The h Bar - 2023-12-17 (page 1 of 2)

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Obliv

02:10

so in general coherent beams are characterized by constant phase relationships

what does this mean for the envelope of the wave?

Like, does this imply that envelopes of coherent beams have a "smooth" shape?

nvm i get it, like for two beams to be coherent they need to have a constant phase difference

what it means to have a constant phase difference, I'm not entirely certain. If it varies by some position dependent function, then it's still coherent but perhaps not trivially?

"In the superposition of in-phase coherent beams, individual amplitudes add together, whereas in the superposition of incoherent beams, individual irradiances add together."

huh

2 hours later…

Obliv

03:54

so dead in here

@naturallyInconsistent are you out partying this fine saturday night

1 hour later…

naturallyInconsistent

05:10

@Obliv yeppuu miahahaha

1 hour later…

Jagerber48

06:16

Are the following statements true:
- SO(n) has countably infinitely many irreducible representations
- Spin(n) has only 1 or 2 irreducible representations

nickbros123

06:54

sad day- PDF Drive stopped working :'(

Jagerber48

07:27

No my last statement is wrong. en.wikipedia.org/wiki/…. Spin(3) = SU(2) has irreducible representations basically with j=0, 1/2, 1, 3/2, 2, ... as I would expect from Griffiths intro to quantum angular momentum.

Any irrep of SO(3) is also an irrep of Spin(3).

Mr. Feynman

07:45

@Jagerber48 $\mathrm{Spin}(3)$ is the universal cover of $\mathrm{SO}(3)$. For this reason they have the same (up to isomorphism) algebras and thus the same algebra representation. As you know irreps are labelled by "angular momenta" (highest weights), but since the universal cover is simply connected every algebra representation descends from $\mathrm{Spin}(3)$ representation,

while the same isn't true for $\mathrm{SO}(3)$, for which only even-dimensional representation descend from a group representation

Jagerber48

08:13

I'm trying to reconcile the "spinors as minimal left ideals of Clifford algebras" definition of spinors with the "spinors as representation spaces of Spin(n) groups" definition. I can see how minimal left ideals are related to irreducible representations. But the dimensionality of an ideal is < the dimensionality of the space. So for a Clifford algebra (e.g. Cl(3)) the ideal must have dimension < 2^n. But we discussed that there are representations of Spin(3) of arbitrarily large dimension.

Perhaps the "spinors as minimal left ideals of Clifford algebras" definition only captures the fundamental irreducible representations, (e.g. the complex 2D Pauli Spinors or 4D Dirac spinors) but does not capture the higher dimensional spinors, such as those corresponding to j=11/2?

Qmechanic

09:02

@Mr.Feynman : Unfortunately, no.

1 hour later…

ACuriousMind

10:27

@Jagerber48 Sometimes people mean "any representation of the rotation algebra that does not lift to a representation of the rotation group" by "spinor", sometimes they specifically mean a Dirac or Weyl spinor on which $\gamma$-matrices can act (and which therefore is a representation of the Clifford algebra)

there's nothing to reconcile, it's just two different meanings of the word

Ryder Rude

10:42

hello. starting from $|p\rangle \rightarrow |\Lambda p\rangle$, how do u derive the action of boosts on the wavefunctions in this basis?

i get : $U\psi = U(\int dp \psi(p) |p\rangle)= \int dp \psi (p) |\Lambda p\rangle = \int dp \psi (p) \sqrt{2\omega _p } \delta (\Lambda p - p')$

user587860

@ACuriousMind What are possible ways in which shifting boundary conditions of a string affect its mass spectrum?

user587860

By shifting the boundary conditions, I do not refer to the Lorentz transformation $X^\mu (\sigma,\tau) \to \Lambda^{\mu}_{\nu}X^\nu (\sigma,\tau)$, as Lorentzian symmetry is a symmetry of the theory $S[\gamma, X] = \frac{T}{2}\int \mathrm{d}^{2}\sigma{\sqrt{-\gamma}}\gamma^{ab}\partial _{a}X^{\mu}(\sigma)\partial_{b}X^{\nu }(\sigma)$. On the other hand, the mass spectrum will depend on the vibrational modes $\tilde{a}_{\mu}, \tilde{a}_{\mu}$. Which are determined by the boundary conditions (such as Dirichlet/Von-Neumann), i.e in Dirichlet condition it holds that $alpha_{\mu} = -\alpha_{\mu}…

Ryder Rude

10:59

physics.stackexchange.com/a/575928 according to this post, $\delta (Lambda p - p') =\frac{1}{\gamma} \delta (p-p')$

ACuriousMind

@Supersymmetry What difficulty do you have in figuring this out from the standard quantization of the open string? There's more or less explicit formulae for the masses in terms of the boundary conditions and the length of the string

Ryder Rude

so we get : $\int dp \psi (p) \frac{1}{\gamma} \sqrt{2\omega _p} \delta (p-p')=\int dp \frac{\psi (p)}{\gamma} |\Lambda p\rangle$

so $\psi (p)$ transforms to $\frac{1}{\gamma} \psi (\Lambda p)$?

user587860

@ACuriousMind Thank you for your response. Yes, I know how a boundary condition can affect its mass spectrum (obvious), and this has to invariant under Lorentz action but other local actions might not be leaving boundary conditions invariant. So I was trying to come up with examples of it

user587860

For example, if I am not mistaken, twist operator might affect the boundary conditions by lifting the requirement that a closed string returns to the point it started off at, creating a twisted sector

