Spinors for Beginners 15: Nilpotents, Fermions, and Maximally Isotropic Subspaces
HTML-код
- Опубликовано: 9 июл 2024
- Full spinors playlist: • Spinors for Beginners
Leave me a tip: ko-fi.com/eigenchris
Powerpoint slide files + Exercise answers: github.com/eigenchris/MathNot...
Professor M Does Science's playlist on multi-particle quantum mechanics with creation and annihilation operators (second quantization): • Second quantization
Cohl Furey's playlist on Clifford Algebras and particle physics: • Division algebras and ...
0:00 - Introduction
0:53 - Creation and Annihilation Operators (Bosons)
2:14 - Fermions
4:58 - Nilpotents
7:14 - Projectors
10:46 - Example in Cl(1,3)
14:18 - More Nilpotents
16:11 - Maximally Isotropic Subspaces
21:02 - Generalizing to C(n,0) and Cl(p,q)
23:57 - Example in Cl(2,0)
26:20 - Conclusion
There's a bad typo at 1:30, the raising operator should have a - sign instead of a + sign. Sorry about that.
Can’t wait for the Lie groups!
Amazing so far, can't wait for you to get into the lie algebra stuff
Your doing great, i am looking forward to christmas vacation where i will properly watch these and take notes❤
Very good and impressive work! Keep on going …
Nice! Another video so soon.
17:00 When building a descending chain of ideals using projectors, you need to be careful each projector you add is *not* orthogonal to the product you already have, else you get the trivial ideal (you "skipped" the minimal ideals). It's kinda obvious when you think about it, as this is basically the definition of orthogonal projectors: they multiply to zero, and the zero projector is the one taking everything to zero when multiplied.
EDIT: a slightly less destructive situation that also can occur is that multiplying the projector doesn't change the product, but you haven't reached a minimal ideal yet.
At 1:30 , the raising and lowering operators are shown to equal each other. I would guess they should differ in the sign of their respective imaginary parts. Am I wrong?
Edit: corrected "operations" (should have been "operators" but somehow my fingers mistyped it.)
The raising operstor should have a minus sign in front of the complex i. My bad.
@@eigenchrisThank you.
Me too!! I realy hope he will take it to higher level!! I even purchased some books on about Lie ...which I can't even flip few pages ..feels like out of my element 😊
Got super excited at the notification and then saw it was down, did you have to make a correction? :P Glad it's back, this made my weekend!
I often have multiple drafts online at any given time, set to private, as I'm making final tweaks to the video. Last night I accidentally took an old version and set it to public. It was missing the short discussion at 19:05 about proving all vectors in the space are isotropic.
Minor correction:
11:25 defines U = \gamma_{t} + \gamma_{z}, but at 11:30 you have UU = (\gamma_{t} + \gamma_{x})(\gamma_{t} + \gamma_{x}) instead.
Sorry. That's a typo.
24:26 If you just take 1=1, i=I, σ1 = σx, σ2=σy, the other mappings immediately fall into place by multiplication. The main thing to remember is that you need to consider V and iV separately to honour that real algebras don't allow multiplying by i (you need to use the pseudoscalar).
Lol. Just as I started typing comment about this video obviously being influenced by Cohl Furey works, you mentioned her explicitly.
Any plans of making a video on her Cl(8) model?
Yeah, I watched her playlist about 2 years ago, but didn't understand anything after video 6 or so, once ideals and projectors came in. I decided to give it another shot last month, because I had since studied ideals and projectors, and suddenly it made sense. I think there's a big jump in video 6, where she expects the viewer to be already familiar with spinors and projectors and Fock spaces. I tried to cover a lot of the same content, but more gradually.
I don't plan on doing much else on Clifford Algebras, other than an appendix video to explain a few details I missed in this playlist. I don't really understand the strong or weak forces at this point anyway, so I don't think I can confidently explain the standard model algebra she's working on.
Love you dad
1:30 Nice half step for the ground state
Can you build a nilpotent in Cl(3,0,R)? If not, then is it more useful to use is Cl(2,0,C)?
i can't wait for the presentation on "Spinors in Particle Physics" 😂
Very elegant video as always! :)
I was wondering how this formalism of nilpotent vectors relates to Newman-Penrose formalism in GR, because in that formalism you are also required to choose a tetrad of null vector fields on your manifold (organized in two pairs), which have to be cross normalized in a similar fashion to what you're doing here. In that formalism you also have to complexify your tangent bundle in order to allow to write down two pairs that work, exactly what you're doing here. I also recall that the Newman-Penrose formalism is how one can think about spinors on a manifold, so the connection seems clear to me. What do you think?
I'm not familiar with that formalism. The wikipedia page says "Their notation is an effort to treat general relativity in terms of spinor notation", so I guess so? I'm afraid I can't say anything more because I haven't learned it.
Oh, sorry, thank you anyway then!
