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

enter image description here

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 $
  • 1
    $\begingroup$ Moreover, what is $\langle \phi\mid\phi\mid\xi\rangle$? $\endgroup$ – Berci Mar 2 at 21:59
  • $\begingroup$ @Berci I am searching that definition but unsuccessfully so far. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.