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[Q]Analytic function of operator A

  1. Nov 4, 2008 #1
    Hi, 5.27 problem in liboff says that if [itex] g(A)f(\varphi) = g(a)f(\varphi),where A\varphi = a\varphi [/itex]

    I tried to solve this problem with tylar expansion.

    [itex] f(\varphi) = f(0) + f^{'}(0)\varphi + \frac{f^{''}(0)}{2!}\varphi^2 + \frac{f^{(3)}(0)}{3!}\varphi^3 + ...[/itex]

    [itex] g(A) = g(0) + g^{'}(0)A + \frac{g^{''}(0)}{2!}A^2 + \frac{g^{(3)}(0)}{3!}A^3 + ...[/itex]

    But when i applied g(A) to f([itex]\varphi[/itex]) i can not get right hand side of equality

    written above because i don't know how can i deal with unseen term such as

    [itex] A\varphi^2 [/itex]or [itex]A^2\varphi [/itex] or etc

    please assist to me.
     
  2. jcsd
  3. Nov 4, 2008 #2

    clem

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    Don't expand f.
    Show that A^n f=a^n f, which is easy to show by repeated use of Af=af.
     
  4. Nov 5, 2008 #3
    Ya I now your way, but How can i deal with [itex] A\varphi^n [/itex] showed in product

    of [itex] f(\varphi) [/itex] expansion and g(A) expansion
     
  5. Nov 5, 2008 #4

    clem

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    As you show, expanding f is useless, because you don't know [tex]A\phi^2[/tex].
     
  6. Nov 5, 2008 #5

    cks

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    I think in this way, in Dirac notation

    [tex]
    A(\|phi>)^2=(A\|phi>)|phi>=(a\|phi>)|phi>=a|phi>^2

    [/tex]
    Similarly,
    [tex]

    A^2\|phi>=A(A\|phi>)=A(a\|phi>)=a(A|phi>)=a^2|phi>
    [/tex]

    The second equation is straightforward.
    The first equation sounds a bit weird to me. I feel like I maybe wrong.
     
  7. Nov 5, 2008 #6

    Hurkyl

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    Just checking -- what do you think [itex]f''(0)[/itex] and [itex]\varphi^2[/itex] mean?
     
  8. Nov 5, 2008 #7

    cks

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    Like Hurkyl pointed out,

    one single operator acts on [tex] \psi^2 [/tex] , acts on two wavefunction, sounds weird.I don't know how to explain it.
     
  9. Nov 5, 2008 #8

    clem

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    Science Advisor

    It is wrong.
     
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