Misleading Phrase: Collapse of wave function

Discussion in 'Physics & Math' started by Dinosaur, Mar 30, 2016.

  1. Dinosaur Rational Skeptic Valued Senior Member

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    The wave function provides probabilities for various possible events.

    The collapse phrase seems like weird semantics.

    You can construct a list of probabilities for dice throws. When a pair of dice is thrown & the result is 7, does anyone state the following?
    The collapse phrase seems semantically weird, suggesting activity stranger than that implied by
    The collapse phrase implies that the wave function is some physical process rather than being analogous to a list of probabilities.
     
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  3. QuarkHead Remedial Math Student Valued Senior Member

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    Actually it is the square of the wave function that provides these probabilities

    Yes, but get this.......

    The so-called "wave function" is a vector - QM-ists call it a state vector. And as such it is entitled to be acted upon by an operator (like any vector). This operator, in QM, is taken to be one of the (maybe many) measurable properties of the state of a QM system.

    The actual set of all possible measurements that can be made on the system are given by the spectrum of possible eigenvalues for the operator(s) that act on the state vector.

    So yes, I agree - the semantics is rubbish and, except in the hands of experts the phrase obscures woolly thinking at best, and lack of comprehension at second best
     
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  5. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I liked all but the last of that. It is about as simple as it can correctly be said.
    Fact that many don't know about eigenfunctions and values is their defect, not QM's.

    I think it was Einstein that said:
    "Every thing should be said as simply as possible, but not more so." Ergo: it is not "rubbish."
     
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  7. arfa brane call me arf Valued Senior Member

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    I think it's more to do with complementarity. When a wavefunction collapses it looks like a particle.
    Or when you measure which slit a particle went through, the wavefunction 'collapses' and no wavelike interference pattern emerges.
     
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  8. danshawen Valued Senior Member

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    Indeed it should. Thanks, area brane. That actually makes a lot of sense now.
     
  9. Farsight

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    Like Quarkhead said, it's the square of the wave function. But take a look at this by Jeff Lundeen:

    "With weak measurements, it’s possible to learn something about the wavefunction without completely destroying it. As the measurement becomes very weak, you learn very little about the wavefunction, but leave it largely unchanged. This is the technique that we’ve used in our experiment. We have developed a methodology for measuring the wavefunction directly, by repeating many weak measurements on a group of systems that have been prepared with identical wavefunctions. By repeating the measurements, the knowledge of the wavefunction accumulates to the point where high precision can be restored. So what does this mean? We hope that the scientific community can now improve upon the Copenhagen Interpretation, and redefine the wavefunction so that it is no longer just a mathematical tool, but rather something that can be directly measured in the laboratory."

    Also see physicsworld. He's saying wavefunction is something real that can be measured in a laboratory.

    Maybe it does, but if wavefunction is real, and if you detect a photon with an electron, and both are comprised of real wavefunction, then the interaction will involve one wavefunction interacting with the other, which makes the square of the wave function sound more plausible. And as for the collapse, you've seen something similar in the Optical Fourier transform. See Steven Lehar's web page.

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    No. You threw a dice, it rolled, it stopped, the 7 showed on top.

    I don't think it needs to be weird. In fact I'd say what is weird is the Copenhagen Interpretation along with "it surpasseth all human understanding".

    Yes. IMHO the simplest way of looking at this is that the photon is an extended entity that takes many-paths, like a seismic wave does. So it goes through both slits in the dual-slit experiment. Then when you detect it at the screen something akin to an Optical Fourier transform occurs, so you see a dot on the screen. It's similar if you detect it at one of the slits - it gets converted into something pointlike so it goes through that slit only, and you lose the interference pattern.
     
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  10. danshawen Valued Senior Member

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    Or another way of looking at (slit experiments with waves) is that the pattern that eventually finds its way to the screen consists of both photons and a lack thereof. Whenever photons cancel each other out, they never do so completely unless they happen to be virtual photons created and completely destroyed in the quantum foam. Those virtual wave functions collapse all the time, even around the moons of Jupiter, and no one takes any particular notice.

    I doubt that the contribution would make much of a difference (and it's about 20 light minutes away), but it is possible in principle to view a double slit experiment performed on Earth from one of the moons of Jupiter, unless the moon eclipses the faint image on the screen. And even in that case, femtosecond cameras have demonstrated that it is possible to see around corners with waves. It makes perfect sense. The image can still be reconstructed from light reflecting off of the sunny side of Jupiter.

    In a universe of energy transfer events, such waves interfere with each other continuously and indefinitely, whether there is math or computer simulations fast enough to keep pace with predicting all of the outcomes, waveform collapses or not.

    The thing I like best about your posts Farsight is that you continue to look for understanding of physical aspects of a situation where someoone like rpenner (or even rfeynman) have given up on that approach and prefer instead to compute line integrals around the moons of Jupiter in order to renormalize or collapse the wave function.

    Feynman never adequately explained his remarks. Until now, no one else had either.
     
    Last edited: Mar 31, 2016
  11. PhysBang Valued Senior Member

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    The problem with Farsight's posts is that he ignores the mathematics that accurately describes how the physical systems actually work. He also ignores the statements of the scientists he quotes when their direct statements contradict the cherry-picked quotations that Farsight likes. This kind of behavior is important to note.
     
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  12. QuarkHead Remedial Math Student Valued Senior Member

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    But surely, the so-called quantum weirdness comes from the fact that individual light quanta self-interfere - i.e. may, under some circumstances, "cancel" themselves out?

    PS by edit: My understanding of Einstein's purely theoretical 1905 paper on the light quantum is that whatever property you can assign to what he called a "light ray" applies equally to the individual quanta of light.

