# Question about performing VQE with an embedded Hamiltonian

Say I have a physical Hamiltonian $$\mathcal{H}$$ which is $$D$$-dimensional and I encode it into a larger matrix $$M$$ which is $$\Delta$$-dimensional. In the cases I care about $$\Delta$$ is strictly greater than $$D$$. Therefore there are $$(\Delta-D)$$ non-physical junk states'' in the basis of $$M$$. Also suppose that I have an ansatz for VQE which is only capable of producing physical states for any configuration of the ansatz parameters.

For VQE: is it strictly required that the entire matrix $$M$$ is Hermitian or just that the embedded physical Hamiltonian $$\mathcal{H}$$ is Hermitian?

For example: If the encoded physical basis states are $$|00>$$, $$|10>$$ and $$|11>$$, and the output of the ansatz circuit is always some superposition of those three, could $$M$$ be something like

$$$$M=\begin{pmatrix} a & \alpha & b & 0 \\ 0 & \beta & 0 & 0 \\ b & 0 & c & d \\ 0 & \gamma & d & f \\ \end{pmatrix}\quad\text{where}\quad \mathcal{H}= \begin{pmatrix} a & b & 0 \\ b & c & d \\ 0 & d & f \\ \end{pmatrix}.$$$$

I've come up with several examples where $$M\neq M^\dagger$$ gives the correct answers with VQE but can't find anything on why that works. Any citations on this would be extremely appreciated.

As a side note: this non-Hermitian-ness in $$M$$ comes from the encoded creation operators not being the adjoint of the encoded annihilation operators unless you restrict yourself to the space spanned by the physical basis states.