yields the exact solution. Generally, whenever the original model $H$ is exactly solvable, the choice $H'=H$ will do.

For a non-trivial model $H$, the choice $H'=H$ is useless for a practical computation. Any simplified reference system, however, yields a consistent approximation. The case of the DMFT discussed in Sec. \[sec:dmft\] is an illustrative example. On the other hand, in the context of the DMFT actually both, $H$ and $H'$, are highly non-trivial models, and further approximations or large-scale numerics are needed to treat the reference system $H'$. More simple approximations for the Hubbard model which are still constistent are generated by considering reference systems with a [*finite*]{} number of degrees of freedom. The reference system of Fig. 1c with $n_{\rm s} < \infty$ is an interesting example which shall be discussed in the following. For small $n_{\rm s}$ one can easily obtain numerical results as a complete diagonalization of $H'$ is feasible.

![ Grand potential $\Omega$ (per lattice site) and the different contributions (per lattice site) according to Eq. (\[eq:vvv\]): $\Omega'$, ${\rm Tr} \ln (- ({\bf G}_0^{-1} - {\bf \Sigma}({\bf t}'))^{-1})$, and ${\rm Tr} \ln (-{\bf G}')$ for $U=4$, $T=0$, and $\mu=U/2$ (half-filling) as functions of $V$. Stationary points (arrows) 