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Expected value in quantum

  1. Oct 13, 2013 #1
    1.GIF
     
  2. jcsd
  3. Oct 13, 2013 #2
    quantum expected value

    1. why <x> is squeez between ψ* and ψ what we doing this?
    2. for <p> [h/i d/dx ] is sqeeze between ψ* and ψ why is that?
    3. if you put latter operator between ψ* and ψ what is going to happen?

    thank you
     
  4. Oct 13, 2013 #3

    bhobba

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    Its easier to see in the Dirac notation - E(A) = <u|A|u>

    But that is only a special case valid for so called pure states - the full rule is E(A) = Trace(PA) where P is a positive operator of unit trace which is the correct definition of a quantum state - pure states |u><u| are a special case. For pure states Trace(|u><u| A) = <u|A|u>

    As to why that formula check out Gleason's Theorem:
    http://kof.physto.se/theses/helena-master.pdf [Broken]

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  5. Oct 14, 2013 #4
    The operators can be represented by matrices. Although usually we think of operators as operating on everything to the right, this is just a matter of convention. The physical observables are self-adjoint operators so it makes no physical difference if they act on the right or the left. This is good because the distinction between "bra" and "ket" wavefunctions is really just an artificial notational part of the model, and can't have real significance. We put the operator in the middle so it can act on the left or the right (but only once).

    The sign of the complex part of the wavefunction also has no real significance. Multiplying a number by its complex conjugate gives the absolute value squared. This is a handy mathematical trick, but there's no reason other than convention to put the conjugate on the left factor or the right factor.
     
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