The h Bar

Davyz2

09:00

ahahahahahahah

ACuriousMind

@NairitSahoo @naturallyInconsistent please, both of you, be a little bit more charitable towards each other

Davyz2

ACM literally being Elohim from the Talos Principle

naturallyInconsistent

@Davyz2 The opposite is the usual situation. The professors tend to know how little they actually know about QFT, and while they are still sloppy in the presentation, they know they cannot deviate from the known path. They wont be openly being reckless

Davyz2

the "known path" decided by the pope of science I guess

naturallyInconsistent

@NairitSahoo in so doing, you insist that everybody else waste their time?

@Davyz2 actually, this is usually the path that they themselves learnt the stuff. We do not need to resort to a conspiracy for this. It is pretty reasonable, if you think about this

Davyz2

09:03

Ah I understand, so instead of changing the path we should just carry the leftover path from the previous teachers, sounds a bit lazy to me

Sanjana

@naturallyInconsistent I just realised that this is that "Raman please" guy!

Comparing him to his usual standards, he is being quite reasonable nowadays :p

naturallyInconsistent

@Davyz2 I have always asserted to students that the laziest person in the room always ought to be the professor...

Davyz2

Yes as long as the lazy person is also smart

naturallyInconsistent

Are professors in physics not smart?

Davyz2

You would be surprised

naturallyInconsistent

09:06

@Sanjana omg lol ok

@Davyz2 I dont automatically respect people; the vast majority of the physics profs that I have seen, command the respect naturally from conversations. Notable exceptions obviously exist, but they are rare from my PoV

Davyz2

I mean, to be totally honest, my college background was a disaster and my teachers were horrible, so I have a confirmation bias going on

I guess on average teachers are alright

it's just that you can get unlucky

naturallyInconsistent

@Davyz2 ... I'd not assume that either. I mean, the average teacher is crushed by bureaucracy and finances. The average prof doesnt really know how to teach, not least since their main job is to research

but good instructors are available. You just learn to cherish them when you find them.

Nobody owes anybody an education

Davyz2

09:25

Nobody owes anybody a free education, you mean

if college was free, i would have zero problems

naturallyInconsistent

@Davyz2 didnt you do college? When ever did profs actually think that they owe you an education?

Davyz2

they never think that, but people do pay their salary, so the least they could do is not waste money

and provide a good education, not some copy-pasted rethoric translated from books which are free online anyway

what is the point of going to the professor for QED, when he just repeats QED from Peskin?

I already have QED from Peskin at home

naturallyInconsistent

@Davyz2 but it works. Oftentimes a crappy professor regurgitating from textbook is sufficient to magically diffuse some arcane knowledge into the minds of resistant youngsters. Like, I already hate the guy with a vengeance, yet I cannot take away that achievement

@Davyz2 nah, it does not have to be good in any way shape or form. The bar is extremely low.

Davyz2

I guess I was just naive before going to college

Sanjana

I have seen people complaining about their profs blindly following Peskin in QFT courses. Why don't people like it? What would people expect to learn about an age old subject which is not in the book? Do people want to learn new stuff or what? I mean... what can the profs do? Present research material or something? what do u have in mind?

Davyz2

09:34

Peskin is free -> The professor is not -> If professor = Peskin I have a problem

Sanjana

@Davyz2 But if the Prof. is not actually producing new results, then he must be copying it from some resource (if not Peskin), and everything is free... right?

Davyz2

Then what entitles him to be a professor?

I can study Peskin back to front, and I can hold a QFT course, right?

naturallyInconsistent

@Sanjana Pesky and Shredder is impossible to understand at the front. The renormalisation bits are remarkably readable. But their experiment to quickly generate students able to compute stuff is a failure.

Davyz2

a professor should teach you how to think and give unique insights and research problems and discuss these problems, that are difficult to find somewhere else

Sanjana

@Davyz2 Don't ask me :p I am just asking for your opinion of what an ideal prof. should be

@Davyz2 ah.

