Відео

Hi, these are very great questions, and looking back, I think I definitely could have done a better job addressing them in the video itself. I think the main points to answer your questions are around 6:36 and 9:58 in the video, respectively, but I'll try to answer both of your questions in some more depth, because they definitely deserve to be expanded upon. (1) This is a bit tricky mainly due to the fact that the same "intuition" for a ball on a spring doesn't really apply for a quantum harmonic oscillator. Probably a better way to think of a quantum harmonic oscillator is something like a piggy bank that only accepts dimes: it can only hold an exactly integer number of dimes, and since the hole is only large enough to fit dimes if you try to put in a different amount of money, it won't fit. In the case of a quantum harmonic oscillator, the dimes are the quanta of energy: you can only add or take away integer multiples of the exact value of energy accepted by the oscillator. Now, when we couple together multiple identical oscillators, any one oscillator can still only accept a discrete number of "dimes" at a time, but they are allowed to pass them back and forth between each other. When we increase the number of oscillators in the system, it gets easier and easier for neighbors to pass these quanta between each other, just like the neighboring masses respond more quickly to pushes and pulls in the the classical case, but you have to handle it a bit carefully to avoid things from blowing up (just like replacing the discrete masses and spring constants with a continuous mass density and tension). The end result is a line of continuous sites where these quanta of energy can live, but it is free to move from one to the next. The other key point is that, due to the uncertainty principle, we can't know exactly where these quanta of energy are living; we have to describe it by a probability distribution for them to be at any one site. The "spreading" effect that you are looking for an analogy for from the classical case is the spreading of this probability of the quantum to be at each site. However, it's very important to not get this probability mixed up with the location of the quantum: the unit of energy lives *at a single site* at any one time (remember, the oscillators can only accept/give up an integer multiple of this energy at a time, so if there is only one unit of energy in the system, it cannot be split up between sites), we just can't know exactly which site it is living at at any one time, so the best we can do is a probability distribution. (2) You're completely right that when we go to higher dimensions, the probability distribution will no longer "look" particle-like. In fact, in d-dimensions, instead of points, you would see the probability of locating the quantum in the system as a (d-1)-spherical shell expanding outward at the speed of sound. This is because we initially put the quantum of energy in a single site, so we have to have infinite uncertainty in momentum. Since the particle is massless, though, it has to travel at the speed of sound (or light in the case of a vacuum theory), and that is why you see a shell instead of a more "dispersive" effect where the probability spreads out over the full space. What you are really seeing is the natural Lorentz-invariance of the field theory! Again, I need to reiterate that the particle-like behavior isn't coming from the behavior of the probability distribution, and in fact this distribution looks very classical in higher dimensions: it is essentially a wave expanding at the speed of sound. The particle-like behavior is coming from the discrete nature of the quantum harmonic oscillators that make up the system. Consider we surround the site where we place the initial displacement (classical) or unit of energy (quantum) with a spherical detector. In both cases, the perturbation travels outward at the speed of sound until it hits our detector. In the classical case, the spherical wave originating from the displacement hits the *entire* detector, i.e. the full detector sees an excitation at once. On the other hand, in the quantum case, the chunk of energy travels outward at the speed of sound, but since there is only a single unit of energy in the system and this unit of energy can only live at one site at a time, the detector will only see a "hit" at *one, specific point*. Of course, we can't predict where that point will be, since all points have equal probability (the spherical probability distribution), but the main takeaway is that there is a fundamentally different behavior between the classical and quantum cases, and this "single-hit" behavior that we see in the quantum case is exactly what we expect from a particle! Hopefully that clarifies your questions a bit more!

