Measuring effective (renormalized) coupling constants

[center o bass]

How is the effective (or renormalized) coupling constant at a given momentum scale scale measured?

If one wants a definition which makes it easy to measure I would think it would be natural to use the LSZ formula which connects the measurable Feynman amplitude with the amputated greens function. For example in phi-four theory

$$i \mathcal{M} = Z^{(n/2)} \Gamma^{(n)}_{\text{on shell}}$$

where $\mathcal{M}$ is the feynman amplitude and $\Gamma^{(n)}$ is the vertex function and is identical to the amputated greens function. Since $M$ should be easy to measure (and in some sense indicates the strength of the interaction) I would guess natural definition of the effective coupling would be that for example $g = Z^{(2)} \Gamma^{(4)}(p_i = \mu)$ at some momentum scale $\mu$. This is exactly the definition of the effective coupling used in conventional renormalization of phi-four theory.

This would indeed correspond to $i\mathcal{M}$ for a scalar particle scattering if $\Gamma$ was evaluated on shell in the definition of $g$. However for an arbitrary scale $\mu$ this does not longer correspond to the LSZ formula. And I no longer see how the effective coupling is easily measured, nor how it is connected to the 'strength' of the interaction.

A similar thing also happens in QED where

$$ie \equiv Z_2 \sqrt{Z_3}ie_0\Gamma^{\mu}|_{\text{on shell}} = \frac{Z_2 \sqrt{Z_3}}{Z_1} ie_0 = i \sqrt{Z_3}e_0$$

and $Z_3$ is related to the vacuum polarization through $Z_3^{-1} = 1 - \Pi(q^2 = 0)$. Now at an arbitrary scale I've seen the definition of $\alpha = e^2/(4\pi)$

$$\alpha(\mu) = \alpha_0 (1 + \Pi(q^2 = - \mu^2))$$

where the photon with momentum $q$ is no longer on shell. Now I do not longer see how this is easily measured or how it connects to the strength of the interaction.

So my question is how is the effective coupling measured in a theory (for example QED) at an arbitrary momentum scale and how does the definition of the coupling then correspond to something we can call the 'strength' of the interaction?

 

[andrien]

So my question is how is the effective coupling measured in a theory (for example QED) at an arbitrary momentum scale and how does the definition of the coupling then correspond to something we can call the 'strength' of the interaction?

you will have to draw all feynman diagrams with vacuum polarizations included to higher orders,when you will write the amplitude then you will find a sum which can be evaluated using some identity of
1/A+B with noncommutating A and B(they may be commuting but not necessarily).When you will sum up then you will be able to replace the fine structure constant by α(μ) type thing which will already take the account of vacuum polarization.The onshell condition is not followed in this procedure apparently.

 

[The_Duck]

How is the effective (or renormalized) coupling constant at a given momentum scale scale measured?

You compute a scattering amplitude or something in terms of the renormalized coupling, then measure the scattering amplitude in an experiment, then solve for the numerical value of of the renormalized coupling given the measured numerical value of the scattering amplitude. A renormalization scheme where the renormalized coupling is defined to be equal to some on-shell scattering amplitude makes this particularly easy, but you can carry out the same procedure in any renormalization scheme.

 

[center o bass]

You compute a scattering amplitude or something in terms of the renormalized coupling, then measure the scattering amplitude in an experiment, then solve for the numerical value of of the renormalized coupling given the measured numerical value of the scattering amplitude. A renormalization scheme where the renormalized coupling is defined to be equal to some on-shell scattering amplitude makes this particularly easy, but you can carry out the same procedure in any renormalization scheme.

Ah, thanks! That was enlightening.
Would you say that, regardless of the renormalization scheme, one could identify the renormalized coupling as something that has to with the 'strength' of the interaction?

The reason I wonder is that one tends to plot $g(\mu)$; is it correct to infer something else than when perturbation theory is a good or bad approximation from these plots? I.e is it, regardless of the renormalization scheme, always correct to infer that a big $g(\mu)$ corresponds to a 'strong' interaction?

 

[The_Duck]

I think in any reasonable renormalization scheme, $g(\mu)$ is indeed measuring the strength of the interaction. To check this you can compute a scattering amplitude in perturbation theory and look at its dependence on $g(\mu)$. Typically we find scattering amplitudes of the form

$g(\mu) + C g(\mu)^2 \ln(E/\mu) + ...$

where E is a typical energy of the scattering particles. If $g(\mu)$ is small, then the first term dominates and the scattering amplitude--which is a measure of the strength of the interaction--is just $g(\mu)$. So we can be sure that $g(\mu)$ is measuring the strength of the interaction if our scattering amplitude has this form.

We might worry that the logarithm in the second term might get large. Then, even if $g(\mu)$ is small, the first term might not dominate anymore. However if we use the coupling renormalized at a scale $\mu \approx E$ then the logarithm will not be large and the first term will indeed dominate. So to quantify the strength of the interaction at an energy $E$ we should use $g(E)$.

So suppose $g(\mu)$ starts small but increases as we move $\mu$ in some direction. At first we can trust perturbation theory and it tells us that the strength of the interaction is increasing. At some point $g(\mu)$ becomes too large for perturbation theory to be reliable. Then we have to either guess what happens at these energies or else do some sort of nonperturbative calculation to see what actually happens. Note that once we get into this regime, the beta function that tell us how $g(\mu)$ runs with $\mu$ can no longer be trusted, if it was computed in perturbation theory.