# In bra-ket notation, what axiom allows "distributing" the conjugate ($^*$) across a sum of bras or sum of kets?

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

1. They are linear:

$$\langle C|\{|A\rangle+\langle B\rangle\}=\langle C|A\rangle+\langle C|B\rangle$$

1. 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:

1. [The] bra corresponding to $$|A\rangle+|B\rangle$$ is $$\langle A|+\langle B|$$.

2. 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"?

• It seems to me that the distibutivity of the complex conjugate is used... I assume it wasn't mentioned as an axiom because properties of complex numbers are known at this point already. Apr 23, 2023 at 19:04
• Yep that makes sense. I'm in Chapter 1 of the book and it painstakingly spelled out 7 axioms for bras and kets plus a review of basic linear algebra, so I expected to be able to solve the problem using only what was explicitly given as axioms and properties. It's weird that the text would explain how to add 2 column vectors but expect distributivity of the complex conjugate to be a given. Apr 23, 2023 at 19:09

If $$z$$ and $$w$$ are complex numbers then:
$$(z+w)^*=z^* +w^*$$
Since the $$\langle C|A\rangle$$ and $$\langle C|B\rangle$$ are complex numbers the statement follows.