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Under what condition <AB>=<A><B> stands?

  1. Jul 10, 2014 #1
    The same thing as title.
     
  2. jcsd
  3. Jul 10, 2014 #2

    Meir Achuz

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    The < has to be an eigenstate of A, and the > an eigenstate of B.
     
  4. Jul 10, 2014 #3
    Can you tell me why?
     
  5. Jul 10, 2014 #4
    [A,B]=0? Am i right?
     
  6. Jul 10, 2014 #5

    Matterwave

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    In some normalized state |f> you want to know when is <f|AB|f>=<f|A|f><f|B|f> (if the state is not normalized, then we have to add some factors in the denominator, which is annoying notationally, but easy to do).

    If A|f>=a|f> and B|f>=b|f> then we can see that <f|AB|f>=ba<f|f>=ab and <f|A|f><f|B|f>=ab<f|f><f|f>=ab. So in the case that |f> is an eigenstate of both A and B, then we will have this equality hold. This condition is therefore sufficient, but is it necessary? Actually it's late right now, and off the top of my head, I am unsure if this condition is necessary, perhaps you can finish the other half of the proof.

    Do we need [A,B]=0? Well, in the above sufficiency argument we required that |f> be an eigenstate of both A and B. So we only required that the state in which we take the expectation value to be an eigenstate of both A and B. We did not require that A and B share a complete set of eigenstates. So although [A,B]=0 is a sufficient condition, since commuting operators share eigenstates, it is not strictly speaking necessary. There could be exotic conditions where non-commuting operators share 1 eigenstate in common for example.
     
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