The h Bar

DIRAC1930

00:02

@ACuriousMind I don't see how stating that there exists only infinite-dimensional unitary representations is useful when a finite dimensional rep won't be able to describe any useful state space

Mr. Feynman

And now representations too

Amit

00:19

The spinors 'cause of Weinstein?? ^_^

Mr. Feynman

@Amit You need to know that some months ago there were several people talking about spinors here everyday

Amit

Ah, yeah, that can scar you dude

Mr. Feynman

For a couple of weeks or even more

Amit

In a less than insightful way? lol, or just with an excess of ACM pinging?

Mr. Feynman

It depends, iirc they were mostly rep theory question (mine were, I don't really remember the others)

Generally I avoid pinging directly with a question, I did that a couple of times maybe

Amit

00:28

I see. Well anything gets boring after a while. That's why school teachers start often with tremendous motivation and... (no need to complete the sentence)

Can't generalize of course. Some get a kick from teaching the same thing over and over again, and some are creative enough to constantly change and improve their methods

Silly Goose

05:51

so when we talk about say 2 identical systems in quantum mechanics

all we are saying is that $H_1 = H_2$ where 1 and 2 indicate system 1 and system 2, respectively. In particular, we are not constraining anything to do with what states system 1 and 2 can occupy (other than what the hilbert space allows)

ACuriousMind

09:36

@SillyGoose 1. What is the Hilbert space if not the space of states the system can occupy? 2. As Hilbert spaces, all separable infinite-dimensional spaces are isomorphic, so you need to be careful what you mean by $H_1 = H_2$

Mr. Feynman

11:00

I'm stuck on an exercise on Carroll. The problem wants me to prove that given dust ($T^{\mu\nu}=\rho u^\mu u^\nu$, where $u^\mu u_\mu=-1$ is the 4-velocity.), the solution of EFE can only be static if the 4-velocity is parallel to the timelike Killing vector

I've been playing around with the conservation law $\nabla_\mu T^{\mu\nu}=0$ without getting much, any hint?

For reference, I'm doing point b)

Amit

11:30

I'm not sure @Mr.Feynman because I haven't understood the Killing stuff well enough yet, but does it have something to do maybe with $$g_{\mu\nu} v^\mu v^\nu = v_\nu v^\nu$$
?

if this entire expression is constant & $v$ is constant also as a vector does it imply that the metric is constant?

Ah sorry, $v$ should be called $u$, more standard for $4$-velocity.

I think it's true but again, I'm not sure about the consequence of being parallel to the timelike Killing vector. If it implies that $u$ is constant wrt time then what I wrote can be relevant

Mr. Feynman

@Amit static metric means that there exists a timelike killing vector orthogonal to spacelike hypersurfaces (as in the text). It is not constant

Anyway, your idea would lead to the geodesic equation $u^\mu\nabla_\mu u^\nu=0$

Amit

You are given that the 4 velocity is parallel to this vector

Mr. Feynman

Which I think plays some role in this stuff

Amit

This is the velocity of the gravitating matter

Mr. Feynman

@Amit I have to prove that it has to be so

If the killing vector exists

Amit

11:44

Ahh, yes but I am suggesting to start from the end :)

See what is the consequence of that, and see if it implies something you know to be true $\leftrightarrow$ style logic :)

But anyway, I was only asking if assuming a constant velocity implies such a vector exists. But I don't want to talk too much nonsense because my recollection about what Killing vectors are approaches zero :)

Mr. Feynman

I've been talking nonsense about that exercise for a bunch of hours now :P

This is some bad way to employ one's time

Amit

Timelike killing of time? ^_^

(I know that Killing is a german surname , that part I remember lol)

Mr. Feynman

12:02

> I read stories by H. P. Lovecraft. Then I can return to trying to show how the construction of a Lie Algebra from a Lie Group can be regarded as a functor safe in the knowledge that whatever a category might be it is, mercifully, neither squamous nor rugose.

