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How to perceive probability value in QM?

  1. Jul 10, 2010 #1
    Hi all,

    I have a confusion right now. To the truth is, I have been studying QM for some years, but somehow, some of its concepts are still not really clear for me.

    I understand that one of components of QM interpretation that generally accepted now is the Born postulate that [tex]\left|\Psi(x,t)\right|^2dx[/tex] is the probability to find the particle between x and dx.

    Then the uncertainty principle [tex]\Delta x \Delta p_x \geq \frac{1}{2} \hbar[/tex] and [tex]\Delta t \Delta E \geq \frac{1}{2} \hbar[/tex], tells the inability to precisely measure two non-commuting observables simultaneously.

    My confusion is here. For sure that we are able to calculate analytically the energy quantization of system such as H atom, with exact value.

    So my question is like this.

    If during the energy measurement of H atom, we get different spread results over time and also over position. Is this spread values are due to the uncertainty principle, or because the limitation of our tools in measurements (related to equipment accuracy, etc.), or because we are actually approaching the measurement that we treat the H atoms system as ensembles (so instead of pure QM, we actually doing measurement as explained by statistical mechanics)?? Where is the significance of the exact calculated H atom energy?

    Sometimes, I still thinking that if we can calculate exactly the energy of the system, then, in every measurements, we should get the very same identical result.. is this way of thinking is wrong?
  2. jcsd
  3. Jul 12, 2010 #2
    Hi leoneri,

    What really matters is what you mean by [tex]\Delta E and [tex]\Delta t. The best place I've read this was Griffith's book.

    It states that [tex]\Delta t[/tex] and [tex]\Delta E[/tex] are the standard deviations of the time required for the state of the body to change by [tex]\Delta E[/tex].

    In your case, if we know that the body has a given energy, for example the hydrogen atom, then [tex]\Delta E = 0[/tex] and as it is a stable state then it would take an infinite time to change, i.e. [tex]\Delta t = \infty[/tex].

    Take care that this has nothing to do with the measurement apparatus used or with the measurement causing any perturbation to the measurement results.

    This meaning is different than the meaning of [tex] \Delta x \cdot \Delta p \geq \hbar [/tex] which talks about the standard deviation measured for an ensemble of identical experiments.
  4. Jul 20, 2010 #3
    Hi Omar, thanks for your reply. Your information from the Griffith's book is really helpful, and your explanation at least confirmed what I have in my mind, that [tex]\Delta x \Delta p_x \geq \frac{1}{2} \hbar[/tex] is somehow related to ensemble concept.

    But I am still questioning about spread results. I will refine my question. I still do not really understand with this spread results that usually mentioned in QM books. Is every time we do measurement, we will always get spread results after doing the experiments several time, or not? Assuming the apparatus do not causing perturbation.
  5. Jul 20, 2010 #4
    Hi leoneri,

    Thanks to you :).

    For the spread mentioned in books; I get is as the following: If we have an ensemble of experiments and we measure the same parameter in all experiments and form a statistical average we will get a spread, i.e. each experiment will give a different result.

    For your own argument, if we neglect or consider there is no effect of time evolution or repeated measurement process on the result, then the results will not be identical and we will have a spread in them.
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