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: $\langle \psi | \hat{a}_{\updownarrow i}^{\dagger 2} \hat{a}_{\updownarrow i}^{2} | \psi \rangle = \langle \psi | \hat{a}_{\mathord{\leftrightarrow} i}^{\dagger 2} \hat{a}_{\mathord{\leftrightarrow}i}^{2} | \psi \rangle = \langle \psi | \hat{a}_{\mathord{\leftrightarrow} i}^{\dagger} \hat{a}_{\updownarrow i}^{\dagger}\hat{a}_{\mathord{\leftrightarrow}i} \hat{a}_{\updownarrow i} | \psi \rangle = 0$.

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    $\begingroup$ I have several questions about the exact model. First, why are you worrying about polarizations at all (i.e., one photon in each rail)? Wouldn't it be much simpler if you didn't consider polarizations? For the second question, with post-selection, what kind of measurements do you want to consider? Is the origin of this question the theorem that quantum computation with linear optics is possible if you start with single photon states and have photon number measurements? Did you want to figure out what states are used in this model, or was there another reason for the question? $\endgroup$ Nov 21, 2010 at 4:37
  • $\begingroup$ @Peter: I really hope Piotr comes back and addresses your questions so we can continue to get the benefit of your expertise here! $\endgroup$
    – David Z
    Nov 21, 2010 at 11:49
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    $\begingroup$ @Peter: I am working with Decoherence-Free Subspaces (with collective decoherence in the form of $|\psi\rangle \mapsto U^{\otimes n} |\psi\rangle$ for $U\in SU(2)$). So pairing of states (e.g. with polarization) is natural. The actual question is 'Starting with one state from DFS, which DFS states can be obtained with certainty (or more generally: what is success rate for $|\psi\rangle \mapsto |\phi\rangle$)?'. By post-selection I mean counting of photons (i.e. taking into account only that measurements in which certain detectors clicked once/twice/...). $\endgroup$ Nov 21, 2010 at 12:29

1 Answer 1


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.

  • $\begingroup$ Thanks. As I am interested in DFS states, I want to have in and out states with the same (and well-defined) number of photons (so I am not interested in anything near coherent states). Good point with quantum Zeno effect - perhaps obtaining success-rate bounds needs additional assumptions (e.g. 'no additional rails, the only post-selection rule: in each rail there was detected exactly one photon'). $\endgroup$ Nov 21, 2010 at 16:35

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