Changing the zero-point energy I have the following Hamiltonian
$$\mathcal{H}(\{x_i,y_i \})=-l\sqrt{2}\sum_{i=1}^N \mathbf{f}_i \cdot \hat{\mathbf{b}}_i+E_0$$
For calculating things like the partition function it would be extremely beneficial to drop the $E_0$. Is a shifted Hamiltonian ok to use and when are we allowed or not allowed to do this?
 A: The addition of a constant energy to a hamiltonian is not a problem for the computation of the partition function.
For example in the canonical ensemble, taking $H=H'+E_0$ and $Z'$ to be the partition function of the system with Hamiltonian H':
$Z = \mu(e^{-\beta H}) = \mu(e^{- \beta H' - \beta E_0}) = \mu(e^{- \beta H'} e^{ - \beta E_0}) =  e^{ - \beta E_0} \mu(e^{- \beta H'})=  e^{ - \beta E_0} Z'$
Where $\mu$ is the trace function in the quantum mechanical case or the the integral over phase-space in the classical case. Since these are both linear operations its no problem to pull the constant number $e^{-\beta E_0}$ out and obtain the above relation.
For the grand canonical and micro-canonical ensembles the calculations are not much different, but in the grand canonical case the result depends on what you mean by $H = H' + E_0$. If you understand it to be a way of rewriting each $n$ particle Hamiltonian as $H_n=H_n'+E_0$, then the same calculation from above can be used, if you see it as a shift of each particles energy, you have $H_n=H_n' + n\, E_0$, and the result is $Z= Z'(\mu'=\mu - E_0)$, the partition function you get from Hamiltonian $H'$ where the chemical potential $\mu$ is shifted down by $E_0$.
The partition function itself is not a physical object, the physical object is a probability distribution on some phase space. The partition function encodes many of the properties of this distribution in a convenient (and sometimes mysterious) manner. For example in the classical canonical ensemble this distribution is given by $p(x) = e^{-\beta H(x)} / Z$ (where $x$ is a set of coordinates parametrising phase space). If we look at what happens to the distribution when we consider a Hamiltonian $H'(x)=H(x)+E_0$, we get:
$p'(x)=e^{-\beta H'(x)}/\,Z' = e^{-\beta H(x) -\beta E_0}/\, (Z e^{-\beta E_0}) = e^{-\beta H(x)} / Z = p(x)$ 
So the actual physical object, the probability distribution, remains unchanged by an energy shift.
