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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"?

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  • $\begingroup$ 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. $\endgroup$
    – HappyDay
    Apr 23, 2023 at 19:04
  • $\begingroup$ 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. $\endgroup$ Apr 23, 2023 at 19:09

1 Answer 1

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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.

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    $\begingroup$ Ah perfect, thank you. I didn't keep in mind the bit that a complete bra-ket represents a complex number. $\endgroup$ Apr 23, 2023 at 19:04
  • $\begingroup$ @AndrewCheong precisely it represents the inner product in some Hilbert space, which is necessarily a complex number. Glad I could help $\endgroup$
    – Chris
    Apr 23, 2023 at 19:07

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