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POVM measurement

  1. Aug 14, 2012 #1

    Demystifier

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    I need a simple, intuitive, and easy-to-comprehend example of a POVM measurement, which is not a projective measurement. Any suggestions?
     
  2. jcsd
  3. Aug 14, 2012 #2
    Take a system which is a tensor factor of a larger system and perform a projective measurement on that larger system with a basis that does not separate in that tensor factor. The subsystem is then undergoing a POVM measurement.
     
  4. Aug 15, 2012 #3

    Demystifier

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    Jazzdude, it is clear mathematically but not physically. How a measurement in such a mixed basis can be performed in practice? Any simple example?
     
  5. Aug 15, 2012 #4

    bhobba

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    Indeed - mathematically I love that explanation - but physically its meaning is unclear. Physically I think the following is better - see page 14-16:
    http://arxiv.org/pdf/quant-ph/0205039v1.pdf

    Its basically what results when you let a measuring system interact with the system being measured then observe the measuring system.

    Also note the cool almost trivial proof of Gleasons Theorem with POVM's.

    Thanks
    Bill
     
    Last edited: Aug 15, 2012
  6. Aug 15, 2012 #5
    Take a system of two particle with nonzero spin and measure the total spin. Then you can describe the measurement as a POVM measurement for each individual particle.

    In quantum information theory one often considers measuring an entangled system in the Bell basis, which also results in a POVM for each subsystem.

    Does that make it clearer? If you're asking how you actually perform an experiment to measure those I must admit that I have almost no knowledge about that.

    Cheers,
    Jazz
     
  7. Aug 15, 2012 #6

    A. Neumaier

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    The family of coherent states is a POVM of the kind you want, describing a simultaneous but unsharp measurements of position and momentum.
     
  8. Aug 16, 2012 #7

    Demystifier

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    Thank you all. The example by Neumaier satisfies my criteria the best.
     
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