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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.

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  • $\begingroup$ Maybe for scalar nonmassive fields i can write but I am not sure: $$L=\partial_{\mu}\varphi\partial_{\mu}\overline{\varphi}=\dfrac{1}{\lambda^2}\partial_{\mu}(\lambda^{-\Delta}\varphi)\partial_{\mu}(\lambda^{-\Delta}\varphi)\Rightarrow \Delta=1$$ $\endgroup$ Jan 21, 2022 at 0:04

1 Answer 1

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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.

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  • $\begingroup$ So, this is true for nonmassive field? $$ \begin{array}{l} S=\int \mathcal{L}\left(\Phi(x), \partial_{\mu} \Phi(x)\right) d^{4} x=\int \mathcal{L}^{\prime}\left(\Phi^{\prime}\left(x^{\prime}\right), \partial_{\mu} \Phi^{\prime}\left(x^{\prime}\right)\right) d^{4} x^{\prime} \\ L=\partial_{\mu} \varphi \partial_{\mu} \bar{\varphi}=\lambda^{4} L^{\prime}=\lambda^{4} \frac{1}{\lambda^{2}} \partial_{\mu}\left(\lambda^{-\Delta} \varphi\right) \partial_{\mu}\left(\lambda^{-\Delta} \varphi\right) \Rightarrow \Delta=1 \end{array} $$ $\endgroup$ Jan 21, 2022 at 17:52

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