Ryder Rude

11:27

i think the derivation shud b : $U (\int dp \psi (p) |p\rangle) = \int dp \psi (p) |\Lambda p\rangle$. then we do a change of variables $p'=\Lambda ^{-1} p$. dp --> dp'/gamma. so we get $\int dp' \frac{1}{\gamma}( \psi (\Lambda ^{-1}p') |p'\rangle $

so $\psi (p)$ is not a scalar, but $\frac{1}{2\omega _p} \psi(p)$ is. is this correct

Ryder Rude

12:23

which one do u like among quantum, classical and stat mech

i think quantum>classical>stat

but GR is its own tier. it has CTCs so not like a milquetoast classical theory

Mr. Feynman

The question is kind of meaningless. Statistical mechanics can be done both in a classical or quantum context

Ryder Rude

yes

lets say quantum minus stat, classical minus stat, and everything stat

Mr. Feynman

I hate all of them

Ryder Rude

what do u love

i feel it's the principle of least action :P

Mr. Feynman

@RyderRude nothing

Ryder Rude

12:28

u will love nihilism :)

Mr. Feynman

Philosophy is not my cup of tea, I don't understand it :P

Ryder Rude

nor do i. i just do my own philosophy :P

Mr. Feynman

I would think so too, then I realized I might be saying naive or idiotic stuff, so I stopped

(not implying that you're doing so of course)

Ryder Rude

thank u for writing that bracket :)

Mr. Feynman

On the web one has to be even more careful not to be misunderstood :P

Ryder Rude

12:32

i feel im just doing practical philosophy. the philosophy i read is far too removed from anything practical

both in vocabulary and in application

u can be simple and practical instead of needlessly complicating

this philosophy is practical tho plato.stanford.edu/entries/structural-realism

Mr. Feynman

Even then, I don't like speaking when I'm not well aware of what I'm speaking and that goes for physics too. That's also why I don't try anymore answering questions on the site

Ryder Rude

i sometimes speak with half-understanding admittedly

but i try to avoid it

but i too only answer stuff i know

-2 is the worst ive got on my answers :P

i just gave my thoughts on CTCs

it was this answer physics.stackexchange.com/questions/191538/… No one actually commented why it's wrong

user587860

13:32

Does it also happen to you that the more you learn, the more anxious you are that you do not know anything?

Ryder Rude

13:52

@Supersymmetry i get more satisfied when i learn more

@Supersymmetry but i feel anxious when i encounter a new counter intuitive thing

user587860

14:14

@RyderRude Exactly

naturallyInconsistent

14:25

@Jagerber48 You will have a much better time considering a massive photon, where s=1, to compare with an electron, s=1/2, as opposed to orbital angular momentum, both of which can have any $\ell\in\mathbb Z^+_0$

naturallyInconsistent

14:38

@Obliv $$\text{when coherent}\ :\qquad\vec E=\vec E_1+\vec E_2\\\text{when incoherent}\ :\qquad|\vec E|^2=|\vec E_1|^2+|\vec E_2|^2$$which is sometimes written as $I=I_1+I_2$

Jagerber48

@ACuriousMind As I see. That is clarifying. I feel like you've probably said before, but how exactly are Dirac and Weyl spinors related to the infinite set of representations of Spin(n) in even and odd dimensions?

nickbros123

is it normal for people to learn Special Theory of Relativity first, perhaps after mechanics and then learn Electrodynamics, directly in language of STR, instead of going through the non relativistic version first (like it is done in griffiths)?

Jagerber48

My hypothesis: In odd dimension the fundamental representation of Spin(n) is a Dirac spinor (dimension 2^{(n-1)/2}) while in even dimension the fundamental representation is a Pauli spinor (dimension 2^{n/2}). Not sure what exactly is captured by the Clifford algebra minimal left ideal in each case. Is it always the fundamental representation whether that is Dirac or Pauli spinors?

ACuriousMind

@Jagerber48 not sure what exactly you mean by the "fundamental" in this case, but any representation of the Clifford algebra is also a representation of Spin(n)

Jagerber48

@ACuriousMind I don't know what I mean by "fundamental" either. I'm just regurgitating some stuff I read, including posts from you. What I've taken from some of your posts (I may be mistaken) is that, in physics, the fundamental representation is defined to mean the lowest dimension faithful irreducible representation of the group.

ACuriousMind

14:50

@Jagerber48 but "lowest-dimensional" is not unique, there can be different representations with the same dimension!

like there are for the Weyl spinors - the left and right handed Weyl spinors are distinct representations with the same dimension

Jagerber48

I see

naturallyInconsistent

@nickbros123 very unusual. A person should learn CM, STR, EM(NR), EM(SR). Jumping straight is kinda awkward.

Jagerber48

My question is, of the infinite set of irreps of Spin(n), which ones do you collect by looking at minimal left ideals of Cl(p, n-p)?

ACuriousMind

@Jagerber48 exactly one

and in general it's some kind of generalization of the "spin-1/2" reps in 3d and 4d

again: You get stuff that can star as the wavefunction/field in the Dirac equation

Jagerber48

Yeah, in 3D (Spin(3)) you get something like the Pauli spinors. Would you call those Dirac or Weyl spinors?