Davyz2

09:37

instead they pretend physics is easy business, and give you a QFT book to follow

Sanjana

Btw I was playing the devil's advocate here tbh. I myself hate one particular course on the internet where the prof. just copies from P&S

But then I thought. Oh it's QED. What new thing they can say :)

@naturallyInconsistent I never asked you this: Do you have a favourite QFT book?

Davyz2

You asked the forbidden question.

You shall be de-allocated from the Universe.

QFT books are nightmare-fuel T_T

naturallyInconsistent

@Sanjana Yes, I do. I am the kind of person whereby if a presentation does not give a reasoning for the route that they are taking, then miao miao will deviate and get thrown off. The wonderful book by Anthony Duncan, explains the basic ingredients extremely well.

Arjun

@naturallyInconsistent pesky and shredder lmao..why do have beef with Schroeder XD?(Tbh I hated his thermal physics book,it was way too hand wavy)

naturallyInconsistent

@Arjun For a long while I didnt know that Shredder is the same guy who wrote the thermal textbook. He is one guy that should just stop writing.

Davyz2

09:43

Peskin be like "alright we are going to sum over the bubbles and pretend it's a number"

naturallyInconsistent

By that I meant that miao miao came to the conclusion that he is no good as an author independently from those two books.

Davyz2

e^{sum_over_drawings}

naturallyInconsistent

@Sanjana Actually, there are quite many reinventions of subjects. For example, technically, the entire set of non-relativistic mechanics is one subject. That means everything from Newtonian mechanics, to Lagrangian and Hamiltonian, to symplectic manifolds, are all just one subject. Needless to say, the treatments are sooooo different.

Similarly, QM by Dirac and QM by Feynman-Hibbs basically dont intersect

Davyz2

Arnold for the win

Arjun

I totally agree with miao..shredder should prolly stop writing books

naturallyInconsistent

09:45

There is no good reason why QED should be one single treatment.

Sanjana

@naturallyInconsistent Hmm. This name is cropping up quite a few times...

naturallyInconsistent

@Davyz2 My QFT prof said that there was once that he used a book that had a mistake on a very important part. Another time he defined something backwards from what it should be. And then basically nothing works. In fact, he said that most QFT texts, if they give a "derivation" of something, only the first line and the last line are trustworthy. The middle steps can easily get something wrong.

And to that, he recommended Sterman because the mistakes are kept at a minimum. He used Srednicki for the teaching, though.

ACuriousMind

@Davyz2 Again, this is just the issue of renormalization, not a unique failing of any particular presentation

naturallyInconsistent

@Sanjana It is extremely different, and pretty wonderful. Then one day I found that a semi-student of mine actually learnt something else from him. Absolutely made my day

Davyz2

i know i know, it's just funny to me how they really write e^{bubbles}

I thought in elementary school they taught not to sum pears and apples, and then in QFT this makes a comeback

naturallyInconsistent

09:50

@Davyz2 Did you know that this is actually a diss? And it is from Heaviside. He had many things to diss

Sanjana

There is one note (book?) on QFT: the one by Siegel. It is very toxic. You want to get rid of it because it is so handwavy at parts, but then you realise that there are so many things in QFT that the usual books don't even bother to talk about.

Davyz2

I could write a dissing anthology on QFT

Sanjana

It is toxic because you love it and hate it at the same time: a perfect textbook for quantum field theory.

Davyz2

perfection

naturallyInconsistent

lol

@Sanjana "the usual books don't even bother to talk about" then you really need to check out Anthony Duncan one day. Like, I originally had it in PDF form, and then I decided I needed it in hardcover.

Davyz2

09:55

I think i'm going to check out this Duncan book

naturallyInconsistent

yay

Even in the very first chapter, i.e. introductory bits, Duncan talked about how Einstein, Dirac, Heisenberg, Jordan, Pauli, and friends, wrestled with the no-sources QED, i.e. just the empty cavity filled only with photons, and derived some amazing results. The pioneers were just so ridiculously smart.

