In dimensional regularization, we must shift the dimensionless coupling $g$ by the renormalization scale $\mu$ (which has unit mass dimension): \begin{equation} g \rightarrow \mu^{4-d} g \tag{1} \end{equation} This is done to ensure that the action will have the correct dimensions.

Apparently, $\mu$ will not affect any observables. But I don't understand how this can be proven.

Edit: I am aware that we do not want to "true" theories to depends on the cut-off $\Lambda \rightarrow \infty$ or $\epsilon=4-d \rightarrow 0$, but this doesn't tell me what will happen with $\mu$.

Edit 2: Ok, maybe I understand it. We introduce the renormalization scale $\mu$ and this is an unphysical parameter. Therefore, we demand that that physical observable are not dependent on $\mu$. This procedure results in the renormalization group equations. Is this correct?

  • 2
    $\begingroup$ your second edit is correct, it is made independant by construction. If it is not the case, then the theory is ill-defined. $\endgroup$
    – bendaizer
    Commented Mar 5, 2014 at 10:27


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