Ryder Rude

14:54

note that the one u collect is a faithful one, so it's not one of the reps of so(3)

ACuriousMind

@Jagerber48 Dirac

The Weyl spinors happen in even dimensions where the unique irrep of the Clifford algebra decomposes into two irreps of the spin group

Jagerber48

and in 4D spacetime (Spin(3, 1) or Spin(1, 3)), based on the Eigenchris videos, I think you get something like typicall 4D dirac spinors from the Dirac equation

ekardnam_

15:09

i was studying spinors in more detail last year's summer and i remember reading a lot of stuff from @ACuriousMind on this website

Mr. Feynman

IIRC reps of Clifford algebras are one of his favorite topics

ekardnam_

it looks like that ahah

Ryder Rude

@ekardnam_ r u learning physics as hobby

ACuriousMind

@Mr.Feynman I try to enjoy all algebra equally, but reps of Clifford algebras are one of these topics that just come up a lot in physics but are poorly explained by standard texts (like most representation theory)

Mr. Feynman

@ACuriousMind I mean, is there a physics book not failing with even Lie algebras reps? :P

ACuriousMind

15:19

no

ekardnam_

> @ekardnam_ r u learning physics as hobby

no

Mr. Feynman

Even worse with Clifford algebras, in my experience. At least for Lie algebras they give some context sloppy as they are

ekardnam_

im in a theoretical physics masters, done all exams just finishing to write my thesis

the main reason i was studying on my own these stuff it is because as they are saying it is not treated decently anywhere in standard books

Mr. Feynman

@ekardnam_ you can reply to messages directly instead of quoting manually

Like I just did

ekardnam_

@Mr.Feynman right

Ryder Rude

15:24

oh

Mr. Feynman

ah

Ryder Rude

are u interested in qft or gr?

ekardnam_

eh

@RyderRude both

mainly QFT though

non-perturbative

Mr. Feynman

RG stuff?

ekardnam_

integrability

Ryder Rude

15:25

nice

is ur masters related to qft

do u have any opinion on whether particles or fields r more fundamental

ekardnam_

@RyderRude yeah, but also to GR

Mr. Feynman

@RyderRude I think it's just fundamental interaction curriculum in Bologna

ekardnam_

mainly it is on the ODE/IM correspondence

Mr. Feynman

There is not much choice in Italy :P

Ryder Rude

i feel Weinberg says particles but many physicists i spoke to believe fields these days

ekardnam_

15:28

@Mr.Feynman no its not

it theoretical physics

its mixed on all physics

i did a lot of condensed matter too

Mr. Feynman

Oh ok sorry

@ekardnam_ unfortunately so did I :P

From my perspective 3-4 courses are a lot

Ryder Rude

r u also interested in quantum gravity

Jagerber48

@RyderRude You didn't ask but I think fundamentally it's all fields. Non-relativistic QM is like classical particle theory, QFT is like classical field theory. That's why it's called quantum field theory.

But, classically, we're used to vector fields. The electron field isn't a quantum vector field, it's apparently a quantum spinor field. That's why I want to understand spinors.

Ryder Rude

@Jagerber48 yeah. i hear this everywhere. but Weinberg takes another approach for some reason

but the more i learn, the less general Weinberg's approach is

ekardnam_

@RyderRude i am indeed

i actually like LQG quite a lot but nobody does that so i drifted away

Jagerber48

15:32

My grad E&M professor was pretty disparaging against Weinberg, really didn't like his approach I guess. I had one textbook by Weinberg and hated it. Haven't looked more into his stuff than that.

ekardnam_

AdS/CFT is also cool too

Ryder Rude

@Jagerber48 yes

Mr. Feynman

bolbteppa has joined the chat

ekardnam_

@Mr.Feynman LQG is vixra level too

Mr. Feynman

@ekardnam_ did you study these on your own or during your master's?

Oh my, what is Vixra? The anti Arxiv?

ekardnam_

15:34

@RyderRude its fields imho, for example think a CFT there are no particle states there (except for the free CFTs like the free massless boson)

ACuriousMind

@Jagerber48 and yet the only field from QFT which really makes classical sense is the EM field - the one field for which the particle picture is most questionable

ekardnam_

@Mr.Feynman apparently it is, bolbteppa was mentioning every two seconds the other night

Ryder Rude

@ekardnam_ also, the particle approach is very bad for curved spacertimes. fields provide a general approach to get Hamiltonians in any spacetime and any frame

Mr. Feynman

@ACuriousMind why the most questionable?

Ryder Rude

and then u can diagonalise that Hamiltonian to see if there r particles

ekardnam_

15:36

@Mr.Feynman the AdS/CFT correspondence is featured in my thesis so i did that for that reason. LQG I was reading on my own about two years ago

ACuriousMind

@Mr.Feynman localization issues, for one; it's pretty hard to say in what sense a photon is really "like" a particle except in the notion that it's a discrete "packet of energy"

@Mr.Feynman it's an "alternative" to arXiv started by Phil Gibbs after no one wanted to publish his non-mainstream views on conservation of energy in GR anymore

Mr. Feynman

@ACuriousMind wouldn't that also involve other massless gauge fields like gluons?

