- #1
center o bass
- 560
- 2
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?
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?