Observation and Reality

Discussion in 'Physics & Math' started by Bowser, Dec 30, 2016.

  1. exchemist Valued Senior Member

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    If I were reading a new theory about that I think I would start by seeing how it deals with the uncertainty principle, because as I say that seems to me to be the place where the two complementary aspects of matter are most elegantly shown as facets of the same thing.
     
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  3. geordief Registered Senior Member

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    The uncertainty principle is where the an increase in the accuracy of position* is balanced by a decrease in the accuracy of a measurement of its momentum? (fair description?)

    Is there a mathematical relationship between that pair of accuracies?


    Is there a "conversion factor" between them?

    *of a particle of course.
     
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  5. quantum_wave Contemplating the "as yet" unknown Valued Senior Member

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    Your are right, and I will start there in regard to QM, but the wave-particle doesn't really exist in QM, and/or can be discounted as a popularization of QM. I think also, you have to take QM as it is described in its postulates, and be aware that the numerous interpretations each take a different slant on the implications of the postulates.

    Never-the-less, the speculative description of a wave-particle that I would start with is not consistent with the postulates of QM, but does have a decided quantum level description. It is not a QM compatible description.

    My ideas of a wave-particle are not based on mathematical probability functions. It features the concept that a particle can be modeled as having both wave and particle characteristics at the same time, and is consistent with the uncertainty you face knowing both measurements at the same time.

    If you are a little generous considering some of the logic I use, the particle can be said to have internal composition, but it diverges immediately from the standard particle model and QM, which have fundamental particles that have no internal composition. There is no getting around that it is speculation, and I acknowledge that these speculations would set physics on end, so I am not presenting it as something that the scientific community will have any reason to address, except if the topic is speculation of what a wave-particle might be.

    The speculation is intended to address the lack of a consensus on what a wave-particle may look like, and how the accompanying Quantum Action Process may work to establish the presence of particles, particle interactions, and a layman version of a quantum gravity solution.

    I'm not sure if it will be allowed to begin here, and if it is, I'm not certain that it won't find its way out the Fringe where I have been discussing it, essentially alone, for a long time.
     
    Last edited: Jan 13, 2017
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  7. exchemist Valued Senior Member

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    Yes indeed, it is one of the most famous relations in QM: Δp.Δx >= h/4π, the two deltas being the uncertainty in momentum and position, on some "x" axis, respectively and h being Planck's Constant.

    It arises because in QM momentum is proportional to the frequency of the wavefunction, while probability density (the likelihood of the particle being detected within a given volume of space) is the squared amplitude of the wavefunction. For a particle localised in a small area of space, you need a wave that only has amplitude in that small area. The only way this can occur with a wave is if the wavefunction is made up of a sum of lots of different components, all with different frequencies (i.e. different momenta) so that they interfere constructively only in a small region.
    Graphic of this here: https://en.wikipedia.org/wiki/Uncer...e:Sequential_superposition_of_plane_waves.gif

    I understand that radio engineers are very familiar with this sort of thing as you get a related phenomenon in radio signals - but don't ask me the details.
     
  8. geordief Registered Senior Member

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    So Planck's constant is a conversion factor? What factors "go into" its make up?I can see that it defines /is defined by position and momentum,so can I say that mass is one factor ,time is another and position another?*

    Are position and time treated as separate "entities" or are they unified as in Relativity?

    EDIT: Is it possible to regard h in QM as the counterpart of c in GR?

    * did I list them all?
     
    Last edited: Jan 13, 2017
  9. exchemist Valued Senior Member

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    6,296
    h is one of the universal constants and relates energy to frequency, as in the famous relation E=hν, which determines the energy of a photon of a given frequency. As far as I know it is not made up from any combination of other constants but is just what we observe to be the proportionality factor in such expressions. So it is fundamental.

    h has the dimensions of "action" viz energy x time, so erg-seconds, or kgm²/sec or equivalent.

