What is a d + id superconductor and why does it break time reversal symmetry? There are a lot of publications dealing with d-wave and d + id superconductivity, but I found no satisfying answer what exactly makes a superconductor d + id and why they break time reversal symmetry. Is it the same with p + ip superconductors? 
Also, what do these tearms (p + ip, d + id) mean? Do the wavefunction of the Cooper pairs in real space or the superconducting gap in k-space have corresponding symmetries?
 A: A $d+id$ superconductor breaks time reversal symmetry just like a $p+ip$ one does. 
In a $d+id$ superconductor, there are two coexisting $d$-wave order parameter. For example, there are two representations of $D_{4h}$ point group that have four nodes (directions where the order parameter vanishes), commonly noted as $\Delta_{d_{x^2-y^2}}\sim \cos(k_x)-\cos(k_y)$ and $\Delta_{d_{xy}}\sim \sin(k_x)\sin(k_y)$. By $d+id$, one means that both $\Delta_{d_{x^2-y^2}}$ and $\Delta_{d_{xy}}$ are nonzero, and moreover the relative phase between $\Delta_{d_{x^2-y^2}}$ and $\Delta_{d_{xy}}$ (since SC order parameters are in general complex) is $\pi/2$. Under time reversal, the relative phase becomes $-\pi/2$, and such a state can be denoted as $d-id$ state. Therefore time-reversal is broken.
Yes, in real space a $d+id$ superconductor is odd under $\pi/2$ rotations, just like a regular $d$-wave superconductor. 
A: At present, the mechanism of the superconductivity in systems like Na$_{0.35}$CoO$_2$·1.3H$_2$O are of high interest. The CoO$_2$ layers have been modeled as a spin 1/2 antiferromagnetic Mott insulator on a triangular lattice. By using resonate valence bond mean-field analysis, this supports the view that the superconducting order parameter has the spin-singlet broken-
time-reversal symmetry chiral $d_{x^2−y^2} \pm id_{xy}$ pairing symmetry.
