# Continuous evolution from 1 eigenstate of O to another O-eigenstate?

1. Sep 12, 2013

### James MC

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)?

2. Sep 12, 2013

### wotanub

If a particle is in an energy eigenstate (aka stationary state) of the Hamiltonian, it won't evolve with time. Observables that commute with the Hamiltonian will not be altered by time evolution of a stationary state. But if it is in some superposition of energy eigenstates, time evolution does come into play allowing the particle to be in potentially different states at time passes. Time evolution is described by Schrodinger's equation.

3. Sep 12, 2013

### James MC

Are you trying to state a condition under which a particle can continuously evolve from an eigenstate of operator O to a different eigenstate of that same operator O? (Because my question is whether this is possible.) In particular, are you saying: if a particle is in a superposition of energy eigenstates then that particle is in an eigenstate of a commuting operator that continuously evolves to another eigenstate of that commuting operator?

If that's what your saying, an example would help. If that's not what you're saying, then I didn't understand your comment.

4. Sep 12, 2013

### wotanub

I'm saying that a particle can change to a different eigenstate of an operator under time evolution, but it will not always change depending on the initial state and the Hamiltonian that is generating the time evolution.

For example, a particle in a stable circular orbit due to a central potential. The energy of the particle is fixed in an energy eigenstate. The angular momentum operator commutes with this Hamiltonian, so the particle's energy eigenstate is in a simultaneous eigenstate with L^2, therefore L^2 will not change as time evolves.

However, operators like X and P do not commute with H, so the particle must be in a superposition of the eigenstates of these operators and the observables associated with them will change as time evolves.

Some may point out that it may be difficult or impossible to prepare a system with a definite energy eigenstate, but my example was only for illustration.

5. Sep 12, 2013

### James MC

Okay, then I would like to learn about examples that illustrate this, and about what the continuous evolution depends on if not initial-state+Schrodinger-evolution...

So the energy of the particle is fixed in one energy eigenstate while the angular momentum of the particle is simultaneously fixed in one angular momentum eigenstate ("will not change as time evolves"). If they're both fixed then this does not illustrate continuous evolution from one eigenstate to another eigenstate of the same observable.

Or am I missing something?

6. Sep 12, 2013

### Jano L.

Check Rabi oscillations for two-state system, with resonant frequency of external radiation. The expansion coefficient for the two states will oscillate back and forth from 0 to 1.

7. Sep 12, 2013

### James MC

Thanks Jano! I've started to research the example and I found a very helpful paper.

This paper states that it actually depends on how long the external radiation is applied to the two-level system. (See section 2.3 (p5).) But the kind of example I'm looking for is illustrated in equations (2.11) and (2.12) so that's really helpful.

I'm now thinking about a way to define in general how this is possible. On page 1. it states "If the system begins in an energy eigenstate, it only acquires a phase as it evolves." And when discussing the key example (2.12) it says "Since an overall phase does not affect probabilities, this (radiation affected) state is equivalent to |2> (the orthogonal state)."

So in terms of stating what in general gives rise to the possibility of continuous eigenstate-to-eigenstate evolution; is it just where the dynamics only has the effect of rotating the state vector in the complex plane in such a way that it gets the vector from one orthonormal eigenvector to the other while avoiding being a superposition of those eigenvectors?

8. Sep 12, 2013

### Staff: Mentor

The wave function evolves according to the time-dependent Schrodinger equation. That evolution doesn't turn the $x=0$ eigenstate of a particle moving at speed $v$ at time zero into the $x=vt$ eigenstate after a time $t$ has passed; it turns it into a somewhat dispersed wave packet with a high probability that a position measurement will yield the result $vt$. To get it into the actual $x=vt$ eigenstate you have to perform a position measurement.

(BTW, this behavior makes sense in a collapse interpretation, which explains both the early success of Copenhagen and the persistence of collapse interpretations in introductory classes today).

9. Sep 12, 2013

### wotanub

That's right, but think about operators like X and Y. If the particle is in an orbit around the z-axis, the x and y position of a particle changes as time evolves.

And as Nugatory has said, the particle is really never in a sharply defined eigenstate for things like position operators.