Let us consider the Dirac complex
\begin{equation}
D_{\rm Dirac}:S^+\to S^-
\end{equation}
where $S^{\pm}$ are the chiral-spinor bundles on $\mathbb{R}^4$.
Using the fact that the bundle $S^+$ is given by $\Omega^{0,0} \oplus \Omega^{0,2}$ twisted by $K^{1/2}$ while $S^-$ is given by $\Omega^{0,1}$ twisted by $K^{1/2}$ where $K$ is the canonical bundle, the equivariant index of the Dirac complex with respect to $T=U(1)_1\times U(1)_2$ action $(z_1,z_2)\mapsto (t_1 z_1,t_2,z_2)$ can be computed by

\begin{eqnarray}
{\rm ind} D_{\rm Dirac}&=& \frac{ t_1^{1/2} t_2^{1/2} + t_1^{-1/2} t_2
^{-1/2}
- ( t_1^{1/2} t_2^{-1/2} + t_1^{-1/2} t_2^{1/2})}
{ (1-t_1)(1 -t_1^{-1})(1-t_2)(1-t_2^{-1})} \cr
&=& \frac { t_1^{1/2} t_2^{1/2}}{ (1 -t_1)(1-t_2)}
\end{eqnarray}
I would like to know the reason why the spinor bundles are equivalent to the Dolbeault complex twisted by the square root $K^{1/2}$ of the canonical bundle. Why is the index of the Dirac operator equal to the one of the twisted Dolbeault operator?

This question comes from the computation of one-loop determinant done by Pestun. (see p.35-36 in the paper and p.34 in the paper) It was shown in those papers that, to compute one-loop determinant \begin{equation} \frac{\det_{{\rm Coker} D} T}{\det_{{\rm Ker} D} T} \ , \end{equation} one can use the Atiyah-Singer index theorem for transversally elliptic operators.