Anomalous field dimension in the on-shell scheme

[aqualone]

In learning QFT, I've found that the renormalization group is often introduced with the MS scheme. I noticed that if one uses the on-shell scheme instead and calculates the anomalous field dimension using
$\gamma_{\phi} = \mu \frac{\partial \ln Z_{\phi}}{\partial \mu}$
one finds that $\gamma_{\phi} = 0$. Is this correct or am I making a mistake? What is the physical interpretation?

I've found this to be the case for both $\phi^3$ and $\phi^4$ theory. I haven't learned QED or anything about QFTs in particle physics so I'm not sure how it plays out there.

I wasn't sure if this is the place to post, or the homework or high-energy section; if this is not the right place, apologies and please feel free to move this thread.

 

[king vitamin]

In massless $\phi^4$ theory in 4 dimensions,

$$\gamma_{\phi} = O(g^2)$$

That is, there's no correction until 2-loop, so if you just do the one-loop calculation, you should get zero.

In massless $\phi^3$ theory in 6 dimensions (where the interaction is renormalizable), Srednicki does the calculation in his book and he does find a one-loop correction. (Note that the one-loop contribution is proportional to the square of the coupling still, unlike the phi4 case).