The internet is wonderful

Amit

lol, it encourages distractions!! ^_^

distractions for me are destructive... when I find that I procrastinate something I put a timer on, and until it beeps I am 100% focused on this thing

I suppose you can call it a spacetime barrier, lol, it will make it sound cooler than it is

But srsly, this method has a nice side effect: if by the time the timer beeps I am immersed in this thing, I often don't wanna leave it, and will go much beyond the prescribed time frame. So it only sets a minimum threshold, not a maximum one

Mr. Feynman

12:19

@Amit My weak points are: having an enormous to-do list, making me basically a Buridan's ass and obsessing over something I don't understand without getting any progress, which leads to a huge waste of time

Amit

I think I went through both of these symptoms and in my case, which I can't generalize to anyone else, I realized that they both originate in exaggerated expectations of what doing this that or the other can "give me"

I think I learned to relax a bit, I don't expect more than just, fun :)

Mr. Feynman

It sounds like you went through this too :P

Amit

I think yes, I think it may even be rather common whenever you are greatly motivated by any subject of study. But actually the obsessing over something you don't understand is useful, the thing is more about how you manage that one... e.g. accepting that you don't have to: 1. understand 100% of anything now 2. get distracted by reading an entire textbook to understand every teeny weeny mathematical detail. Lol, point 2. happens a lot for "rigorously" leaning physics students

But I think it's better to ignore all advice, and make up your own advice, that other people can in their turn ignore... lol

DIRAC1930

13:28

Is this infinite-dimensional unitary rep have anything to do with there being an infinite number of particles allowed? However, this can't be right because for $SU(2)$ we have a $2 s +1$ dim. rep where $s$ is the number of particles but we never say that we have an infinite unitary rep of $SU(2)$ when describing spin

Someone help

ACuriousMind

@DIRAC1930 the irreducible unitary rep is on the 1-particle space, I don't know what you're talking about

also, $s$ is the spin, not the number of particles

DIRAC1930

What do we mean by infinite unitary representation then?

What are we saying about the operator $U$ in $U \phi(X) U^{-1}=\phi(\Lambda^{-1}X)$

$\phi(X)$ here is $1$ dimensional

ACuriousMind

@DIRAC1930 no, it isn't

DIRAC1930

Doesn't the dimension of $U$ have to match for the above equation to make any sense?

ACuriousMind

$\phi(X)$ is an operator-valued distribution acting on an infinite-dimensional Hilbert space (the space of states of the quantumtheory)

you'Re thinking about the dimension of the target space of the classical field

the two have nothing to do with each other

DIRAC1930

13:38

Yes I got confused thanks

ACuriousMind

see physics.stackexchange.com/a/174908/50583 for this particular confusion

DIRAC1930

Above you said that $U$ acts on the $1$ particle space but the field operator acts on the full Hilbert space

ACuriousMind

I said that it acts irreducibly on the 1-particle space

its representation on the full Hilbert space (Fock space) is reducible since it's just the Fock powers of the 1-particle representation

DIRAC1930

So $U$ is really $U=U_1 \oplus (U_1 \otimes U_1) \oplus \dots$?

ACuriousMind

of course

DIRAC1930

13:43

How do I explicitely show how $U$ acts on $\hat{\phi}(X)$ when $\hat{\phi}(X)$ doesn't manifestly have that structure?

Won't $\hat{\phi}(X)$ only act on $\bigotimes^{\infty} U_1$?

for a free field anyway

due to the integral in the field operator producing an infinite number of modes

$\int \mathrm{d}k \hat{a}^\dagger_k \dots$

ACuriousMind

I don't understand the question

both $U$ and $\phi$ are operators on the full Hilbert space

so $U\phi U^\dagger$ makes sense

Ryder Rude

@Amit i will try this too as i suffer from Buridan's ass

DIRAC1930

How do I know that $U$ acts irreducibly on the 1 particle Hilbert space?

