Are there special advantages in this representation of sl2?

Question

In Mathematical Physics I often find representations of $sl(2)$ that instead of being based on the usual vectors $e,f,h$ are based on based on these vectors $$J_{3}=-\frac{1}{2}h,\,\,\,\,J_{+}=-f,\,\,\,\,J_{-}=-e,$$ with these relations $$\left[J_{3},\,J_{+}\right]=J_{+},\,\,\,\,\left[J_{3},\,J_{-}\right]=-J_{-},\,\,\,\left[J_{+},J_{-}\right]=2J_{3}.$$ And a special base on the vector space $V$ that results in these matrices (here $j$ is $n/2$) $$\pi_{n}\left(J_{3}\right) =\left(\begin{array}{ccccc} -j & 0\\ 0 & -j+1 & \ddots\\ & 0 & \ddots & 0\\ & & \ddots & j-1 & 0\\ & & & 0 & j \end{array}\right),$$ $$\pi_{n}\left(J_{+}\right) =\left(\begin{array}{cccccc} 0 & 0\\ \sqrt{2j} & 0 & \ddots\\ & \sqrt{2\left(2j-1\right)} & \ddots & 0\\ & & \ddots & 0 & 0\\ & & & \sqrt{2\left(2j-1\right)} & 0 & 0\\ & & & & \sqrt{2j} & 0 \end{array}\right),$$ $$\pi_{n}\left(J_{-}\right) =\left(\begin{array}{cccccc} 0 & \sqrt{2j}\\ 0 & 0 & \sqrt{2\left(2j-1\right)}\\ & 0 & 0\\ & & 0 & \ddots\\ & & & \ddots & \sqrt{2\left(2j-1\right)}\\ & & & \ddots & 0 & \sqrt{2j}\\ & & & & 0 & 0 \end{array}\right).$$ I see here that that $J_{-}$ and $J_{+}$ are the transpose one of the other. Are there special analytical or logical advantages in this formulation? Why is it so often chosen in mathematical physics? I'm interested in mathematical advantages and not historical or sociological advantages... Thanks in advance

Answer 1

This is the „ladder operator“ representation of the $\mathfrak{sl}(2,\mathbb{C})$. It is particularly important since it is used to classify the irreducible representations of the Lie Algebra, mainly due do the fact that the minimal/maximal eigenvalue of $J_3$ (that is $j$ in your notation) determines the dimension of the representation (2$j$ +1). For more information see Humphrey’s book on Lie Algebras. Futhermore, there is a physical interpretation of $J^\pm$ corresponding to annihilation- and creation-operators of ‚spin quanta‘.