Mean lifetime of a particle (quantum)

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K.J.Healey
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(*NOT HOMEWORK :) *)

If I have a particle that oscillates between states A and B, and I calculate the probability of transition (or of rather being in state B after an IC of being in A @ t=0)
and that probability goes like:
Pb = C1 Sin^2 (C2 t)

And I want to find the average/mean lifetime of a particle that starts in A what do I do?

It seems like I could consider only C1, since on average the Sin^2 term goes to (1/2)
So <Pb> == C1 (1/2)
And therefor <Pa> = 1-C1 *(1/2)

but I feel like I'm making an error in not considering the frequency of oscillation. If this were a particle-antiparticle system oscillating between the two, wouldn't the oscillation frequency matter?
Or do you say that since in a part-antipart system, the period of oscillation must be very long (as in a neutron osc), therefor in any given realistic time interval we can say that the Sin term goes like:
Sin[C2 t]^2 -->> C2^2 t^2
the ol' sin[x]->x for small x series expansion.

So the approximate probability
Pb == C1 C2^2 t^2 for t<<periodicity
then Pa = 1-Pb

But now what would the mean lifetime of a particle in A be thanks to one of these probabilities?

Or am I doing this all wrong?

Thanks
-KJH
 
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This problem is very strangely worded. If it is true that the system OSCILLATES between the two states, then "lifetime" doesn't make sense (nothing is decaying). But the TRANSITION time from a to b is just 1/C2 in your notation, and C1 == 1! This follows from either the uncertainty principle, or Schroedinger's equation.

Perhaps I'm misunderstanding?
 
Yeah I figured it out. It was much more complicated than I had assumed but I ended up with what I wanted by bypassing this train of thought all together.

I was wondering the transition time, BUT they don't oscillate between 0 and 1. The Hamiltonian has off-diagonal elements which makes the oscillation suppressed for the b state. And I was trying to relate this suppression of oscillation to a change in the probability of finding a particle in the B state.

Thanks for the attempt, but I got it.