The free Dirac operator is the differential operator of the following form $$ T_0 = i \alpha \nabla + \beta,$$
where $\alpha$ and $\beta$ are Hermitian $4 \times 4$ matrices, and $T_0$ is selfadjoint on some domain $D(T_0)$ on infinite-dimensional Hilbert space. Dirac operator acts on $C^4-$valued functions $\psi(r) = (\psi_i(r))^4_{i=1}$ in Hilbert space $(L^2)^4.$
If we want to include the interaction in the differential equation (which is often the case, since the free Dirac operator is not so physically interesting) one could just use the following type of the interaction:
$$V(r) = \frac{Ze^2}{r^2},$$
where $Z$ and $e$ are just numbers. This is the important (Coulomb) type of the potential in physics.
Now the new 'interacting' differential operator is defined as
$$T = T_0 + V(r).$$
My questions: Could this type of interaction $V$ somehow 'destroy' the selfadjointness of the operator $T$, even if $T_0$ is selfadjoint on some domain? If yes, could this be simply avoided by choosing the new domain for $T=T_0 + V$ without the $r=0$, since $V$ is singular there? Is there a reason why would selfadjointness be ruined only for large coefficients Z, but not for small ones?
More information about $V$: $V$ is a $4 \times 4$ symmetric matrix, whose elements $V_{ij}$ are (complex valued) measurable almost everywhere finite functions on $R^3$.