For a sharp cut-off regularization, we introduce the UV cutoff $\Lambda$. When we need to do momentum integral, we integrate the momentum ball with radius $\Lambda$. This $\Lambda$ has the explicit physical meaning of UV cutoff.

For $\phi^4$ in $4$-dim, when we use dimensional regularization, we introduce an arbitrary mass scale $\mu$ $$S= \int d^Dx \frac{1}{2} (\partial_\mu \phi)^2 + \frac{m^2}{2} \phi^2+ \frac{\lambda \mu^\epsilon}{4!} \phi^4 $$ with $\epsilon = 4-D$.

Up to now, the introduction of the arbitrary mass scale $\mu$ is just to keep parameter $\lambda$ dimensionless. It can be any number. But when we write the RG equation and beta function, we give the $\mu$ physical meaning of a UV cutoff. However textbook doesn't explain why.

My question:

  1. Why does this arbitrary mass scale $\mu$ has the physical meaning of UV cutoff?
  • $\begingroup$ This wont answer your question, but I believe you have to perform that integral before you use dimensional regularization, otherwise that integral doesnt make sense. $\endgroup$
    – Craig
    Commented Dec 19, 2018 at 4:56
  • $\begingroup$ Related questions here and here. $\endgroup$
    – knzhou
    Commented Dec 19, 2018 at 20:40

1 Answer 1


Dude, you are confused! In the dimensional renormalization scheme (Feynman used to call the shell game of renormalization dippy Hocus-Pocus), it's the $\epsilon = 4-d$ which plays the role of UV cutoff.

The renormalization scale $\mu$ is the energy scale ($p^2 \sim \mu^2$) you anchor your renormalized parameters, such as coupling $g_{renor}|_{p^2=\mu^2}$, Usually, $\mu$ is chosen at the physical process scale in concern, rather than an UV cutoff scale $\Lambda$ which is assumed to be close to the Planck scale $M_p$.


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