# Scale invariance in (2+1)D nonrelativistic field theory

Context: I am reading a paper named 'Nonrelativistic field-theoretic scale anomaly' on scale invariance in nonrelativistic field theory.

The Lagrangian density for the scalar field is given by, $$\mathcal{L} = i\phi^\dagger\partial_t\phi+\frac{1}{2}\phi^\dagger\nabla^2\phi-\frac{v_0}{4}\phi^\dagger\phi^\dagger\phi\phi.\tag{2.5}$$

Scale transformation: The scale transformation is given by, \begin{align} \boldsymbol{x}&\rightarrow e^\alpha \boldsymbol{x}\\ t&\rightarrow e^{2\alpha}t\\ \phi&\rightarrow e^{-\alpha}\phi. \end{align}\tag{3.1}

With these scale transformations, the paper states that \begin{align} \delta\phi &= (1+\boldsymbol{x}\cdot\nabla+2t\partial_t)\phi\tag{3.2}\\ \delta\mathcal{L}&=(4+\boldsymbol{x}\cdot\nabla+2t\partial_t)\mathcal{L}.\tag{3.4} \end{align}

Question: My question is about deriving the equations for $$\delta\phi$$ and $$\delta\mathcal{L}$$. I assume the equations have omitted the scale parameter $$\alpha$$ which is common in most papers and books. I have been able to derive the first equation but ended with a negative sign in front. A short derivation for that: \begin{align}\delta \phi &= \phi'(x,t)-\phi(x,t)\\ &= \phi'(x',t')-\phi(x,t)-[\phi'(x',t')-\phi'(x,t)]\\ &= -\alpha(1+\boldsymbol{x}\cdot\nabla+2t\partial_t)\phi \end{align} where I have kept terms upto first order in $$\alpha$$ and used the scale transformation equations. However my question is how to get the equation for $$\delta \mathcal{L}$$? Is there a short way to do it? I have tried $$\delta \mathcal{L}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta(\partial_\mu\phi)+\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi$$ and used the transformations but I do not get the desired result. Any help is appreciated.

Your missing sign could be the difference between a passive or active transformation. As for finding $$\delta L$$, your equation should work but it's hard to see where you've gone wrong without more detail of the calculation...