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< f | A f >^2 = < f | A^2 f >

  1. May 12, 2005 #1
    where A is arbitrary operator (not only ermiton), f is function, that < f | f > = 1. How to prove, that f is eigenfunction of operator A?
     
  2. jcsd
  3. May 12, 2005 #2

    Galileo

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    Try to rewrite the equation.
     
  4. May 12, 2005 #3
    i've prooved it when operator is ermiton, but i can't proove for arbitrary operator...
     
  5. May 12, 2005 #4

    dextercioby

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    What do you know about this baby [itex] |f\rangle\langle f| [/itex]...?

    Daniel.
     
  6. May 12, 2005 #5
    i've never used it before.
     
  7. May 12, 2005 #6

    dextercioby

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    [tex] \langle f|f\rangle =1 \Rightarrow |f\rangle\langle f|f\rangle =|f\rangle \Rightarrow |f\rangle\langle f|=\hat{1} [/tex]

    Does that help...?

    Daniel.
     
  8. May 12, 2005 #7
    Thank you. I've prooved it with your help. The puzzle is solved :smile:
     
  9. May 12, 2005 #8

    dextercioby

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    Some details:

    1.[itex] |f\rangle\langle f| [/itex] is called the projector onto the vector [itex] |f\rangle [/itex].I assumed the Hilbert subspace is unidimensional and that the basis is formed by this vector [itex] |f\rangle [/itex].Therefore,the projector is generaly ~ to the unit operator on this subspace,but in this case,due to the orthonormalization,it coincides with the unit operator.

    2.It's called HERMITEAN (or symmetric) operator,after the name of the 19-th cent.French mathematician Charles Hérmite.:wink:

    Daniel.
     
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