# Definition of outer product

I am trying to understand the concept of outer product in quantum mechanics. I read "Quantum Computing explained" of David MacMahon.

I can understand the transition in (3.12): $$(|\psi\rangle \langle \phi | )|\chi\rangle \rightarrow |\psi\rangle \langle \phi |\chi\rangle$$

But how to get $$(\langle \phi | \phi | \chi \rangle ) | \psi \rangle$$ ?

Why it is possible to get through such steps?

1. $$|\psi\rangle \langle \phi | \chi\rangle$$
2. $$|\psi\rangle \langle \phi | \phi | \chi\rangle$$
3. $$\langle \phi | \phi | \chi\rangle |\psi\rangle$$
• Moreover, what is $\langle \phi\mid\phi\mid\xi\rangle$? – Berci Mar 2 at 21:59
• @Berci I am searching that definition but unsuccessfully so far. – Roma Karageorgievich Mar 2 at 22:05

I'm thinking that it's a typo, and all the author wanted in the last term was to write $$(\langle \phi|\chi\rangle)\,|\psi\rangle.$$ The proof uses that you have a kind of associativity in the first equality $$(|\psi\rangle\langle\phi|)\,|\chi\rangle= |\psi\rangle\,\langle\phi|\chi\rangle$$ which I think is brought out of the blue if you introduce bras and kets out of nowhere.
The equality is obvious if you notice that kets as simple column vectors in $$\mathbb C^n$$, and bras are their adjoints (conjugate transpose). In that setting your equality is $$(\psi\phi^*)\,\chi=\psi\,(\phi^*\chi)=(\phi^*\chi)\,\psi,$$ where the associativity is that of the product of matrices.