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Electromagnetic waves experience dispersion and these result in a "chirp" in frequency after traversing some distance. This chirp can be heard when waves from a lightning strike in one magnetic hemisphere of the Earth propagate along magnetic Earth's magnetic field lines through charged particles trapped in them and are received in the other magnetic hemisphere, and are used to estimate distances to Fast Radio Bursts (Just how fast is a Fast Radio Burst thought to be?) since dispersion by the interstellar medium is well-characterized.

Question: Are there conditions under which matter waves of bound systems like molecules experience dispersion? If so, how would it manifest itself experimentally?

I don't know if this is a "for example" or a separate question, but it helps reflect my current lack of understanding of the problem. Atoms and molecules have been demonstrated to exhibit diffraction and even Bragg scattering from electromagnetic standing waves1 when there is a dipole moment. Since diffraction is dispersive, do different "frequency components of the molecule's wave" (whatever that means) get spread out in a sort of a spectrum?

1recent news "Bragg Diffraction of Large Organic Molecules"

uhoh
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2 Answers2

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Can matter waves of bound systems experience dispersion? If so, how would it manifest itself experimentally

Electromagnetic waves have variations in energy as a function of (x,y,z,t)

The quantum mechanical waves that describe particles and molecules are not energy waves. Each particle/molecule is described by a wavefunction $Ψ$ , a complex function which is a solution of the quantum mechanical equation, BUT is not measureable. What is measured in experiments is $Ψ^*Ψ$ which , according to the postulates of quantum mechanics, is a probability to find the particle/molecule at (x,y,z,t). See the double slit experiment one electron at a time

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Each electron is a dot seemingly random, the wave nature appears with the accumulation of events which is a probability distribution.

So any type of interference is possible ,but it will manifest itself in the probability distributions , in no sense in matter waves that can be timed, as far as I can see.

As far as "chirps" concerned: we are in the QM regime when talking of bound states. They are quantum mechanically bound, and described mathematically by QM wavefunctions. This means that any single measurement can show no chirps. Any chirps will be seen in the probability distributions.

anna v
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  • Okay so in the electromagnetic examples (whistlers and dispersion measurements of radioastronomical signals) there are zillions of photons in each pulse and the radio reception and frequency vs time analysis does the equivalent of building up a histogram of them, whereas in an experiment based on counting individual molecules no matter what happens in a dispersive media we're still counting individuals and each one collapses into a single count upon measurement. – uhoh Jul 26 '20 at 00:24
  • So actually I don't understand drawing the distinction between electromagnetic waves and matter waves as one being an "energy wave" and the other being complex and quantum mechanical. Maybe it's more that the electromagnetic examples I gave are collections of a very large number of photons -- we usually don't count radio photons experimentally due to their individual very low energies -- while we often do count individual electrons, or alpha particles or in this case molecules. – uhoh Jul 26 '20 at 00:27
  • As always there are different ways to look at the same problem, I guess for me it's better to think of the radio waves as a zillion individual photons, with each completely quantum mechanical in nature than it is a single "radio wave packet" treated classically. The QM treatment should always give the right answer, it's just that it's harder sometimes. Thank you very much for your answer! – uhoh Jul 26 '20 at 00:32
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    Single photns behave the same as the electrons https://www.google.gr/search?q=swiss+double+slit+single+photon&ie=utf-8&oe=utf-8&client=firefox-b&gfe_rd=cr&ei=sGFmWYrFCO_v8AfrtLGADQ . The classical em and its mathematcs emerges from a superposition of the photon wavefunctions, not a simple addition. and yes it is very complicated https://motls.blogspot.com/2011/11/how-classical-fields-particles-emerge.html – anna v Jul 26 '20 at 03:44
  • Would the answer to "Can matter waves of bound systems experience dispersion?" be "No"? or just "Not in this case?" If it's the latter, then I can accept this answer and then ask a new, more general question, something along the lines of the concept of matter wave dispersion making sense in any context ever, or if there are any proposed methods for its detection. I'll have to think about exactly how to ask that though. – uhoh Aug 06 '20 at 01:49
  • The answer is the same as for electrons. there are no matter waves, it is probability distributions that show wave like behavior. Matter waves occur only classically in water waves and fluids in general where atoms and molecules are displaced and can be modeled by wave equations – anna v Aug 06 '20 at 04:17
  • Wikipedia is not the final word in physics, but the first sentence to the article says Matter waves are a central part of the theory of quantum mechanics, being an example of wave–particle duality.. If you would like to add your assertion that "there are no matter waves" where it can be voted on, that would be great! – uhoh Aug 06 '20 at 15:37
  • @uhoh this is simply old history of quantum mechanics before it became a solid theory with the quntum mechanical equations, which ,by postulates http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qm.html have wave behavior in the accumulated probability distributions, as my answer simply illustrates. Individual particles are not waves. I do not intend to get involved in wikipedia arguments. – anna v Aug 06 '20 at 18:55
  • Thank you for your continued replies! I will go and think about this further. For single particles (e.g. electrons or photons) a dispersive medium or scatting from a diffracting structure will behave as you've illustrated, and the uncertainty in the final position, or width of the spread of "spots" when many particles have been tested will depend on the uncertainty in their initial spread in energy or momentum, whichever is being dispersed. But for the diffraction of a bound system like an atom or molecule, where observables may be more complex than a spot where it hits on a screen... – uhoh Aug 06 '20 at 22:20
  • ...I'm still pretty sure the answer to my question as asked is yes this is possible because the bound system has internal degrees of freedom. So I've added a bounty. – uhoh Aug 06 '20 at 22:21
  • we are in the QM regmie when talking of bound states. They are quantum mechnically bound. This means that any singlel measurement can show no chirps. Any chirps will be seen in the probability distributions – anna v Aug 07 '20 at 03:44
  • Can't a bound QM system like a molecule can have several observables? Even in the case of single point particle like an electron can not we measure both its spin and its direction or velocity i.e. use an ExB velocity filter for speed, a magnetic gradient for spin, and a position-sensitive channel plate for the measurement? For a molecule certainly there are many more observables available besides its center of mass motion and though computationally expensive, molecules can certainly be treated as QM systems. – uhoh Aug 07 '20 at 04:23
  • For an extremely weakly bound, rotating system, perhaps a dispersive effect might even split them into two separate particles. I'm pretty confident that the effect of dispersion on bound QM systems will have some observable effect, and will need some solid math and/or reliable supporting source to be convinced otherwise. – uhoh Aug 07 '20 at 04:30
  • all observables obey the probabilistic law of QM. If a molecule splits, it means an extra interaction/differen wavefunction and still the results of the split will follow probabilistic laws. – anna v Aug 07 '20 at 05:08
  • In that particular example then, that the split occurred would be detectable and therefore an experimental manifestation of the dispersive field. I'm giving this as a counterexample that you would need a large number of repeats. – uhoh Aug 07 '20 at 05:32
  • there would just be two random looking points in the equivalent screen as in the example above. An accumulation would be needed to show any signals due to the wavefunction – anna v Aug 07 '20 at 05:46
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As stated in the comment by @ChiralAnomaly, all wave packets corresponding to particles with nonzero mass should experience dispersion, because velocity is (classically) proportional to momentum and all wave packets have a spread of momentum due to the uncertainty principle.

