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Homework Help: QM - observables

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data
    If A and B were observables, and say the simultaneous eigenkets of A and B [tex]{|a',b'>}[/tex] form a complete orthonormal set of base ket. Can we conclude that [tex][A,B]=0[/tex]?

    2. The attempt at a solution

    Assume [tex]{|a',b'>}[/tex] is incompatible:

    [tex]AB|a',b'>=a'b'|a',b'>[/tex] <-- skipped several steps
    [tex]BA|a',b'>=a'b'|a',b'>[/tex]

    [tex]AB|a',b'>-BA|a',b'>=0[/tex]
    [tex][AB-BA]|a',b'>=0[/tex]
    [tex][A,B]|a',b'>=0[/tex]
    [tex][A,B]=0[/tex]

    and so we reach a contradiction. Therefore, we conclude that [tex][A,B]=0[/tex] assuming the simultaneous eigenkets of A and B [tex]{|a',b'>}[/tex] form a complete orthonormal set of base ket.

    was this thought process correct?
     
  2. jcsd
  3. Sep 15, 2007 #2

    dextercioby

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    Yes, but the observables are incompatible, not the vectors. And your LaTex would look better by using "[tex] \rangle [/tex]" instead of ">".
     
  4. Sep 15, 2007 #3
    It is important to mention that vectors [itex] |a',b' \rangle [/itex] form a full basis. Therefore any vector in the Hilbert space can be represented as a linear combination of these basis vectors. Therefore, by linearity, your eq. (1) is valid for any vector [itex] |\psi \rangle [/itex]

    [tex][A,B] |\psi \rangle =0[/tex]

    Then you can conclude that eq. (2) holds.

    Eugene.
     
  5. Sep 15, 2007 #4
    Sorry, you mean I should assume that A and B are incompatible and the do the proof by contradiction right?
     
  6. Sep 15, 2007 #5

    dextercioby

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    Nope, compatibility of the observables is no issue here. And you don't need to use any "proof by contradiction". A direct proof is enough. And that's what you did in post #1 in this thread.
     
  7. Sep 15, 2007 #6
    Then I'm not sure what you mean by this:
     
  8. Sep 15, 2007 #7

    dextercioby

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    I was correcting your expression: the observables are compatible/incompatible, and not the vectors, since you cannot measure vectors, but only observables.
     
  9. Sep 16, 2007 #8
    just wonder, how would this change if [tex]|a' , b' \rangle[/tex] did not form a complete orthonormal set of base ket
     
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