@@zapphysics those are fantastic answers, and I so appreciate the time you took to respond! I am much clearer on the first question, and your answer on the second question covers my question well. The remaining questions I have is less about your video and more general questions about quantum field theory. I know it is a mistake to take the metaphors used in the visual depiction of a theory and overextend them, and I know slogans like "the electron is an excitation in the electron field" are gross simplifications. So the many UA-cam videos of multiple interacting two dimensional fields with high (or low) points representing particles can only be taken so far. But they all rely on there being "particle-like" localization of the probability density field that endures for some duration of time, and for an unconstrained excitation, I don't understand how that is possible. Everything gets smeared out at the speed of sound. (The other question I have is about how you START with a localized distribution, but that is just the preparation end of the measurement problem, so also not a question directly related to your video!) I am also confused about the actual quantum field - is there THE electron field for the universe, or separate ones for each system/preparation? How many dimensions does it have in multiple-particle systems? I guess I don't understand what entanglement looks like in the field picture. Don't feel obligated to respond, but if any future videos come down the pipeline addressing these kinds of issues... put me down as excited!

@andrewmilne9535 Yes, I certainly took the easy way out in this video by doing a non-interacting QFT :) I think one must always be extremely wary of anyone trying to present a "true" visual representation of an interacting quantum field theory. The main issue is that, outside of extremely special theories (e.g. superconformal field theories), it is not known how to find exact solutions for these probability distributions, whereas non-interacting QFTs are actually quite straightforward to solve, though they aren't really all that interesting for describing nature. In fact, it isn't even known whether or not interacting quantum fields in general are consistent mathematical objects (if you're interested in learning more, this is a good place to start: en.wikipedia.org/wiki/Quantum_field_theory#Mathematical_rigor). What's typically done in particle physics is that one starts with the non-interacting case, which we can solve, and then add in small perturbations caused by interactions. In this case, one can get reasonable answers and predictions (though the convergence of the resulting series at extremely high orders is questionable), but the mathematics behind it quickly becomes substantially more difficult than the free case, so visual representations of these on UA-cam, particularly of realistic scenarios like QED, are immediately going to be a bit suspicious. Not saying that you shouldn't ever trust these, but I'm just trying to get the point across that it becomes significantly more difficult to get the answer, especially beyond first order in perturbation theory. Going beyond, if someone is saying that the theory they are visualizing is non-perturbative, they would need some sort of lattice calculations, which are unbelievably computationally expensive, or an unrealistic theory with extremely high degrees of symmetry that is going to be somewhat difficult to extrapolate to reality. I have seen some videos of the sort that you are describing and some are honestly a bit horrifying that they are being passed off as true. One that jumps to mind essentially just took Feynman diagrams, spread out the lines and called them excitations of the corresponding fields. So they ended up with things like virtual photon field excitations in their visualization, which is just nonsensical: we know that everything must be described in terms of probability densities of some observation (in this case, it would be a position observation), and we cannot observe "virtual particles" because they are unphysical. In some sense, these virtual particles never even exist and are really an intermediate tool we use to do perturbation theory; if you were able to solve the theory exactly, you would never need to talk about virtual particles. I think that your intuition is good on the matter: if something is showing a completely non-dispersive, but point-like probability density in any spatial dimension higher than one (and even in one spatial dimension, you should at least see left-right dispersion like in the video), it certainly calls into question the validity of what they are showing. One could always start with some Gaussian distribution which minimizes both spatial and momentum uncertainty, but at the end of the day, there is no getting around the uncertainty principle if you have a quantum theory. Calling something e.g. the electron field only makes sense in this perturbative picture. Really, only in the free picture, but in perturbation theory, we assume that the fields are close to their non-interacting counterparts. In this case, yes, single units of energy that you add to the field will always result in particles of that type (really, they will be slightly different than the free cases, altered by the interactions). The fields themselves are infinitely expansive and fill the whole universe. That is why all electrons have the same mass, charge, etc. As soon as we go away from this nice, perturbative case, though, everything goes downhill: for example, if I try to add a single unit of energy to the Yang-Mills field, I won't end up with a gluon, even though the fundamental degrees of freedom that describe this field are the gluons. (I can't tell you what you would actually get, since again, nobody knows how to solve this problem, but the likely answer is that you would end up with some *massive* glueball which is uncharged under the Yang-Mills interactions.) If I'm honest, I'm not exactly sure how to answer all of your other questions unfortunately. I do, however, want to make a follow up to this video that addresses some of these points eventually (along with some other misconceptions that I am seeing a lot in the comments). I am even toying with the idea of actually trying to do a similar setup with some (perturbative) interacting theory, but like I said, it is a lot of work to make sure that it is done properly, so we will see.

