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Using bra-ket notation, how do I prove that the inner product of a ket with itself (ie, $\langle A | A \rangle$) is a real number?

I understand that the rule of inner products states that: $\langle B | A \rangle = \overline{\langle A | B \rangle}$.

Therefore $|A\rangle$ is equal to it's own complex conjugate, and hence must have no imaginary part, but I'm having trouble writing the proof in a formal way.

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I don't understand your doubt. Since $\langle A|A\rangle=\overline{\langle A|A\rangle}$, $\langle A|A\rangle\in\mathbb R$ because, among the complex numbers, those which are real numbers are those which are equal to their own conjugate.

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  • $\begingroup$ Oh, it's just that simple? I think I was over complicating it in my mind. I thought there would be some long, rigorous proof. Thank you! $\endgroup$ – Matt Hill Sep 4 '17 at 18:59
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By definition, we know that: $$\langle A|B\rangle \in \mathbb{C} $$

and that

$$\overline{\langle A|B\rangle} = \langle B|A\rangle$$

$\forall A, B$.

Then, also $\langle A|A\rangle \in \mathbb{C} $, and $\langle A|A\rangle = \overline{\langle A|A\rangle}. $

The only numbers $x \in \mathbb{C}$ such that $x = \overline{x}$ are the real numbers. Hence, $\langle A|A\rangle$ is real.

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If you want/need more detail you could set $\langle A|A \rangle = x + iy$ and then use $\overline{\langle A|A \rangle} = \langle A|A \rangle$ to conclude that $-y = y$, which is true only for $y=0$.

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The argument form

$x = \bar{x}$ therefore $x$ is real

is a standard one; generally, nothing more needs to be said.

If you were in a setting where you felt that you need to include a proof of this rather than just using it, a simple way is

$$ \mathrm{im}(x) = \mathrm{im}(\bar{x}) = - \mathrm{im}(x) $$

and thus

$$ 2 \mathrm{im}(x) = 0 $$ $$ \mathrm{im}(x) = 0$$

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