Yes. In physics, we think of a Dirac operator \(D\) as a first-order differential operator such that \(D^2=\Delta\) is the Laplacian. (Mathematically, its an operator acting on the sections of a spinor bundle, see here for details). In the case of supersymmetry, the interesting question to ask is about the possible supersymmetric decompositions of the Dirac operator, i.e., decompositions that satisfy an appropriate SUSY algebra.

For the sake of pedagogy, consider \(d=4\), where we have 2 SUSY decompositions of \(D\), one of which is called chiral SUSY, and the other complex SUSY. Here I will consider only the first, since it's well-known and a little easier to construct. For the complex case cf. the ref. below.

Let \((\gamma^m)\) be our Clifford algebra generators (i.e., \(\{\gamma^m,\gamma^n\}=2g^{mn}1\)), \(D_m=\partial_m-iA_m\) the covariant derivative for a vector potential \(A_m\), and \({\not}D=\gamma^m D_m\). Then Dirac operator can be decomposed as \[{\not}D = Q_+ + Q_-,\]

where \[Q_+=\frac{1}{2}(1+\gamma^5)\gamma^m D_m, \quad Q_-=\frac{1}{2}(1-\gamma^5)\gamma^m D_m.\]

As an easy exercise, you can see that \(Q_+\) and \(Q_-\) are nilpotent operators, and letting \(H=-(\gamma^m D_m)^2\), the usual \(\mathcal{N}=1\) SUSY algebra is satisfied: \[H = \{Q_+,Q_-\}, \quad [Q_+, H] = [Q_-,H]=0,\]

as you learn in a SUSY QM course.

For details and a discussion of the complex SUSY case, cf.