Colorful Quantum Mechanics (Standard Model Part 5)
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- čas přidán 16. 07. 2024
- Introducing quarks is great, but it seems to introduce more questions than it answers. In this video, we will take a look at how to resolve some of the seeming problems that arise in the quark model by introducing color charges, gluons and the quantum chromodynamics!
0:00 Questions from Quarks
1:27 Dubious Delta Baryons
2:45 New Quantum Numbers
9:58 A Brand New Force!
14:50 Curious Coupling Constants
18:30 Recap
Thank you so much for making these videos. The bridge between popular science and graduate-level courses is severely lacking in the literature. This provides useful intuition to ground us as we go deeper into the math.
6:02 Non-trivial elements of U(1) don't have determinant 1. SU(1) is just the trivial group.
Great work! Have you considered starting a Patreon? I would gladly give a couple dollars a month to support these videos. You are uniquely talented at explaining the basics without dumbing it down.
@Dreeev Thank you, that means a lot to me! I have considered Patreon, and I am not opposed to it, I just am unsure what all else I can offer for e.g. different Patron tiers and things like that and I wouldn't feel great about taking people's money and only giving them what they could have for free. Also, as you are probably aware, I'm not the most consistent with my upload schedule, and I don't really want it to turn into an obligation to crank out videos so my Patrons are getting their value, etc. I've tried to limit myself to a schedule in the past and I felt like the content I produced really took a hit. So I am kind of torn on the idea, but like I said, I am not opposed, it just will take some extra thought before I go through with it!
@@zapphysics I personally wouldn't expect any reward tiers or regular uploads, it would just be a way to support these videos more directly. I think most people would be quite understanding, especially towards an educational channel. However, I understand how it could create some perceived pressure on your end anyway.
Either way, thanks for the videos!
You are one of the best quantum physics teacher on CZcams, you make great videos, you deserve mor subs.
The boi drops another banger
Really interesting and amazing video Zap ⚡
loveittt
Pls talk about the skyrmion model and holographic QCD which make very good predictions
Can we make multiple cosmic microwave background images & measurements? +is there anything that could alter red shift? Say tons of gas cloud's? Black holes, galaxies, the type of Light the star is giving off? So how would you find a "reliable measuring system from super nova's? It seems variable?" I don't think we fully understand Gravity Yet. ~i know that's random here but this made me think of our current state of physics. If u look up "what is Gravity" right now, you get all sorts of science videos explaining that gravity is not a force, & they talk about time dilation. ~i think there is more we have yet to understand & perceive about Gravity. Sorta like something we've all seen a million times, but finally someone finds a new way to look at it and notices things most people have overlooked. *I in no way am saying this because I think I have the answers. No. This is just a gut feeling that I hope people way smarter than me discover. I got this guy feeling after observing the state physics has been in for a long time now. Something is telling me we have to look back on things and see if there is anything we may have overlooked? It happens all the times in discoveries and progression. There's Nothing embarrassing or to be ashamed of. It's natural. It's a part of growth and inventions.
My gut tells me that gravity is greatly intertwined with electrodynamics, density, and all the factors that make up many things in our Cosmo's. Just that gravity is this unique property of our universe in multiple scales. Small form, mid form, mega form. Think of large Filaments throughout the Cosmo's/multiple galaxies/black holes/nebula's that span many light-years across • then • Stars/planets/ asteroid's/solar systems/orbits/atmosphere's/magnetosphere/cosmic bubble around our solar system/then all the effects on our planet that gravity plays a role in. •Lastly• the small factors of matter/static charges that cause things to start sticking to each other/ particles/magnetism/density/temperature/velocity/pressure (probably many more things I'm leaving out but hopefully u get the point, these factors all play such a crucial role in our Cosmo's and I really think we have further to learn about it.) *But that's just my personal opinion
6:08 Actually, no; ðe only member of SU(1) wið det = 0 is ðe identity map, corresponding to ðe 1×1 identity matrix: (1). Could you please set ðis right? 😅
Oðerwise, a deeply informative video! 👍
I guess I still dont fundamentally understand how charges interact seemingly at a distance. Are they radiating gauge bosons in all directions or just at the nearby particle?