@ekardnam_ Oh, I see. I'm a bit late :P

@ACuriousMind god that sounds kinda childish :P

ACuriousMind

@Mr.Feynman well, gluons appear neither as particles nor as fields in what we observe due to confinement :P

ekardnam_

@Mr.Feynman uhm low energy QCD is very different than low energy QED

@ACuriousMind yeah this

Ryder Rude

y not fields

is confinement not modelled by qft

Mr. Feynman

15:41

Sure and that's why gluons look more questionable to me :P

ACuriousMind

@Mr.Feynman generally most physicists will assume that if your paper is on vixra it's crackpottery, and generally this is true

ekardnam_

@Mr.Feynman they make sense because of asymptotic freedom

naturallyInconsistent

oh, it is a party here tonight

Ryder Rude

wasnt Oppenheim's paper on vixra

Mr. Feynman

I can't take this Vixra seriously

ekardnam_

15:42

in fact it is believed that theories which arent asymptotically free or asymptotically conformal cant be quantized consistently

Mr. Feynman

It sounds like Alucard and Dracula

ekardnam_

@RyderRude it wasnt i think

Mr. Feynman

I assume some of you know Hellsing

Ryder Rude

oh. i thought i read vixra

Mr. Feynman

What do you think of studying physics perturbatively? :P

ekardnam_

15:44

@Mr.Feynman bad :P

but most importantly

not fun

Mr. Feynman

Just to be clear, I'm not talking about perturbation theory

Ryder Rude

perturbative series r asymptotic in qed

@Mr.Feynman it's the only way

Mr. Feynman

I mean perturbative as a way of learning: you read without caring much about details first (zeroth order), then you re-read that same material to expand a bit etc.

Ryder Rude

but it doesnt always converge

Mr. Feynman

Maybe it is more proficient that spending ages on a single line

@RyderRude it does because the cutoff is death

Ryder Rude

15:47

lol

ekardnam_

@Mr.Feynman then yes

this is the way

ACuriousMind

@Mr.Feynman I think for almost every technical text that is the correct way to approach it (in particular because it spares you a lot of time if it turns out the text doesn't actually show at the end what you wanted to read it for)

ekardnam_

my friends laughed because my strategy in studying is often starting from the last chapter and going backwards

ACuriousMind

reading the conclusion first is a perfectly reasonable approach

Ryder Rude

i feel the last chapters r like spoilers.

like when u accidentally read some spoiler line in a novel

similar for spoiler eqns

Mr. Feynman

15:49

@ACuriousMind and for physics I agree, but according to my experience math is highly non perturbative (at least in the way books are written :P)

Either that or math pages are much denser than physics pages

Ryder Rude

i spoiled the Dirac eqn for myself

Mr. Feynman

@RyderRude this attitude only got me to slow down without results, so now I want spoilers :D

Ryder Rude

but the build up to the Dirac eqn isnt very interesting anyway :P

just some bs about negative probabilities

ekardnam_

i hate reading maths because instead of typying and naming stuff they would say "definition 10.3 and theorem 3.5 together yield to lemma 16.3" now i have to go back how many pages to recall what that definition or theorem was

Ryder Rude

@Mr.Feynman yes. i also love spoilers in physics now

naturallyInconsistent

15:52

I think it is just that maths stuff love to define a lot of silly terms that are not obviously used a lot of the time, and so you have to memorise a lot of technical terms that are not helpful to understanding stuff, until you realise that the theorems that they are busy proving is actually a lot less confusing if they simply stated what the defined terms meant.

ACuriousMind

@Mr.Feynman I think mathematics has a lot more variation in style than the typical physics paper

Ryder Rude

it's maybe becuz math is a much broader field

in terms of range of topics

physics is more focused. like every paper gonna talk about experiments

ACuriousMind

so it's not quite as easy to develop a general approach to reading those papers as you need to get a feel for the style first: Is this one of these Bourbaki-style expositions where the first chapters will contain abstract definition that only make sense later? Or is this one of these "state the big result up front and then explain the proof bit by bit" pieces?

naturallyInconsistent

I'd also point out that there is a tendency for maths texts to like to have a very tiny book as the intro, so then they cover very concisely just the relations. However, then they like to have a counterexamples book as a separate text, and applications in yet another separate text, etc.

Mr. Feynman

@ekardnam_ that is something I like. My problem is that with math I feel almost forced to do things like the source does - except minor details - so knowing just the ideas is not enough. For physics I feel like I can just learn the ideas and do everything myself

naturallyInconsistent

15:57

They think that this is the best way to structure and write texts. I mean, sure, strictly speaking, in maths, the originating idea/history/applications isn't always the focus of a maths exposition, but that often makes it incredibly difficult to read, and they don't think that it is difficult to read stuff that way.

Mr. Feynman

@ACuriousMind at the moment I'm only reading math books, not papers

ACuriousMind

There's a lot of famous math papers that are not so abstractly structured at all

ekardnam_

@Mr.Feynman yeah i get you

naturallyInconsistent

Whereas in physics we tend to merge applications and history and counterexamples into an introductory text, so that the text gets bigger, but it is much more accessible to students that way

ACuriousMind

@Mr.Feynman well, textbooks are a different thing - it doesn't make that much sense to try to skim-read textbooks because the point of the book is to teach you how stuff is done, not necessarily to get you to the results the quickest

ekardnam_

15:58

@ACuriousMind there are also physicists that do all in their means to make simpler stuff the most abstract possible

conversely

ACuriousMind

sure, sure

naturallyInconsistent

@ekardnam_ and like, sometimes it is so... noooooooo...

Mr. Feynman

@ACuriousMind yes and it wouldn't work as you'd lose sight of previous definitions. Yet, learning from books seems to be an even longer challenge at this point

naturallyInconsistent

@Mr.Feynman Like, not every little thing needs a defined name, dammit!