Davyz2

The golden age indeed

Face reveal of ACM:

Sanjana

11:27

@naturallyInconsistent I went offline for reading that, right after your text. I feel you :3

naturallyInconsistent

11:53

yay~~

Nairit Sahoo

12:53

@ACuriousMind Am I right about the following? The Noetherian sense of transformations you talked about (which acts only on the fields) is what people call "active" diffeos and the coordinate transformations which acts upon the coordinates is what people call "passive" transformations. I can see now that they are different. But sometimes people say that they are the same. What do they mean?

@ACuriousMind c.f. Will's comment on your answer here.

ACuriousMind

@NairitSahoo I don't like the active/passive language and no, this distinction is not the same. Mathematically, a coordinate change is something conceptually completely different from what I've called "transformations in the sense of Noether's theorem", in that you can explain the latter fully without coordinates (cf. my first answer I linked above), but obviously you can't explain what a coordinate change is without coordinates.

Nairit Sahoo

@ACuriousMind But why? I thought, Active transformations are just like your "transformations in the sense of Noether's theorem". I thought, it is not a coordinate transformation. In active transformations you don't change the coordinates, but the object itself.

ACuriousMind

I'm not sure what you mean by "why". Transformations and coordinate changes are distinct operations. But people say active and passive transformations are the same. So obviously "active" and "passive" can't mean the same as "transformation" and "coordinate change" in our context.

And again, I don't like the active/passive terminology to begin with - it's silly and I've never seen any actual mathematicians talk that way about transformations or coordinate changes. I cannot explain its details and will not defend it.

Nairit Sahoo

13:09

@ACuriousMind You said "So obviously "active" and "passive" can't mean the same as "transformation" and "coordinate change" in our context." Can't "active" transformation mean transformations in the sense of Noether's theorem and "passive" transformations mean coordinate transformations?

@ACuriousMind Okay

Ryder Rude

I have a post about this

This physics.stackexchange.com/questions/759863/…

ACuriousMind

@NairitSahoo Huh? The argument I'm making here is simply transitive logic: We have "active" = "passive" (where '=' means equivalent in some sense) but "transformation" != "coordinate change". So we cannot have "active" = "transformation" and "passive" = "coordinate change", as this would imply "transformation" = "coordinate change", resulting in a contradiction.

Ryder Rude

I call the transforms of Noether's theorem as "active", but its not necessary ofc

u can call it whatever u want

The important thing is that Noether's theorem considers a family of trajectories, so it's not changing the co ordinates

ACuriousMind

@RyderRude ...note that I have already linked my own answer to that question earlier, and that is the answer accepted by the OP and most upvoted.

I am not sure what you think you add by always linking your own posts even when they're the worst answers on a thread by all objective measures

I don't link my own answers when they're not the best answers

Ryder Rude

@ACuriousMind i dont think my answer is the worst. It's the shortest and written in a non abstract language. Ur answer is more abstract. They serve different purposes

Answers don't have to be best or worst. they can say the same thing differently

Nairit Sahoo

13:18

@ACuriousMind Yeah I understood that without this cute set of steps too :p What I am saying is that can't transformations be of two types: one acting on coordinates (passive ones), another on fields (active ones)??

ACuriousMind

@RyderRude Your answer is at +4, mine is at +24 and accepted. You don't even define what you mean by the nebulous words "active" or "passive", let alone demonstrate that this is really the problem. The difference is not the level of abstraction, the difference is that I actually give a mathematically rigorous answer to a question that asked for mathematical rigour.

@NairitSahoo I do not understand the question.

Nairit Sahoo

@ACuriousMind ok. no problem

Ryder Rude

but i have explained the meaning of "active transforms" here : Let's say you have an action that is a fixed function of the field trajectory S[ϕ(x,t)]
. Noether's theorem looks for a family of field trajectories indexed by a parameter α
, ϕα(x,t)
, such that the action is invariant for the family. Now, note that not just any family of field trajectories will have an invariant action.

But again, it doesnt matter if we choose to name the above as "active", but many people do. It is standard terminology

ACuriousMind

@RyderRude That's not a definition, and a passive view of a transformation parametrized by some $\alpha$ would likewise produce a family of expressions $\phi_\alpha$.

