I have two abstract orthogonal vectors $\mid a\rangle$ and $\mid b\rangle$: $\langle a\mid b\rangle=0$, but I don't know the lengths $\mid a\mid=\sqrt{\langle a\mid a\rangle}$ and $\mid b\mid=\sqrt{\langle b\mid b\rangle}$.
I would like to find two rotated vectors
$\mid a'\rangle=c_1\mid a\rangle+c_2\mid b\rangle$
$\mid b'\rangle=c_3\mid a\rangle+c_3\mid b\rangle$
that are still orthogonal, in terms of the coefficients $c_i$. Is this possible without knowing the lengths?
If they were of unit length I could use
$\mid a'\rangle=\cos\theta\mid a\rangle+\sin\theta\mid b\rangle$
$\mid b'\rangle=-\sin\theta\mid a\rangle+\cos\theta\mid b\rangle$
$\ \ \ \ \ \ \Longrightarrow \langle a'\mid b' \rangle=0$
And if I knew their lengths I could use
$\mid a'\rangle=\cos\theta\mid a\rangle+\frac{\mid a\mid}{\mid b\mid}\sin\theta\mid b\rangle$
$\mid b'\rangle=-\frac{\mid b\mid}{\mid a\mid}\sin\theta\mid a\rangle+\cos\theta\mid b\rangle$
$\ \ \ \ \ \ \Longrightarrow \langle a'\mid b' \rangle=0$
But what about my case?
Note: I don't actually need the general rotation, just some rotated still orthogonal set; ideally rotated $90^{\circ}$.
EDIT: I realised that I can set up the equations for $c_i$, and I get
$c_1^*c_3\mid a\mid^2+c_2^*c_4\mid b\mid^2=0$
which I can't solve; does that mean that what I need is impossible or could there still be some trick?
For an arbitrary angle $\theta$, we have $$ \langle a'\mid b'\rangle=\\ -\sin\theta\cos\theta\langle a\mid a\rangle+\cos\theta\sin\theta\langle b\mid b\rangle = \\ \cos \theta \sin \theta (\langle a\mid a\rangle - \langle b\mid b\rangle) =\\ \frac 12 \sin (2 \theta) (\langle a\mid a\rangle - \langle b\mid b\rangle) $$ So, in order to have $\langle a'\mid b'\rangle = 0$, we must either have $\langle a\mid a\rangle = \langle b\mid b\rangle$, or $\sin (2 \theta) = 0$.