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I Quantum Zeno effect to influence a particle's movement

  1. May 14, 2018 #1
    Hi,

    I've had an idea I would like feed-back.

    Let's imagine a particle on an x axis. On it's initial state it is localized in segment [0,1] as a purely symmetric wave.

    The quantum Zeno effects tells us that if we measure wether the particle is still in [0,1] n times during one second, the probability that we find the particle outside of [0,1] tends to 0 when n tends to infinity.

    Now suppose that during 1 second we measure n times wether the particle is in [-0.5,1] and then we do a final measurement to know wether it is in [-0.5,0.5].

    When n tends to infinity the probability that we find the particle outside of [-0.5,1] tends to 0, but during the second of intensive measurements the wave still spreads normally to [-0.5,0]. Therefore the probability that the particle is found in [-0.5,0] is strictly greater than 0, and the probability that it is found in [-0.5,0.5] is obviously strictly greater that 0.5.

    Therefore by this rather simple (and improvable) measurement scheme, we seem to be able to influence the direction the particles goes towards, and not only freeze it.

    What do you think? Is that right or do I make a logical mistake? Or is that an already know effect?

    Thank's in advance!
    Pierre.
     
  2. jcsd
  3. May 14, 2018 #2

    PeterDonis

    Staff: Mentor

    What does "purely symmetric wave" mean? What Hamiltonian are you using, and what state do you think the particle is in? You need to be able to write down mathematical expressions for these things in order to reason properly about this experiment.
     
  4. May 15, 2018 #3
    Thanks for the reply.

    Sorry I meant symmetric around 0.5. For example you can take the initial state in the form Ψ(x,0)=a*e^(i*b*(x-0.5)) on [0,1] .

    Ok I will try to go further with the math (it's been a long time I did no get deeply into quatum physics math), but I thought it was intuitive enough to ask the question without it.
     
    Last edited: May 15, 2018
  5. May 15, 2018 #4

    PeterDonis

    Staff: Mentor

    This state isn't localized to [0, 1]; it has support everywhere on the real line. You can't just arbitrarily say it vanishes outside [0, 1].
     
  6. May 15, 2018 #5
    @PeteDonis

    After measuring that the particle is in [0,1] the wave vanishes outside of [0,1] by
    Wave function collapse right? So you may consider that a first measurment occurs at time t=0 and provoke that vanishing outside of [0,1]

    To visualize what I'm saying in both this post and the first one you can have a look at that (see at 3'50):
     
    Last edited: May 15, 2018
  7. May 15, 2018 #6

    PeterDonis

    Staff: Mentor

    Assuming you make such a measurement, what is the particle's state following the measurement? You can use the collapse postulate to determine this state, that's fine: but you still need to figure out what that state looks like, mathematically. You can't just take a state that has support on the entire real line and truncate it outside [0, 1].
     
  8. May 15, 2018 #7
    The state after the first measurement (that determines the particle is in [0,1]), i.e this initial states, would be as I said Ψ(x,0)=1[0,1]*a*e(i*b*(x-0.5)) ∀x∈ℝ . (This time expressed with the indicator function (1[0,1]), a is chosen so that ∫|Ψ(x,0)|²dx=1). This first measurement determines the initial state, as I said "On it's initial state it is localized in segment [0,1]"

    I'm not sure to understand where you want to go. Maybe you think my wave is not physically realistic because has a constant amplitude, then fine we can let's add an attenuating factor (Gaussian)
    Ψ(x,0)=1[0,1]*a*e(i*b*(x-0.5))*e(-(x-0.5)^2) .

    But it does not change anything to my reasoning on first post, which should work with pretty much any wave symmetric around 0.5.
     
    Last edited: May 15, 2018
  9. May 15, 2018 #8

    PeterDonis

    Staff: Mentor

    No, it isn't. Or at least, you can't just assume it is. You have to show that it is. And you won't be able to. See below.

    No, I think it's not physically possible because it's not an eigenstate of whatever operator corresponds to measuring the position and finding it to be in the interval [0, 1]. It's an eigenstate of the momentum operator for a free particle over the entire real line.

    That doesn't help because you're still arbitrarily truncating the state at the endpoints [0, 1]. So it still won't be an eigenstate of any operator that measures position and finds it to be in the interval [0, 1].

    A hint: any state that is an eigenstate of such an operator will have to, at the very least, have ##\Psi = 0## at the endpoints of the interval.
     