@Andrew Ferris This is a phenomenal question. The idea that these particles are spitting out gauge bosons is _strongly_ rooted in perturbation theory, which of course is just an approximation as to what is really going on. Now, I can't really paint a picture of what is actually happening because, theories like the standard model can't be solved exactly, so there isn't really a good picture to paint. However, if we remember that these are all quantum _field theories_ we might be able to get some sort of intuition about what's going on by looking at the more familiar classical field theories.
So, say I have two fields, A and B, and I allow them to "talk" to each other at each point, just meaning that stuff that is happening in A makes stuff happen in B and vice versa. This is all that we mean when we say that two things are "coupled:" effects in one influence effects in the other. So, if I poke field A to make some small perturbation, this will also have effects in B. But since these are fields, the effects travel through the fields to other points as well (think of just hitting a sheet of fabric: the effects don't "stay" at the place where you hit it). So, when I perturb A, that creates effects in B and those effects travel through B as well as A. But remember that B also induces effects back in A. So, if I am sitting at some different point and looking at A, I will see extra effects other than the ones from the initial "poke" in A, these extra effects, of course, coming from the field B. This is all that is really happening when we have particle interactions: the "poke" in a perturbative QFT is just a particle. So if two quantum fields are coupled, then it shouldn't be too surprising that there can be long-distance effects from this coupling! In principle, that is all this sort of interaction really is.
@@zapphysics your AB picture reminded me of the Reeh-Schlieder theorem, where action at a distance is described as a consequence of entanglement between different patches of (fields in) spacetime.
With, specifically, close patches being more entangled, than distant ones; but never 0.
That made me realize all of a sudden that "pokes" in the fields don't need to send actual/physical "messages" (like virtual particles), but they may influence each other simply because they're all entangled!
This lefts "messaging" part as a still reliable mathematical representation I believe, but makes much more physical sence to me.
Even if the baryon has a spin 1/2, at least two quarks have the same same spin state.
Banger tweet bestie
at 8:35 on combining three 3's, I think there is a small error. You won't get a 3 representation. There would be a Decouplet (10) , two adjoint (8), and a singlet (1). Although that doesn't change anything else in this otherwise excellent video.
What now? The Charged Fermion Propagator does not have the same mathematical formula as in a previous Part! See timestamp 13:59.
thanks. what are the laws of conservation that come from these SU symmetries? is it law of color charge conservation? also what law of conservation comes from SU 2
Interesting video! These are getting deep :)
Does it make sense to talk about the location of the quarks "in" or the size of the hadrons if we use the exclusion principle to say they can't be in that same state? Or is it an energy state rather than a position they would have to share?
On a related note I've been thinking about how fermions can band together to form bosons (in particular Cooper pairs). Does the exclusion principle then only care about the composite boson, or does each individual fermion still obey it? Basically; can Cooper pairs stack?
I was thinking about this in relation to very compact stars, like the interior of neutron stars but i suppose this could also happen inside of hadrons?
@Narf Whals It makes as much sense to talk about the locations of quarks in hadrons as it does to talk about the locations of electrons in atoms. The big difference is that we don't know the potential felt between the quarks like we do between e.g. the electron and proton in the Hydrogen atom (again, this just comes down to the fact that low-energy QCD is non-perturbative), so we don't really know the wavefunctions corresponding to quarks in hadrons.
The "size" of a hadron is a bit tricky to pin down, mostly because almost all of the hadrons aside from the proton are unstable (and the proton might still be unstable with a crazy long half-life). Typically the size of the proton is determined by its "charge radius." However, there is a slight issue in that different measurements yield different results (the so-called proton radius puzzle).
So the Pauli exclusion principle just forbids us from putting multiple identical fermions into the same quantum state. For a bound state, that doesn't have to be restricted to a single position in space (again, think of the orbitals of an atom). So, once fermions bind together to form total bosonic states, the combination will now always behave like a boson: if I try to stack two e.g. Cooper pairs, I can do so because the fermions live in different bound states. Perhaps the most famous example of this is superfluid helium-4: the atom itself is made up of fermions, but it is in a total bosonic state. When it is cooled down so that only long-range effects are present, the bosonic nature allows the atoms to "stack" so that they flow past each other with zero viscosity. This is also how they have formed Bose-Einstein condensates in labs. They do it using atoms which are total bosonic states, so that at very low temperatures, when you can't see the internal structure (i.e. that they are really made up of fermions), their behavior will be totally bosonic, allowing them to all pile into the ground state together.
