Formally we have $$Cl_{1,3}(\mathbb{R})\otimes_\mathbb{R} \mathbb{C} \cong Cl_4(\mathbb{C})\cong M_4(\mathbb{C})\cong \text{End}(S)$$ and on a more explicit level (see wiki) we can exhibit the spinor module as
$$S =\textstyle \bigwedge^{\text{even}}W \oplus \bigwedge^{\text{odd}} W$$ where $W$ is the isotropic space with respect to the standard $\mathbb{R}^{1,3}$ quadratic form extending to $\mathbb{C}$
Also, I know that $$\text{Spin}(1,3):=\{v_1\cdots v_{2k} : \vert v_i \vert =1\}$$ can be seen as a 2 to 1 cover of $SO(1,3)$ by the adjoint action on the Cliford algebra $C^{\times}_{1,3}$
Now my question is,
Is it even possible to derive the spinor module as
without resorting to some contrived identification such as viewing the hermitian matrix
$$X = \begin{pmatrix} x_0 + x_1 & x_2 + ix_3 \\ x_2 - ix_3 & x_0 - x_1 \end{pmatrix}$$
as $\mathbb{R}^{1,3}$ by the matrix norm $-\det(X) = -x_0^2 + x_1^2 + x^2_2 + x^2_3$ or whatsoever ?
[say, how can I construct this from the viewpoint of $\text{Spin}(1,3)\cong SL(2,\mathbb{C})$ adjoint action above? or is it even related?]
I believe this is related to how I exhibit the real structure $Cl_{1,3}(\mathbb{R})$ in its complex algebra (since the question about the hermitian matrix is related to its real algebra but I cannot see how to do this)