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So I am reading this book where they say this:

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My understanding of how operators work is that : $$\langle n|\hat{a}|\alpha \rangle = \langle n|\hat{a}\alpha \rangle = \langle n\hat{a}^\dagger |\alpha \rangle $$

but it seems that they are saying $$(\langle n|\hat{a} )|\alpha \rangle$$ then defining $\langle n|\hat{a}$ . But this seems incorrect given my understanding. because isn't it the adjoint of $\hat{a}$ acting on the bra?

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  • $\begingroup$ What does this have to do with expectation values? $\endgroup$
    – Sandejo
    Commented Oct 16, 2022 at 4:52
  • $\begingroup$ Which book? IMHO Dirac notation is just confusing in the very most cases. The adjoint is still an operator on the Hilbert space and is not acting on bras, which are elements of the dual space. Possibly related. $\endgroup$ Commented Oct 16, 2022 at 6:27

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Your understanding of how the adjoints of operators work is correct. Recall that the notation of $\langle n\hat{a}^\dagger|$ is essentially a shorthand for the adjoint of the ket $\hat{a}^\dagger| n \rangle$: $$ \langle n\hat{a}^\dagger | = (\hat{a}^\dagger| n \rangle)^\dagger = \langle n | \hat{a}. $$ This allows you to define the action of $\hat{a}$ on the bra $\langle n |$ via the adjoint of $\hat{a}^\dagger |n \rangle$.

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  • $\begingroup$ Why can't one just say that $\langle n \hat{a}^\dagger| = \langle n| \hat{a}^\dagger ?$ $\endgroup$ Commented Oct 16, 2022 at 2:02
  • $\begingroup$ @realanswers because then the notation for the adjoint wouldn't work. You wouldn't be able to say $\langle n | \hat{a} | \alpha \rangle = \langle n \hat{a}^\dagger| \alpha \rangle$ as in your post. $\endgroup$ Commented Oct 16, 2022 at 2:11
  • $\begingroup$ Then how is this notation illuminating in using the dagger why not just say it's $\langle n a |\alpha \rangle$ ? it seems just confusing $\endgroup$ Commented Oct 16, 2022 at 3:10
  • $\begingroup$ @realanswers For an inner product $\langle \cdot,\cdot\rangle:H\times H \longrightarrow \mathbb C$, which by convention is anti-linear in the first and linear in the second argument here, we have $\langle \psi,A\,\phi\rangle=\langle A^\dagger \,\psi,\phi\rangle$, which follows from the definition of the adjoint (one has to be careful for which vectors this holds in infinite dimensions, but this shouldn't matter for this question), cf. this $\endgroup$ Commented Oct 16, 2022 at 6:34

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