Nairit Sahoo

@ACuriousMind Just to be clear: when you say proper symmetries, in the context of this statement, gauge symmetries fall within the class of "proper" symmetries, right?

ACuriousMind

13:24

@NairitSahoo Yes - I mean symmetries in the sense of Noether's theorem, both first and second.

@RyderRude If this is indeed "standard terminology", you should have no difficulty producing a reference that agrees with you.

Ryder Rude

@ACuriousMind that is correct. but note that my family are all in the $(x,t)$ co ordinate system, otherwise i would have to use some notation to describe the change of co ordinates. I also state that the action function is fixed here, which means no change of co ordinates

I think it is clear that i am considering a family after fixing the co ordinate system

But ur answer is more accurate maybe because it completely gets rid of co ordinatea

ACuriousMind

if indeed "Noether's theorem is only about active transformations" in "standard terminology", it should be easy to find this stated in the literature

Ryder Rude

I can indeed find a physics reference defiining active transforms. The old saying that "active = passive" comes from geometry discussions where they talk about moving the point vs moving the axes

The geometry discussion indeed confuses these terms. but they r clear in physics

lemme find the reference

ACuriousMind

@RyderRude I'm not asking for a reference for the definition of "active", I'm asking for a reference that agrees with your specific statement that Noether's theorem is about active transformations. As a preliminary data point, the word "active" occurs 0 times in the Wikipedia article for Noether's theorem. Why do you think that is?

Ryder Rude

@ACuriousMind note that i only said that the active terminology was standard. My statement is something i made while writing that answer. But when I googled noether theorem and active transform, i got many hits: physics.stackexchange.com/questions/158398/…

@ACuriousMind also see this answer by QMechanic math.stackexchange.com/a/4612850

Nairit Sahoo

13:40

@ACuriousMind One thing. Particle mechanics can be considered as field theory where the role of fields is taken by coordinates. So true Noetherian symmetries are transformations of fields (and not coordinate changes) which keep the action invariant, would seem to imply that in particle mechanics Noetherian symmetries do act on coordinates because fields are just coordinates here.

Ryder Rude

Now, it is possible that wiki never used this term.. but I'm just saying that it's a valid use of the terminology of "active transforms" @ACuriousMind

Nairit Sahoo

Wouldn't that imply that point particle mechanics is invariant under all diffeos. in the sense of Noether?

ACuriousMind

@RyderRude ...so you claim that this is "standard terminology" and then all you can do to prove this is coming up with other SE answers instead of referring to, well, standard literature? What do you think "standard terminology" means?

When someone claims that something is standard, I expect them to be able to back this up with more than trying to google references in the moment

@NairitSahoo No, that's not how "particle mechanics as field mechanics" works. The "coordinates" in the sense of field theory in the case of particle dynamics is just time -that's why we call it 0+1 dimensional field theory.

Ryder Rude

I only said that the active transform terminology was standard...i can give a reference for that...about my phrasing of Noether's theorem, i got only stackexchange hits...but i will try to find something

I didn't say that my phrasing of Noether's theorem was popular. But only that it is correct

U can see qmechanic using the exact same terminology, i think it is at least a correct phrasing...idk if it's popular.

I will have to check

ACuriousMind

The "coordinate changes" are just new choices of the time coordinate, and in fact you can already see the same kind of distinction in this case: Of course the integral $\int S(t) \mathrm{d}t$ is invariant under coordinate substitutions $t'(t)$ like any integral, but only a special class of theories is "invariant under time reparametrizations" such that they are a symmetry in the sense of Noether.

It's a well-known (in the relevant literature, intro courses tend to not discuss this) result that the Hamiltonians of such theories are generically zero; these are toy models for the diffeomorphism invariance of GR.

