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Nielsen and Chuang define $A^{\dagger}$ as the unique linear operator on $V$ such that for all vectors $|v\rangle, |w\rangle$ in $V$, $$(|v\rangle, A|w\rangle) = (A^{\dagger}|v\rangle, |w\rangle).$$

I'm having trouble deducing why $(AB)^{\dagger} = B^{\dagger}A^{\dagger}$ using only this definition.

I apply the definition to get $(|v\rangle, AB|w\rangle) = ((AB)^+|v\rangle, |w\rangle)$, but I thinking there's some other property that I can't think of, or am unaware of to prove this from scratch.

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    $\begingroup$ From $(|v \rangle, AB | w \rangle)$, move $A$ to get $(A^+ |v \rangle, B | w \rangle)$. Then, let $A^+ |v \rangle = |u \rangle$, and try to move $B$ over as you did with $A$. $\endgroup$
    – Muphrid
    Oct 15, 2016 at 2:05
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    $\begingroup$ Btw. you're using a suboptimal way to display the ket notation in Latex. The normal commands are \langle and \rangle. Using the bra-ket notation in such a redundant way (writing scalar products with normal brackets) seems cumbersome to me. You could write it as follows: Let $V$ be a Hilbert space and $A$ be a linear operator on $V$ (one may want to add details concerning domains of definition). Then there is a unique linear operator $A^{\dagger}$ (\dagger. Other common symbols are H, $*$) called the adjoint, such that for $v,w$ in $V$ $$\langle v, Aw\rangle=\langle A^{\dagger}v, w\rangle$$. $\endgroup$ Oct 15, 2016 at 2:13
  • $\begingroup$ Thanks. I see it now. I'm also having trouble seeing why $(A|v \rangle)^+ = \langle v| A^+$ using the definition that $|v \rangle^+ = \langle v|$. $\endgroup$
    – theQman
    Oct 15, 2016 at 2:17
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    $\begingroup$ It breaks my heart to say this, but one way to see what's going on is to choose a basis. $\endgroup$
    – WillO
    Oct 15, 2016 at 2:41
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    $\begingroup$ @Muphrid that should be an answer $\endgroup$
    – David Z
    Oct 15, 2016 at 9:44

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Let's just do the process in two steps. Each time we just apply the rule $$(|v\rangle, A|w\rangle) = (A^{\dagger}|v\rangle, |w\rangle) . $$ We want to apply this to $(|v\rangle, AB|w\rangle)$. Let's start by substituting $B|w\rangle\rightarrow |u\rangle$. So then we have an expression for which we can apply the rule: $$ (|v\rangle, A|u\rangle) = (A^{\dagger}|v\rangle, |u\rangle). $$ Now we replace $|w\rangle$ and define $A^{\dagger}|v\rangle\rightarrow |t\rangle$. So now we have another expression for which we can apply the rule $$ (|t\rangle, B|w\rangle) = (B^{\dagger}|t\rangle, |w\rangle). $$ Finally we replace $|v\rangle$. So, starting form the original expression, we get the result $$ (|v\rangle, AB|w\rangle) = (B^{\dagger}A^{\dagger}|v\rangle, |w\rangle), $$ which shows that the order of the operators changed. We didn't really need to do the substitutions, but it makes it obvious how to apply the rule.

Addition:

Why does $$ \tag{1} (A|v\rangle)^{\dagger}=\langle v|A^{\dagger} , $$ if we use $(|v\rangle)^{\dagger}=\langle v|$ as given?

There are probably various ways to show this. The way I can think of now is to say that it follows from $$ (\langle w|A|v\rangle)^{\dagger}=\langle v|A^{\dagger}|w\rangle . $$ To see that the latter should be true one can use the trace say that $$ \langle w|v\rangle={\rm tr}\{|v\rangle\langle w|\} $$ Now $|w\rangle\langle v|$ acts like an operator. So we can use the rule about operators to say that $$ (A|v\rangle\langle w|)^{\dagger}=(|v\rangle\langle w|)^{\dagger}A^{\dagger} =|w\rangle\langle v|A^{\dagger} $$ This shows how the operators transform together wth the states. I could probably have gone straight to this last expression by just appending the $\langle w|$ at the end in (1).

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  • $\begingroup$ Thanks, I see it now. I'm still confused about this bit of text that comes right after: using the definition that $|v\rangle+=\langle v|$ we get $(A|v\rangle) = \langle v|A^+$ $\endgroup$
    – theQman
    Oct 15, 2016 at 12:22
  • $\begingroup$ Oops, that should be $(A|v\rangle)^+ = \langle v|A^+$. $\endgroup$
    – theQman
    Oct 15, 2016 at 12:40

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