1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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...

Of course you need to sum the contributions from the corresponding variations in the conjugate field.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.