I’m trying to figure out if $\left< a \middle| b \right> = \left< b \middle| a \right> $ when $\left| a \right>$ and $\left| b \right>$ are eigenfunctions with the same eigenvalue $\lambda$. I tried a few vectors and it seems to be correct, but is it general? If not, is there any extra conditions needed for this to be true?
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$\begingroup$ Eigenfunctions of what? $\endgroup$– Tobias FünkeCommented Apr 11, 2023 at 22:21
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5$\begingroup$ Take $|b\rangle=c |a\rangle$ with $c\in \mathbb C$ and $c\neq 0$. Thus $\langle a|b\rangle = c \langle a|a\rangle$, but $ \langle b|a\rangle = \bar c \langle a|a\rangle$. This even works for normalized vectors, i.e. you can take $|c|=1$. And it is obvious that $|b\rangle$ is an eigenfunction of some (linear) operator iff $|a\rangle$ is (with the same eigenvalue). Your claim is thus false. $\endgroup$– Tobias FünkeCommented Apr 11, 2023 at 22:24
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2 Answers
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By using the notation $\langle . \lvert . \rangle$, you are referring to a complex inner product.
Let $\mathcal{H}$ be a complex vector space. A complex inner product is a map $\langle . \lvert . \rangle: \mathcal{H}\times \mathcal{H} \rightarrow \mathbb{C}$ such that $\langle . \lvert . \rangle$ is bilinear (or just linear in one of the arguments), positive-definite, and conjugate symmetric. Conjugate symmetry gives us for any $a,b \in \mathcal{H}$: $$ \langle a \lvert b \rangle = \langle b \lvert a \rangle^*.$$ Hence the necessary condition is that $\langle a \lvert b \rangle \in \mathbb{R} \iff \langle b \lvert a \rangle \in \mathbb{R}$.
answered Apr 11, 2023 at 23:11
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$\left< a \middle| b \right> = \left< b \middle| a \right> $ if and only if either side (and therefore both) is real.
The property that $\left< a \middle| b \right> = \left< b \middle| a \right>^\ast $ (with the asterisk denoting complex conjugation) is completely general and one of the foundational axioms of the inner product. The rest follows from that.
answered Apr 11, 2023 at 22:43