Which qubit states are accessible with linear optics operations? Given a quantum state of $n$ qubits, and being restricted to linear optics (that is, the output annihilation operators are linear combinations of the input annihilation operators):


*

*Which states are accessible, that is can be made from $|\psi \rangle$ with certainty? (Are there any simple criteria?)

*If we want to get (any) $|\phi \rangle$ from $|\psi \rangle$ and we allow post-selection, are there any known bounds on the success rate? 


If it is going to simplify anything, one may assume that we consider states having exactly one photon in each rail:
$\langle \psi | \hat{a}_{\updownarrow i}^{\dagger 2} \hat{a}_{\updownarrow i}^{2} | \psi \rangle
= \langle \psi | \hat{a}_{\mathord{\leftrightarrow} i}^{\dagger 2} \hat{a}_{\mathord{\leftrightarrow}i}^{2} | \psi \rangle
= \langle \psi | \hat{a}_{\mathord{\leftrightarrow} i}^{\dagger} \hat{a}_{\updownarrow i}^{\dagger}\hat{a}_{\mathord{\leftrightarrow}i} \hat{a}_{\updownarrow i} | \psi \rangle
= 0$.
 A: Thanks for the clarification. Your question makes sense to me now. I'm not really going to be able to answer it. In general, if you start with a photon number state, and put it through linear optics, I believe the state you get looks like a big, ugly mess if you try to write it down in any reasonable basis. 
I don't think you'll be able to get most quantum states without postselection. An argument for this is: suppose you could get all quantum states. Then you could get coherent states. Reversing the circuit, you would then be able to get your start state $|{\psi}\rangle$ by starting with some coherent state and using linear optics. But if you start with coherent  states, and use linear optics, all you'll get is coherent states. 
Maybe a better path is to investigate is what happens when you do a small linear optics transformation, and then project back into your DFS. What you would then want are a set of transformations which keep you in the DFS (using the quantum Zeno effect) but where the quantum Zeno effect doesn't end up keeping your original state exactly the same. I don't know whether such linear optics transformations exist.
