Consider a QED scattering process $e^-+e^-\rightarrow e^-+e^-$. The scattering crosssection at the tree-level depends on the square of the fine-structure constant $\alpha$ (apart from the electron mass and the CM energy). But $\alpha$ is a running coupling. My question is what value of $\alpha$ i.e., the value at which scale is to be substituted in this expression to obtain a numerical prediction? What is that value?

If instead, the scattering amplitude is calculated upto various higher orders, should one use a different value of $\alpha$ than used for tree-level result?

  • 1
    $\begingroup$ A shortened explanations is here: youtube.com/watch?v=oI-Fw-eyccI $\endgroup$ Jul 16, 2016 at 19:49
  • $\begingroup$ I trust that you are familiar with the Mandelstam variables? $\endgroup$ Jul 16, 2016 at 20:18
  • $\begingroup$ @dmckee: Do you address your question to me or to SRS? $\endgroup$ Jul 17, 2016 at 9:52
  • $\begingroup$ Using a running coupling is like the old tradition of using a velocity dependent relativistic mass. Sure, it usually helps to easily and quickly arrive at the correct expression, but one should not use this concept systematically: it doesn't always work. In the same way you cannot take a non-relativistic formula and substitute $m\to m(v)$ to get the relativistic correction, you neither can take a tree level amplitude and substitute $\alpha\to\alpha(s)$ to account for loop corrections. This sometimes fails (as in light-by-light scattering, where the tree amplitude vanishes). $\endgroup$ Dec 5, 2016 at 14:03

3 Answers 3


The quick and simple answer is : the value of $\alpha$ you measure at the scale for which you want to do predictions. Now, practically speaking, we don't measure $\alpha$ at each scale we do an experiment, but rather use renormalization group flow.

The explanation to the running of the coupling constant comes from renormalization and a simple way to picture renormalization in this case is the following. The diagram related to the running of $\alpha$ is the basic fermion-photon interaction diagram with two fermion legs and one photon leg. Consider now the series made by adding photon edges connecting the two fermion legs. It happens that the series is divergent instead of convergent, meaning that treating bare electromagnetic interaction with perturbation would be meaningless: you have to consider all the orders of this divergent series. The trick is now to tie the infinity this series diverges to to a physical constant you can measure. For that purpose, suppose that I rewrite $\alpha$ as $\tilde{\alpha}$ which takes into account the whole divergent series so that the Feynman diagram basic interaction block at the first order already includes the divergent series. Formally speaking, this coupling parameter is now infinite, compared to the bare value. One then measures this constant at the relevant scale and finds a finite value. The formally infinite parameter now has a finite value.

Finally, the running of this coupling parameter with a change of scale can be computed using the machinery of the renormalization group. In practice, we measure the coupling constant $\alpha$ at some energy scale and and "run" it using the renormalization group equation (with the $\beta$ function of QED, see this).


In quantum electrodynamics, we are provided with the Lagrangian, $$\mathcal L = \bar{\psi}(i\gamma^\mu D_\mu -m)\psi - \frac14 F_{\mu\nu}F^{\mu\nu}$$

where $D_\mu$ is the gauge-covariant derivative. In order to renormalise the theory, the standard practice is to go to renormalised perturbation theory, by rescaling the fields and constants, and expressing the Lagrangian as the sum of renormalised quantities and counter-terms.

For the electron charge, we usually take something like $e= Z_e^{-1}\mu^{-\epsilon/2}e^{(0)}$, and so the renormalised charge is related to the bare charge through a factor $Z_e$ and the renormalisation scale $\mu$ in dimensions $d= 4-\epsilon$, to keep the coupling dimensionless.

When we impose the condition that correlation functions of our theory are finite, we can derive expressions for $Z_e$ and the other renormalisations, which depend on the scale $\mu$. We can set up a differential equation then relating $e$ and the scale $\mu$:

$$\mu \frac{d}{d\mu} e = \frac{1}{12\pi^2}e^3.$$

Noting $\alpha = \frac{1}{4\pi}e^2$, we can use this to relate two couplings at different scales, that is,

$$\alpha(\mu_2) = \frac{\alpha(\mu_1)}{1 - \frac{1}{3\pi}\alpha(\mu_1)\ln \frac{\mu_2^2}{\mu_1^2}}.$$

This demonstrates that the coupling changes with scale, and is known as renormalisation group flow, though there are two main approaches to the RG and multiple interpretations. With this explained, to finally answer your question, you choose a coupling appropriate for the renormalisation scale of the experiment.

Higher order loop corrections will change the beta function, which in turn means the coupling at different scales will be predicted differently, and normally it is practice to stay consistent in perturbation theory and use the same order for all quantities.


Consider a QED scattering process $e^-+e^-\rightarrow e^-+e^- $. The scattering cross-section at the tree-level depends on the square of the fine-structure constant α (apart from the electron mass and the CM energy).

Yes, see NIST. The fine structure constant is typically given as: $$\alpha ={\frac {e^{2}}{(4\pi \varepsilon _{0})\hbar c}}= 0.0072973525664(17)$$ It's circa 1/137. However as NIST says:

"Thus α depends upon the energy at which it is measured, increasing with increasing energy, and is considered an effective or running coupling constant. Indeed, due to e+ e- and other vacuum polarization processes, at an energy corresponding to the mass of the W boson (approximately 81 GeV, equivalent to a distance of approximately 2 x 10-17 m), α(mW) is approximately 1/128 compared with its zero-energy value of approximately 1/137. Thus the famous number 1/137 is not unique or especially fundamental".

But α is a running coupling. My question is what value of α i.e., the value at which scale is to be substituted in this expression to obtain a numerical prediction? What is that value?

Somewhere between 1/137 and 1/128.

If instead, the scattering amplitude is calculated up to various higher orders, should one use a different value of α than used for tree-level result?

Yes. But life isn't so simple. The NIST article also says this:

"According to quantum electrodynamics (QED), the relativistic quantum field theory of the interaction of charged particles and photons, an electron can emit virtual photons that can then emit virtual electron-positron pairs (e+, e-). The virtual positrons are attracted to the original or "bare" electron while the virtual electrons are repelled from it. The bare electron is therefore screened due to this polarization".

This is misleading. Virtual particles are virtual. They only exist in the mathematics of the model. Moreover conservation of charge tells you that e doesn't vary, and so you should know that if α varies, it has to be because ε0 or ħ or c varies.

That's a surprise to some. But see the second paragraph here where Einstein talks about a gravitational field being a place where the speed of light is spatially variable. That's from 1920. Also see Clock Test of Relativity at Four Solar Radii. Alpha is thought to vary with gravitational potential. To muddy the waters further note how dmckee mentioned Mandelstam variables, and the Wikipedia article mentions invariant mass. Well, guess what? When you drop an electron some of its mass-energy is converted into kinetic energy. This is dissipated when the electron hits the ground, and you're left with a mass-deficit. (When you lift the electron you do work on it and add energy to it, increasing its mass). So the fine structure constant is not constant, invariant mass varies, NIST optical clocks going slower when they're lower and $c_0={1\over\sqrt{\mu_0\varepsilon_0}}$ suggests that both ε0 and c varies, and some even say h varies. So good luck with those calculations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.