# What distinguish a non-interacting two fermions gate from an interacting one?

We start with a Hilbert space consists of two fermions(a and b), which is 4-dimensional. The basis would be $$|vec>$$,$$b^\dagger|vac>$$,$$a^\dagger|vac>$$,$$a^\dagger b^\dagger |vac>$$.

Denote all the unitary transformation on this space as $$U^{tol}$$. One of the subspace of great interest would be the unitary transformation that can be realized as a Hamiltonian evolution. $$U(t) = Texp[-i\int_0^t H(t)dt]$$. I intuitively think all the unitary in $$U^{tol}$$ can be written as hamiltonian evolution, i.e. connected with the identity. But I am not sure.

We can further put constraints on the evolution Hamiltonian. If the Hamiltonian respect parity, i.e. it only mix state with the same parity, the matrix representation of the operator would be like: $$\begin{bmatrix} u_{11}&0&0&u_{12} \\ 0&v_{11}&v_{12}&0 \\ 0&v_{21}&v_{22}&0 \\ u_{21}&0&0&u{22} \end{bmatrix}$$

We can denote this subset of unitary as $$U^p$$. If we put another constraint of the Hamiltonian being non-interacting terms, i.e. $$a^\dagger a$$,$$b^\dagger b$$,$$a^\dagger b+b^\dagger a$$, $$a^\dagger b^\dagger + ba$$, the allowed unitary is further suppresed. Denote this subset as $$U^{NI}$$. An example of a non-interacting unitary would be: $$\begin{bmatrix} 1&0&0&0&\\ 0&0&i&0\\ 0&i&0&0 \\ 0&0&0&1 \\ \end{bmatrix}$$ while an interacting unitary would be: $$\begin{bmatrix} 1&0&0&0 \\ 0&0&1&0\\ 0&1&0&0 \\ 0&0&0&1 \end{bmatrix}$$.

Now my question is, is there anyway to tell whether a unitary is connected with identity operator with a non-interacting Hamiltonian? Or is there any invariant that can be computed from the unitary matrix to signify this?

For two-qubit unitaries, these gates are known as "matchgates". They encompass all gates which have the block-diagonal form you list, and where the determinant of the even-parity $$2\times 2$$ matrix $$(u_{ij})$$ is the same as the one of the odd-parity $$2\times 2$$ submatrix $$(v_{ij})$$.