An exercise says,
Using the axioms for inner products, prove
$$\{\langle A| + \langle B|\}|C\rangle = \langle A|C\rangle+\langle B|C\rangle.$$
The two axioms I've been given are
- They are linear:
$$\langle C|\{|A\rangle+\langle B\rangle\}=\langle C|A\rangle+\langle C|B\rangle$$
- Interchanging bras and kets corresponds to complex conjugation:
$$\langle B|A\rangle=\langle A|B\rangle^*$$
One of the solutions I've found starts...
\begin{align} \{\langle A| + \langle B|\}|C\rangle &= [\langle C|\{|A\rangle+|B\rangle\}]^*\hspace{2em}&\text{Axiom 2}\\ &= \langle C|A\rangle^*+\langle C|B\rangle^*&\text{???}\\ &=\ ... \end{align}
I see that it distributed the $\langle C|$ using Axiom 1, but what axiom justifies "distributing" the conjugate, $^*$?
It's clearly not one of the two inner product axioms. Earlier in the chapter we were given some additional definitions:
[The] bra corresponding to $|A\rangle+|B\rangle$ is $\langle A|+\langle B|$.
If $z$ is a complex number, then [...] the bra corresponding to $z|A\rangle$ is $\langle A|z^*$.
I'm wondering if the second definition about $z$ is coming into play. Can $z$ stand in for a bra, e.g. $\langle C|$? But even so, it doesn't explain the "distributing" of the conjugate. Do I maybe need to expand the bras and kets to row and column vectors and do some recombination to arrive at this "distribution"?
This is just a property of complex numbers, which you can prove easily.
If $z$ and $w$ are complex numbers then:
$$(z+w)^*=z^* +w^*$$
You can prove this by writing the complex numbers in their real and imaginary part.
Since the $\langle C|A\rangle$ and $\langle C|B\rangle$ are complex numbers the statement follows.