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Average value of observable

  1. Mar 2, 2014 #1
    Why we define that average value of some observable ##\hat{A}## in state ##\psi## is
    ##(\psi,\hat{A}\psi)##
    Why this isnot perhaps
    ##26(\psi,\hat{A}\psi)##?
     
  2. jcsd
  3. Mar 2, 2014 #2

    Doc Al

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    Staff: Mentor

    What if the state ψ was an eigenstate of the observable A?
     
  4. Mar 2, 2014 #3
    In that case ##\hat{A}\psi=a\psi##,
    so
    ##(\psi,\hat{A}\psi)=a##.
     
  5. Mar 2, 2014 #4

    Doc Al

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    Right, where ##a## is the eigenvalue of the observable. Not ##26a##!
     
  6. Mar 2, 2014 #5
    Tnx.
     
  7. Mar 16, 2014 #6
    But still after I think. Ok. When the system is in some eigen- state ##\varphi_n##, we measure energy ##E_n##. So
    ##(\varphi_n,\hat{H}\varphi_n)=E_n##
    but if system is in some other state, perhaps, ##\phi(x)## which is not eigenstate of observable ##\hat{H}## how I can be sure that average value of energy is
    ##(\phi(x),\hat{H}\phi(x))##?
     
  8. Mar 16, 2014 #7

    kith

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    The expectation value can be defined as <H>=∑npnEn. If you use Born's rule for the pn you get <ψ|H|ψ>.
     
  9. Mar 16, 2014 #8

    Doc Al

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    Staff: Mentor

    You can express the general state ##\phi(x)## in terms of the eigenfunctions: ##\phi(x)## = ##a_1\varphi_1## + ##a_2\varphi_2## ...
    where ##a_n^*a_n## represents the probability of measuring ##E_n##. Thus the average value will be ##(\phi(x),\hat{H}\phi(x))##.

    (This is equivalent to what kith just said about using the Born rule.)
     
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