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I know that in general the following statement is true: $$\langle\phi|\chi\rangle = \langle\chi|\phi\rangle^* $$

And for the operator $A$ then the following identity also holds: $$ \langle \psi| A|\phi\rangle = \langle\phi|A^\dagger |\psi \rangle$$

Does this mean that (1) implies (2)

$$| \psi\rangle = |\chi\rangle|\phi\rangle\tag1$$ $$\langle \psi|= \langle\phi|\langle\chi|\tag2$$

since I assume$$ \langle\psi|\psi\rangle = 1?$$

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  • $\begingroup$ Yes, my apologies, I've fixed it. $\endgroup$
    – DJA
    Commented Oct 18, 2020 at 16:56

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If you are dealing with a composite system as it seems, you don’t need to change the order of $\psi$ and $\phi$ from (1) to (2), since the first (second) ket/bra always refers to the first (second) subsystem of your larger system. Along the same reasoning, you don’t need to change the order if the two kets/bras refer to two different degrees of freedom of the same system

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  • $\begingroup$ So you mean the convention already suggests that only the bra/kets that are in the same Hilbert space will interact with one another? $\endgroup$
    – DJA
    Commented Oct 18, 2020 at 16:51
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    $\begingroup$ Yes, you can say it like that $\endgroup$
    – Milarepa
    Commented Oct 18, 2020 at 17:14

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