Ryder Rude

13:50

I can't find much...i think all the sources find the "active" terminology redundant in the phrasing of Noether's theorem, because they already specify a family of trajectories... So it's already precise @ACuriousMind

ACuriousMind

It's a bit hard to convey these differences in the abstract, it is a nice exercise to look at one reparametrization-invariant system and one that isn't and figure out exactly how they differ, even while the overall integral is still "invariant" under the substitution

Ryder Rude

i think it is just redundant there. If u specify a family, u don't have to say active

My answer says both active and proceeds to give a family. active is correct there but redundant

Slereah

14:12

taps the sign

Nairit Sahoo

15:19

@ACuriousMind Yeah I know that. I wasn't saying that

@ACuriousMind I know. I have done for the relativistic free particle

I was saying something more trivial I believe... Since Noetherian symmetries act on fields, and since the $x$ are the fields here, which we usually call coordinates in a more elementary level course. So in this special case of point particle mechanics, Noetherian symmetries also act on coordinates...

@RyderRude Carroll's book tries to explain diffeomorphism invariance via active vs passive. But yeah I don't think explicitly talks about any direct relation to Noetherian symmetries which I would love to "speculate", but that would only be a speculation atmost (for me atleast as of now...)

On the other hand, ACM's explanation is clean. I also liked Prahar's answer.

@RyderRude You could say in some sense I was trying to bridge ACM & Prahar's answer to yours.

@Slereah neat

@Slereah Where is this from?

Slereah

@NairitSahoo arxiv.org/abs/gr-qc/0603087

Sanjana

@Slereah Minor nitprick on a point which is not probably relevant for this discussion: sourcefree Maxwell equations are not invariant under conformal transformations in generic dimensions. It is merely scale invariant (apart from being obviously Poincare invariant)

ACuriousMind

15:35

@NairitSahoo yes, but only if you use a colloquial and not the technical meaning of "coordinates" :P

Nairit Sahoo

@Sanjana What? But I derived that it is true in our EM class.

Slereah

@Sanjana Also called the conformal invariance :p

Or Weyl group, if you're being fancy I guess

"Conformal group" is one of those ambiguous word

ACuriousMind

@NairitSahoo No, you derived scale (Weyl) invariance. People often confuse the two.

Sanjana

@NairitSahoo Can't be. Scale invariance vs. conformal invariance is a big thing. But maybe as Slereah is suggesting it is a fight over terminology. Anyway the point is that Maxwell equations in generic dimensions are not invariant under SCTs.

Yeah... that

Or maybe... wait you are in UG, right? Maybe you are in 4D that's why it just works.

In 4D it is invariant under the full conformal group iirc.

Nairit Sahoo

Yes we derived it for 4D only!

ACuriousMind

15:37

see physics.stackexchange.com/q/38138/50583 and its linked questions

Nairit Sahoo

@ACuriousMind Yeah I know the difference wordwise, but I didn't pay much attention to this in the context of source free EM.

@ACuriousMind thanks

Slereah

Wasn't there some famous big historical fight over this

And one guy said FIFTEEN GENERATORS

I vaguely remember the anecdote

Nairit Sahoo

@Sanjana Does this work only for 4D or some other dimension also?

Sanjana

@NairitSahoo You have full conformal invariance only in 4D. I mean you can just transform the action according to conformal transformations acting on fields and then you precisely see that $\delta S \propto (d-4)$.

This I read from Zee's GR book

Nowadays scale invariant but not conformally invariant theories are plenty. The prototypical example if you allow NR field theories is elasticity

As Slereah is saying there was some big fights. Polchinski helped solve the problem to some extent in 2D. Roughly Scaling symmetry+Unitarity $\implies$ conformal invariance in 2D

Slereah

Found it

Sanjana

15:48

I also forgot. Was it Coleman?

Slereah

> It happened so, that there was an International Conference in Dubna, the main topic of which was so-called scale symmetry, promoted with great fanfare by the Bogoliubov School. This scale symmetry was mostly a political slogan, good for dissertations and career moves but not for any practical applications in a world of Physics.
After the Plenary Session devoted to the Scale Symmetry, one of the Western Physicists asked the speaker: “What is the difference between Scale Symmetry and Conformal Symmetry?” Apparently, the rumors about new symmetry were already spread, so this was what KGB use…

FIFTEEN PARAMETERS

naturallyInconsistent

is there a chloroform symmetry?

