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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 $\Delta$?

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

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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).

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