    Plus I believe the actual experiment has been performed in a Young-like set up - it makes no difference to the result whether you fire a beam of light through an appropriately positioned double slit, or whether you fire a single photon.
     
    Last edited: Mar 31, 2016
  13. arfa brane call me arf Valued Senior Member

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    It's about superposition; photons don't disappear if there is a cancellation of "probability waves" at some point, the photon will still appear somewhere else.
    Not that I'm implying you might think that, but some here probably do have that misconception.
     
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  14. Farsight

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    Virtual photons are "field quanta" Dan. It's like they're abstract chunks of field. When the electron and proton attract one another the resultant hydrogen atom doesn't have much in the way of an electromagnetic field. So you can say the electron and photon have exchanged field. But they haven't been throwing photons at one another. There are no short-lived real or virtual photons popping in and out of existence in this scenario, or in any other scenario. Virtual photons are not the same thing as vacuum fluctuations.

    That they do. And whilst waves are extended-entity things, they can look pointlike.

    Thanks Dan. I will never accept the surpasseth all human understanding baggage that has grown up around quantum physics.
     
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  15. danshawen Valued Senior Member

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    on "field quanta", I found this:

    http://physics.stackexchange.com/questions/3038/what-are-field-quanta

    Which makes sense mathematically iff you buy a point in spacetime possessing inertia all by itself.

    I don't subscribe to that interpretation of QFT, but I am permanently banned from talking about that here, evidently. I will strictly observe this restriction.

    But, thanks, Farsight.
     
  16. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Well we (some proficient with the math, anyway) do know how to descrirbe QM very well. If I recall correctly they can calculate the value of the fine structure constant to 16 significant figures (probably as many as you like, but measurements can only go that precise).

    As far as QM very being consistent with human experience, no that ain't gona happen. I know as I have made single photos in the two paths of a Mach Zehner interferometer be more than a meter from itself, when measuring their length as 35 cm.
     
  17. Farsight

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    Noted Billy.

    I wasn't keen on the answers there. Have a look at this one, where retired particle physicist anna v said virtual particles only exist in the mathematics of the model. They aren't short-lived real particles that pop in and out of existence. That's a pop-science myth.

    My pleasure Dan.
     
  18. PhysBang Valued Senior Member

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    Farsight does not accept that mathematics can be used to describe physics. He only accepts vague, textual descriptions of physical systems that seem to make some sort of sense to him.
     
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  19. arfa brane call me arf Valued Senior Member

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    Quantum mechanics doesn't "surpass all understanding"; experiments are available to ordinary people who don't work or study at universities and research centres like CERN.

    Understanding how to use the mathematical side of it isn't that hard either. What is difficult is reconciling what appears to happen in quantum experiments with our ordinary sense of reality; obviously quantum logic is extraordinary and requires (or appears to) some revision of our native understanding of space and time--we have somehow evolved the "wrong" ideas about them.

    And another comment about mathematics of the physical variety: I personally have found it quite easy to finish a 2nd year university course on solid state physics and get my head around the Hilbert spaces of probability amplitudes, but I have a set of notes (not mine) about mathematical physics, and it's a lot harder to understand.

    Then again, I have no idea why the Hilbert space has two complex dimensions (for each particle), and I haven't found anything definitive.
     
    Last edited: Apr 3, 2016
  20. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    For you perhaps the math is not hard, but 99.99% of the population can not even read with understanding a tensor equation, in part because where the sub and super scripts are and whether or not they are repeated, (and perhaps other conventions of that extremely compact notation, I have forgotten) completely control the meaning of the equation.

    And secondly, it is not the modern POV about space/time that is so tough to get one's head around, it is little things humans believe like:

    Any one discrete object like an electron or photon can not be far from its self in a multi-path interferometer. Etc. (but they can be far from themselves!)

    Or

    An observation in Boston can not instantaneously determine the results of a related observation in NYC. Most believe that invalidates the speed of light limits.
     
  21. arfa brane call me arf Valued Senior Member

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    If you can understand a 2-bit register as a tensor of two 1-bit registers, tensors aren't really all that puzzling. This is how an online pdf about quantum computing kicks off, eventually replacing registers with 0 or 1 in them with qubit registers.

    So how many dimensions does a 2-bit, or n-bit register have?
     
  22. arfa brane call me arf Valued Senior Member

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    Ok, can we do that with a register of decimal digits? Yes, we can!

    That's what any decimal number is. A 1-digit register of decimal numbers can hold ten distinct values, what we call the "place for units", then a 2-digit register has a place for tens, and so on. So, when you "tensor" two digits in base 10, you get ten times as many possible values. In general you get k times as many values for digits in base k.

    And you write these "tensor products" as simply concatenated elements. However, you can be really pedantic and say that you have column vectors instead of digits, and you use Dirac notation, so if you have two digits like \( \{d_1,d_2\} \). their tensor product is \( |d_1\rangle \otimes |d_2\rangle \), but we can (and we do) simplify all that to just \( d_1d_2 \). The indices tell us which digit is "the first", so we don't get something like 58 is eighty five.

    In QM, the column vectors are 2-vectors. like in binary numbers, but each component is a complex number.
     
    Last edited: Apr 4, 2016
  23. Farsight

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    This is "point-particle thinking", and IMHO it's the source of a lot of problems. If you remember that we're dealing with quantum field theory rather than quantum point-particle theory, you can soon start thinking of a photon as something more like a seismic wave than a billiard ball. It doesn't have any surface, it takes many-paths. When you then put it through pair production to create an electron (and a positron) the electron still isn't some little billiard-ball thing. It isn't some little round thing that has a field. It is field, this field has a spherical symmetry, and it doesn't have a surface.
     
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