Relation for Inner Product with States from a Complete Set

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  • #1
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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.
 

Answers and Replies

  • #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]?
 
  • #3
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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.
 
  • #4
George Jones
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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.
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]?
 
  • #5
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thanks a lot.
 

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