# Simple calculation on coordinate transformation of Lagrangian (Qualls' CFT lecture note)

I have a question while reading "Lectures on conformal field theory" by Qualls (https://arxiv.org/abs/1511.04074). $$^1$$

Question. I cannot find that the transformation (1.12) makes the action invariant. My attempt is as follows. Since $$Q=(ct+d)Q'$$, it follows that $$\frac{dQ}{dt}=cQ'+(ct+d)\frac{dt'}{dt}\frac{dQ'}{dt'}=\cdots=cQ'+\frac{1}{ct+d}\frac{dQ'}{dt'}.$$ Also, we have $$t=\frac{dt'-b}{-ct'+a}$$ by inverting the transformation, and finally we obtain

$$L'=\frac12 \left( cQ'+\frac{1}{\left(c\cdot \frac{dt'-b}{-ct'+a}+d\right)^2} \left(\frac{dQ'}{dt'}\right)\right)^2 -\frac{g}{2\left(c\cdot \frac{dt'-b}{-ct'+a}+d\right)^2Q'^2}.$$ How this transformed Lagrangian related to the original one?

$$^1$$ Please ignore the red "Why?" in the picture, which is answered in this related Phys.SE post.

TL;DR: The conformal transformation (1.12) is only a quasi-symmetry of the action (1.11), i.e. it only preserves the action modulo boundary terms.

Sketched proof:

1. The conformal transformation (1.12) leads to $$\mathrm{d}t^{\prime}~\stackrel{(1.12)}{=}~\frac{\mathrm{d}t}{(ct+d)^2}, \tag{A}$$ $$\mathrm{d}Q^{\prime}~\stackrel{(1.12)}{=}~\frac{\mathrm{d}Q}{ct+d}-\frac{cQ\mathrm{d}t}{(ct+d)^2}.\tag{B}$$

2. The velocity transforms as $$\frac{dQ^{\prime}}{dt^{\prime}}~\stackrel{(A)+(B)}{=}~ (ct+d)\frac{dQ}{dt}-cQ.\tag{C}$$

3. The potential term is strictly invariant: $$\frac{\mathrm{d}t^{\prime}}{Q^{\prime 2}}~\stackrel{(A)+(1.12)}{=}~\frac{\mathrm{d}t}{Q^2}. \tag{D}$$

4. The kinetic term changes with a total time derivative term: $$\left(\frac{dQ^{\prime}}{dt^{\prime}}\right)^2 \mathrm{d}t^{\prime}-\left(\frac{dQ}{dt}\right)^2 \mathrm{d}t ~\stackrel{(A)+(C)}{=}~-\left(\frac{d}{dt}\frac{cQ^2}{ct+d}\right)\mathrm{d}t.\tag{E}$$ $$\Box$$