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Cloning quantum states

  1. May 16, 2008 #1
    Suppose you had a devide which, upon input of one of two non-orthogonal quantum states [tex]\left|\psi\right>[/tex] or [tex]\left|\phi\right>[/tex] correctly identified the state. How could you use this device to clone these states (in violation of the no-cloning theorem)?
     
  2. jcsd
  3. May 16, 2008 #2
    Depends what you mean by "identify".
    Usually the state is coupled to another system, say, [tex]\left|\rho\right\rangle_\mathrm{copy}[/tex] which then gets changed through some interaction to reflect the original state. If you destroy the original state but still want two identical copies, you'll need a third system, but the details are the same.

    The way you phrased the question, you could use a polarizer to correctly identify which orthogonal state a photon was in (assuming you restrict your photons to either state). Of course, you can't make a cloning device out of this.
     
  4. May 16, 2008 #3
    I quoted the question exactly. I guess by "identify" he means "outputs 1 or 0" depending on the input. Which means that it can't take an arbitrary input.
     
  5. May 16, 2008 #4
    Ok, then depends what you mean by "output".

    I know I sound pedantic but it's for a reason. One can spend forever looking for an answer to an ill-posed question.

    However, I misread the question about the states being non-orthogonal. I will ponder.
     
  6. May 16, 2008 #5
    Ok so the simple answer is that, if you've somehow evaded the no-cloning theorem, your device will say either [tex]\psi[/tex] or [tex]\phi[/tex]. Then it's in the domain of state preparation to clone the state. If you "know" the exact state, then you can create as many copies of it that you like. It just depends on what you're measuring.

    For instance, if the two states are photons polarized at [tex]30^\circ[/tex] and [tex]45^\circ[/tex], then simply produce more photons and align your polarizers in one of those two orientations.
     
  7. May 16, 2008 #6
    But if you "know" the state, it's trivial!
     
  8. May 16, 2008 #7
    I think the unwritten part of the question goes like this: a bunch of particles are randomly prepared as either psi or phi. psi and phi are not orthogonal states. they are sent to an analyzer. explain why the analyzer can't tell which paticle was prepared in what state (even given that the analyzer posesses knowlege of the set of preparation states). Feel free to beat me if im wrong.
     
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