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Eigenvectors associated with distinct values of an observable are orthogonal, according to quantum mechanics.

Does this entail that a quantum system cannot continuously evolve from one eigenstate into another, for ANY observable?

At first, that seems strange: it seems like a particle should be able to "travel" in the sense of continuously moving from one eigenstate of position to the next.

Another example: can't a particle "speed up" (i.e. go continuously go from one velocity eigenstate to the next)?

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Neutrino oscillation. –  dmckee Sep 12 '13 at 17:21

1 Answer 1

Time evolution is dictated by the Hamiltonian. What is certain is that a state $|E\rangle$ with energy $E$ will after any time $t$ still be orthogonal to a state with energy $E'$. But for other operators $O$, this is not the case, as long as $[O,H] \neq 0.$ The latter condition is important: if $[O,H] = 0$ and at time $t=0$ a state $|\psi\rangle$ has eigenvalue $\lambda$ under $O$, then for all times $t$ it will have this eigenvalue. If $[O,H] \neq 0$, the state will walk through different sectors in $O$-space under time evolution.

A single, free particle (with Hamiltonian $H \propto P^2$) cannot speed up for the reason that $[P^2,H] = 0,$ or in other words, because of conservation of energy. If you have a multi-particle system, then you can definitely have non-vanishing matrix elements between states with different momenta (as long as the total momentum is conserved). This is the basis of scattering theory.

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