ACuriousMind

@DIRAC1930 1. That it has to have a meaningful restriction to the 1-particle space is intuitively obvious: Lorentz transformations neither create nor destroy particles. 2. That this restricted action is irreducible follows from the representation theory of the Lorentz group, in this case known as Wigner's classification

the best physics exposition (modulo horrible notation) of this representation theory is once again in Weinberg, the mathematical treatments rely on Mackey's theory of induced representations applied to the little groups

DIRAC1930

Ah okay, thanks. Why would a reducible rep have the possibility of creating or destroying particles?

Ryder Rude

13:54

How to have peace in life in ur opinion?

ACuriousMind

@DIRAC1930 that's not what I said

DIRAC1930

Oh yeag

ACuriousMind

do you understand what an irreducible representation is?

DIRAC1930

yeah*

That there are no invariant subspaces

Ryder Rude

Is the term "reducible rep" used for direct sums of irreducible reps, or for tensor products of irreducible reps? @ACuriousMind

ACuriousMind

13:57

@DIRAC1930 right, and so my argument is that since Lorentz transformations can't change particle number, an irreducible representation can't contain states with different numbers of particles since the subspace generated as the orbit of a state with $n$ particles can't contain a state with $m\neq n$ particles, hence would be an invariant subspace

Ryder Rude

Like, both of these seem built from irreducible reps

ACuriousMind

@RyderRude neither

physicists may play a bit loose with these terms, but a reducible representation is just any representation that contains an invariant non-trivial subspace

reducible representations that are sums of irreducible representations are decomposable

that every reducible representation is decomposable is not true in general, and is a non-trivial statement that needs to be proved e.g. for semi-simple Lie algebras

Ryder Rude

I can understand y the Dirac field wud b called a reducible rep. But the multi-particle states r tensor products of the one-particle space. Then y r these called reducible reps? @ACuriousMind

Is the term "reducible rep" used for tensor products too?

ACuriousMind

I just gave the definition of "reducible" and I don't know what is unclear about it

Ryder Rude

Does this definition cover tensor product stuff too?

Ooh yeah

ACuriousMind

14:03

a tensor product of representations is itself a representation

Ryder Rude

Sorry

ACuriousMind

this representation can be reducible or not

DIRAC1930

The Fock space is a direct sum of tensor products

Is there a way to explain the constant particle number without using orbits?

ACuriousMind

what do you mean, "without using orbits"?

an orbit is a very basic notion in group and representation theory

it's just "the set of all states you can reach by applying all the group/algebra operators to this state"

it's not a complicated or exotic notion

DIRAC1930

Oh thanks

I've always found rep theory too abstract

For me anyway

Ryder Rude

14:09

How to have peace in life in ur opinion?

@ACuriousMind @Slereah @DIRAC1930 @Amit @Mr.Feynman

ACuriousMind

with whom are you currently at war that you don't have peace?

DIRAC1930

By being able to be your true self in this world

Ryder Rude

@ACuriousMind i feel rush-y instead of calm

@DIRAC1930 very concise and to the point :)

ACuriousMind

What does that mean? Why do you think you should feel calm?

Ryder Rude

I feel better when calm. I'd rather spend whole life being calm @ACuriousMind

DIRAC1930

14:13

ACuriousMind feels at peace when no one asks him about asymptotic states lol

Ryder Rude

Lol

But i guess rush has its use too @ACuriousMind

ACuriousMind

@RyderRude so why don't you?

Ryder Rude

I cant. I need some advice to b able to do it

How do u guys find peace?

ACuriousMind

@RyderRude What I'm trying to say is that the answer depends on why you specifically don't feel as you'd like to

Ryder Rude

I've become used to rush i think

ACuriousMind

14:17

what does "rush" mean? You can't focus on a single topic for any length of time? You're restless and you move around a lot? You get impatient and annoyed when you don't make visible progress?

