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Continuous and discreet parts of an electromagnetic wave

  1. Oct 11, 2005 #1
    I'm just a hobbyist in things quantum and in the course of my reading, I have found it a bit confusing figuring out which parts of quantum theory deal with finite numbers of discreet values and which parts require continuums.

    For example: Last night I was reading up on qbits and in the course of the definition the author mentioned the phase, polarization and frequency of wave functions. Which of these three wave properties are continuous and which are discreet?

    Or put another way, are phase and polarization restricted to a finite number of quantized values? Or can a wave adopt any possible phase between 0 and 90 degrees? Even ones that requlre irrational numbers?

    Similarly, can the polarization of a electromagnetic wave be any possible angle or can they only be polarized at angles whose sides are the integer solutions of a^2 + b^2 = c^2?

    Lastly, if an electron in an atom falls from a higher energy state to a lower one, emitting a wave packet of red light, is the shape of that wave sinusoidal? Or more of a sawtooth or square wave due to quantization?

    Any explanations greatly appreciated. (and please be kind as I have no training in quantum physics. I have a decent handle on trig and algebra but no experience with calculus)

    Thanks

    Ken
     
  2. jcsd
  3. Oct 12, 2005 #2

    vanesch

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    Quantum mechanics is a very confusing subject indeed, and what it doesn't share with other matters of study is, the more you study it, the more confusing it gets :smile:.
    I think that a lot of your questions arise from something which is indeed puzzling in quantum theory: there is a difference between what one can measure (and obtain as a result of measurement), and what "is". It can lead very far to discuss "what is" but let us for the moment take it that the wave function describes the state (what "is") of the system. Well, that wavefunction can be in quite some continuous ways be dependent on several parameters, such as phase, polarization, frequency..., so in this respect, these quantities are continuous parameters describing the wavefunction (or the possible wavefunctions).
    However, when you go and MEASURE a property, sometimes the outcomes have to be discrete. That doesn't mean that the wavefunction had that discrete property, but it is just so that you can only obtain a few discrete values. The most obvious one is polarization.
    If you measure polarization IN A CERTAIN DIRECTION (it is fundamentally impossible to measure it in several directions at the same time), then the answers you can obtain are discrete. For instance, if it is a light particle, you can only get "up" or "down". But that doesn't mean that the incoming wavefunction was "up" or "down". In fact, the incoming wavefunction could be continuously "distributed" amongst "up" and "down", and this mixture just gave you the PROBABILITIES of getting "up" and "down".
    It's one of those weird properties of quantum mechanics, which is different from classical theory. There are many more of these weird properties, it is just a beginning !
    Short answer: the wave function can take on any polarization angle, with just any real number. But the result of measurement cannot !
    There are different ways to answer the question. An approximate answer (which is fundamentally not 100% correct) is that you'll get a "wave packet" out, which is a "sine wave in a bumpy envellope".
    I tried as much to give intuitive pictures, but this has its limits of course. QM can only really be understood using a good dose of sophisticated math. Given QM's intuitive weirdness, this clinging to the math is even more important than in other fields of physics where intuition can go much further.
    cheers,
    Patrick.
     
  4. Oct 12, 2005 #3
    Would the time a measurement was taken have some effect on the measured/quantized value?
    With these constraints, how does one determine circular polarization? It would seem to be impossible to determine this from one measurement.
    Is the situation similar to the projection of a 3D object on a 2D surface? For example: If you shine a light on a stretched slinky and rotate the slinky along the long axis, the shadow on the wall will trace a wave pattern but the movement of each point will be either up or down even though the point on the slinky is moving in 3D space.
    Is this because the act of choosing a sampling angle, in essence, preselects what discreet polarization angles you will measure?
    This brings me to my real question. I just didn;t know how to phrase it before.
    Even though the only values we can measure will be quantized, it is assumed that what is going on "under the hood" is actually continuous. Correct?
    If I got that part right, are there any limits either found or intuited as to the precision of the "floating point" values used by the under the hood continuous properties? Or is the resolution assumed to be infinite?
    If the sine waves in the "lumpy envelope" can truly have any mathematically definable phase, wavelength and polarization, wouldn't that indicate that every point in the universe is infinitely complex? Since it would require "floating point" numbers of infinite length and precision to define the envelope states exactly?
    If this is true, it would imply that not just every physical point, but every mathematical point in the universe can contain an infinite amount of information.
    I most likely took a wrong turn somewhere in there but if not, you're right. Quantum stuff is bizarre!
    Time for a beer and mindless tv show : )
    Thanks for replying
    Ken
     
  5. Oct 12, 2005 #4
    Followup

    There's something strange here...

    Since the act of measuring requires an instrument to perform the measurement, and since the instrument itself is comprised of "particles" that would imply that the discreet "up" or "down" values are imposed by the particles within the instrument. The particles themselves are blind to the underlying continuum.

    If quantum entities themselves can only detect "up" or "down", is there any compelling reason to assume what lies "under the hood" is in fact a continuum?
     