    Historically it gained traction as a result of Max Planck's success in using it to overcome the Ultraviolet Catastrophe: https://en.wikipedia.org/wiki/Ultraviolet_catastrophe , which was one of several breakdowns of classical physics at the end of the c.19th that ushered in the quantum era. Einstein (yes him again) was also heavily involved, as it was he that first proposed the quantisation of light, using Planck's constant.

    QM does not rely on a concept of spacetime, as GR does.
     
  10. geordief Registered Senior Member

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    537
    Thanks,that is new territory for me....
     
  11. exchemist Valued Senior Member

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    This whole area was what I found most exciting in the 6th form and at university. The revolution in understanding between about 1880 and 1930 was amazing. Of course it still has not entirely settled down, due to the difficulty we still have with the apparent implications of some QM concepts. But it works, and I still get a kick out of the apparently disparate phenomena that QM was able to explain. Things like atomic spectra, the temperature-dependence of the heat capacity of gases, refraction of light, the Periodic Table, chemical bonding.......
     
  12. quantum_wave Contemplating the "as yet" unknown Valued Senior Member

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    6,583
    You guys have covered it pretty well. The up ending that my wave-particle composition and quantum action mechanics would require seems hardly worth stiring up concerns in P&M. It is really just speculation that should remain a Fringe topic.

    I have a thread elsewhere asking, "Does a single particle display both its wave nature and its particle nature?" There has been some interesting discussion, though much less content. I just wanted to pass on this link, which I found interesting in support of the idea that the photon can show both a wave and a particle at the same time:
    https://en.m.wikipedia.org/wiki/Afshar_experiment
     
    Last edited: Jan 13, 2017
  13. exchemist Valued Senior Member

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    6,296
    Anyway, revenons à nos moutons, as they say in France.........

    I thought this summary of Quantum Mysticism might be a good short read for those readers still with the original topic of the thread: https://en.wikipedia.org/wiki/Quantum_mysticism

    According to this, some of the originators of QM, such as Schroedinger, came to wonder if consciousness of the observer indeed played a role somehow, but this idea then fell from favour is now not generally speaking embraced by physicists.

    It has however enjoyed a new lease of life in various forms of New Age pseudoscience.

    Please Register or Log in to view the hidden image!

     
  14. QuarkHead Remedial Math Student Valued Senior Member

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    Hmm....

    You are stating the Heisenberg Uncertainty principle by assuming uncertainty in position and momentum. Sort of circular.

    Moreover, using operator calculus, I get a slightly different result.

    Suppose a position operator \(\hat{x} \) and a momentum operator \(\hat{p}= i\hbar\frac{d}{dx}\). (I have justified this latter equality [here=http://www.sciforums.com/threads/do-the-particles-ever-collide-in-qed.144077/page-12#post-3272383]).
    Then if \(\hat{x}\) and \(\hat{p}\) commute - that is if makes no difference whether we know the position first or the momentum first, we say these operators commute. That would be written as \( [\hat{p}, \hat{x}] = \hat{p} \hat{x}- \hat{x}\hat{p}=0\).

    The Uncertainty Principle states this is not the case. Rather that \([\hat{p}, \hat{x}]=[i\hbar\frac{d}{dx},\hat{x}]=i \hbar\) where \(\hbar = \frac{h}{2 \pi}\).

    Which differs from your assertion by a factor of 2. Should I care about this?
     
    Last edited: Jan 14, 2017
  15. Q-reeus Valued Senior Member

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    2,497
    For case of ground state of a particle-in-a-box, with infinite potential at walls, wavefunction is a pure half-sinusoidal function of one frequency, for a given box width L:
    http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html
    Scarcely different for case of finite well potential: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pfbox.html#c1
    For 3D rectangular box case, there are 3 such orthogonal ground state solutions in general. No more.
    I'm puzzled as to how you arrived at that highlighted in red, given above.
    In #86, you write: "h has the dimensions of "action" viz energy x time, so erg-seconds, or kgm²/sec or equivalent."
    The last expression naturally interprets as energy x velocity, not energy x time. A typo presumably.
     
