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Some basic misunderstanding in Quantum Mechanics

  1. Jun 3, 2012 #1
    Hi, guys,

    As a beginner, I have some problems to understand QM.
    Thanks a lot in advance for your response !

    1. According to QM interpretation, the measured particles, they become spike, do they have chances to back normal status (the status before becoming spikes)? If yes, how? if no, why?

    2. If I understand correctly, any kinds of particles, after being measured, become same spikes(described by Dirac function), or not? it sounds very artificial / weird if it is. Naively, how (god?) could make the particles become same spikes just after being measured? Or the same spikes postulation is just a temporary successful theory/model which has sustained experimental test for a few decades?

    3. According to QM, the expectation value <x> is the average of repeated measurements on an ensemble of identically prepared systems. Then when calculating momentum p = mv, what's the "m" ? It's the average mass of those "an ensemble of identically prepared systems" ? If yes, then how to know the number of systems were involved when measurement was processed? If not, it would be the sum of mass of all measured systems? So, if this interpretation is correct, then the momentum is a kind of "hybrid" quantity in the sense of the <v> = d<x>/dt come from average value while the "m" come from sum of all systems. Or, I missed everything?

    BTW, I'm mainly reading Griffiths QM, 2nd version, international edition.

    Thanks again !
  2. jcsd
  3. Jun 3, 2012 #2
    1. The "spikes" become non-"spikes". This is because the Schrödinger equation tends to widen it (if you start with something delta function-esque, it evolves into a Gaussian that keeps getting wider; at least in vacuum (i.e. V = 0)).

    2. It depends, in the most common interpretation the "spikening" is a postulate yes. And it is quite weird. In another interpretation (pilot-wave theory a.k.a. de broglie bohm interpretation) the spikening is deduced/it's a result, having to do with an effect called "decoherence".

    3. The m is just the ordinary mass. But if you want, it's equal to the average of the ensemble, since all identically prepared systems have the same mass, but I don't think looking at it like this is helpful and you likely have some sort of misconception about this notion of ensemble averaging, but it's hard to say what exactly.
  4. Jun 4, 2012 #3
    Dear mr. vodka,

    Thanks a lot for your response !

    Still, I would like to discuss a little bit more on those topics.

    1. As to the "spike" and "non-spike" transformation, it's a kind of "double-way" transformation, right? I'd like to make water wave as an example to understand it, make a comment on this if possible please.
    Imaging the water waves in sea, they usually have small fluctuation(tide) - this kind of "quiet" fluctuation(tide) due to the gravitation between earth and moon. Well, once a bomb was threw down from a plane to the quiet sea surface(corresponding to a measurement performed on a particle), a water "spike" produced. Later, the "spike" get wider and wider, finally back to original status - small fluctuation(tide).
    The point is if no external perturbation, particle always keep fluctuating like water wave, right? Then, why particle "select" this kind of way as their "life style"? Does the gravitation also the "power source" of particle's fluctuation? But in the scale of particle, gravitons might be "seen" only in very high energy scale(> 1000Tev ?).

    2. OK.

    3. Thanks a lot for your statement of "all identically prepared systems have the same mass".
    I doubted the m because in particle physics, there are many unbelievable things. For instance, the mass of proton is ~938 Mev, while its composed quarks u and d are 2.4 and 4.8 Mev separately. Which means the sum of three quarks u + u + d ~=10 Mev. The variation is too big(938 Mev vs 10Mev), right?
    So, I think I might missed some fundamental concepts on mass in QM. And the mass definition in quantum mechanics is not so clear(at least to me).

    Thanks again
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