Indeed there is. Apologies for tooting my own horn, but you can find it in this paper, cowritten with Chris Koehl and Bill Spence. Instead of repeating the explanation, I refer you to this MathOverflow answer from a decade ago.
Added after the comment
From the point of view espoused in that paper, this is what remains when one takes a supersymmetric field theory in 4-dimensional Minkowski spacetime and dimensionally reduces to a line. Taking that line to be spacelike, the supersymmetry algebra breaks to the centraliser of the three-dimensional complement of that line. Schematically, the Poincaré superalgebra is $$\mathfrak{sp}= \mathfrak{so}(3,1) \oplus S \oplus V $$ where $V$ is the 4-dimensional real vector representation of $\mathfrak{so}(3,1)$ and $S$ is the 4-dimensional real spinorial representation of $\mathfrak{so}(3,1)$. Both $V$ and $S$ are irreducible representations. The Lie superalgebra is $\mathbb{Z}$-graded with $\mathfrak{so}(3,1)$ in degree $0$, $S$ in degree $-1$ and $V$ in degree $-2$. The only non-obvious bracket is $$ [S,S] = V $$ Now pick a spacelike vector $v \in V$. Then the centraliser of $v^\perp$ in $\mathfrak{sp}$ is the Lie subalgebra $$\mathfrak{so}(2,1) \oplus S \oplus V $$
Geometrically, $v^\perp$ acts trivially, $v$ is central and acts like the Laplacian, the $\mathfrak{so}(2,1) \cong \mathfrak{sl}(2,\mathbb{R})$ subalgebra is spanned by $L,\Lambda,H$ and $S$ is spanned by $\partial$, $\partial^*$, $\bar\partial$ and $\bar\partial^*$.
(You may have to complexify everything, by the way.)