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I am having trouble understanding how an operator actS when it is between a ket and a bra:

For example, let $\hat{a}$ be a ladder operator, as in a simple harmonic oscillator case, and let $\phi_n$ an energy eigenstate associated to the energy $E_n$. Then I want to compute $\langle \phi|\hat a^2|\phi \rangle$

  • Option 1 $$\langle \phi_n| (\hat{a} (\hat{a} | \phi_n \rangle)) \propto \langle \phi_n | \phi_{n-2} \rangle = 0 $$

  • Option 2 $$(\langle \phi_n|a)(a| \phi \rangle)$$

I mean, in the first case it gives zero, but what about the second case?

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    $\begingroup$ Can you define your system? Why do you say it is zero in the first case? For a generic state (not the ground state) you don't get zero $\endgroup$
    – ohneVal
    Commented Feb 2, 2021 at 16:28
  • $\begingroup$ @ohneVal the action of a in this energy eigenstate will be; $$a|\phi> = \sqrt{\phi|}|\phi-1>$$, acting twice will give us $$ \sqrt{\phi(\phi-1)}|\phi-2>$$, this is orthogonal to $$<\phi|$$ $\endgroup$
    – El ele
    Commented Feb 2, 2021 at 16:49
  • $\begingroup$ Let us add all of that to the question to make it complete, by the way by the same logic both give 0. $\endgroup$
    – ohneVal
    Commented Feb 2, 2021 at 16:56
  • $\begingroup$ You are currently still mixing up notation. When you write $|\phi\rangle$, do you always mean energy eigenstates $|\phi_n\rangle$ or are you interested in statements about general vectors $|\phi\rangle$? I assume from the way that the question is posed that all $|\phi\rangle$ should be replaced by $|\phi_n\rangle$ (and similarly for the bras) $\endgroup$
    – TBissinger
    Commented Feb 2, 2021 at 17:06

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The second option also gives zero, operators can act to the left via adjoints, that is $$\hat{a}|\psi\rangle = (\langle \psi | \hat{a}^\dagger)^\dagger$$

or applied to your case $$\langle \phi_n | \hat{a} = (\hat{a}^\dagger|\phi_n\rangle)^\dagger \propto \langle \phi_{n+1}|$$

Using the above result, you have $$(\langle \phi_n |\hat{a})( \hat{a} |\phi_n\rangle) \propto \langle \phi_{n+1} | \phi_{n-1}\rangle = 0$$

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