Ryder Rude

I'm restless and i move a lot

I dont feel like doing nothing

I always want to do the next thing

ACuriousMind

I mean...nothing inherently wrong with that

some people are more driven than others

Mr. Feynman

@RyderRude if you're talking about being at peace with yourself, that is an issue I'm trying to figure out too :P

Ryder Rude

But i love doing nothing whenever i sort of force myself to do it

But mostly, i cant convince myself to do that

I love doing nothing more than being restless

@Mr.Feynman we're in this together :)

ACuriousMind

@RyderRude so do it more! block time explicitly for doing the kind of nothing that you enjoy! turn off all the distractions that could remind you of other things you could be doing in that time! if you're motivated by checking off goals from a list, set explicit goals at the start of a week for how many hours you want to take for yourself that week and measure your progress. If the problem is that you feel driven to do stuff, make "relaxing and doing nothing" a task like everything else ;)

Ryder Rude

14:22

I'd like to more often b able to start doing nothing

ACuriousMind

I don't really have personal experience there but I know people who do this and as strange as it sounds it seems to work for them :P

Ryder Rude

I will try this then. But I've been procrastinating this

Mr. Feynman

@RyderRude Is there a specific reason why you can't just do it? Do you feel forced to study?

Ryder Rude

I feel like doing anything, not necessarily studying @Mr.Feynman

I can feel like walking too

Mr. Feynman

I think that "nothing" is like "vacuum" :P

Ryder Rude

14:26

@ACuriousMind What do u do to get ur sort of peace?

Mr. Feynman

It's not really empty

@RyderRude He throws darts on targets with the usernames of the regular h bar users pinging him (?) :P

Ryder Rude

Lol

@Mr.Feynman What is ur definition of getting peace?

DIRAC's is :being urself

Mine is : doing nothing

ACuriousMind

@RyderRude I don't really have that kind of problem, I don't get restless easily. Like, most of my friends talk about how hard it was to mostly stay isolated during the early days of the pandemic and feeling restless because they couldn't do a lot of things but I...didn't really have a big problem with that? I just got a bit annoyed. If anything my problem is that I'm too lethargic.

Mr. Feynman

@RyderRude Getting better to the point I can finally be satisfied and proud of myself

Of course I'm talking about Physics, I'm already at peace with the rest of my life

Ryder Rude

Oh.. This definition can get a bit problematic.. Becuz u may never b proud of urself

When u learn something new, it just becomes obvious after a few days

So the proudness doesnt last

Mr. Feynman

14:32

12 mins ago, by Mr. Feynman

@RyderRude if you're talking about being at peace with yourself, that is an issue I'm trying to figure out too :P

Ryder Rude

Sorry. Prize stuff is not important in this discussion

We only chase understanding. No one is chasing prizes

Mr. Feynman

@RyderRude Also, I'm a very negative person and mistakes along with comparison with others are quite troublesome

But maybe that's not the question you asked and this definition of peace is too self-dependent

So I'll make my definition more human-invariant by reformulating it as "being happy with what you are"

Ryder Rude

This is also a gr8 definition of peace

@ACuriousMind What is ur definition of the peace u desire? U r not restless, but is there some other peace u desire?

Something not defined in terms of restlessness

ACuriousMind

I'm quite content with the way my life is right now

Ryder Rude

U r living the dream then :)

Amit

14:43

The search for peace is warfare :) But maybe I'll say something less hyperbolic when I get back home lol

Mr. Feynman

@RyderRude Don't shake his hand, he's our antiparticle

Amit

@RyderRude Okay here goes a less hyperbolic version: peace is 1. Figuring out what you want and either 2. Devoting yourself to it completely or 3. Giving it up completely. That's the easier part! The harder part is managing conflicting wants/desires. I think for example a person who only wants one thing is a very lucky person :)

Ryder Rude

i've been engaging in all my interests half-assedly @Amit

I will try devoting myself to smthing then

@Amit Hav u devoted urself to physics?

