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Take the Mach-Zehnder interferometer as an example. A photon passes through a beam splitter, is reflected off mirrors, and interferes with itself at another half-silvered mirror. No measurements or disturbances are made in between. The preferred basis of the photon is given by which path it takes after passing the second beam splitter. The problem is, this preferred basis is nonlocal in between.

Is there a problem with locality here, or is silly ol' me just plain confused?

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What exactly do you mean with "preferred basis"? There is no measurement as far as I can see that would make one basis stand out. –  A.O.Tell Oct 19 '12 at 14:09
    
Generally, in scattering theory, the outgoing momentum state are considered to be the "preferred basis", since it is unlikely that someone will make a huge spherical mirror to recohere all the outgoing photons. –  Ron Maimon Oct 19 '12 at 14:14
    
If you start with input into only one port of the first beamsplitter then the second beamsplitter should reconstruct that state and the photons come out at one of the output ports of the second beamsplitter. This is probably what he means by the preferred basis. But the "basis" is really only generated through the virtue of inputting photons into only a single port of the first beamsplitter. If that did not happen you would need a measurement somewhere to generate the basis. –  SMeznaric Oct 19 '12 at 16:15

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