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Quantum states transition

  1. Mar 19, 2015 #1
    Hello everyone.
    Can someone explains me the meaning of quantum state transition?
    For example consider an electron which is in the superposition of two energy eigenstates of a given hamiltonian, now, if no one perturbs the state with a measure, nothing happens and the superposition remains the same for all the time. Suppose you had been able to measure one of the states, this means that the electron decays in that state and it remains there if no perturbation is applied on the system and you have all the probability on that state, am I right? If so, I think this is described by the fact that the two states are orthogonal.
    Consider the case where the wave function is a dirac delta in 0 with the hamiltonian of the free particle, what happens theorically after you made the measure of the position? The state remains the same or it evolves spontaneously?
     
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  3. Mar 19, 2015 #2

    DrClaude

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    A position eigenstate of a free particle is not an eigenstate of the Hamiltonian of a free particle. Therefore, the state will evolve. You can see this with Gaussian wave packets that spread out with time.
     
  4. Mar 19, 2015 #3
    So I conlude that there is a transition probability between the position eigenstates, does it have any sense?
    Does the probability transition have anything to do with time?
     
  5. Mar 19, 2015 #4

    DrClaude

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    I've never heard "transition" used in that way, as it usually refers to discrete states.
     
  6. Mar 19, 2015 #5
    What is meant by transition probability between two sates? Physically speaking.
     
  7. Mar 19, 2015 #6

    DrClaude

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    The transition probability between states ##|i\rangle## and ##|f\rangle## is the probability per unit of time that a system initially in state ##|i\rangle## will be found in state ##|f\rangle##. It is proportional to ##\left| \langle f | \hat{W} i \rangle \right|^2##, where ##\hat{W}## is the perturbation responsible for the transition.
     
  8. Mar 19, 2015 #7

    atyy

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    Just to add to DrClaude's remarks - when a probability is used in quantum mechanics, it is ultimately always in the context of measurement. If no measurement is made, the system evolves deterministically according to the Shroedinger equation.
     
  9. Mar 20, 2015 #8
    Does the fact that there is a non zero transition probability between two states imply that the state of the system is a superposition of them?
     
    Last edited: Mar 20, 2015
  10. Mar 20, 2015 #9

    DrClaude

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    Yes. The original (initial and final) states that are considered are those of the unperturbed Hamiltonian. But the initial and final states are superpositions of eigenstates of the Hamiltonian including the perturbation. Often, the perturbation applies for a finite amount of time, hence it makes sense to talk about the problem in terms of unperturbed states.
     
  11. Mar 20, 2015 #10

    bhobba

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    Just to add a bit more detail to Dr Claude's excellent reply, this is what's called Fermi's Golden Rule:
    http://moleng.physics.upatras.gr/personnel/Koukaras/download/FermiGR.pdf [Broken]

    Of interest is the interpretive aspect of it and the quote from Ballentine.

    Don't agree with Ballentine adopts a view that's different to the orthodox one (presumably Copenhagen) but that would be a whole new thread and I don't want to derail this one.

    Thanks
    Bill
     
    Last edited by a moderator: May 7, 2017
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