Amit

15:00

@RyderRude No, I devoted myself to peace, then gave it up completely because I realized I don't really want total peace. So Im completely content half assing around things too now, lol. Hope that makes sense

Ryder Rude

Lol

Yes, according to some philosophies, peace may not b the right thing to want in life

Amit

That's the thing yeah... we want peace up to a point, then it becomes boring. But real peace doesn't work that way :) I think it's a case of "careful what you wish for..."

Ryder Rude

@Amit oooh but u actually did get peace. U just got it in a "liar paradox" method

U r at peace with half assing and with ditching peace

Amit

No idk, I don't bother about calling it peace or suffering... well Im pretty sure I still suffer sometimes lol

Ryder Rude

@Amit this is y i think that Buddhists r missng out

Amit

15:04

I do agree that perhaps the only possible peace is giving up on peace, which redefines a new kind of peace

Ryder Rude

But I also feel Buddhists r doing betr than everyone else

Amit

A wise buddhist will mention the fourfold or eightfold, i cant recall, "way", and that peace is attained differently by different types of people

Ryder Rude

I feel conflicted about whether monks r missing out or they r living the dream

@Amit i will google it then

What do all of u think? R monks missing out on good stuff or r they living the dream life everyone wants?

Amit

Eh I don't endorse this stuff

They make good beer sometimes don't they?

Ryder Rude

I guess so. They must have their niche way of partying

They probably drink and meditate

Amit

15:10

No I don't think they are allowed to consume it

Ryder Rude

Oh. Then maybe it's some other monks than Buddhists

Some of them r very weird

Amit

en.m.wikipedia.org/wiki/Trappist_beer

Example

Amit

15:23

Tom Baker, past dr who actor has very funny stories about being a catholic monk for seven years before he became an actor

Funny / mortifying.. apparently the discipline was often quite cruel

Ryder Rude

15:34

Ok i think monks just falsely believe this cruel stuff will lead to peace, but it never does. Monks just end up missing out on the good stuff

Amit

Some of them are just escaping their life, and possibly it can save someone's life too... even if he takes a time out only for a few years

So I can't generically dismiss this

I guess that some people are more devoted to stuff by their nature. And if you have to decide between devoting yourself to enlightenment or to alcohol say... then probably the former is less harmful lol

Ryder Rude

Hav they shared their enlightment? Can it be read and learned or must it be experienced?

Amit

I don't think it exists, but it's a way to spend your time nonetheless

Ryder Rude

"In Buddhism, Nirvana is the highest state one can achieve and it is also considered by Buddhist monks in Buddhism. According to Buddhist tradition, one who attained Nirvana will be free from worldly desires and suffering of life and will also be free from the Wheel of Life, Bhavachakra."

The above seems like the monk stuff is a scheme to bypass the programming of evolution

Evolution has programmed us to b like this. But monks want to be something else

user 85795

If those were monks at the coronation; they looked like they were free from worldly "desires," but who knows...

ACuriousMind

15:48

@user85795 ...why would there be Catholic monks at an Anglican ceremony?

user 85795

Corrected, 🙏 thanks.

When was the last time Germany had a coronation of a king?

Amit

@RyderRude I can only say good luck lol

Loong

16:08

@user85795 1888?

or maybe 1913 if you include Bavaria

user 85795

Thanks.

Mr. Feynman

17:51

There is one thing bugging me about the infinite vacuum energy in QFT. When we ignore gravity we can gleefully neglect the zero-point energy and everything works fine because only energy differences matter.

Now, as soon as we try to put GR into the game, the absolute value of energy is relevant since $G=8\pi G T$. I wonder, why would we say that the "correct" energy is the infinite one? What would the absolute energy even be? When gravity is absent, neither choice of energy is more "correct" than the other

ACuriousMind

@Mr.Feynman who claims the "correct"energy is the infinite one?

usually we say that in gravitational theories, we are no longer free to renormalize the vacuum energy to 0, but we must instead renormalize it to the observed value of dark energy

Mr. Feynman

@ACuriousMind I haven't seen that claim directly but I have heard that taking gravity into account we can't neglect the infinite energy term

ACuriousMind

yes, what they mean is what I said in my second message - you can't just "ignore" the energy and set it to zero

Mr. Feynman

I see, so the claim that there is something mysterious about it and that there is some compensating mechanism killing off the infinity is just incorrect

ACuriousMind

where did you see that claim?

