# Relation for Inner Product with States from a Complete Set

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.

## Answers and Replies

George Jones
Staff Emeritus
Science Advisor
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?
thanks for the quick reply.

George Jones
Staff Emeritus
Science Advisor
Gold Member
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?
thanks for the quick reply.

Yes.

Now, what is the only possibility for $|\Psi>$ if $<\mathbf{p}|\Psi>$ is zero for every $\mathbf{p}$?

thanks a lot.