# Relation for Inner Product with States from a Complete Set

## Main Question or Discussion Point

Hi.

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

$$<\Omega^{\pm}^{\dagger} \Psi_n|p>= ... = 0$$

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

How can this formally be shown?

thanky you.

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George Jones
Staff Emeritus
Gold Member
First, something a little more general.

How is a ket $\left| \psi \right>$ expressed with respect to the complete set of states $\left| p \right>$?

This isn't mentioned in the book. But i assume $$|\Psi>=\int |\mathbf{p}><\mathbf{p}|\Psi> d\mathbf{p}$$ as usual. Or what you mean?

George Jones
Staff Emeritus
This isn't mentioned in the book. But i assume $$|\Psi>=\int |\mathbf{p}><\mathbf{p}|\Psi> d\mathbf{p}$$ as usual. Or what you mean?
Now, what is the only possibility for $|\Psi>$ if $<\mathbf{p}|\Psi>$ is zero for every $\mathbf{p}$?