@@zapphysics But given all other properties are fixed and equal, position _is_ restricted. And superfluid helium still has and fills a volume in space, so the atoms don't all sit on top of each other. Is that just the electromagnetic repulsion between the electrons?
If we go with the electrons again if we try to add an electron with the same spin and energy we get psi=C12+C21=0 (trying to adopt the notation from your fermion/boson video). This is a mixed state?
But if they hypothetically formed a cooper pair we can't use this two particle wave function anymore because the cooper pair is now one boson and asking what happens when we exchange the electrons doesn't make sense? Is this a pure state?
Could we then stack as many cooper pairs into the lowest energy level as we wanted?
I suppose not, because they repel each other and we'd need to add energy and break the pairs.
But why I initially thought about this is actually neutron stars. Because my pet theory about what happens when a black hole forms is that the neutrons form cooper pairs and then all happily stack on top of each other in what may or may not be a singularity.
And a recent PBS spacetime video told me that _this actually happens_ in the core. But if that happens, and cooper pairs will stack in position space, why does the core still exert an outwards pressure?
@@narfwhals7843 I think we have to be a bit careful about what we are talking about. In particular, when we talk about cramming a bunch of bosons into the same ground state, then that isn't the same as saying we are putting them all in the same position. The ground state will have some spatial extent to it and I believe that the idea for putting these bosonic composite states into the the same ground state is that this spatial extent (or characteristic wavelength) of the ground state is large enough that it can't "see" the internal structure of the composite. The same goes for the types of states where the Pauli exclusion principle applies. Think of the orbitals of the Hydrogen atom: these orbitals are not just a single location, but are spread out in space. Despite this, we can only put a single spin up and a single spin down electron in each orbital, otherwise we violate the Pauli exclusion principle.
We have to be a bit wary whenever we talk about the exact location of something in quantum mechanics, since we know that we can't exactly know this information by the uncertainty principle. However, we _can_ think of the "state" of a free particle as occupying some volume in phase space, as long as the "width" of this volume satisfies the uncertainty principle (think of segmenting phase space into little boxes of size hbar/2). But keep in mind what this is saying: there is absolutely no problem piling identical fermions into the same space *as long as they have different momentum states*. So really, there is no issue with black hole formation coming from piling neutrons into the same spatial region. Once you have achieved a dense enough region (through e.g. stellar collapse) to form a black hole, it doesn't matter how high of a momentum the neutrons have; they are going towards the singularity no matter what.
Just as another note: I would be a bit skeptical of someone saying that something "actually happens" on the interior of a neutron star. In reality, we have no idea what happens at the core of a neutron star. This is again because QCD is so tricky, and actually even more so in this case, since it involves identifying the thermodynamic properties of QCD. Essentially, the goal is to find a phase diagram corresponding to QCD, but many thermodynamic techniques involve perturbation theory (which we can't use) and even lattice QCD has serious problems due to computers not really knowing how to handle complex phases which show up in the calculation. I think that things like the quark-gluon plasma and color superconductivity are expected to occur, but we certainly don't know for sure. Some approximations of the calculation are given on the wikipedia page for color superconductivity (en.wikipedia.org/wiki/Color_superconductivity), but another interesting one that I would add is that there is work being done in this field using the AdS/CFT correspondence (I am not an expert here, but I have talked with people who are working on this).
damn particle physics is hugeeeeee
Does the following quantum model agree with the Spinor Theory of Roger Penrose?
Quantum Entangled Twisted Tubules: "A theory that you can't explain to a bartender is probably no damn good." Ernest Rutherford
When we draw a sine wave on a blackboard, we are representing spatial curvature. Does a photon transfer spatial curvature from one location to another? Wrap a piece of wire around a pencil and it can produce a 3D coil of wire, much like a spring. When viewed from the side it can look like a two-dimensional sine wave. You could coil the wire with either a right-hand twist, or with a left-hand twist. Could Planck's Constant be proportional to the twist cycles. A photon with a higher frequency has more energy. (More spatial curvature). What if gluons are actually made up of these twisted tubes which become entangled with other tubes to produce quarks. (In the same way twisted electrical extension cords can become entangled.) Therefore, the gluons are a part of the quarks. Quarks cannot exist without gluons, and vice-versa. Mesons are made up of two entangled tubes (Quarks/Gluons), while protons and neutrons would be made up of three entangled tubes. (Quarks/Gluons) The "Color Force" would be related to the XYZ coordinates (orientation) of entanglement. "Asymptotic Freedom", and "flux tubes" are logically based on this concept.
