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Is there a proof for this either way?

For the normalized kets $\left|a \right\rangle, \left|b\right \rangle, \left|c\right \rangle $

If $$ \left\langle a\middle| b \right\rangle = 0 \quad\text{and}\quad \left\langle a \middle|c \right\rangle = 0, $$ are $\left|b\right \rangle$ and $\left|c\right \rangle$ orthogonal to each other? That is, must $\left\langle b \middle|c\right \rangle = 0$?

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    $\begingroup$ No, e.g. $b=c$. $\endgroup$
    – lemon
    May 12, 2015 at 12:18
  • $\begingroup$ are a,b and c are supposed to be e.g. linearly independent? $\endgroup$
    – innisfree
    May 12, 2015 at 13:59
  • $\begingroup$ @lemon That is an answer, not a comment. $\endgroup$
    – rob ♦
    May 12, 2015 at 21:05

3 Answers 3

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No. Just let $|b\rangle = |c \rangle$.

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In general no. If you think of $\left|a\right>$, $\left|b\right>$ and $\left|c\right>$ as vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$, you can say that $\vec{a} = \vec{b} \times \vec{c}$, so that $\vec{a}$ is orthogonal to both $\vec{b}$ and $\vec{c}$, but this doesn't imply that $\vec{b} \cdot \vec{c} = 0$.

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(1) Answer is Yes in this example:

Consider complete set of orthonormal energy eigenfunctions of "particle in a box problem (infinite potential well)" in quantum mechanics.

Choose any 3 of them. If any one of these chosen 3 is orthogonal to remaining two, then these two are orthogonal to each other.

(2) Answer is No in the example of cross product of spatial vectors (as answered by MFH above).

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