  6. Oct 13, 2005 #5

    vanesch

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    Hehe, I like your reasoning :biggrin:
    There is a serious "problem" in quantum theory, which is bugging people since the very beginning in the 1920ies. It shows up under very different faces, like "Schroedinger's cat", "EPR", "two-slit experiment", "quantum erasure"....
    Much ink has flown over the issue, and in my opinion, the issue still stands. But it is going to be extremely hard to explain this without being able to refer to the mathematical machinery ; I'll try but I'll loose some accuracy along the way.
    But let me first say this: the problem is a "foundational" one, not a "practical" one. In practice, things work out fine. But we sneak in something odd in order to do so, which makes the calculations work, but is not "right" on a fundamental level.
    If you do, in practice, a quantum calculation, it goes in two steps. First, you do the calculation with what's called "wave functions" and "wave equations" (like Schroedinger's equation), to go from the initial state "prepared by the experiment" to the final state "observed by the measurement apparatus".
    This part is called "the unitary part of quantum theory" in math slang. You could call it "continuous" if you want to.
    And then we take the second step: we declare part of the setup to be "measurement apparatus", do as if it didn't consist of just more atoms and stuff we are studying, and we apply a TOTALLY DIFFERENT procedure: we calculate probabilities of outcomes using something that goes under different names: "projection postulate" ; "Born rule" etc...
    (in fact we already used this in the setting up of the initial state too). This is the "discrete" part. This step doesn't really make sense but makes the calculations work out fine. Now you could think that this is some kind of shortcut to a more complicated calculation using the "unitary" part if you'd apply it to the complicated system of the measurement apparatus, but one can easily show that both mathematical procedures are incompatible: the "unitary" part can never give rise to the "discrete" part.
    So that's the puzzle that's with us since the beginning of quantum theory, and many people have many ways of "resolving" the issue (me too), but they are all weird in one way or another (and when I say, weird, it really means weird!).
    For instance, Schroedinger himself pointed out the difficulty with his cat: according to the "continuous" evolution, the cat would be BOTH alive and dead (that's continuous) in varying amounts, while if we look at it, of course it is one or the other (discrete).
    I cannot discuss the different ways of resolving the issue without introducing some of the mathematics of quantum theory. But I hope you get the gist of it.
     
  7. Oct 13, 2005 #6

    vanesch

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    Depends on the situation. A VERY weird situation is this: particle oscillation. It is known in a certain type of particle (K_S), and it is now strongly suspected (ok, I think, generally accepted) for neutrinos.
    It goes like this:
    You make particles of type "A". You can say that, because if you measure their type immediately after they've been produced, they all are of type A.
    But now you leave them there for a certain time and you measure them again. Surprise: half of them are of type A and half of them are of type B.
    You redo the experiment and wait for a longer time now between the production and the measurement. Surprise: they are all of type B now !
    You redo the experiment again and still wait longer: they're again all of type A!
    (this is what's happening to solar neutrinos).
    Again, one of the strange aspects: you could use ANOTHER measurement apparatus that tells you LEFT or RIGHT polarized particles. Again a discrete answer. This time you'd have lost the linear polarization information.
    YES !
    Good question. We don't know. In standard quantum mechanics, it's continuous as can be, using real (in fact, complex) numbers.
    If you want to, you can say that, yes. A quantum state is a way more complex beast than a classical state.
    It's a subtle issue. Let us say that the information you can *extract* from a quantum state is limited. It is not clear what it means for a state to "contain" information if you cannot extract it.
    But you already have that problem in Newtonian mechanics. The precise position of a particle, given by 3 real numbers x y and z, already contains "infinite information". A real number contains "infinite information" but you cannot extract it, because you can only measure x, y and z up to a finite precision, hence you only get out a finite amount of info.
    cheers,
    patrick.
     
  8. Oct 13, 2005 #7
    Followup

    I think I meant "precisely define" rather than "contain."

    True. Hadn't thought about it that way.

    After further pondering this stuff, I think the essential question can be boiled down to this: Is space itself quantized? Or is it continuous?

    If there exists some unit that represents the minimum distance a particle or quanta or whatever may travel then there would be a resolution at which continuous functions like sine, cosine, log etc would start to introduce minute errors. Even formulas like the pythagorean theorum would break down.

    Added at 7:50: Just following these thoughts where they lead...

    In order for space to be quantized, one possibility would be that it is built from units with a fixed tetrahedral structure as that would ensure that the distance between a vertex or node and any of it's nearest neighbors is identical.

    Quantized space introduces another problem though, namely: What happens when a wave/signal/particle is in transition between one vertex node and another? Since the quantized distance is by definition the smallest allowed distance, would waves/signals/quanta disappear from the universe and reappear in their new location?

    Quantized space would necessitate quantized time as well otherwise, it spawns the question: Where is the signal when it has only been traveling 1/2, 1/4, 1/8 the time it takes to leave one node and appear at a neighboring node? (and what is it traveling through)

    Quantized space and time would imply quantized "force" or "energy" as well. Since the only allowed travel time between node points is either zero or the quantized time unit (1), that would mean, at this scale, all signals travel at the same speed (between nodes) and therefore have the same energy. Higher energies would be the result of either multiple signals traveling similar paths (parallel) or multiple signals traveling the same path at different times (serial) and would be a simple multiple of the base quantized energy (Planck's constant?)

    If quantized space did have a tetrahedral structure, signals would travel along some vectors faster than others as, for some vectors, the path would consist of straight lines. Other arbitrary vectors would require some "sawtoothing" to get from point "a" to point "b" and the sawtoothing would increase the travel time. (red shift?) This would bring up the interesting possibility that two points that are mathematically equidistant from a source point could detect signals at different times due to the different sawtoothing required to reach them.

    Anyhoo... Not trying to prove anything here, just playing with ideas

    Ken
     
    Last edited: Oct 13, 2005
  9. Oct 15, 2005 #8
    This might be forever unanswerable (untestable). The form in which we get our information (electromagnetic phenomena) is quantized. The physical structure of the undisturbed medium is unknown.
     
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