Mr. Feynman

Doesn't it? :P

ekardnam_

16:02

@naturallyInconsistent i do prefer a name to, say, property 3.6

at least i can remember the name easier than the number

naturallyInconsistent

@ekardnam_ That is definitely the case. But then there are a million different names and like, just go away

ekardnam_

i'll go now guys and girls

bye

naturallyInconsistent

Like, if we keep using something, then it will definitely get a name. It is the silly little stuff that seriously stop putting names on them allllll

Mr. Feynman

I had this analysis Prof that used to number theorems on the blackboard too

naturallyInconsistent

@Mr.Feynman If they are numbered the same as the referenced text, then that is really good, tbh

Mr. Feynman

16:03

But she usually got confused and we had many "theorem 2" in the same lecture

naturallyInconsistent

oh no

Mr. Feynman

@naturallyInconsistent the counter reset each lecture

naturallyInconsistent

It should be Theorem (book author, number)

Mr. Feynman

@naturallyInconsistent but at least she never used the numbers to reference stuff :P

"As we proved in a previous theorem"

naturallyInconsistent

@Mr.Feynman Quantum Physics module ended with Legendre polynomials. Atomic & Molecular Physics module picked up with "As seen in Quantum Physics, the radial solutions of the Hydrogen atom goes as Laguerre polynomials..."

nickbros123

16:07

@Mr.Feynman im guilty of this when I take notes while reading books- references only pertain to equations used in the very same proof. but I also use a hardbound diary with dates atop, so its easy to just write " trivial to see, from result in page 11th feb, combined with theorem in page 3rd march"

naturallyInconsistent

@nickbros123 That's not bad.

ACuriousMind

@Mr.Feynman I feel this comes back to my favourite von Neumann quote that we don't learn things in mathematics, we just get used to them. Sure, the first time you see an unfamiliar definition, you have to go back to the definition every time it's used. But as you get used to it, there should be some internalizing going on where you understand what the definition does without having to formally spell it out every time

Mr. Feynman

@nickbros123 I do that too when I'm not writing latex

@naturallyInconsistent lol

naturallyInconsistent

@ACuriousMind But then this is precisely also the kind of evidence that tells us that naming should be an organic thing that rises from repeated exposure to the same thing over and over, until it becomes clear to the student's mind that hey, this needs to be memorised. It should not be stated up-front and expected to be remembered the entire time afterwards.

Mr. Feynman

@ACuriousMind now, that's a nice way to interpret the quote. I always thought of it as "Math is so difficult that even math people don't really understand it"

So did you just get used to rep theory or gauge theory? :P

ACuriousMind

16:09

@Mr.Feynman yes

a simple example is that when people start learning about representation theory, they tend to be very cross about people just saying "let's have a representation $V$" when the definition clearly says that a representation is a tuple of a vector space and a map, and they get lost by people not spelling out the maps

nickbros123

@naturallyInconsistent i think it is an organic thing more often then not: like for eg: defining contractive sequences- they literally contract if you add them!!! also, squeeze or sandwich theorem!! brilliant naming!

ACuriousMind

but once you've seen this often enough, you start doing this yourself because it's just quicker and everyone understands how it is supposed to work

Mr. Feynman

I've been through that phase :P

bolbteppa

@Jagerber48 the wikipedia explains this

In geometry and physics, spinors are elements of a complex number-based vector space that can be associated with Euclidean space. A spinor transforms linearly when the Euclidean space is subjected to a slight (infinitesimal) rotation, but unlike geometric vectors and tensors, a spinor transforms to its negative when the space rotates through 360° (see picture). It takes a rotation of 720° for a spinor to go back to its original state. This property characterizes spinors: spinors can be viewed as the "square roots" of vectors (although this is inaccurate and may be misleading; they are better...

naturallyInconsistent

@nickbros123 Of course; you remember them well because the names are amazingly good. And then you also remember them when they are incredibly bad, e.g. clopen sets.

Mr. Feynman

16:12

I've also had my "write explictly actions"-phase

$\theta(g,p)$ instead of $g\cdot p$

And with compositions it gets messed up :P

nickbros123

@naturallyInconsistent lol, no one should use clopen sets

just say open and closed

ACuriousMind

"clopen" is a great name and I'll die on this hill

Jagerber48

@bolbteppa I've spent tons of time on that Wikipedia page and even in that section. I can't figure out what $W$ or $W'$ is in that section so I can't understand it.

bolbteppa

$\omega$ there is the analog of a vacuum state $|0\rangle$, but we're creating this vacuum inside the Clifford algebra using elements from the Clifford algebra directly, as far as I can tell it's supposed to be a product of annihilation operators made from the gamma's. This way when you apply another annihilation operator to it it gets annihilated

ACuriousMind

just 10/10, no notes

Mr. Feynman

16:13

@nickbros123 it is quite more common than you'd imagine

bolbteppa

You then act on it with creation operators to build up states, that's what the end of the section is saying

Jagerber48

I like having lots of clear definitions for things in a "Bourbaki" style. I find that my confusions are typically resolved by better understanding a series of definitions of the things I am studying.

nickbros123

I live in R^n so the only open and closed subset of R^n is itself, I rarely use it tbh

Mr. Feynman

I once read a couple of comments on MO about "Frobenius" becoming a verb, an adjective etc

Frobenate

In the context of Frobenius theorem in DG

nickbros123

ive seen an argument like that but for "cantor"

for arguments that follow along cantors theorem lines

bolbteppa

16:16

In even dimensions, by acting with all creation operators on the vacuum, you are thus going to get $|0>, \gamma_{\mu}|0>,\gamma_{\mu_1 \mu_2}|0>,...$, which I explained gives a Dirac spinor, this is the minimal left ideal. This is the exact same thing I discussed in my spinor answer, a column vector expressed as a matrix with one non-zero column in the Clifford algebra of matrices. You can then project onto the even and odd subspaces getting Left and Right Weyl spinors

