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Relation for Inner Product with States from a Complete Set

  1. May 24, 2009 #1
    Hi.

    I've found the following relation (in a book about the qm 3-body scattering theory):

    [tex]<\Omega^{\pm}^{\dagger} \Psi_n|p>= ... = 0[/tex]

    where [tex]|p>[/tex] is a momentum eigenstate.
    So it is shown, that the inner Product is zero. Then they conclude that [tex]\Omega^{\pm}^{\dagger}|\Psi_n> = 0[/tex] because the p-states form a complete set.

    How can this formally be shown?

    thanky you.
     
  2. jcsd
  3. May 24, 2009 #2

    George Jones

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    First, something a little more general.

    How is a ket [itex]\left| \psi \right>[/itex] expressed with respect to the complete set of states [itex]\left| p \right>[/itex]?
     
  4. May 24, 2009 #3
    This isn't mentioned in the book. But i assume [tex] |\Psi>=\int |\mathbf{p}><\mathbf{p}|\Psi> d\mathbf{p}[/tex] as usual. Or what you mean?
    thanks for the quick reply.
     
  5. May 24, 2009 #4

    George Jones

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    Yes.

    Now, what is the only possibility for [itex]|\Psi>[/itex] if [itex]<\mathbf{p}|\Psi>[/itex] is zero for every [itex]\mathbf{p}[/itex]?
     
  6. May 24, 2009 #5
    thanks a lot.
     
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