Modern physicists say the Strong Force is mediated by a constant exchange of Mesons. The diagrams produced by some modern physicists actually represent the Strong Force like a spring connecting the two quarks. Asymptotic Freedom acts like real springs. Their drawing is actually more correct than their theory and matches perfectly to what I am saying in this model. You cannot separate the Gluons from the Quarks because they are a part of the same thing. The Quarks are the places where the Gluons are entangled with each other. It is almost as if they cannot see the forest for the trees.
Neutrinos would be made up of a twisted torus (like a twisted donut) within this model. Gravity is a result of a very small curvature imbalance within atoms. (This is why the force of gravity is so small.) Instead of attempting to explain matter as "particles", this concept attempts to explain matter more in the manner of our current understanding of the space-time curvature of gravity. If an electron has qualities of both a particle and a wave, it cannot be either one. It must be something else. Therefore, a "particle" is actually a structure which stores spatial curvature. Can an electron-positron pair (which are made up of opposite directions of twist) annihilate each other by unwinding into each other producing Gamma Ray photons?
Does an electron travel through space like a threaded nut traveling down a threaded rod, with each twist cycle proportional to Planck’s Constant? Does it wind up on one end, while unwinding on the other end? Is this related to the Higgs field? Does this help explain the strange ½ spin of many subatomic particles? Does the 720 degree rotation of a 1/2 spin particle require at least one extra dimension?
Alpha decay occurs when the two protons and two neutrons (which are bound together by entangled tubes), become un-entangled from the rest of the nucleons
. Beta decay occurs when the tube of a down quark/gluon in a neutron becomes overtwisted and breaks producing a twisted torus (neutrino) and an up quark, and the ejected electron. The phenomenon of Supercoiling involving twist and writhe cycles may reveal how overtwisted quarks can produce these new particles. The conversion of twists into writhes, and vice-versa, is an interesting process.
Gamma photons are produced when a tube unwinds producing electromagnetic waves.
>>>>>>>>>>>>>>>>>>>>>>
Within this model a black hole could represent a quantum of gravity, because it is one cycle of spatial gravitational curvature. Therefore, instead of a graviton being a subatomic particle it could be considered to be a black hole. The overall gravitational attraction would be caused by a very tiny curvature imbalance within atoms.
>>>>>>>>>>>>>>>>>>>>>>
In this model Alpha equals the compactification ratio within the twistor cone. 1/137
1= Hypertubule diameter at 4D interface
137= Cone’s larger end diameter at 3D interface
A Hypertubule gets longer or shorter as twisting occurs. 720 degrees per twist cycle.
>>>>>>>>>>>>>>>>>>>>>>>
How many neutrinos are left over from the Big Bang? They have a small mass, but they could be very large in number. Could this help explain Dark Matter?
You say "If an electron has qualities of both a particle and a wave, it cannot be either one." .
When physicists call an electron a particle, they do not necessarily mean that it has all the properties that *you* think of a "particle" as having. Similarly, when they call it a wave, or say it acts like a wave.
"Particle" and "wave" are only words/names. They are used in ways that their users fine useful.
Electrons are "particles" in the sense of "particles" that they mean when they say that electrons are "particles". The preceding statement is not about what electrons are, but about what they mean when they say "particle".
(Also, as you have recognize, electrons are *not* "particles" in the sense of "particles" which is used when people say that "electrons have some aspects of particles but not others.".)
Do not overestimate the power of words.
You also say "Alpha decay occurs when the two protons and two neutrons (which are bound together by entangled tubes), become un-entangled from the rest of the nucleons". This seems incorrect to me. Suppose you have some nucleons bound together, which is going off in some direction, starting in a configuration where the position and momentum both have relatively low uncertainty. The uncertainty of the position will increase over time, but the uncertainty in the relative positions of the nucleons (relative to one-another) should remain fairly small. If it then decays (I suppose post-conditioning on when it decays?) emitting an alpha particle, the position of the alpha particle and the position of the rest of the nucleus should still be entangled, and as such, the positions of the nucleons in the alpha particle should still be entangled with those of the nucleons in the nucleus. (At least, until the position of either has been observed).