Mr. Feynman

17:57

Oh, I heard that during a QFT lecture (that was actually a break)

ACuriousMind

there is a strange argument that people sometimes make that the smallness of dark energy in reality is "mysterious" because "naturalness arguments" would suggest it to lie around the Planck scale

I suppose you misheard a variant of that argument

Mr. Feynman

I guess you are right because that divergence with an appropriate cutoff gives rise to the cosmological constant problem, which is the infamous discrepancy

So they could have meant some compensating mechanism causing $\Lambda_{\text{theoretical}}\gg\Lambda_\text{measured}$

It was really a bad choice to use $\Lambda$ when talking about a cutoff lol, of course I meant the cosmological constant. The theoretical value would be the one with the Planck scale cutoff

DIRAC1930

18:24

@ACuriousMind Does this mean I can form a Casimir operator for particle number out of the Lorentz generator operators?

Eg. is $s^2$ effectively particle number in non-rel?

ACuriousMind

@DIRAC1930 Why would it be? If you combine two 1/2-particles of SO(3), you get spin-1 rep + spin-0 rep. Where do you get a "particle number" operator from here?

DIRAC1930

Oh yeah

I'm confused about why a reducible rep can give the possibility of Lorentz transformations changing particle number

ACuriousMind

5 hours ago, by ACuriousMind

@DIRAC1930 that's not what I said

5 hours ago, by ACuriousMind

@DIRAC1930 right, and so my argument is that since Lorentz transformations can't change particle number, an irreducible representation can't contain states with different numbers of particles since the subspace generated as the orbit of a state with $n$ particles can't contain a state with $m\neq n$ particles, hence would be an invariant subspace

there is no implication here that a "reducible rep can give the possibility of Lorentz transformations changing particle number"

DIRAC1930

Ah I read it wrong the first time round

So the irrep must contain states with a fixed particle number

So we need to find an irrep of the Lorentz group on the Hilbert space that is unitary. Physically Lorentz transformations can't change particle number. And since the Fock space is built entirely from direct sum/product representations of the 1 particle Hilbert space, enforcing it for the one particle Hilbert space is enough to make everything work out

So we look for a unitary irrep of the Lorentz group on the one particle Hilbert space

And we find only infinite ones are possible

Is this somewhat correct?

@ACuriousMind

Silly Goose

18:54

Hm I guess I saw that you set $H_1 = H_2 = H$ earlier when talking about identical systems and I was mimicking that @ACuriousMind

But using equality as opposed to isomorphic would just make things more convenient when dealing with identical systems right?

or what I am thinking is that it allows us to label the "same" state among different identical systems with one ket

err okay maybe just skirt around the second thing you brought up by limiting to finite dimensional systems :P.

hm okay wait so a system is not determined by a hilbert space right?

(i am asking the above to answer the question "is it wrong to say that a hilbert space represents a system")

PM 2Ring

19:12

@SillyGoose "A new study from researchers at Northeastern University, in collaboration with scientists from MIT and the University of Glasgow, investigated what happened when a group of domesticated birds were taught to call one another on tablets and smartphones". news.northeastern.edu/2023/04/21/parrots-talking-video-calls

Silly Goose

omg XD

the internet revolution for birds has come

DIRAC1930

It's difficult to tell what parts of group theory/rep theory are needed versus what is there to help mathematicians understand physical concepts

I'm a firm beliver in the group theory pest ncatlab.org/nlab/show/Gruppenpest lol

