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I'm learning about operations on qubits, and I came across this statement:

Suppose $|w\rangle = U |v\rangle$, and we want $U$ to preserve state norms. Then $\langle w|=\langle v|\bar U^{\operatorname{T}}$. What can we say about $\langle w | w \rangle$? It ought to equal $1$ (norms are preserved). Note that we have

$$\langle w | w \rangle = \langle v | \bar U^{\operatorname{T}} U | v \rangle$$ but since $\langle v | v \rangle = 1$, we must have that $\bar U^{\operatorname{T}} U=I$ and we call $U$ satisfying this criterion unitary.

What's unclear to me is how we were able to "take out" $\langle v | v \rangle$ in order to simplify the multiplication. Is there some notion of commutativity that allows us to draw this conclusion?

actinidia
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    Note: the common notation for $\bar U^T$ is $U^\dagger$ in QM literature, $U^H$ in some of the numerical linear algebra literature, and $U^*$ in the wider mathematical literature. – Ben Grossmann Sep 23 '20 at 06:15

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Perhaps this will help: $$ \langle w| w \rangle = \langle w| \ |w \rangle = \left( \langle u | U^\dagger\right)\cdot \big(U | u \rangle \big) = \langle u| \left( U^\dagger U\right) | u \rangle = \langle u | \ | u \rangle = \langle u|u \rangle. $$ The relevant notion is not that of "commutativity" but of associativity.

Ben Grossmann
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  • If I understand you correctly, this shows why $\langle w | w \rangle = 1$ if $U$ is unitary. How would we use $\langle v | v \rangle = 1$ to show that $\langle w | w \rangle = 1$ implies $U$ is unitary? – actinidia Sep 23 '20 at 06:19
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    @Tiwa Be careful with that statement: it is not true that for a unit vector $|v \rangle$ and $|w\rangle = U|v \rangle$, if $\langle w|w \rangle = 1$ then $U$ is unitary. What is true is that if for every unit vector $v$ and associated $w = U |v \rangle$ we have $\langle w| w\rangle = 1$, then $U$ is unitary. – Ben Grossmann Sep 23 '20 at 06:22
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    @Tiwa If you want to prove this second (true) result, then this fact is useful – Ben Grossmann Sep 23 '20 at 06:30