In addition, you should know that the fine structure constant is not exactly 1/137.
There other things to correct, but I don't want to.
Where is the variational part before q fields, please if you could tell me to watch it? 🖖🙂
The Photon Propagator is also not the same!
One thing I don't get: why should a uuu baryon be subject to the PEP when the three fermions are all in different positions? Different position is a different state, right?
Another PEP issue I can't get past: if I have two particles with the same momentum, their relative position remains constant, so you can't get one right on top of the other unless they were already superimposed. I don't see why I need a special arbitrary law for fermions randomly follow in absence of any force to explain the necessary action required to modify the path of the particle.
I'm having trouble with the relationship between symmetry and charge.
If I understand correctly, when we have a local symmetry, we need the gauge field to accommodate for the local transformation.
To me, this indicates that _every_ particle that has this *local* symmetry must couple to the gauge field, because that's the point of the gauge field.
This would mean that the charge of the theory is the same property as having the local symmetry.
In QCD this seems to be the case as SU(3) is just the symmetry of the color space. Can we have particles that have global symmetry in this space but not local? Is that what confinement is?
In QED we usually talk of the charge separately. Like we can have particles that have local phase invariance but do not need the gauge field to make up for it. Is that the case? Do neutral particles have local phase invariance?
In terms of the covariant derivative, with the gauge as the connection term, we have this term multiplying the charge, which is zero for neutral particles. But that seems like its just a convenient way to write it.
In your QED video, (if I understand correctly) you write this term as psibar*psi times the gauge derivative. Will this turn out to be zero, if the field is not locally symmetric?
@narfwhals7843 I think most of your intuition is correct, but I will add a couple of comments to hopefully clarify some things.
> If I understand correctly, when we have a local symmetry, we need the gauge field to accommodate for the local transformation.
> To me, this indicates that every particle that has this local symmetry must couple to the gauge field, because that's the point of the gauge field.
> This would mean that the charge of the theory is the same property as having the local symmetry.
You're exactly correct that any field that transforms under the local symmetry must couple to the gauge field. Keeping in mind that the gauge field is really the connection between two points which are now allowed to transform differently under the symmetry transformation, this is very easy to see: suppose I don't couple such a field to the gauge connection. Then, I am not accounting for this extra local transformation as I compare two different points, so two field configurations which are related by the local symmetry transformation will actually "look" different according to the theory and the symmetry is explicitly broken (two fields which are related by a symmetry transformation should always look the same).
The charge is essentially how much the field transforms under the symmetry, and is always going to be proportional to the generators of the symmetry group. In QED (and any U(1) theory in general), the generators are any real numbers, so fields can have any charge under a U(1) symmetry. For a more complicated symmetry group like SU(3), fields are restricted to only transform as representations under the group (this is true for U(1) as well, it's just that any real number corresponds to a valid representation).
> In QCD this seems to be the case as SU(3) is just the symmetry of the color space. Can we have particles that have global symmetry in this space but not local? Is that what confinement is?
Not quite. When we make a symmetry transformation, the transformation has a set number of transformation parameters (the same as the number of generators), i.e. "how much" you are transforming by. These transformation parameters can vary on spacetime, but they will be the same for any field we are transforming. A global symmetry would then need to be completely disconnected from the local one since it would have totally different transformation parameters, so it would probably make more sense to just consider them as separate symmetries. Confinement just has to do with the behavior of couplings/potentials of a theory. However, you can do the opposite, where you have a large global symmetry and only gauge a sub-group of it. For example, I could have a global U(2) symmetry where I gauge the U(1) sub-group of the U(2).
> In QED we usually talk of the charge separately. Like we can have particles that have local phase invariance but do not need the gauge field to make up for it. Is that the case? Do neutral particles have local phase invariance?
> In terms of the covariant derivative, with the gauge as the connection term, we have this term multiplying the charge, which is zero for neutral particles. But that seems like its just a convenient way to write it.
