# Invariant quantity from commutators

I am reading this paper on conformal quantum mechanics by De Alfaro, Fubini, and Furlan. There, they find the algebra of the generators of $$(0+1)$$-D conformal transformations (Eq. 2.23) $$[H,D] = iH\;, \qquad [K,D] = -iK\;, \qquad [H,K] = 2iD\;.$$ Here $$H$$ is the Hamiltonian operator, $$D$$ is the dilatation generator, and $$K$$ is the special conformal operator in $$(0+1)$$-D. On the next page, they define an operator $$G = uH+vD+wK$$ where $$u, v,w$$ are constants. Next, they say that from the commutator relation it is easy to see that the quantity $$\Delta = v^2-4uw$$ is invariant with respect to any general conformal transformation $$G\rightarrow U^{-1}GU$$.

I can check that this is true using any $$U$$. But what I do not understand is how one can "easily" determine what the expression of $$\Delta$$ should be just from looking at the commutation relations? How do they get this $$\Delta$$?

• Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. Commented Jul 31, 2021 at 19:40

Perhaps de Alfaro et al note that$$\left[G,\,\left(\begin{array}{c} D\\ H\\ K \end{array}\right)\right]=iM\left(\begin{array}{c} D\\ H\\ K \end{array}\right),\,M:=\left(\begin{array}{ccc} 0 & u & -w\\ -2w & -v & 0\\ 2u & 0 & v \end{array}\right).$$Since $$M$$ is singular and traceless, its eigenvalues are of the form $$0,\,\pm\lambda$$. If we diagonalize to isolate the $$0$$ eigenvalue, the others' product should be a homogeneous quadratic, so $$\lambda$$ is a square root thereof. And it comes as no surprise, given how quadratic equations work, that $$\lambda=\sqrt{v^2-4uw}$$.
Note in particular we don't need to compute $$M$$ to realize that, nor to realize it will be singular ($$G$$ gives an obvious kernel element) or traceless (that just requires the "self-interacting" commutators $$[H,\,D],\,[K,\,D]$$ to have opposite coefficients).