Mr. Feynman

There is a guy in my dep whose name became a verb with the meaning "nitpick"

naturallyInconsistent

@Jagerber48 I'd have to warn you that, oftentimes, in science, you adopt a tentative definition at the start, and realise that you have to keep changing definitions later.

bolbteppa

In matrix language, this is a Dirac spinor written as a left and right Weyl spinor in the column vector, and the Clifford algebra gamma matrices look like $\gamma_{\mu} = \begin{bmatrix} 0 & \sigma_{\mu} \\ \overline{\sigma}_{\mu} & 0 \end{bmatrix}$ (bar may be in the wrong place off the top of my head) corresonding to this left/right split

nickbros123

@naturallyInconsistent , do u have any other example of such unnecessary naming schemes

Jagerber48

@naturallyInconsistent if a theory (that models experiments) can be described mathematically then in the end you should be able to write it down with a consistent set of definitions. Maybe as you're learning and figuring things out you need to take tentative definitions and move things around as you get a better understanding.

naturallyInconsistent

16:19

@nickbros123 There are a whole lot. However, I am not anywhere as sure that they are useless as the standard comical few.

Jagerber48

But I find this strategy to be confusing and a waste of time. I'd rather learn the consistent final theory from the get-go, even if it's hard.

Mr. Feynman

Did you people get any book for Christmas?

naturallyInconsistent

@Jagerber48 That's insane, tbh. Consider Maxwell's theory of EM. First of all, final theory, in diff geo form? In GA form? In vector form? In potentials form? In covariant form? In gauge form? In QED? In GR?

bolbteppa

A spin 11/2 representation is a direct product of 11 spin 1/2 representations, this Clifford algebra procedure is studying the spin 1/2 representation, not direct products related to it

Jagerber48

@naturallyInconsistent Any form that is internally consistent

So vector form is fine at first as long as your clear about what the terms mean as you go

nickbros123

16:22

@Mr.Feynman purchased Greub- Linear Algebra , Rudin-PMA , Erwin Kreysig- Differential Geometry

Jagerber48

And you learn that there are equivalent (also self-consistent) other presentations.

naturallyInconsistent

@Jagerber48 The vector E and B fields form is internally consistent. It is the standard textbook choice. But then it is so inimical to any later development, that you essentially have to redo the entire subject in some other form later anyway

Jagerber48

@naturallyInconsistent Sorry, I don't see the point you're making here

All I'm saying is I prefer a fairly high level of mathematical rigor when learning theoretical physics topics.

Maxwell's equations can be presented with rigor in vector form or diff geo form.

naturallyInconsistent

@Jagerber48 If you consider E and B fields as $\vec E$ and $\vec B$, then yes, you can get the physics correct, but then it will be utterly disconnected from whatever is taught later on using gauge theory, or using differential forms, or whatnot.

ACuriousMind

@Jagerber48 in that case spinors should be your least concern in QFT :P

naturallyInconsistent

16:25

It is like, yes, you can do a lot of physics using Newtonian mechanics, but anybody who has learnt about Lagrangians and Hamiltonians will be able to tell you how little you actually know if you only know about Newtonian mechanics.

bolbteppa

As far as I can tell, $W$ is the subspace of creation operators, and $W'$ is the subspace of annihilation operators, so multiplying all products in $W'$ is just products of anti-symmetric annihilation operators which is the exterior algebra of these annihilation operators, and the product of all of them is then the $w$, the analog of a vacuum state annihilated by all annihilation operators

Jagerber48

@ACuriousMind So I hear. But if I want to understand the complications in QFT (I'd like to one day) I feel like I won't have a hope if I don't even have a good understanding of what spinors are.

ACuriousMind

the mathematical theory of spinors in terms of Clifford algebras is perfectly well-developed and even rather short, if you're willing to let go of this strange fixation of not using representation-theoretic formulations

naturallyInconsistent

@Jagerber48 No, you can do renormalisation on the basic spinless scalar field.

Jagerber48

@naturallyInconsistent This is orthogonal to my original point somehow. Let me rephrase what I was saying at first: "I find that my theoretical confusions are typically resolved by introducing more mathematical rigor". I know this isn't true for everyone.

@naturallyInconsistent Maybe I should try that

bolbteppa

16:27

If your clifford algebra is written in terms of creation and annihilation operators, and $\omega$ is a product of all annihilation operators, then, in even dimensions, $CL(V)\omega$ is all creation operators acting on the vacuum, which is exactly the Dirac spinor as I've explained multiple times, thus it reduces to two Weyl spinors, which the wiki also says

naturallyInconsistent

@Jagerber48 I am actually on board with this: I have been reminding many profs that, it is not the profs that need the rigour, it is that students have a better chance of understanding things if the presentation has some rigour to prevent the misunderstandings.

bolbteppa

All we're doing is trying to reproduce the idea of creation operators acting on a vacuum directly from the elements of the Clifford algebra themselves, i.e. yes ignoring representation theory is a mistake that will only confuse you as it is currently doing

naturallyInconsistent

Now I bring in another point, though. You might simultaneously be looking for the most general way to do subjects, and just teaching directly the final thing. e.g. teaching multi-dimensional differential forms right away instead of building it up from basic calculus. This, however, immediately loses stuff---the Fundamental Theorem of Calculus taught in 1D basic stuff has a wider range of applicability than the generalised Stokes's theorem that generalises FTC to n-D.

