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Jan 11, 2016 at 18:41 history edited Martin CC BY-SA 3.0
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Jan 10, 2016 at 7:59 history tweeted twitter.com/StackPhysics/status/686095089261752320
Jan 10, 2016 at 3:46 comment added Andrew @AleksandrSokolov That's why you need to pick a scheme to fix the ambiguity in what you choose for $Z$. The point of MSbar is that you only subtract the divergent piece (so there are no finite pieces in $Z$ as you send $\epsilon\rightarrow 0$). This is a choice. The scheme dependence drops out of any physical quantity though, which is guaranteed by setting up RG equations. Also, you are right the form of $Z$ is not obvious without doing some work, they have omitted the calculation of the relevant loop diagrams (which is perfectly acceptable since there are no new ideas in that calculation).
Jan 10, 2016 at 3:08 answer added Nogueira timeline score: 1
Jan 7, 2016 at 12:59 comment added Yossarian @ACuriousMind but can we define it this way? I mean, can we choose whatever we like for the $Z$?
Jan 6, 2016 at 13:35 comment added ACuriousMind They explain it directly after that equation in a footnote: "Physical quantities can be made finite even if we include terms of order $\epsilon^0,\epsilon^1,\epsilon^2$ in $Z_i$. This corresponds to taking other renormalization prescription." and "The reason why the MS scheme is used more often than other schemes is that empirically perturbative series for various physical quantities exhibit good convergence behaviors.", i.e. this form of the constant is the definition of the MS scheme.
S Jan 5, 2016 at 19:21 history bounty started Yossarian
S Jan 5, 2016 at 19:21 history notice added Yossarian Draw attention
Dec 30, 2015 at 17:19 history edited Qmechanic CC BY-SA 3.0
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Dec 30, 2015 at 16:11 history asked Yossarian CC BY-SA 3.0