Why in quantum mechanics must orthogonal states stay orthogonal?

Question

Given two states $|A(t)\rangle$ and $|B(t)\rangle$. If $\langle A(0)|B(0)\rangle=0$ then for all $t$, $\langle A(t)|B(t)\rangle=0$.

This is a fundamental rule of quantum mechanics. And we can imply that states evolve unitary with $|A(t)\rangle = U(t)|A(0)\rangle$.

Which is equivalent(?) to saying that states evolve with linearly.

One can think of this as two arrows on a circle. And they evolve by going round the circle, keeping at right angles from each other.

But one could imagine an evolution where the speed of the arrow on a circle depends on it's position on the circle. Then the arrows would not stay orthogonal.

It would be replacing the unitary group with the holomorphic-diffeomophism group.

States would evolve with operators $\psi'(t)=iH(\psi(t))\psi(t)$. i.e. non-linearly. But would always remain distinguishable.

Would this be against some physical principle?

Edit: Although the arrows would move around the circle at different speeds they would stay on the circle and hence the evolution is still unitary and hence states would always stay the same length. You would simply have a unitary operator dependent on the state e.g. $U(\psi(t))$. e.g. $\langle A(t)|A(t)\rangle$ would always stay the same value. But $\langle A(t)|B(t)\rangle$ would not.

Answer 1

I would say that the physical principle is linearity. The whole basis of quantum mechanics is the notion that things like time evolution are linear. It is the fact that time evolution is linear that allows double-slit interference, for example. Once you combine linearity with norm-preservation, you get unitarity as a mathematical consequence.

A linear operator $U$ that preserves the norm of all states must be unitary. Proof: $U$ must preserve the norm of the state $|A\rangle + e^{i\phi}|B\rangle$ for all $A,B,\phi$. Since $U$ also preserves the norms of $|A\rangle$ and $|B\rangle$ we have $$e^{i\phi} (\langle UA|UB\rangle-\langle A|B\rangle)+e^{-i\phi} (\langle UB|UA\rangle-\langle B|A\rangle) = 0$$ Since $e^{i\phi}$ and $e^{-i\phi}$ are linearly independent functions of $\phi$ this implies $\langle U A| UB\rangle = \langle A | B\rangle$ for all $A,B$. This is the definition of a unitary operator.