How is this equivalence for flipping an equation of ket-vectors to an equation of bra-vectors, true? (Complex conjugate "distributes" across product?)

Question

I'm given, in the setup of an exercise (not even the exercise itself):

$$|i\rangle=\alpha|u\rangle+\beta|d\rangle \implies \langle i|=\langle u|\alpha^*+\langle d|\beta^*$$

I'm confused. Why is that self-evident? I was given few axioms, the seemingly relevant one here being:

$$\langle A|B\rangle=\langle B|A\rangle^*$$

Suppose I started with a placeholder $X$...

$$\langle X|i\rangle=\alpha\langle X|u\rangle+\beta\langle X|d\rangle$$

...and then did the conjugation...

$$\langle i|X\rangle=(\alpha\langle X|u\rangle)^*+(\beta\langle X|d\rangle)^*$$

Then I'm tempted to write

$$\langle i|X\rangle=\alpha^*\langle u|X\rangle+\beta^*\langle d|X\rangle$$

But then I'd have to ask, how come we can "distribute" the $^*$ to both terms in a product (as opposed to a sum). If that is legal, then can someone show me a proof? (I have been given axioms relating to associativity, commutativity, etc. for bra- and ket-vectors.)

And supposing that was legal, the next step would also be sketchy to me:

$$\langle i|=\alpha^*\langle u|+\beta^*\langle d|$$

What, we can just strip all ket-vectors on both sides? (I guess though I should have had a problem with this step from the very beginning, when I did the inverse and added bra-vectors on both sides.)

Answer 1

Recall that $\alpha$ and $\langle X|u\rangle$ are just complex numbers, as $\langle\cdot|\cdot\rangle$ is a map from $\mathcal H\times\mathcal H$ ($\mathcal H$ being the Hilbert space, e.g., $\mathbb C^n$) to the complex numbers. But the complex conjugate ${}^*$ of a product of complex numbers is the product of the complex conjugates so $$ (\alpha\langle X|u\rangle)^*=\alpha^*(\langle X|u\rangle)^*\overset{\text{Axiom}}=\alpha^*\langle u|X\rangle\,. $$ The second step then relies on the equivalence of the following statements:

1. $|x\rangle=|y\rangle$
2. $\langle v|x\rangle=\langle v|y\rangle$ for all $v$

(the non-trivial direction being 2. $\Rightarrow$ 1., of course). This is a variant of the statement that two vectors coincide if and only if all their coordinates coincide.

A common yet slightly advanced way of proving this is to use the basis formula for Hilbert spaces: given any orthonormal basis $(|n\rangle)_n$ of your Hilbert space---an object which always exists---then $$ |x\rangle=\sum_n\langle n|x\rangle\,|n\rangle\overset{\text{2.}}=\sum_n\langle n|y\rangle\,|n\rangle=|y\rangle\,, $$ hence 1. holds.

Answer 2

You can also use different notation: if $v$ is a vector in Hilbert space $H$, denote by $v^\dagger$ the map $H \rightarrow \mathbb{C}$ such that for all $w \in H$, $v^\dagger(w) := \langle v, w\rangle$.

It is then easy to prove that $(\lambda v)^\dagger = \lambda^* v^\dagger$. Indeed, let $w \in H$. We then have $(\lambda v)^\dagger (w) = \langle \lambda v, w\rangle = \lambda^* \langle v,w\rangle = \lambda^* v^\dagger (w)$. This being true for all $w$, we then have the announced statement. You can similarly show that $(v_1 + v_2)^\dagger = v^\dagger_1 + v^\dagger_2$.

The last step is to realize that physicists denote $v$ by $\vert v \rangle$ and $v^\dagger$ by $\langle v \vert$.