The h Bar

12:52 AM

Is it possible to know when a close vote was given? I have one on a recent question of mine, and I want to know if it occurred before or after I edited the question (due to a comment pointing out a mistake).

John Rennie

4:49 AM

@user400188 Look at your reputation and it will show the time you got the downvote next to the -2 reputation change.

@user400188 Which question was it? I cannot see a downvote on either of your two recent questions.

user400188

5:02 AM

I wasn’t downvoted, I only got a close vote on the question (I guess they were nice)

John Rennie

Oops, sorry, yes.

user400188

So, is it the case that you can't know when a close vote occurred? (unless you watch it happen).

Also, it just occurred to me that this question would appear in the close vote queue, once it's given its first close vote (If I am understanding how that works properly).

John Rennie

The question did appear in the closed vote queue, but all that shows is the two reviews:

It doesn't show when the original close vote was.

user400188

Oh, ok then. Thanks for the help :)

Slereah

7:23 AM

Morning

Slereah

8:57 AM

I need some help regarding closed timelike curves

I need to build a time machine so that Angular was never invented

ACuriousMind

9:40 AM

@Slereah but how will you keep people from inventing it again?

Manas Dogra

"The error bars suggest a $3\sigma$ discrepancy between the two measurements" what does this statement mean? $\sigma$ means the standard deviation(error) usually but what should this mean when two measurements with two different errors are compared like this?

I don't think it's relevant but the measurements are of Hubble constant from different experiments. One is $H_0=74.0\pm 1.4$ and another is $H_0=67.4\pm 0.5$ in the same units

Slereah

ACuriousMind

9:55 AM

@ManasDogra It means that you need to add/subtract more than three times their respective standard deviation to them to make them compatible - in your example, 74-3*1.4 = 69.8 and 67.4+3*0.5 = 68.9. It's not $4\sigma$ because subtracting 1.4 from the first value again lands you at 68.4, which is below the second value (hence they are compatible at $4\sigma$).

Manas Dogra

Thanks

More Anonymous

11:47 AM

@Slereah Whats angular?

Nihar Karve

12:05 PM

AngularJS, presumably

Yashas

Why is the Hamiltonian density conserved? If you plug zero into both indices in eq. 2.30, you will get $\Theta_{00}$ to be conserved with time. Eq 2.31 shows that it is nothing but Hamiltonian density.

ACuriousMind

...why did you ask a question and then answer it yourself in the same message?

Yashas

@ACuriousMind Sorry, I didn't word it correctly. The question is I don't understand why the density is conserved. It makes intuitive sense if Hamiltonian is conserved but why is the density also conserved?

ACuriousMind

@Yashas what do you mean "why"? Didn't that text derive its eq. 2.30 via Noether's theorem?

Yashas

Yes, but it was surprising. I am accustomed to seeing Hamiltonian being conserved but was a bit surprised that to learn that the density is conserved too.

Why can't energy move from one place to another? Or could it not spread out or concentrate in one place? The density changes in these cases, right? But the total energy can still remain conserved.

Semiclassical

12:34 PM

@ACuriousMind use the closed timelike curve to go back to the Big Bang and embed the message “don’t create Angular” in the CMB

either that or grandfather-paradox anyone who decides to invent it

but the former has the advantage of messing with cosmologists

”God exists, and He hates Angular”

ACuriousMind

@Yashas you need to look closer what the equations actually say

the conservation law is not $\partial^0\Theta_{00} = 0$, but $\partial^0\Theta_{00} = \partial^j\Theta_{j0}$

i.e. the claim is not that energy density is constant, just that its change with time is associated with a spatial flux of energy $\Theta_{j0}$

Yashas

12:59 PM

Very embarassing. I picked out a single element instead of the sum. Still, the author says that density is conserved right below eq 2.32. Is that a typo?

ACuriousMind

yeah, that's a bad formulation - they mean "the integral of the Hamiltonian density over $V$ is conserved if there's no flux through its boundary $S$"

DIRAC1930

2:12 PM

When we canonically quantize through $\Phi \rightarrow \hat{\Phi}$ what pictures are we going from and to?

e.g. is it the Heisenberg picture?

ACuriousMind

Why would there be a picture at all there?

you only have to decide what picture you're in once you talk about time evolution

DIRAC1930

That's what I was thinking

Okay thanks

What is meant by a 'dense subspace of the Hilbert space' in more physical terms?

