Can we numerically find ground-state of a 1D tight binding Hamiltonain with odd sites at half filling? We can numerically find ground state energy and wavefunction of a 1D Hamiltonian at half-filling ($L = \#$ of sites and $N = \# $ of particles) using exact diagonalization. i.e at $L = 10$ and $N = 5$, we can digonalize Hamiltonian.
My question is, can we find ground state of a Hamiltonian with $L = $ (any odd number let's say 11) at half-filling?
I asked this question to a professor and he said yes, we can! there is a way to reach grand canonical from canonical; but he also said that he does not properly remember how to do it. He said one guess is that 
$L$ with $11$ sites $= \frac{1}{2} (L$ with $10$ sites + $L$ with $12$ sites$)$
Can someone please clarify me how is it possible?
Edit: system consists of spinless fermions with periodic boundary conditions.
 A: In the canonical ensemble this is clearly impossible. In this ensemble we keep the number of particle fixed, which means we work in the subspace where
$$
\hat{N} = N, \ \ \ \ (1)
$$
Here $\hat{N}$ is the particle number operator:
$$
\hat{N} = \sum_j a^\dagger_j a_j
$$
Since the eigenvalues of $\hat{N}$ are the non-negative integers, there are no states fulfilling 
$$
\hat{N}  =\frac{L}{2}
$$
when $L$ is odd. 
You can of course obtain a solution when you work in the grandcanonical ensemble. This amounts to impose the contstraint on average: i.e. we impose, 
$$
\langle \hat{N} \rangle_\mu = \frac{L}{2},  \ \ \ \ (2)
$$
where $\langle \bullet \rangle_\mu $ is the grandcanonical average which depends on the chemical potential $\mu$.
There is indeed a way to go from the grandcanonical to the canonical. This is probably what your professor was alluding to. I think to remember that it amounts to project back the grandcanonical solution onto the relevant $\hat{N} = N $ subspace. However the problem here is that again, that subspace is empty. 
The procedure you outlined may work but there seems to be no reason why we should limit to averaging over $N=10$ and $N=12$. In principle you have all possible value of $N$ (of course the relative weights will be small for large $N$). 