Sanjana

Oh it was Migdal...

@naturallyInconsistent What would you use that for :p

naturallyInconsistent

miao gonna keep that a secret miehehe

Slereah

Slightly confusing because in math, the scale symmetry is called the conformal orthogonal group

Sanjana

15:58

You are the history guy here. Do you know who used "conformal" first?

I mean atleast which community?

Slereah

Not sure

I know the original paper on the topic was in the 1850's

Let's see what it was

didn't seem to use the word in the 1850's

the first guy to bring the conformal group to physics (outside of the observation that Maxwell equations were invariant under it from some guy) was Weyl IIRC*

Thomas used the word in 1926 tho

In "On conformal differential geometry"

Nairit Sahoo

16:15

Are there any non-trivial diffeomorphism invariant theories where by invariance I mean true Noetherian symmetries. So that for each of these diffeomorphisms we have conserved charges?

Slereah

Being invariant under every diffeomorphisms is a pretty tall order

I guess the Lagrangian $L = 1$ would be diffeomorphism invariant in that sense

Maybe some topological theories idk

Sanjana

Apart from topological theories, one trivial example I can think of is $CFT_1$. All transformations in 1D are conformal.

Ryder Rude

@NairitSahoo u don't need to bridge :) my answer doesn't say anything different...

@NairitSahoo the important point is just that Noether's theorem considers a family of trajectories while keeping the co ordinate system unchanged. whether or not u call this an active transform is irrelevant

Although it is an active transform in the standard definition, as it amounts to changing the object (the field)

Nairit Sahoo

@Sanjana No I want a theory which is invariant under all diffeomorphisms, and each of these should give me conserved charges according to Noether's theorem

Ryder Rude

Also, $active \neq passive$ in this context, as the functional form of the Lagrangian is fixed in active, but changeable in passive

Slereah

16:27

I should really read up on topological QFT

Nairit Sahoo

I am sure it is overly restrictive. But, then is it completely trivial or there is atleast some little amount of non-triviality in these...

Ryder Rude

The idea that active = passive comes from geometry discussions where they just have axes and a point, and there's no equivalent of a lagrangian

Sanjana

@NairitSahoo Related: but not exactly what you are looking for. There are these gauge transformations which are not connected to the identity or equivalently don't die off at infinity. So when you define the charge from current you can't kill surface terms. These gauge transformations although "gauge", they act non-trivially at infinity and give you non-zero charges

Now... the surprising fact: these are symmetries of well known systems like Maxwell ED, Einstein gravity, etc. They were just discovered recently (post 2013 atleast iirc). These are called asymptotic symmetries.

You might have heard of soft theorems (very handwavingly scattering amplitude with no photon is proportional to scattering amplitude with low energy photon)... These can be rewritten as Ward identities (conservation laws). Which symmetries do they correspond to??

Ryder Rude

@Sanjana really interesting

Slereah

arxiv.org/pdf/1407.6364

^this seems to have info on the topic

Sanjana

16:31

guess what... asymptotic symmetries!

Nairit Sahoo

@Sanjana so?

Sanjana

Oh sorry I got carried away, and missed what you were asking...

Okay so these symmetries are infinite in number and are local diffeomorphisms in the sense that they act on each point on the "celestial" spheres at infinity (recall Penrose diagram for flat space has suppressed spheres...), and the action can be independently specified for each point on the sphere!

So there you go infinite diffeomorphisms giving infinitely many conserved charges.

Also see that paper by Slereah. I don't specifically know what's that about but they also use the same formalism (Covariant Phase space formalism), to derive the charges.

It is really sleek.

Nairit Sahoo

That's a lot of nice information

@Slereah This looks nice too

@RyderRude Hmm I understand

@ACuriousMind Just pinging you in case you also know of some crazy examples where the theory is truly diffeomorphism invariant.

ACuriousMind

@NairitSahoo they won't - the diffeomorphisms are spacetime-dependent symmetries, hence gauge, so it is Noether's second theorem that applies, not the first - you do not get conserved charges from that

look at GR - GR has this "diffeomorphism-invariance", but it has no corresponding conserved charges, the conserved quantities in GR come specifically from Killing vectors/isometries

Nairit Sahoo

@ACuriousMind Yeah that I understand. But I am asking for theories with global diffeomorphism invariance

Is there anything like that?