Jagerber48

@bolbteppa I feel like my questions have roughly been answered. It seems like the minimal left ideals in clifford algebras reproduce the Dirac spinors which are the lowest dimensional irreps of Spin(2n+1) and are like... almost the lowest dimension irreps of Spin(2n).

Minimal left ideals in Clifford algebras do NOT reproduce the infinite series of irreps of Spin(n).

naturallyInconsistent

Education is a very difficult topic. Pedagogy is a field of study in itself.

bolbteppa

16:33

Yes I think that's good enough

Jagerber48

Actually... would it be correct to say that Dirac spinors are defined to be the specific irrep of Spin(n) that you get from taking minimal left ideals of the Clifford algebra?

That is, Dirac spinors are specifically the irrep of Spin(n) that corresponds to the irrep of Cl(p, n-p)?

ACuriousMind

Yes

because the point of Dirac spinors is that you want to use them in the Dirac equation, where $\gamma$s act on them

Jagerber48

Ok cool

Is there an analogous story for vectors actually?

bolbteppa

and people only even thought of this minimal left ideal business by noticing when you put a spinor into a column vector of a matrix of the same size as the gammas, you get another spinor back of the same kind when you apply the gammas to it, as the history section of that wiki explains

Jagerber48

Like the vectors V can be thought of as a specific one of the infinitely many irreps of SO(n)?

bolbteppa

16:37

I explained the difference between the vector and spinor representations in my answer that was removed

ACuriousMind

@Jagerber48 I mean, this is the wrong way around: The reason SO(n) is interesting is because it's the group of rotations of vectors

and vectors are interesting because they're the natural notion of "direction" in space

bolbteppa

On a simple level, under commutation with the generators of the spin group i.e. infinitesimal rotations (represented via gamma's), the gamma's transform as vectors

ACuriousMind

but you already know the story there of defining vectors via directional derivatives

bolbteppa

That follows by studying 'relativistic invariance of the Dirac equation' which is where people usually first see it

@ekardnam_ Shame

30

A: Why is Standard Model + Loop Quantum Gravity usually not listed as a theory of everything

One can pinpoint the technical error in LQG explicitly: To recall, the starting point of LQG is to encode the Riemannian metric in terms of the parallel transport of the affine connection that it induces. This parallel transport is an assignment to each smooth curve in the manifold between point...

Jagerber48

@ACuriousMind right I know this is the wrong way around for vectors. But I think requiring representations for spinors is also the wrong way around! So im

wondering

bolbteppa

16:44

22

A: Can Loop Quantum Gravity connect in any way with string theory?

Dear Lawrence, an equivalence between LQG and string theory - or an LQG-like description of string theory physics - has surely been an attractive idea for many physicists (myself included) but it is impossible because of fundamental differences in virtually all general features and predictions of...

Jagerber48

if I can at least treat vectors and sponsors the same.

ACuriousMind

@Jagerber48 no you can't

as I keep saying, spinors are more specific than vectors - all manifolds have vectors, not all have spinors, and the obstruction is specifically the ability to choose the spinorial representations at every point consistently

bolbteppa

Again my answer explained why you can't treat vectors and tensors vs spinors the same, under $S_{ij} = i[\gamma_i,\gamma_j]/4$ the vector representation transforms as $|k> \to [S_{ij},\gamma_k]|0>$, while in the spinor rep we have $|k> \to S_{ij}|k>$ which can give $|ijk>$ (a rank 3 tensor) or an $|i>$ depending on the indices

ACuriousMind

and the whole underlying reason spinors appear in quantum theories but not classical theories is representation-theoretic, too: All the proper linear representations of SO(n) arise as tensor products of vectors, so you don't need anything except vectors to build all the classical objects, but in the quantum theory, we have to admit projective representations, which includes spinors

naturallyInconsistent

@Jagerber48 You can at most identify a spin-1 with spherical vectors. Not even the usual vectors, but vectors as seen in spherical harmonics way.

Jagerber48

16:49

I guess I’m thinking about accidentalfouriertransforms answer which seemed to imply vectors could also be defined as irreps

naturallyInconsistent

@Jagerber48 which I just told you...

bolbteppa

Vectors, tensors, and spinors are irreducible representations of $so(n)$, they give different 'fundamental representations'

Fundamental representation

In representation theory of Lie groups and Lie algebras, a fundamental representation is an irreducible finite-dimensional representation of a semisimple Lie group or Lie algebra whose highest weight is a fundamental weight. For example, the defining module of a classical Lie group is a fundamental representation. Any finite-dimensional irreducible representation of a semisimple Lie group or Lie algebra can be constructed from the fundamental representations by a procedure due to Élie Cartan. Thus in a certain sense, the fundamental representations are the elementary building blocks for arbitrary...

Jagerber48

Yeah sorry typing on my phone niw

bolbteppa

I think that's enough to mull over

naturallyInconsistent

Anyway, I think this is part of the sadness of life: Quite a lot of stuff have multiple incompatible generalisations, and so anybody seeking a unified single way to see them all, is doomed.

