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I'm trying to understand charge renormalization in QED. I know that one can write the full photon propagator as


where $\Pi$ is regular at $0$. Obviously this leads to a running coupling.

However I don't see why we also have to renormalize

$$e \mapsto \sqrt{Z}e$$

where $Z$ is the residue of the propagator pole at $0$. Peskin and Schroeder say that this renomalization is valid for low-$q^2$ scattering but that just confuses me even more!

So far as I understand it $Z$ renormalization is a direct result of the LSZ formula, which says that scattering processes with $n$ external legs have to be scaled by $(\sqrt{Z})^n$. But the renormalization here is on an internal photon propagator, so seems to be unrelated to LSZ.

What am I missing here?

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up vote 7 down vote accepted

There is a ward identity that links the charge renormalization to the photon's wave function renormalization. Ward identities are relationships between correlation functions that follow from the quantum theory having a symmetry. In this case the gauge invariance of QED relates (among other things) the electron's two point function (propagator) to the $eA\bar{e}$ three point vertex.

If we write out the lagrangian including arbitrary scalings for $A$ and $\psi$ and I also put in a constant $Z_e$ to let the charge scale \begin{equation} \mathcal{L} =-\frac{1}{4} Z_1 F_{\mu\nu} F^{\mu\nu} + i Z_2\bar{\psi}\gamma^\mu\partial_\mu \psi + \sqrt{Z_1}Z_2 Z_e e \bar{\psi} \gamma^\mu A_\mu \psi \end{equation}

$Z_1,Z_2$ and $Z_e$ are all fixed by renormalization to cancel the divergent parts of the loop integrals.

The ward identity says that $Z_2=\sqrt{Z_1}Z_2 Z_e$, or in other words

\begin{equation} Z_e=\frac{1}{\sqrt{Z_1}} \end{equation}

Since $Z_1$ as I have defined it leads to a residue $1/\sqrt{Z_1}$ in the photon propagator, this is this is equivalent to the equation you wrote above. (by the way this factor of $1/\sqrt{Z_1}$ will appear on all photon propagators, not just the on shell ones).

Note that as a consequence of the ward identity I can rewrite the last two terms in the lagrangian as

\begin{equation} i Z_2\bar{\psi}\gamma^\mu\partial_\mu \psi + \sqrt{Z_1}Z_2 Z_e e \bar{\psi} \gamma^\mu A_\mu \psi= Z_2 i\bar{\psi}\gamma^\mu(\partial_\mu -i e A_\mu)\psi \end{equation}

The right hand side is the gauge covariant derivative.

So when you adjust $Z_1$ to fix the norm of the photon's propagator to 1 (to match the LSZ formula and so forth), you also must adjust the electric charge by an appropriate amount. Alternatively you could go and compute the $eA\bar{e}$ three point function (using a regulator that preserves the gauge invariance, such as dim reg) and you would find that you had to renormalize it by this amount (which amounts to a of the ward identity at 1-loop).

Bonus comment: The constant $Z_1$ which appears in the lagrangian is a 'wave function renormalization', its just a rescaling of the field $A$ by $A\rightarrow \sqrt{Z_1} A$. How do we know what the right value for $Z_1$ is? It's a convention, and the convention is fixed by the LSZ formula. The LSZ formula tells you how to compute observables, and its based on a convention where the photon propagator has residue 1. So if there were no quantum corrections we would set $Z_1=1$. Loops correct the action, so we have to pick a value of $Z_1$ to cancel off the loop contributions. The total $Z$, $Z=Z_1+Z_{loops}$, will end up equaling 1, but we pick $Z_1$ to cancel out the loop contributions. However, we are using $Z_1$ in our definition of the free photon theory around which we are perturbing, and so we have to use $Z_1$ consistently every time we use a photon propagator. (There are actually many conventions for exactly where you put things, this is just one way to think of it.) However, worrying about putting factors of $Z_1$ on photon propagators (or choosing a convention where you put those factors somewhere else) only really starts mattering if you do higher loops, because $Z_1-1$ is already $O(\hbar)$. At your level the main point to realize is what's going on conceptually: the $Z_1$ in the action sets the size of ALL photon propagators (because its really the overall normalization of the photon field). We use the LSZ formula to fix the normalization, but that fixes the normalization for all propagators, not just the external ones.

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thanks for your answer. So Peskin and Schroeder are wrong when they say that the renormalization is valid for low $q^2$ only? Also, why do you adjust $Z_1$ to fix the norm of the photon propagator to $1$? I don't see how this relates to the LSZ formula, which is all about external on-shell legs not internal propagators. Is there a good resource you can point me to about this? – Edward Hughes Aug 3 '13 at 22:53
I like Srednicki as a text. If you don't have it then you can see his lecture notes here: Peskin and Schroder aren't wrong. You always fix the renormalization constants at one scale, and then that's all the fixing you get to do, everything else at every other scale is determined. Setting the electron charge by renormalizing at $q^2=0$ is a convenient choice experimentally. As an aside, note that if there was a derivative $\partial A \bar{\psi} \psi$ then a diagram with that vertex would vanish at $q^2=0$. So that choice won't always work. – Andrew Aug 3 '13 at 22:59
Ah right okay - so they just use a $q^2 = 0$ prescription for convenience then. I misread their meaning and thought they meant that it was only valid at that scale. – Edward Hughes Aug 3 '13 at 23:02
Oh, and is there a quick explanation for why you choose $Z_1$ to make the norm of the photon propagator equal to $1$? Is it a mathematically motivated thing (like LSZ) or just a physical convenience? Many thanks for your help - I'll certainly accept your answer now! – Edward Hughes Aug 3 '13 at 23:03
I added a comment about $Z_1$ to the answer, see if you like it. As for the $q^2$ issue: You do fix the value of the charge by evaluating the $eA\bar{e}$ vertex at $q^2=0$. Now everything is finite. But now if you change the values of $q^2$, you will find that the charge depends on q^2 (renormalization group). But those are all finite effects--you subtract the infinities off at one energy scale, and input your physically observed value for the charge at that scale. Then the theory predicts what you get at other energy scales. – Andrew Aug 3 '13 at 23:19

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