Silly Goose

as in you do not think the group theory way is good :0

DIRAC1930

I just find it too difficult lol

Silly Goose

i have been recommended to work through brian c. hall's lie groups, lie algebras, and representation theory since it takes a more concrete approach (deals only with matrix groups, algebras, etc.) while still covering some important business but i haven't really read much of it yet. but idk it would be hard to work through if one hasn't taken introductory group theory and real analysis (at least from what i read)

DIRAC1930

19:23

Peter Woit's book is good

There is also a book on QM group theory by Van der Waerden that goes through non-rel spin perfectly

Silly Goose

i was wondering if future theoretical physicist will just have to be mathematicians :P

you already have the slew people in string theory who are trained as mathematicians it seems

DIRAC1930

I think the best ones can do both very well

I immediately get stuck when I can't attribute anything physical to the math

Mr. Feynman

@SillyGoose That day will be the death of Physics

@SillyGoose You don't need group theory to study Lie Groups

Really, knowing what a group is and some other basic definitions is enough

Silly Goose

ah i should've said like the basic definitions and stuff @Mr.Feynman like what a group is what a representation is and so on

Mr. Feynman

Representations are fully explained in the book

DIRAC1930

19:29

That book is quite complicated I found

Silly Goose

it helps to have a basic idea of what you are doing before doing it though :D

it took me almost a whole semester to understand (at least i think) that a representation is a mapping (satisfying blah blah blah) :P from Dummit and Foote

they use the term so fast and loose

Mr. Feynman

So what you really need is some linear algebra and the basic definitions

@DIRAC1930 Mhh, I agree that sometimes the author takes a long way to prove things that using differential geometry would be straightforward

Silly Goose

is partt of the point of the book not to use diffe g though? or i thought it was to like give a presentation of lie theory with the least preliminary background pretty much

Mr. Feynman

But that's also complementary to the approach he's taking of defining Matrix Lie groups as subgroups of $\mathrm{GL}(n,\mathhb{C})$ satisfying some conditions and only later it turns out that matrix lie groups are lie groups

And then you can read all of it with the machinery of DG

Silly Goose

my real mathematical weakness (all of which are weaknesses anyways) is analysis :P i abhor the subject

maybe later on topology/diffe g will make me warmer towards it though

Mr. Feynman

19:33

Well, Lie Groups are a DG topic. Basically they are differentiable manifolds endowed with a group structure such that the operations are smooth; When dealing with matrix groups you can take a less DG geometric approach

As it boils down to calculus/analysis in $\mathbb{R}^{n}$

Amit

@SillyGoose did you study complex analysis?

Silly Goose

just real analysis and not much of it @Amit up to continuity and limits of functions (ch 4 of rudin) so not even differentiability and so on

i realized that the course setup at the university i took analysis is strange (or maybe this is the usual case?) it covers topics from constructing the reals to limits/continuity of functions in analysis I. then it covers differentiability, integrability and some complex analysis in analysis II. then there are two graduate level analysis courses

Mr. Feynman

@SillyGoose I'm not sure Hall's book requires to be grounded in Real Analysis (also, serious Real Analysis can be pretty scary afaik, the Real Analysis I know is that of my freshman and sophomore years), only something regarding power series in chapter 2 as far as I recall

@SillyGoose What you have described is a standard analysis I syllabus

Amit

Ah. There's a book called "Visual Complex Analysis"... some like it very much. In general I think complex analysis may be a more fun way to study real analysis (I'm assuming you're one of the people that can start there, because you know enough already about real analysis even if not "rigorously")

Silly Goose

at least in the first chapter it helped to know what continuity and compact and so on meant @Mr.Feynman but yeah it doesn't seem like one needs to know real analysis

Mr. Feynman

19:37

Well, yes. Althought in this case we're working in the standard real topology, those are not specific concepts of analysis

Silly Goose

it would be nice to see what complex analysis is like :D @Amit maybe i will try taking another analysis course next year

Mr. Feynman

I hate analysis since last October

Amit

I never took such a course but I heard many people fell in love with complex analysis, and I think it translates quite well to real analysis knowledge as well