  10. May 15, 2018 #9
    Hi,

    Thanks a lot for your reply and your time.

    I'm not sure to follow you, are saying that we should replace [0,1] by ]0,1[ everywhere?

    Do we agree, about the feasibility (at least theoretical) of making a measurement like in the video above; that is making a quantum measurement that measures whether the particle is in [0,1] (]0,1[?) or not, but does not determine where the particle is precisely in this interval?

    Do we agree that doing such a measurement, assuming that the particle is measured inside the interval, it zeros out the wave outside of [0,1] (]0,1[ ? ), whatever the wave is at this time?

    Pierre.
     
  11. May 15, 2018 #10

    PeterDonis

    Staff: Mentor

    No.

    As a reasonable idealization, sure, I have no problem with this.

    Not if by "zeros out" you mean "leaves the wave function exactly it was before the measurement inside the interval, but makes it zero outside the interval". The measurement will change the wave function inside the interval; it won't just erase it outside the interval.
     
  12. May 15, 2018 #11
    >As a reasonable idealization, sure, I have no problem with this.
    I should have precised that the effect I wanted to study in the first post, is a potential theoretical effect deriving from quantum physics, not necessarily something we can do in practice.

    >The measurement will change the wave function inside the interval;

    What do you mean by change? At the exact time of measurement t=0 the remaining wave inside of the interval is just multiplied by a constant in order to keep ∫|ψ(x,0)|²dx=1 right? (then of course it evolves later according to the Schrödinger equation). If not what change will happen according to you at time t=0?
     
  13. May 15, 2018 #12

    PeterDonis

    Staff: Mentor

    For any x in the interval [0, 1], the value of ##\Psi(x)## after the measurement will, in general, be different than the value of ##\Psi(x)## before the measurement.

    Wrong.

    The only way to figure that out is to do the math: find the operator that describes the process of measuring the position and finding the result to be in the interval [0, 1], then apply that operator to the state ##\Psi(x)## before the measurement to obtain the state ##\Psi(x)## after the measurement.
     
  14. May 15, 2018 #13
    Last edited: May 15, 2018
  15. May 15, 2018 #14

    PeterDonis

    Staff: Mentor

    If you're looking at snulty's answer, I'm not sure it's correct. Bobak Hashemi's answer raises some issues with it.
     
  16. May 16, 2018 #15
    To my mind everything works, snulty's just adding normalization (multiplying by a constant in order to keep ∫|ψ(x,0)|²dx=1).
    Bobak Hashemi's just presenting the discrete version of this operator.

    Anyways I think my assumption on first post is correct, I will keep researching and I may write a more concrete proof.
     
  17. May 16, 2018 #16

    PeterDonis

    Staff: Mentor

    No, he's pointing out that, if you are measuring position within an interval, you are making a discrete measurement, not a continuous measurement. A continuous measurement of position would be measuring position to infinite accuracy: exactly one point.

    The issue with snulty's answer is that he is assuming that the operator that realizes "measure position within a discrete interval" is of the same form as the continuous position operator, which in the coordinate representation is just "multiply by x". But the position operator snulty came up with is just "multiply by 1 inside the interval, and 0 outside the interval", which is not the same as "multiply by x", nor is it the discrete analogue of "multiply by x". That is what Bobak Hashemi is getting at when he says "##\chi_I## is a number, not an operator".

    The proper way to model this kind of measurement is to discretize position, i.e., to find an orthonormal basis of states ##\psi_I##, indexed by discrete, disjoint intervals ##I##. One of those intervals would be [0, 1] (possibly with one or both endpoints open, to ensure that no two intervals overlap at the endpoints). But those states will not, in general, be the same as the continuous wave function ##\Psi (x)## restricted to each interval. The only thing you can say for sure about them is that the amplitude for each discrete state will be equal to the integral of the continous wave function ##\Psi (x)## over the interval that discrete state covers.

    This is the sort of model Bobak Hashemi is describing. And with that kind of discrete model, you can do what you have been saying when a measurement occurs: if the measurement fixes the position within the interval [0, 1], then the output state is the discrete state ##\psi_{[0, 1]}##, renormalized to unit amplitude. The probability of getting this result will be equal to the integral of ##|\Psi(x)|^2## over the interval [0, 1]; but that is not the same as saying that the discrete state ##\psi_{[0, 1]}## is just the continuous wave function ##\Psi(x)## truncated to only be nonzero in the interval [0, 1].
     
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