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[Q]Eigenfunction of inverse opreator and another question.

  1. Nov 7, 2008 #1
    Hi.

    Do you know eigenfunction of inverse operator, for instance [itex] \hat{A^{-1}} [/itex] given that [itex] \hat{A}\varphi = a\varphi[/itex]

    textbook said eigenfunction of inverse operator A is the same as [itex] \varphi [/itex]

    which eigenvalue is [itex] \frac{1}{a} [/itex]

    Can you prove that?

    And is it really that [itex] [A,A^{-1}] = 0 [/itex] so both opreatator have a common
    eigenfunction if eigenvalue is not degenerate, this theorem is called commutator theorem?
     
  2. jcsd
  3. Nov 7, 2008 #2

    clem

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    Just operate with A^-1 so you get [tex]A^{-1}A\phi =\phi =a A^{-1}\phi[/tex].
    No need to commute A with its inverse.
    Then proving the commutator=0 follows.
     
    Last edited: Nov 7, 2008
  4. Nov 7, 2008 #3

    clem

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    The latex doesn't seem to work. The last term is \phi=a A^{-1} \phi.
     
  5. Nov 7, 2008 #4
    I don't understand why [itex] AA^{-1}\varphi = \varphi [/itex] It is absolute true that

    [itex] AA^{-1} = 1 [/itex] But applying [itex] AA^{-1} [/itex] to some function is

    different matter. for example we assume A and its inverse can be matrix, function f is also matrix.

    [itex] A(A^{-1}f) [/itex] is not [itex] (AA^{-1})f [/itex] right?
     
  6. Nov 8, 2008 #5
    Hm, yes, the law of association holds for matrices, so those two last expressions are equal.
     
  7. Nov 8, 2008 #6

    Fredrik

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    No, it can't be a different matter. To say that two operators X and Y are equal means that Xf=Yf for all functions f. This is no different from saying that two functions f and g are equal if f(x)=g(x) for all x.
     
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