Jagerber48

16:54

Bolbteppa yeah I think there’s something about fundamental representations that I need to understand

ACuriousMind

careful: Physicists use the word differently from mathematicians :P

Jagerber48

@ACuriousMind I need to understand this also

ACuriousMind

to a physicist, "fundamental" usually just means the "obvious" or "defining" representations (such as vectors in $\mathbb{R}^n$ for SO(n))

to a mathematician, the fundamental representations (plural!) are the ones from which every other representation can be constructed via tensor products

the confusion arises probably from the defining representation usually being fundamental and e.g. for SO(3) there not being any other fundamental representation

Jagerber48

I mean I guess V is always a representation of SO(n), right?

ACuriousMind

what is V

Jagerber48

16:59

R^n

ACuriousMind

that's what I mean by the defining representation

Jagerber48

Right yeah

This helpful. I have to go now but thanks so much all for the conversation

user587860

@ACuriousMind Your assertion that "reps can be constructed from fundamental reps via tensor product" is only true when G is semi-simple and finite-dimensional Lie group

ACuriousMind

@Supersymmetry Sure, but Lie groups are always finite-dimensional in the standard definition and, like, all of Lie theory usually assumes semi-simplicity

I'm not sure what you think this nitpick adds

user587860

@ACuriousMind No, that's surely a good and precise description but I wanted to add that defining fundamental reps that way cannot be general enough, as there are subtleties (that you know a lot better) in infinite-dimensional cases and non-semisimplicity

ACuriousMind

17:10

see, this is one of these cases where I don't think hitting someone just encountering the topic for the first time with some fully general definition is useful - why would I care about fundamental representations in the cases where this statement about constructions doesn't hold?

sure, the definition of fundamental in terms of special weights generalizes, but there is not a lot of use you can get out of that definition in those more general cases

user587860

You're right, I apologize. But your answer, as usual, contains lots of deep insights that are all textbook-worthy! :)

Sanjana

17:57

Why does the Fourier series of $|\sin(x)|$ contain only even multiples of the argument in $\cos(nx)$? I am not asking why it should not contain any sine...I am asking why does the sum on $n$ appear for only even numbers. Sure, I can check it via explicit calculation, but can I infer that from some symmetry or something?

For $|x|$ the sum on $n$ is over odd numbers...same for a square wave.

ACuriousMind

@Sanjana Because if $\sin(x)$ is periodic inside some interval, then $\lvert \sin(x)\rvert$ is period inside half of that interval

Sanjana

Okay that explains why one can get rid of half of them, but which half---I mean why keep the even and not the odd like in $|x|$?
And this also happens for half sine wave i.e. $f(x)=0 \text{ for } -\pi<x \le 0 \text{ and } \sin(x) \text{for} 0 \le x \le \pi$

This half sine wave do not have that periodicity property the $|\sin(x)|$ i.e. full sine wave has

ACuriousMind

@Sanjana Think about it the other way around: If you have a function $f(x)$ periodic on $[0,L]$, then it is also periodic on $[0,2L]$, and also has a Fourier series there, and both series have to be the same.

Sanjana

18:13

Ok, thinking done...then?

ACuriousMind

@Sanjana It's the same reason as for $\lvert \sin(x)\rvert$.

@Sanjana if you had thought of what I wanted you to think of, you'd now understand why it's the even terms :P

write down the Fourier series for both $[0,L]$ and $[0,2L]$ and look at them until you see why only the even terms occur in the one for $[0,2L]$

naturallyInconsistent

I would just note that cosine is odd around $\pi/2$ whereas sine is even around there, and that extends to all odd multiples. This guarantees that the surviving cosines are all even.

Sanjana

So what's the general trick? Like whenever I see some odd function I can say $a_n=0$, is there some general rule like this which people has/I can formulate for these kind of situations?

naturallyInconsistent

I think ACM was trying to assert a particular general trick, but I am not fully understanding that either

ACuriousMind

18:32

1. Fourier series are just expanding the space of functions $L^2([0,L])$ in the basis $f_{L,n}(x) = \exp(\frac{2\pi}{L}nx)$. Extending this series from $[0,L]$ to $\mathbb{R}$ produces a function with period $L$. We have a relation between bases for multiples of $L$: $f_{L,n} = f_{2L,2n}$, i.e. the even basis functions on $[0,2L]$ are the basis on $[0,L]$.

By the general nature of how bases work, if the function is $f(x) = \sum_n a_n f_{L,n}$ on $[0,L]$, then in terms of $f(x) = \sum_n b_n f_{2L,n}$, using orthogonality of the basis and that it's still the same function yields directly $b_n = 0$ for $n$ odd and $b_{2n} = a_n$.

naturallyInconsistent

Yes, but remember that she was also telling you that the half-sine, which is not reducible to a smaller period, is also having a even-only cosine series.

Sanjana

Yeah same goes for $|x|$, square wave, etc. ...

I can understand nIS's point but can't generalise enough.

naturallyInconsistent

With |x| the series is odd-only. That falls out of what ACM is saying

As for square wave, I have to point out that you kinda have to move the square wave to a place where the symmetry is obvious, and then note that horizontal translation will only change the phases.

Sanjana

18:56

@naturallyInconsistent For rectangular wave I managed to find a generic trick--- For functions with $f(t)=-f(t-T/2)$ the even $b_n$s and $a_n$s become zero. But this doesn't apply to the cases I mentioned...so I guess, there's something else which maybe ACM is trying to say, but I don't see

Also the above condition on the function is easy to visualize, just shift the upper half cycle by half period to the left/right and reflect along the horizontal axis, if it matches with the lower half cycle then the function satisfies the above condition.

I am wanting some rule like that which tells me immediately about the $|x|, | \sin(x)|$, the half-sine-wave series via symmetry arguments...

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