  16. exchemist Valued Senior Member

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    Good point, I suppose I should have made more clear what I meant by "small" in that context. The wavefunction for a QM object confined by a potential certainly does not have significant amplitude throughout space, though the same principle applies if one tries to measure its position or momentum exactly, within its envelope of non-zero amplitude, e.g. an atomic orbital or something.
     
  17. exchemist Valued Senior Member

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    6,296
    OK, I wasn't trying to derive it, I was merely answering the question by providing the expression for it. (Geordief is not a physicist, just an interested and intelligent layman.)

    As for the factor of 2, hbar/2 is what I have always understood and is what appears in my textbooks and on the Wiki entry: https://en.wikipedia.org/wiki/Uncertainty_principle

    I agree with your statement of the commutation relation [p,x]=i hbar, but what seems to fall out of that (I am rusty on the maths) is the expression I gave. The deltas are rms deviations from mean value of the observables concerned and, from what I recall, getting those out of the commutation relation is a bit of work.
     
  18. QuarkHead Remedial Math Student Valued Senior Member

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    Sure, fair enough

    If by rms you mean root mean squared, then you are using the "deltas" as standard deviations from the expectation value which, roughly speaking is the mean of a large number of observations - eigenvalues for some operator. (I think - my statistics is rubbish!). Whereas I was treating the deltas as errors, or variance around the mean.

    I suppose that within certain confidence limits - say 95% - that these differ by a factor of 2?

    As I say, my statistics is crap, so I may be talking the same.
     
  19. exchemist Valued Senior Member

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    6,296
    Yes that's what I meant by rms. This seems to be how the uncertainty relation is usually expressed.

    My stats is a lot more rusty than my chemistry, but I do recall that Variance is the mean of the squares of the differences from the mean, so the Std Dev is just the square root of the variance, i.e. the rms value of the deviations from the mean. I am not sure exactly what quantity you were calculating, but obviously it has to be some way of quantifying the degree of spread of values around the mean or expectation value.
     
  20. QuarkHead Remedial Math Student Valued Senior Member

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    1,502
    Been thinking about this a bit (but not deeply).

    I suggest that for \(n\) data sets, the mean of means is equal to \(\frac{1}{n}\) the mean over all elements in these sets. (Proof by brute force!).

    Suppose we assume that the expectation value for the operator acting on an Hilbert space of state vectors is in some sense a "mean". Suppose we further assume that the above applies.

    Then, writing \(\hat{x}\) for the position operator and \(\hat{p}\) for the momentum operator, and \(\langle\hat{x}\rangle\) (likewise for the momentum operator) expectation value, and with a flourishing hand-wave I write

    \( \frac{1}{2}\langle [\hat{x},\hat{p}]\rangle\). (Since there are 2 "data sets")

    So if \([\hat {x},\hat{p}] = i \hbar\), then \( \langle\hat{x}\rangle \langle \hat{p}\rangle = \frac{1}{2}i \hbar\).

    Rubbish?
     
  21. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    Dan...I now know what is length, it was written the iron rod!i
     
  22. exchemist Valued Senior Member

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    I think it might be.

    I'm sure there are derivations available on the net - in fact there is one here, in which the factor 2 seems to appear as a consequence of an operation involving self-adjoint matrices: https://courses.physics.illinois.edu/phys580/fa2013/uncertainty.pdf

    I confess I can no longer follow this, as I have not done any matrix mechanics since university in 1975! It would take longer for me to get it back than I have patience for, so I'm afraid I can't help further.
     
  23. QuarkHead Remedial Math Student Valued Senior Member

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    1,502
    I agree, on reflection

    OK, fair enough. Thanks for all the fish (and the link)
     

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