If you are interested in a cool formulation of general relativity I suggest checking it out ;)
With this methods one can for example classify algebraically various GR solutions (through their Petrov Type) and study the decay of gravitational radiation at far infinity (through the peeling theorem). A highly recommended and elegant formulation in my opinion!
one thing you didn't mention (and I suppose it wasn't immediately relevant to what you did) is how useful projectors and nilpotents are for power series.
a nilpotent e²=0 will always kinda behave like a derivative:
(a + b e)^n = sum_k choose(n, k) a^(n-k) b^k e^k = a^n + n a^(n-1) b e | all other terms 0 because e² = 0
and an orthogonal projector basis like a p + b q just ends up ignoring all mixed terms
(a p + b q)^n = = sum_k choose(n, k) a^(n-k) b^k p^(n-k) q^k = a^n p + b^n q | all other terms 0 because p q = 0
I figure these sorts of constructions would be really useful for figuring out functions' behaviors by expressing objects in the corresponding bases
Thank you for these videos.
sure about the raising and lowering operators at 1:33? they are the same.
That's a typo. The raising operator should have a - instead of a +.
Spinnors for beginnors!
Cohl Furey is the one working on connecting particle physics to like R x C x H x O, right? Where O is somehow related to quarks?
Since clifford algebras are associative in the usual geometric, product, they won't directly produce octonions but afaik there is a way to have an octionion style product anyway, right?
I wonder what it, like, "means", that stuff is non-associative. Non-commutativity seems to essentially correspond to interaction / measuring. If stuff commutes, the order doesn't matter, so the involved objects basically don't interact at all. They go through orthogonal processes, basically.
But what does it mean in this sense, when stuff fails to associate?
You can check out her playlist to learn more, but she treats octonions as functions (applied via left-multiplocation), and function composition is always associative. So you can build chains of octonions associatively, but you must always evaluate the multiplication from right-to-left to get a consistent function result.
I honestly don't understand the last few of the videos in her playlist. It works mathematically but it feels a bit "clunky". Not as "pretty" as most other stuff I've seen with Clifford Algebras.
I don't know what non-associativity means intuitively. The cross product is also non-associative, but you can "fix" that by introducing the wedge product and bivectors as an alternative.
@@eigenchris Yeah I have seen the series before (but not armed with the knowledge you conveyed so I was quite lost)
I should probably check it out again
I think the Octonion product is on one hand non-associative in the same way as the cross product, as it's the 7D cross product (plus a scalar), but on the other it's more fundamental as you can't just as easily "complete the space" to fix it?
But if I guess if her definition achieves associativity, there *is* a way around that issue after all
I’m curious if anyone knows, can you always “factor” any projection operator into two nilpotent operators? I understand the video provides proof that you can go from two nilpotent operators to projection operators, but I’m curious if you can always go the other way around.
I believe it's always possible if the projector is of the form (1+U)/2, and you allow yourself to use complex coefficients. You can re-write U as a product of 2 anti-commuting elements, and use them to build 2 nilpotents as I've shown in this video. I'm not sure if there are other projectors in CA that don't follow this form.
Absolutely loving this series. Is there a way to contact you for a relevant idea I had? Not sure why, but YT sometimes marks long comments as spam and hides them or deletes them.
Can you try just replying to this comment? If that doesn't work, I can give you another way to contact me.
Tried again, didn't work.
Real heavy stuff, but I love it. Once we've got our solid foundation, I hope we get to cover the fundamental question: why? Why do these spinors, which have this very particular algebra, roll out of the Klein-Gordon Equation? I know the simple explanation: you can factorize KG into the Dirac Equation by including Clifford elements as part of the solution, but why are we suddenly allowed to use these alien mathematic objects? Or if that question cannot be answered, I'd like to see us playing around with those things a little more, rotating them etc., getting a real feel for how they work.
I'll show more stuff about rotating and boosting spinors in the Lie Algebra section. But I don't really know "why" spinors exist in physics. As I say in this video, they are associated with Fermions, which obey the Pauli Exclusion Principle.
Amazing video! I often see the conversation phrased in terms of projectors or in terms of nilpotents, but I don't think I've ever seen a clear equivalence of these two ways of speaking. I 100% get the complex case for any signature, the problem I am having trouble with is the case of the real algebra. Is there a similar construction of spinors in arbitrary real Clifford algebras Cl(p,q,R) (you mentioned that there would be an appendix on that, will it come before the Lie algebra section or will you end the section on Clifford algebras with this construction?)
I don't have a specific time scheduled for it. I don't have the proofs figured out yet so it will most likely come in a couple months, after a few Lie Algebra videos. You can check out Lounesto's book, as I think it comes the closest to handling spinors in real clifford algebras out of all the sources I've looked at.
@@eigenchris Yeah, I've looked at a couple different texts and most of them skip over the real case in the spinor section, but I'll give this one a try, thank you! Also, any plans on projective/conformal algebras?
