The usually negligibly small thermoelectric effects in superconducting heterostructures can be boosted dramatically due to the simultaneous effect of spin splitting and spin filtering. Building on an idea of our earlier work [Phys. Rev. Lett. $textbf
{110}$, 047002 (2013)], we propose realistic mesoscopic setups to observe thermoelectric effects in superconductor heterostructures with ferromagnetic interfaces or terminals. We focus on the Seebeck effect being a direct measure of the local thermoelectric response and find that a thermopower of the order of $sim200$ $mu V/K$ can be achieved in a transistor-like structure, in which a third terminal allows to drain the thermal current. A measurement of the thermopower can furthermore be used to determine quantitatively the spin-dependent interface parameters that induce the spin splitting. For applications in nanoscale cooling we discuss the figure of merit for which we find enormous values exceeding 1 for temperature $lesssim 1$K.
We have shown by using the exact solutions for the two-electron system in a parabolic confinement and a homogeneous magnetic field [ M.Taut, J Phys.A{bf 27}, 1045 (1994) ] that both exact densities (charge- and the paramagnetic current density) can b
e non-interacting $cal V$-representable (NIVR) only in a few special cases, or equivalently, that an exact Kohn-Sham (KS) system does not always exist. All those states at non-zero $B$ can be NIVR, which are continuously connected to the singlet or triplet ground states at B=0. In more detail, for singlets (total orbital angular momentum $M_L$ is even) both densities can be NIVR if the vorticity of the exact solution vanishes. For $M_L=0$ this is trivially guaranteed because the paramagnetic current density vanishes. The vorticity based on the exact solutions for the higher $|M_L|$ does not vanish, in particular for small r. In the limit $r to 0$ this can even be shown analytically. For triplets ($M_L$ is odd) and if we assume circular symmetry for the KS system (the same symmetry as the real system) then only the exact states with $|M_L|= 1$ can be NIVR with KS states having angular momenta $m_1=0$ and $|m_2|=1$. Without specification of the symmetry of the KS system the condition for NIVR is that the small-r-exponents of the KS states are 0 and 1.