Thank you so much for the kind words, I really appreciate it and I'm glad you enjoyed the series! Yes, I am absolutely planning to cover anomalies and lattice calculations, and want to start a whole separate series on BSM in the future! Also, to answer your question, I actually am mainly a flavor physicist (I do some BSM stuff here and there though), so this video has a special place in my heart!

This has to do with the fact that we can split fermions into two pieces: a left-handed piece and a right-handed piece (the "handedness" is somewhat related to the two spins, up/down, of the fermions, but not exactly). Typically, in order for a complex fermion (like the charged leptons or the quarks) to have a mass in the standard model, it needs to have a left- and a right-handed component, which allows it to couple to the Higgs. However, since neutrinos are neutral under electromagnetism and QCD, they only interact via weak interactions, which only talk to the left-handed pieces of the fermions, so we have only ever observed left-handed neutrinos. Without a right-handed counterpart, neutrinos can't interact with the Higgs in the standard model and so don't get a mass. We know that in nature neutrinos have mass, so you might wonder why don't we just add a right-handed neutrino to the standard model, have it couple to the Higgs, get a mass and call it a day? The issue is that, unlike the other fermions, the neutrino is neutral under all conserved gauge symmetries of the standard model, so there is a possibility that it is a *real* fermion, not complex, typically called a Majorana fermion. If we are allowed to extend the standard model (which we have to do anyway to add a right-handed neutrino), there are ways to give a real, left-handed neutrino a mass WITHOUT adding a right-handed component. So, without knowing for sure which scenario actually describes nature, we don't know yet how to extend the neutrino sector to explain the neutrino masses.

@@zapphysics so there are multiple ways to do it, but we don't know which matches reality?

This is certainly the natural question to ask after the discovery of all of the generations of quarks. In principle, yes, but they would likely have to behave quite differently than the standard model quarks. The issue is that new quarks would have to be quite a bit heavier than the top quark, otherwise we would have seen their effects in e.g. FCNC decays (or perhaps even direct production at the LHC). The main problem with this is that, if they couple to the Higgs (which is the only way in the standard model that particles can get masses to begin with), quantum corrections coming from interactions should drive the Higgs mass up to the scale of the heaviest quark. Since the Higgs mass is already very close to the top mass, it is theoretically very difficult to come up with a model with additional quarks where the new quarks are heavy, but the Higgs keeps its top-like mass. Now, one could just suppose that the new quarks don't talk to the Higgs, and get their mass from some other mechanism, but these new quarks would have to behave very weirdly compared to the ones we have observed. In particular, if we decompose them into left- and right-handed pieces, these pieces would have to interact the same way according to standard model interactions (otherwise the masses would violate the symmetries of the standard model). Compare this to the standard model quarks whose left- and right-handed components behave very differently, particularly with weak decays. Nonetheless, these new quarks can arise in some theories and there are active searches at the LHC for these so-called heavy "vector-like" quarks.

@@zapphysicsand what about additional leptons? ;)

@Firedragon9898 the charged leptons are pretty much the exact same story since the known charged leptons get their masses from the Higgs, and if there were additional charged leptons lighter than the Higgs/top, then we would have seen them in similar decays to those where we measured the tau. So, there isn't too much room for additional charged leptons aside from similar, vector-like leptons where the left- and right-handed pieces are treated the same by the standard model. Neutrinos are a bit of a different story, since they already don't have mass in the standard model. There are several ways that neutrinos can get a mass (which we know they do, in reality, have) and some of these require some additional neutrinos for it to work. But in some sense, the standard model already only has "half" the number of neutrinos as charged leptons since we have only observed left-handed neutrinos, so one way to give neutrinos mass is by adding a right-handed neutrino basically allows the neutrinos to talk to the Higgs in the first place. This isn't the only way to give neutrinos mass, but it is perhaps the simplest, so again there are several experiments working on trying to find evidence for right-handed neutrinos.

Neil turoks work on right handed nuetrinos looks promising, imho.