An experiment to prove this dispersion would be relatively easy: for example, diffraction of a particle beam by a crystal will produce first- and second- order beams with angular spread that depends on the momentum spread of the original wave packets of the particles.

A single detected particle does not exhibit dispersion, because (if its velocity is detected) it can only have one velocity. Dispersion and wave phenomena in general are features of the probability distribution, which can only be measured in the form of a large number of detections of particles associated with identical wave packets.

In the case of a bound particle such as an electron in a hydrogen atom, it's not obvious what "dispersion" would mean. However if the velocity of the electron could be measured at exactly the same point for a large number of identical atoms, the magnitude of the velocity would always be the same because the energy spectrum is discrete. The direction of the velocity would vary randomly. So, if "dispersion" means having a spread of magnitude of velocity, there is no dispersion on a typical bound system. If there are some bound systems without discrete energy spectra, those systems would exhibit "dispersion" in that sense.

S. McGrew
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  • Please see the edit to my answer. – S. McGrew Aug 06 '20 at 23:52
  • OK, please see the further edit, which directly addresses bound particles. – S. McGrew Aug 07 '20 at 01:58
  • Thanks for the edit! While I chose a large organic molecule specifically for its size and complexity, you've chosen a single hydrogen atom for your example; the smallest possible system without going sub-atomic. Intuitively we like to think of the electron as well-bound to the proton and spherical, which minimizes any consideration of the differential effects of the dispersive medium or field on it relative to the proton, but I still see this as avoiding the question. – uhoh Aug 07 '20 at 02:03
  • Can you make a similar argument for a large organic molecule as I've discussed in the question, where there is far less coupling between the wave functions of the individual constituents within the system? – uhoh Aug 07 '20 at 02:03
  • I suspect that a large molecule can look a lot like a system with energy "bands" rather than a discrete energy spectrum - but I don't know. Probably if it is isolated it has a discrete spectrum but with very closely spaced lines that look continuous. – S. McGrew Aug 07 '20 at 02:05
  • Let's take a semiclassical example that the molecule were spinning around an axis within the scattering plane but perpendicular to the direction of motion. The "top" is moving forward and the bottom backward relative to its center of mass. Would not the dispersive medium then generate a torque about an axis parallel to the direction of motion? This is a limited, semi-classical example only and I believe it can be address quantum-mechanically, but it's one that tries to address why we should not think of bound systems as point particles. I suppose that my "rotation" relates to your "bands" :-) – uhoh Aug 07 '20 at 02:08
  • Not sure about that. In QM, the angular momentum spectrum is discrete, for any system. Also not sure what you're referring to by the term "dispersive medium". We can move this to chat if you would like. – S. McGrew Aug 07 '20 at 02:39
  • one example of a "dispersive medium" is given in the question; strong, periodic electromagnetic standing wave in a vacuum. Others might include a strong gradient in some other field. Let's see what other answers are posted. – uhoh Aug 07 '20 at 02:45