Any time that you have a theory which will depend on *differences* (i.e. derivatives) of fields at spacetime points where the fields transform under a local symmetry, you will need a gauge field. Again, this just has to do with the fact that we are really connecting different points of fields that could be transformed differently, and the gauge field automatically accounts for this. It absolutely isn't wrong to say that a neutral field transforms with "charge" zero. This is just because the "zero" representation is always going to be a representation in any group, since this corresponds to the identity operator and by definition, a group must have an identity element. In this case, yes, the gauge term in the covariant derivative vanishes. I don't think that this is just a convenient way to write it, but it is the correct way to think about it.
> In your QED video, (if I understand correctly) you write this term as psibar*psi times the gauge derivative. Will this turn out to be zero, if the field is not locally symmetric?
I'm not sure what you mean here by "not locally symmetric." If you mean that the theory truly isn't invariant under local transformations of this field, then the symmetry is explicitly broken, and you can no longer talk about having a local symmetry.
If you mean that it doesn't transform under the symmetry (or I guess more properly that it transforms as a 0 representation of the group), then the part of the covariant derivative containing the gauge field will vanish when acting on this field, but the standard derivative will still hang around. However, it will only vanish in the coupling to that field, not to any fields which actually transform non-trivially under the symmetry.
Hope that helps!
@@zapphysics Wow, thank you for always giving such extensive answers!
So having a non-zero charge tells us how much the gauge transformation affects the field.
A zero charge means the field "transforms trivially", so no change at all.
I think my confusion was exactly the difference between your last two paragraphs. A trivial transformation is still a local symmetry.
So the "amount" of charge is a distinct property of the fields, which is restricted to representations of the gauge group.
And that is why we (sometimes) need to talk about it separately.
How is that related to the psibar*psi term in the derivative here czcams.com/video/qtf6U3FfDNQ/video.html ? Or am I misunderstanding which derivative that is?
@narfwhals7843 ah, this term comes exactly from trying to take the derivative of the gauge-transformed field. I apologize in advance for the formatting, but when I take psi-> exp(i*e*theta(x))*psi and psibar -> psibar*exp(-i*e*theta(x)), where e is the charge of the field and theta(x) is the spacetime-dependent transformation parameter, the impertant terms with the derivative of psi transform as psibar*dpsi/dx -> psibar*(dpsi/dx + i*e*psi*dtheta/dx). The first term isn't a problem, since this is the term we started with, but the second term is an issue, since it adds a term we didn't have before, which is clearly an issue for a symmetry. However, when we have a gauge-covariant derivative, we also have a coupling like psibar*i*e*A*psi where A is the gauge field. Under a gauge transformation, the gauge field also transforms like A -> A - dtheta/dx. So, when we add this all together, the combined transformation of the gauge field as well as the local transformation of the charged fields exactly cancel, leaving the theory invariant under the combined transformation. Note that this is exactly related to the fact that the charge in the transformation is the same as the charge in the gauge-covariant derivative: the transformation of the gauge field makes no reference to the particular charge of any one field, so in order for the cancellation to happen correctly, the gauge field needs to paired with the appropriate charge in the covariant derivative acting on the field.
Now, this story gets slightly more complicated for a non-abelian group, but the general idea is the same. Just replace the charge with the appropriate generators of the group corresponding to the representation of the given field.
@@zapphysics Thank you, this was very helpful!
Isn't it that gluons are solely virtual particles ? Then can we say that gluons really exist or are they merely an artifact of the perturbation theory ?
@Wittgenstein bro this is a bit of a tricky question that should be approached a bit carefully. The first point to make is that the idea of virtual particles _comes from perturbation theory_ and really should be considered a mathematical tool we use to approximate calculations, not as a comment on anything physically happening. So, if we have a strongly-coupled theory like low-energy QCD where perturbation theory breaks down, it doesn't make sense to talk in terms of virtual particles anymore. Really, it doesn't even make sense to talk about particles in general, since a particle picture of an interacting quantum field theory assumes that an excitation of an interacting quantum field will behave very similar to that of a free field (where excitations look exactly like particles) with corrections coming from the interaction which are calculated using perturbation theory. When we have a strongly-coupled theory, this assumption breaks down and excitations of a strongly-interacting quantum field theory will look very different from the free-field, particle-like approximation.
So, it isn't so much that gluons are always virtual particles as much as the dynamics of the gluon field at low energies does not resemble particle-like behavior anymore.