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Measurement query

  1. May 31, 2005 #1
    Just working through some more quantum information stuff, and have come accross a stumbling block - I'm clearly missing something obvious.

    Consider a source emits states [itex]|\Phi\rangle = \cos\theta |0\rangle + e^{i\phi}\sin\theta |1\rangle[/itex] with fixed [itex]\theta[/itex] and random phases [itex]\phi[/itex], with equal probability for each phase.

    How can I show that a measurement of the operator Z ([itex]Z|0\rangle = |0\rangle , Z|1\rangle = -|1\rangle[/itex]) doesn't yield any information about the state emitted by the source?
     
  2. jcsd
  3. May 31, 2005 #2
    Just by computing the probability values: P(Z=0,|psi>)=<psi||0><O||psi> and P(Z=-1,|psi>)=<psi||1><1||psi> (<0|1>=0)

    Seratend.
     
  4. May 31, 2005 #3
    Thanks, I was pretty sure it was something simple I was overlooking - wood for the trees and all that!

    This leads on to another measurement question: Suppose a source emits two states:

    [tex]|\Phi_1\rangle = \frac{1}{\sqrt{2}}(|0\rangle + e^{i\phi}|1\rangle)[/tex]
    [tex]|\Phi_2\rangle = \frac{1}{\sqrt{2}}(|0\rangle - e^{i\phi}|1\rangle)[/tex]

    Where [itex]\phi[/itex] is an arbitary fixed phase. What measurement can be used to distinguish between the two states? They form an orthonormal set, so clearly can be distinguished, I just can't see 'how' to measure them.
     
  5. May 31, 2005 #4
    Just define another observable Z' with the good set of eigenvectors. I think now you are able to guess what vectors you have to choose.

    Seratend.
     
  6. May 31, 2005 #5
    Ah of course. I was trying to express the states as linear combinations of the eigenvectors of 'standard' operators rather than define my own.

    Thanks for the pointers.
     
  7. May 31, 2005 #6
    Just to confirm, would it be correct to define the measurement

    [tex]M=|\Phi_1\rangle\langle\Phi_1 |-|\Phi_2\rangle\langle\Phi_2 |[/tex] to measure the states above?

    (Edited to change + to -)
     
    Last edited: May 31, 2005
  8. May 31, 2005 #7
    Yes for the eigenvalues +1 and -1 (but you are free to select others).

    Seratend.
     
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