Can an entangled state in general be created by destructive interference in some subspace?

Question

For instance, if we have a general two-qubit state $$|\psi\rangle=\frac{1}{2}(|0\rangle+e^{i\varphi_a}|1\rangle)\otimes(|0\rangle+e^{i\varphi_b}|1\rangle)=\frac{1}{2}(|00\rangle+e^{i\varphi_b}|01\rangle+e^{i\varphi_a}|10\rangle+e^{i(\varphi_a+\varphi_b)}|11\rangle,$$ is there a way to create a Bell state simply by arranging for the states $|01\rangle$ and $|10\rangle$ to destructively interfere with one another so that those states are never observed? This seems exactly like the sort of thing that happens in certain processes in particle physics where two pathways/histories for a particle decay process destructively interfere with one another so neither history is ever actually observed.

If this simple case is possible, is it then possible in general to arrange any arbitrary entangled state of $N$ qubits by taking a state like $$|\psi\rangle=\prod_{j=1}^{N}(a_j|0\rangle_j+|1\rangle)=\sum_{(q_0,q_1,q_2,...q_N)=(0,0,0,...,0)}^{(1,1,1,...,1)}|q_0,q_1,q_2,...,q_N\rangle$$ and arranging for, say, all the states where more than one $q_j$ is in state $|1\rangle$ to destructively interfere with one another so the state becomes the entangled state $$|\psi\rangle=|10000...\rangle+|01000...\rangle+|00100...\rangle+...|00000...1\rangle~?$$

Answer 1

For distinguishable spins, $|01\rangle$ and $|10\rangle$ are two orthogonal vectors of Hilbert space. Therefore, they cannot interfere with each other. Also, a tensor product of states belonging to different subsystems is always separable (and never entangled) independently of the phases $\phi_a$, $\phi_b$ (or the coefficients $a_j$).