Silly Goose

XD

what happened last october

Amit

yes, reals are just complex numbers without imaginary parts ^_^

Mr. Feynman

19:39

I wanted to attend a Functional Analysis course in the math dep but the Professor suggested to take a course that translates to "Foundations of Superior Analysis", which was basically measure theory so I had to give up

Silly Goose

wait i think i am confusing concepts :P

superior analysis 0.0

Mr. Feynman

I don't know how it translates properly in English. It's called "Analisi Superiore" in Italian

That was a literal translation

Amit

lol everything sounds good in Italian

Mr. Feynman

Oh, I found an Italian university that holds this course in English and it's called Higher Analysis

tutturu

Silly Goose

functional analysis would also be cool :P

i have two slots left for math electives. i was thinking to do topology and diffe g though or at least topology

Mr. Feynman

19:42

Take Diff Geometry for sure

Silly Goose

maybe a more relaxed math course though XD i will be taking 3 math electives and a quantum ind study that semester

okay maybe ill do diffe g

do you think topology is too much math and not really related to physics ?

Amit

Related for sure

Mr. Feynman

Topology is important in differential geometry (although I wouldn't say you need to have had a course in topology to learn it) and knowing some can't be a bad thing

Silly Goose

ah i see

Mr. Feynman

Also, I heard that topology is being used in GR but I cannot provide further information about that

Amit

19:45

@Slereah ain't here ^_^

Silly Goose

XD

i cannot tell if people use topology in physics to mean mathematical topology or something else

it seems like the former but yeah. as in "topological quantum computing"

Mr. Feynman

@Amit Don't worry, "GR" evokes him

Amit

lol... like a spell

ACuriousMind

May 18, 2019 at 15:42, by ACuriousMind

@RyanUnger If you say 'GR' three times in front of a mirror, Slereah will appear and talk to you about wormholes

Silly Goose

LOL

Amit

19:47

lol... that's like a jedi fairytale

Mr. Feynman

@SillyGoose What I can tell you is that sometimes you'll hear physicists misusing "topology" as a synonym of "geometry" which is also misused as a synonym of "shape"

Silly Goose

oh mayn

Mr. Feynman

@ACuriousMind I'm slowly becoming you, wait for it

Silly Goose

i should find out what topological matter means but yeah it always just sounds like buzzwords :P

it seems like topological quantum computing may be interesting though, which has led me to read about identical particles :P

Mr. Feynman

Right now I'd like to rush QFT to learn some String Theory

I guess that QED and a little QCD are still too little knowledge

Silly Goose

19:52

are the physical concepts (as opposed to the math) related to QFT and string theory difficult to understand?

Mr. Feynman

> Right now I'd like to rush QFT to learn some String Theory

Disclaimer: it's not like I'm desperately in love with ST, just that I want to get away from QFT asap :P

@SillyGoose I don't know about ST, but my current understanding of QFT is: "shut up and calculate"

Amit

There must be more ways to get away from QFT ^_^... and anyway, isn't QFT formulated in like $10$ different ways? You have to be able to find one that you like... ^_^

Mr. Feynman

Oh, I don't dislike QFT either. I'm just uncomfortable with a lot of conceptual things :P

Amit

I see

DIRAC1930

20:38

@ACuriousMind Is this all incorrect chat.stackexchange.com/transcript/message/63525738#63525738 ?

Charlie

21:43

@DIRAC1930 What you've written seems correct, the fact that the Lorentz group does not admit finite-dimensional unitary representations is a consequence of it's topology (it is non-compact).

DIRAC1930

21:55

How do we know the structure of the Hilbert space? In non-rel QM, the basis states are explicitly stated and then the $n$-body operators are subsequently found through this however in rel. QFT, the commutation relations seem to come first and then it is just assumed to naturally act on a Fock space

I suppose maybe you can infer it from the creation and annihilation operators navigating between the different $n$ particle subspaces

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