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.

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    $\begingroup$ Which is equivalent(?) to saying that states evolve with linearly. It's not equivalent to linearity. You can have linear operators that are not unitary, e.g., projection operators. $\endgroup$ – Ben Crowell Dec 5 '18 at 20:59
  • $\begingroup$ In nonhermitean but PT-symmetric QM, evolution is not unitary. $\endgroup$ – Cosmas Zachos Dec 5 '18 at 22:33
  • $\begingroup$ For what it's worth, if quantum evolution weren't unitary, you could control the future and/or destroy the universe. $\endgroup$ – DanielSank Dec 6 '18 at 0:55

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.

  • $\begingroup$ But we might find out one day that the double slit experiment is only approximately linear. For example when the double split experiment is done in the presence of massive gravitational fields?????. BTW your proof is that linear-->unitary but not unitary-->linear. $\endgroup$ – zooby Dec 5 '18 at 22:11
  • $\begingroup$ Unitarity includes linearity in the usual definition. It is entirely possible that we will some-day find that quantum mechanics is only approximately linear, and there are people that consider such theories. Penrose, for example, has advocated for adding non-linearity to QM to explain collapse. My only point is that if you take linearity as a physical principle then you get unitarity instead of just norm-preservation. $\endgroup$ – Luke Pritchett Dec 6 '18 at 13:55

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