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On page 257 in Peskin's QFT book a qualitative sketch of the QED coupling is given (see the picture below). Why should I expect such a behavior from QED? The QED beta function is $$\beta_{qed}=\mu\frac{dg}{d\mu}=\frac{g^{3}}{12\pi^{2}}$$ So why the fixed point is not zero? Is there a need to go beyond one loop here? Is this figure enough to determine that QED has a bound state of electrons or photons in high enough energies?

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  • $\begingroup$ You have open a bounty, but it seems to us (at least me, and I guess Trimok would agree) that we have answered your question. What is not clear yet ? (For instance, you have not commented my answer.) $\endgroup$ – Adam Jan 18 '14 at 20:12
  • $\begingroup$ if you equate the beta function to zero (in order to get a fixed point) you get a zero according to the beta function mentioned above and NOT 1/m. This part still confuse me. $\endgroup$ – Yair Jan 18 '14 at 20:33
  • $\begingroup$ You can find a (real) fixed point only if $m=0$. Otherwise for energy smaller than $m$ (or, in RG point of view, $\mu<m$), the flow is stopped. (Physically, it's because it cost to much energy to create a electron-positron pair.) $\endgroup$ – Adam Jan 18 '14 at 20:38
  • $\begingroup$ Take the case of m>0. shouldn't I expect zero fixed point according to the above? $\endgroup$ – Yair Jan 18 '14 at 20:48
  • $\begingroup$ Do you know the RG a la Wilson ? It might help you understand what's going on in the flow. For $\mu\gg m$, the flow is dominated by the fixed point (for $\mu\gg m$, the case $m=0$ is a good approximation). Then, when $\mu\simeq m$, the fixed point does not really matter any more and the flow stops. $\endgroup$ – Adam Jan 18 '14 at 20:53
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This is like the method of dominant balance (for solving differential equations). The equation for the beta function you have written down, is only true approximately, when the energy scale is much larger than the electron mass.

At sufficient energy to create a loop with a virtual electron, the evolution of $\alpha(r)$ goes as $$\frac{1}{\alpha(r)} = \frac{1}{\alpha(r_0)} + \lambda \log(\frac{r}{r_0})$$

So the RG evolution goes on as you expect, till about the scale $r < m^{-1}$.

At smaller energies, the other (neglected term) in the beta function becomes more and more important, since $g^3$ is pretty small. Due to this crossover, the coupling stops getting renormalized after this scale (electron mass) and freezes at the value $\alpha^* \sim \alpha (m^{-1})$, and this is the value of the coupling constant for $r > m^{-1}$. So the graph flattens out before the coupling becomes zero.


For $r > m^{-1}$, the potential between two distant static charges is then $V(r) \sim \dfrac{\alpha^*}{r}$, and we are in the "Coulomb phase".

If the electron was massless, then the coupling would keep varying even at larger length scales, and the potential between two charges, for $r \to +\infty$, would look like $$V(r) \sim \dfrac{\alpha(r)}{r} \sim\dfrac{1}{r \log r}$$

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  • $\begingroup$ Why it become "frozen" for r~m^{-1}? this is the fixed point of QED? Also, if QED become strongly coupled why we don't have electrons bound state? $\endgroup$ – Yair Jan 16 '14 at 0:58
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    $\begingroup$ In fact, we have to look, for a energy scale $\mu$, with $\mu \sim \frac{1}{r}$ at the mutual importance, in the RG flow equation, of $\alpha(\mu)$ and $\frac{m(\mu)}{\mu}$ For large $\mu$, the RG flow equation is governed by $\alpha(\mu)$, because $\frac{m(\mu)}{\mu}$ is very small, while for small $\mu$, $\frac{m(\mu)}{\mu}$ is very large, and is governing the RG flow equation. See the interesting discussion in this paper, Chapter $17$, page $31$ (see also the previous chapter $16$ page $29$). $\endgroup$ – Trimok Jan 16 '14 at 11:37
  • $\begingroup$ thanks though it's no so clear to me what you intended to say or how is it related to the three questions I asked. $\endgroup$ – Yair Jan 16 '14 at 12:10
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    $\begingroup$ For the fixed IR point, I think that it is correct. In fact, you have IR divergences in QED, but it could be seen that these IR divergences can be cancelled. For your last question, search on the web for "Bound-state QED". $\endgroup$ – Trimok Jan 16 '14 at 12:31
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    $\begingroup$ @Yair: This is like the method of dominant balance (for solving differential equations), if you will. The equation for the beta function you have written down, is only true approximately, when the energy scale is much larger than the electron mass. So the RG evolution goes on as you expect, till about the scale $r \sim \frac{1}{m_e}$. The other (neglected term) in the beta function then becomes more and more important, since $g^3$ is pretty small. Due to this crossover, the coupling stops getting renormalized after this scale (electron mass). I have just paraphrased Trimok. $\endgroup$ – Siva Jan 19 '14 at 23:01
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I don't have the book with me, but somewhere (maybe in the same chapter), they discuss the potential felt by two electric charges in real space. In momentum space, the potential is roughly $\alpha(p)/p^2$, that you can transform back to real space, for static charges, calling $\alpha(r)$ what is different from $1/r$ in the potential (i.e. $\alpha(r)\equiv r V(r)$).

Then, we can ask the meaning of the graph. From the renormalization flow of $\alpha(\mu)$, we know that $\alpha$ increases when $\mu$ increases, that is, at high energy, the interaction constant is larger and the potential in real space will be larger at short distance (because short distance "=" high energy). The physical picture is that when you look closer than the Compton length of the electron (of order 1/m), the bare charge is less screened by the vacuum fluctuations of electron-positron pairs.

If the electron was massless, the RG flow of $\alpha(\mu)$ would imply that $\alpha$ goes to zero at very large distance. However, the mass of the electron stop the flow at $\mu\simeq m$ and $\alpha(\mu)=1/137$ for $\mu<m$, that is, the potential in real space is the usual Coulomb potential behaving as $1/r$.

(By the way, there is no bound-state of electrons and photons...)

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