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

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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 $
  • 2
    $\begingroup$ Moreover, what is $\langle \phi\mid\phi\mid\xi\rangle$? $\endgroup$
    – Berci
    Mar 2, 2019 at 21:59
  • $\begingroup$ @Berci I am searching that definition but unsuccessfully so far. $\endgroup$ Mar 2, 2019 at 22:05

1 Answer 1


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.


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