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Kitaev's way of exactly solving his honeycomb model described here is to describe the system using Majorana fermions, a description that introduces a lot of unphysical degrees of freedom. The up shot to this though is that the Hamiltonian reduces to a quadratic form, $$H = \frac{i}{4} \sum_{ij} A_{ij} c_i c_j,$$ where $c_i$ is a local Majorana operator at the $i$-th spin site. As he describe in Eq. (14) and the paragraph above it, after we finish a calculation with the Majorana description of the model we should always project back onto the physical subspace, $$|\psi\rangle_{\text{phys}} = \prod_i \frac{1}{2} (1+D_i) |\psi\rangle $$

My question is the following: Suppose I solve for the ground state $|\psi_0\rangle$ of the Hamiltonian above. Is there a possibility that $$ \prod_i \frac{1}{2} (1+D_i) |\psi_0\rangle = 0 $$ Is there a reasons to believe that $|\psi_0\rangle$ must have a component onto the physical space? I don't mean to imply that it mustn't. I'm asking if it's, in principle, possible.

More detail. Maybe I'll more context to the question. I was doing some numerics for the Kitaev model on a torus, see my earlier question here. On a torus with a $2\times 2$ period I calculated the ground state energy using the original spin Hamiltonian (not converting to a fermionic representation) and compared to what I get when using Kitaev method of passing to a fermionic description and working in the standard gauge where all gauge variables are $+1$. To my surprise the two numbers did not match. Not only that, the energy that I get using Kitaev method, doesn't match any excited state energy of the original model. To me at least, this seems to suggest that the state I reach with Kitaev method is unphysical. I should add that when moving to a bigger torus with period $3 \times 3$, I do get both methods to agree on the ground state energy. But still, I'm left with the question above.

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