ACuriousMind

16:42

that doesn't mean anything

what on earth is a "global diffeomorphism" supposed to be?

Nairit Sahoo

@ACuriousMind Okay... what I tried to mean is that just like we got theories which are invariant under rotations, boosts and translations. Is there a theory which is invariant under every diffeomorphism?

ACuriousMind

again, that's - to the extent that that phrase means something - GR

Sanjana

@NairitSahoo That's just GR

ACuriousMind

it's a gauge symmetry and Noether's second theorem applies

Nairit Sahoo

@Sanjana You just spoke of some global diffeomorphism invariance, right?

Sanjana

16:48

Not at all: I should again warn you... firstly, these asymptotic symmetries are very different from the usual symmetries which ACM is talking about. They precisely depend on the choice of boundary conditions which is bit of an art when you want to do "real world physics". They do give charges, but even then they are still very much local.

I also told you precisely how they "avoid" Noether's 2nd theorem.

naturallyInconsistent

I'm having ramen~

Nairit Sahoo

GIVE IT TO ME. RAMEN PLEASE.

Sanjana

here we go again... Is that your trigger-word or something :p

Nairit Sahoo

Raman no. Raman please. Raman ok. Raman you.

Sanjana

ok. Is anybody else seein' this?

Slereah

16:59

@ACuriousMind From what I can see Chern-Simons does have charges from the diffeomorphism group?

Not that confident in my knowledge of topological theories tho so idk

naturallyInconsistent

dangles lone noodle

Nairit Sahoo

Raman will not let you rest. Raman is very mad

Raman pompong chong dhong

naturallyInconsistent

obscures

ACuriousMind

@Slereah I also don't know that much about the weirder edges of CS theory; the annoying thing is that people often don't properly distinguish between classical and quantum effects; what may appear in the quantum theory as a "global" symmetry from the large gauge transformations modulo small gauge transformations may still be full gauge classically, for instance

Nairit Sahoo

@ACuriousMind Are these large gauge transformations what Sanjana was speaking of?

ACuriousMind

17:09

not necessarily

stop trying to learn random topics from throwaway lines in this chat :P

4

Nairit Sahoo

Okay boss!

Nairit Sahoo

18:05

Sometimes it is meant that solving a theory=finding n-point correlators upto arbitrary $n$. I get that this might be the case for perturbation theory. But do n point correlators contain entire information about the non-perturbative spectra of the theory?

ACuriousMind

@NairitSahoo yes, that's the Wightman reconstruction theorem - all QFTs with the same Wightman functions are unitarily equivalent

Obliv

18:18

in ballentine we have inner products satisfying $(\psi,\phi)=(\phi,\psi)^*$, which means $\langle \psi |\phi\rangle = \langle \phi|\psi\rangle^*$. How come $\langle \psi | A |\phi\rangle = \langle \phi|A|\psi\rangle^*$ isn't correct? I know it should be $\langle \psi |A\phi\rangle = \langle A^{\dagger}\phi | \psi\rangle$ (I think?)

by correct, I should say doesn't follow immediately from the definition of our inner product (it's true in the case of hermitian operators)

I guess I should have just expanded the right side.. so $\langle \phi | A|\psi\rangle^* = \langle A^{\dagger}\phi | \psi \rangle^* = \langle \phi | A\psi\rangle^*$

well I don't know where to go from here. If we let $A = A^{\dagger}$ then idk how that last statement is equivalent to the other one $\langle \psi|A\phi\rangle = \langle A^{\dagger}\phi|\psi\rangle$

ACuriousMind

@Obliv Just apply the definitions: $\langle \phi \vert A\vert \psi\rangle^\ast$ is $(\phi, A\psi)^\ast$ in the inner product notation, and so: $(\phi, A\psi)^\ast = (A\psi, \phi) = (\psi, A^\dagger\phi) = \langle \psi \vert A^\dagger \vert \phi\rangle$.

