How to find scaling dimensions of the scalar and gauge vector fields

The problem is

"Find the scaling dimensions of the scalar and gauge vector fields."

As I understand, a scalar field is a field with lagrangian: $$\mathcal{L}=\partial_{\mu} \phi^{*} \partial_{\mu} \phi-m^{2} \phi^{*} \phi \tag{1}$$ And gauge field has lagrangian: $$\mathcal{L}=-\frac{1}{4} B_{\mu \nu} B_{\mu \nu}+\frac{m^{2}}{2} B_{\mu} B_{\mu} \tag{2}$$ So, I need to find the $$\Delta$$ parameter after changing the scale: $$\begin{array}{c} x \mapsto x^{\prime}=\lambda x \\ \varphi(x) \mapsto \varphi^{\prime}\left(x^{\prime}\right)=\lambda^{-\Delta} \varphi(x) \end{array} \tag{3}$$ But I have no idea how to do that, and I am also not sure that these fields have scaling symmetry.

• Maybe for scalar nonmassive fields i can write but I am not sure: $$L=\partial_{\mu}\varphi\partial_{\mu}\overline{\varphi}=\dfrac{1}{\lambda^2}\partial_{\mu}(\lambda^{-\Delta}\varphi)\partial_{\mu}(\lambda^{-\Delta}\varphi)\Rightarrow \Delta=1$$ Jan 21 at 0:04

You have written the Lagrangian for a massive complex scalar field in (1). The classical scaling dimension of $$\phi$$ is just the mass dimension $$[\phi]$$. Considering d=4 dimensions, the action $$S=\int d^4x L$$ is dimensionless. $$[S]=0$$ and $$[d^4x]=-4$$ (length has inverse dimension of mass) mean that $$[L]=4$$. Therefore, (1) implies that $$[\phi]=1$$ and so the classical scaling dimension of the scalar field is 1.
This is also known as the engineering dimension, since when we turn on interaction terms (e.g. adding in $$\phi^4$$), the classical scaling dimension recieves a correction when the theory is renormalised. This correction is called the anomalous dimension. A similar story is true for the theory of a gauge vector field.
• So, this is true for nonmassive field? $$\begin{array}{l} S=\int \mathcal{L}\left(\Phi(x), \partial_{\mu} \Phi(x)\right) d^{4} x=\int \mathcal{L}^{\prime}\left(\Phi^{\prime}\left(x^{\prime}\right), \partial_{\mu} \Phi^{\prime}\left(x^{\prime}\right)\right) d^{4} x^{\prime} \\ L=\partial_{\mu} \varphi \partial_{\mu} \bar{\varphi}=\lambda^{4} L^{\prime}=\lambda^{4} \frac{1}{\lambda^{2}} \partial_{\mu}\left(\lambda^{-\Delta} \varphi\right) \partial_{\mu}\left(\lambda^{-\Delta} \varphi\right) \Rightarrow \Delta=1 \end{array}$$ Jan 21 at 17:52