Getting the conserved quantity corresponding to dilatational symmetry of $L=\frac{1}{2}m\dot{q}^2-\frac{1}{2}m\omega^2q^2.$

Question

Consider the Lagrangian of a $1$D classical linear harmonic oscillator: $$L(q,\dot{q})=\frac{1}{2}m\dot{q}^2-\frac{1}{2}m\omega^2q^2.\tag{1}$$ After a scaling, $$q\to \bar{q}=\alpha q,\quad \text{or}, \delta q=(\alpha-1)q,\tag{2}$$ where $\alpha$ is a real constant, the Lagrangian changes to $$L\to\bar{L}=\alpha^2L,\tag{3}$$ so that $$\delta L=\bar{L}-L=(\alpha^2-1)L.\tag{4}$$ Since an overall scaling of the Lagrangian does not change Lagrange's equation of motion, this is a symmetry of the system. However, the change in the Lagrangian, in this case, cannot be expressed as the total derivative of a function of the form $F=F(q,t)$. Therefore, the standard formula for conserved quantity: $$Q=\sum_j p_j\delta q_j-F,\tag{5}$$ cannot be directly applied here. In this case, how does one apply Noether's theorem to get the conserved quantity?

Answer 1

Well, even though the dilation/scaling (2) is a symmetry of the EOM, it is not a quasisymmetry of the action (1), cf. example 3 in my Phys.SE answer here. So Noether's theorem does not apply.