Probably not. The channels Sudgylacmoe and Bivector channels likely have videos on those topics.
@@eigenchris I wanted to bring the Vaz and Rocha book on Clifford Algebras and Spinors to your attention if you didn't already know about it. In Theorem 4.6 they show that the number of commuting projectors in Cl(p,q,R) is q-r_{q-p} where r_j is the 5th Radon-Hurwitz number. I found this odd because it means that the construction of spinor spaces is fundamentally different in the real and complex case.
In the complex case we have a constructive proof for what the commuting projectors are (in terms of nilpotent as you have here), and they are always of the form (1+bivector)/2.
But in the real case, the closest we can come is an existence proof with the number of commuting projectors that exist. In general there are no nilpotent, and in general the projectors do not look like (1+bivector)/2.
Do you have any thoughts about why the complex case has an algorithm for constructing projectors but the real case is much more complicated?
@@GiordanoGaudioThanks. I actually have a copy of that book but I find it extremely hard to understand. The quotes on the back praise it for being clear, but I find it very hard to get through. I'll give it another look.
I don't have any deep thoughts about the real vs complex case. I think the fact that complex coefficients allow you to turn a symbol that squares to -1 into a symbol that squares to +1 for free makes the constructions a lot easier and algorithmic. With real coefficients only, the counting of commuting projectors becomes a lot more complicated.
Your series on spinors is really great! I've learned a lot thanks to you! Please continue to make detailed videos on advanced topics like this one.
But on this video, I'm starting to get confused. The way you define your creation and annihilation operators seems odd for me . If you consider many particles states in principle there is no "full state" and you can always add more particles to a state so that alpha+ never gives zero when applied on many states particles. And within this framework of many-states, I'm not sure that alpha- is also nilpotent. Can you explain how the example you described of only two ladder operators which enable to transit between a kind of 2-level system is compatible with usual many-states quantum mechanics ?
Okay I've finished the video but I got your point somehow. It's just an analogy to relate to what we know from QM make nilpotent element of the clifford algebra can be useful to build minimal left ideal. But I'm still confused because it's looks like we are trying to build the usual construction of spinors (or fermion state)in QM but I'm having trouble to relate what you've shown us with many states QM and spinors and fermions
Often the creation/annihilation operator comed with an inout variable, usually position or momentum, so that you can create particles with specific positions or moments. I didn't bkther showing this in the video because it wasn't relevant.
Why are the raising and lowering operators the same equation?
typo. Raising operator should have a - sign.
Can you tell me what kind of software, what programs are available to perform the calculations you have outlined?
It's all microsoft powerpoint.
@eigenchris Thanks for the info.
But this is a program for creating various presentations.
I am looking for programs for modeling physical processes, modeling nuclear reactions, modeling at the level of elementary particles.
!!!!! Programs with which I can simulate on the computer the fission of uranium at the level of quarks, leptons, bosons.
@@ko-prometheus As far as I know, no one has managed to build a computer simulation of nuclear fission on the level of quarks so far. The interactions of quarks with each other are _very_ complicated, usual simulations don't include many (valence) quarks, lots of simulations even include no (valence) quarks at all, but only gluons and sea quarks (that's called the "quenched approximation"). For uranium, you would need to simulate about 700 valence quarks!
@bjornfeuerbacher5514
So we can say that the theory of interaction at the level of elementary particles quarks, gluons, etc. has not been developed yet. So all the models that exist do not give answers how interactions occur.And this means that without understanding how the process of nuclear chain reaction occurs at the level of elementary particles, we can not learn to control this process.We can not make a small nuclear reactor on, say, plutonium and control it on the basis of the laws of elementary particles quarks, gluons, etc. We will not be able to install such reactors in airplanes as a source of energy or in cars or in passenger sea liners.....Only by learning the process of nuclear chain reaction, controlled or uncontrolled at the level of fundamental particles, we will be able to control nuclear chain reaction very precisely!!!!!!.
I would like to learn how to do this. It would be good if someone else would keep me company, understanding elementary particle physics.Of course only computers can model such processes. But you have to know what you are modeling and how you are modeling.
Please make more joke videos, I really loved your momentum joke video 😂😂
youll just have to wait until april fools day
Well what about me though?
Initially, Cl(1,3) is presented without a third "parameter" 'C', but in the creation of the beta projectors, we start to have imaginary/complex coefficients. Is there a reason this is allowed?
Also, I recall when looking through the Wikipedia article that in complex Clifford Algebras, you can always form a basis where each basis vector squares to positive 1. If you start with a basis where some vectors square to positive 1, and some square to negative 1, you can do something similar to what you presented in the video and form a new basis where the vectors that squared to positive 1 remain the same, and the vectors that squared to negative 1 "gain" a factor of i, causing them to also square to 1. Since this is just scalar multiplication of vectors that had already formed a basis, the vectors should still form a basis.