic fields correct? If not, why?

Any help or comments will be appreciated.

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    $\begingroup$ Tip: Consider to only ask 1 question per post. $\endgroup$
    – Qmechanic
    Jun 6 at 8:32
  • $\begingroup$ I know that it is strongy adviced to ask 1 question per post, but my questions are heavily related, so I thought I should ask them together... $\endgroup$
    – schris38
    Jun 6 at 8:42

2 Answers 2

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  1. Setting $\Pi(0)$ to any other value also would have been valid. There are infinitely many schemes and all of them can in principle be used to make the same physical predictions. See two previous answers for some explanation of this. Alternatively, you can pick another scale $\Lambda_1$ and prescribe $\Pi(q^2 = \Lambda_1^2)$ as your renormalization condition. To use results at one scale to make predictions at another scale just requires that renormalized observables are finite functions of renormalized couplings. But the on-shell scheme has the further property that the renormalized mass is also the physical mass, similar to what we do at tree level. This makes it a common pedagogical choice.

  2. Replacing $e$ with $e \tilde{\mu}^{\epsilon / 2}$ (where the $4\pi e^{-\gamma}$ represents the bar in MS-bar) ensures that $e$ is a dimensionless coupling in $d = 4 - \epsilon$. However it is indeed part of renormalization so I would've preferred to write $e_0 = Z_3 \tilde{\mu}^{\epsilon / 2} e$ all in one go. The explicit power of $\mu$ in the Lagrangian is the classical contribution to the running coupling while powers of $e$ in $Z_3 - 1$ are the quantum contributions. Recall that the interaction Lagrangian comes with powers of $i/\hbar$. This justifies the common phrasing that $\tilde{\mu}^{\epsilon / 2}$ is classical because you would still need it even if you were not including any powers of the interactions and therefore not seeing any loops or divergences to cancel. As you say, there are other regulators where it never shows up.

  3. The four components of a Dirac fermion are associated with particles and anti-particles of both helicities but these are states of external particles. The propagator has to do with the field itself and there's only one Dirac field in your action. Said another way, the propagator has two indices which both run from 1 to 4 so the effects of particles and anti-particles are both captured. In other reading you might come across some "anti-propagators" but these refer to anti-time ordered fields and do not play a role in this formalism. You can try computing the Fourier transform of \begin{align} \left < T \left [ \psi_\alpha(x) \bar{\psi}_\beta(y) \right ] \right > = \left < T \left [ \bar{\psi}_\beta(y) \psi_\alpha(x) \right ] \right >. \end{align} with a mode expansion and you will see that $a$, $a^\dagger$, $b$ and $b^\dagger$ are all be involved. You can call this a "fermion propagator", "anti-fermion propagator" or "fermion and anti-fermion propagator" as desired. But whatever you call it, it's clear that this is the only one you need from inserting $e A_\mu \bar{\psi} \gamma^\mu \psi$ a bunch of times and using Wick's theorem.

The bonus is basically a "check my work" question but I do not see any problem with the expression you wrote down.

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  • $\begingroup$ Hi @Connor Behan and thank you so much for your reply. I would like to ask some follow-up questions though. 1) From what I understand from your answer and the two posts you cite, setting $\Pi(0)$ to be some arbitrary value corresponds to a choice for the renormalization scheme. Also, you said that performing an experiment at some energy $\Lambda_1$ to measure a cross section, helps us fix the counter terms at that energy scale and having done so, we can fix the counter terms at any energy scale. Does that mean that, in principle, I can set a value for $\Pi(q^2=\Lambda_1^2)$ for instance? $\endgroup$
    – schris38
    Jun 7 at 8:10
  • $\begingroup$ Instead of doing so at $q^2=0$, which corresponds to the mass of the photon? And if so, (a) does this correspond to a choice of renormalization scheme as well? (b) why do we prefer to fix the energy scale such that the photon is on-shell? Also, 2) I realize now that making the substitution $e\rightarrow e\tilde{\mu}^{\varepsilon/2}$, rendering the coupling dimensionless, is a process called dimensional reguralization. This procedure is a part of the renormalization process and it is introduced to control the divergent behavior of the integral, but there are other regularization techniques yes? $\endgroup$
    – schris38
    Jun 7 at 8:11
  • $\begingroup$ Also, why do you say that "the explicit power of $\mu$ in the Lagrangian is the classical contribution to the running coupling, while powers of $e$ in $Z_3−1$ are the quantum contributions". What does make the first contribution classical and the second one quantum? $\endgroup$
    – schris38
    Jun 7 at 8:16
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    $\begingroup$ Please see the bold parts. $\endgroup$ Jun 7 at 11:00
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    $\begingroup$ Pretty much. Even if the propagator is not in a loop, its indices are still contracted with something so they all have an effect. $\endgroup$ Jun 7 at 12:23
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  1. Why do we impose $\Pi(0)=0$? And how is this associated with the fact that we are choosing the "on-shell renormalization scheme"?

In an early chapter, Srednicki also talked about the renormalisation of the propagator, and there he was talking about how we get to choose $\Sigma(m^2)=0=\Sigma^\prime(m^2)$. The discussion there can be just brought over, because for the photon, $m=0$

$1^\prime$. Could this choice be exchanged with some other choice capable of fixing the counter term?

Definitely. For beginners, though, we tend to stick with the conceptually simple. Only later should you learn about the computationally more efficient choices.

  1. Is the substitution $e→e\tilde\mu^{\epsilon/2}$ what we call charge renormalization, i.e. the same as starting with a bare charge $e_0$ in the Lagrangian and simply substituting the bare charge $e_0$ with $e\tilde\mu$?

The fact that it depends upon $\epsilon$ should be a hint that it is an artefact of the dimensional regularisation that we are wanting to use. If you do not do this, then $e$ will stop being dimensionless (read as meaningful). I am not willing to pry my repressed memories to figure out what this substitution is doing other than this accounting hackery. Sorry.

  1. Why is there not an anti-fermion diagram?

The fermion operator $\hat\psi$ already included both the fermion and its anti-fermion creation and annihilation operators inside it.

bonus question: what if my Lagrangian contained another set of spinor fields, that correspond to another species of spin 1/2 particles.

Yes, and it should be trivial, which is why no author decided to cover this.

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