I'm currently reading Schwartz's QFT text and I'm confused on how observables are supposed to be independent of subtraction schemes. In the text it seems that the renormalized loop diagrams are different depending on the subtraction scheme chosen. So is it that the renormalized parameters have different meanings depending on the subtraction scheme and/or the regulator?
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$\begingroup$ Related: physics.stackexchange.com/q/57965/2451 , physics.stackexchange.com/q/73403/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jan 8, 2019 at 0:20
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1 Answer
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It's because we define, for instance, the mass to be the pole of the renormalized propagator. Namely in yukawa theory the propagator will be given by something like
$$G^R = \frac{1}{Z} \frac{i}{\not{p} - m_0 + \Sigma(\not{p})} $$
where $\Sigma_R$ is the sum over all 1PI insertions, and $m_R = \frac{1}{Z_m} m_0$ We can absorb the $Z, Z_m$ and plug in in for $m_0$ and redefine $\Sigma$ so that
$$ G^R = \frac{1}{1+\delta}\frac{i}{\not{p} - m_0 + \Sigma_2(\not{p})} \\ \boxed{G^R= \frac{i}{\not{p} - m_R + \Sigma_R(\not{p}, m_R, Z_m, Z)} } $$
Now the point is that you don't need your counter terms (or $Z_i$s) to be the same as somebody else's, and you don't have to agree on the $m_R$s either. Rather, all you need at the end of the day is to make sure that you have a pole at $\not{p} = m_P$. That is,
$$ \begin{cases} m_P - m_R + \Sigma_R(\not{p} = m_P) = 0 \\ 1 = \lim_{\not{p} \to m_P} \frac{\not{p} - m_P}{ \not{p} - m_R + \Sigma_R(\not{p}, m_R, Z_m, Z)} \end{cases} $$
where a stricter condition on $\Sigma_R$ can be found by using L'Hôpitals rule. Note that this is just the mathematical definition of a simple pole at $m_P$.
The condition that the mass $m_P$ gives a pole to the propagator is enforced in all schemes (it just looks different for different schemes). So no matter how we define $M_R$ or $\Sigma$, as long as we enforce the fact that the pole of the propagator is the physical mass then it doesn't matter how you get there.
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$\begingroup$ Thanks for the reply. How does the one condition nail down the renormalized greens' function seeing as we have multiple free parameters in the renormalization scheme? Am I correct in understanding that the renormalized green's function is supposed to be independent of subtraction schemes across its entire domain? As in literally being the same function for every four-momentum argument? $\endgroup$ Commented Jan 8, 2019 at 0:32
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$\begingroup$ In your original question you asked about the observables. And the only observable here is the physical mass, which is defined by making sure there is a pole in the propagator at the physical mass. I should note that there is one more condition I didn't mention, which I will put in my answer. $\endgroup$ Commented Jan 8, 2019 at 0:34
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$\begingroup$ Okay but doesn't the Green's function contribute to physical effects such as corrections to Coulomb's law? I was under the impression that the renormalized green's functions were strictly physical since they appear in finite results of the theory. $\endgroup$ Commented Jan 8, 2019 at 0:43
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$\begingroup$ It could seem so at first. But the correction to the coulomb potential is $\propto \frac{e_P^2}{m_P^2} \delta^3(x)$, and so it only depends on physical quantities, which are just numbers. $\endgroup$ Commented Jan 8, 2019 at 0:59
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1$\begingroup$ Okay, I see. But don't different schemes alter the momentum depnendecy of the dressed propagator? It doesn't seem obvious to me that these should all drop out when we calculate observable effects. $\endgroup$ Commented Jan 9, 2019 at 10:04