Obliv

hmm.. going from $(\phi,A\psi)^*$ to $(A\psi,\phi)$ means we conjugate the scalar result. When we want the operator to act on the 2nd argument again, we transpose?

so in other words the transpose of the operator on the ket is the same as conjugating the result

ACuriousMind

there's no transpose here

only adjoint

Obliv

oh i forgot what an adjoint was brb

ACuriousMind

I've simply used the property of $(v,w)^\ast = (w,v)$ and the definition of the adjoint, $(v,Aw) = (A^\dagger v, w)$

Obliv

18:30

so what is the adjoint? It's not the inverse of $A$, is it?

it's the map $A^{\dagger}$ satisfying that relation

but what is it actually doing

ACuriousMind

for complex matrices, it's the transpose + complex conjugation of the elements

Obliv

@ACuriousMind so if $v$ were in the dual of the space $w$ is in, $A^{\dagger}$ changes the unique vector by Riesz theorem s.t. $w$ alone corresponds to each functional $A^{\dagger}v=\chi$? Whereas otherwise $(v,Aw)$ meant $v$ was the fixed vector of $V$ that corresponded to each vector $Aw \in V$?

ACuriousMind

I'm not quite sure what you're trying to say with that

in this notation, both $v$ and $w$ are in $V$, not in the dual

IF you want to phrase this in terms of the representation theorem: The map (i.e. element of the dual) that for fixed $v\in V$ sends $A_v : w\mapsto (v,Aw)$ is the same element in the dual as the dual element of $A^\dagger v$.

Obliv

sorry, yeah. I just meant that if we had $\langle v | A | w \rangle$ this meant $v,w \in V$ and for the functionals in $V'$, they map $V\times V \to \textbf{F}$ by $(v,Aw)$ for all $w \in V$. If we have $A$ act on the bra vector instead, (there is only one by Riesz theorem right?) then we have a new vector $\chi = Av$ which now corresponds to the vectors $w \in V$ in the map $V \times V \to \textbf{F}$ with $(\chi,w)$ for all $w \in V$?

@ACuriousMind yeah basically that

ACuriousMind

the functionals in $V'$ are not maps $V\times V\to \mathbb{F}$, they're just maps $V\to \mathbb{F}$

Obliv

18:42

but aren't inner products binary operations?

ACuriousMind

what does that have to do with $V'$?

the dual $V'$ is the space of linear maps $V\to\mathbb{F}$, that's the definition

Obliv

Idk where I'm going wrong in my reasoning because I definitely agree with that, but I also feel like because of Riesz theorem and the way we define a functional as an inner product, it should map $V\times V \to \mathbb{F}$

ACuriousMind

I don't understand where the inner product comes into this

The Riesz theorem defines a map $V\to V', v\mapsto (w\mapsto (v,w))$

The inner product is a map $V\times V\to \mathbb{F}, v,w\mapsto (v,w)$

Obliv

Yeah I get it now, we don't actually need it to be a binary operation because it's using a fixed vector I think

so $F_{\phi}(\cdot) = (\phi,\cdot)$ has one argument

ACuriousMind

if you want to be fancy you can say that the Riesz map is the curried version of the inner product

Obliv

18:50

oh never heard of that, thanks.

Slereah

Do you mean the adjoint object in the dagger symmetric monoidal category

ACuriousMind

it's a concept that mostly comes up in functional programming/type system discussions

but it holds generally: You can turn ("curry") any map $A\times B\to C$ into a map $A\to (B\to C)$

Slereah

People use it a lot for tensors without saying it rly

ACuriousMind

where $(B\to C)$ is the space of maps $B\to C$

Slereah

As part of the tensor-hom adjunction business

Obliv

18:52

@Slereah this guy nlabs

Slereah

Been reading a bunch of categorical QM stuff lately

Fun times

Proving the point of the quantum fourier transform in there is quite a hassle apparently

Obliv

I imagine learning category theory is similar to just learning a new language

Slereah

you get a bit used to it after a while

Slereah