1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: QM - observables
  1. QM: Identity (Replies: 1)

  2. QM: Force (Replies: 4)

  3. QM: Commutator (Replies: 2)

  4. Qm notation (Replies: 6)

  5. QM: spin (Replies: 4)

Loading...