Any time you see a commutator in QM, does that mean a Lie algebra is implicitely involved?

ACuriousMind

those are two very different questions :P

DIRAC1930

They are in neighbouring paragraphs in the book I'm reading lol

ACuriousMind

@DIRAC1930 you can think about a dense subspace as a subspace of vectors with which you can approximate any other vector arbitrarily closely (since the closure of the dense space is the whole space, all vectors not in the dense subspace are at least limit points of it)

DIRAC1930

2:21 PM

'Defining the commutator of two operators $A$ and $B$ on $\mathcal{H}$ is not as innocent as it may seem. The reason is that, in QM, observables are typically described by operators which are not defined on the whole Hilbert space $\mathcal{H}$ but only on a dense subspace. As a consequence, the composition of operators and hence also their commutator is not nessecarily defined'

Why aren't the operators defined on the whole space?

ACuriousMind

@DIRAC1930 the set of linear operators on a vector space is a Lie algebra with the commutator as the Lie bracket, I'm not sure what you mean by a Lie algebra being "involved" in that

DIRAC1930

And then after stating the the definitions of bi-linearity, antisymmetry and the Jacobi Identity, the book then says 'again we can summarize these properties by the statement that the commutator endows the space of linear operators on $\mathcal{H}$ with the structure of a Lie algebra'

ACuriousMind

@DIRAC1930 because there are no self-adjoint unbounded continuous operators defined on the whole space, cf. Hellinger-Toeplitz, but physical observables are often unbounded quantities like position or momentum

DIRAC1930

Is the following just a statement regarding the Complete Set of Commuting observables? 'Finally, the maximal abelian subalgebra provides the quantum numbers of the Hilbert space vectors'

ACuriousMind

CSCO is essentially just a different name for "maximal Abelian subalgebra", yes

DIRAC1930

2:34 PM

If we have two continuous symmetries, do we have to find the maximal Abelian subalgebra of all the generators combines?

ACuriousMind

in order to do what?

DIRAC1930

To classify the states

ACuriousMind

sure

but the Cartan subalgebra of $G\times H$ is just $\mathfrak{c}_G\oplus\mathfrak{c}_H$ if we denote the maximal Abelian subalgebra of $G$ as $\mathfrak{c}_G$, so it's not different from finding them invididually - you don't suddenly have to do something different for two symmetries

DIRAC1930

Why have you done the direct product of 2 algebras?

ACuriousMind

I was thinking in terms of the groups

$G$ is the group, $\mathfrak{g}$ is its algebra - the algebra of $G\times H$ is $\mathfrak{g}\oplus\mathfrak{h}$

DIRAC1930

2:39 PM

But where do you do the direct product of groups?

ACuriousMind

what did you mean by "having two symmetries" if not that you have one symmetry group $G$ and another symmetry group $H$ such that the total symmetry group is $G\times H$?

DIRAC1930

I meant like, for example, non relativistically, rotational symmetry and $SU(2)$ spin symmetry

ACuriousMind

that's not two different symmetries

spin is just rotation

DIRAC1930

I meant angular momentum and spin

ACuriousMind

still the same group

or rather, angular momentum and spin are not separately conserved, there's no two separate symmetries there

DIRAC1930

2:43 PM

Interesting

ACuriousMind

the whole point of spin is that it interacts with e.g. magnetic fields much like angular momentum does, and so you can convert one into the other

DIRAC1930

What are two different symmetries in non-rel physics

CPT and SO(3)

ACuriousMind

but to your original point: of course, in some concrete setting you can have that they are separately conserved, then you have two copies of the $\mathrm{SO}(3)$ rotation group as symmetries, with the full symmetry being $\mathrm{SO}(3)\times \mathrm{SO}(3)$

DIRAC1930

I'm confused

If I had to non relitavistic particles

Would SO(3)XSO(3) be a symmetry?

No that doesn't sound right

ACuriousMind

depends

symmetry is a property of your Hamiltonian/action

DIRAC1930

2:48 PM

$G\times H$ is the same as $G \oplus H$ right?

ACuriousMind

there is no $\oplus$ for groups

DIRAC1930

What is I had two Hamiltonians put together like $H_1 \oplus H_2$

Wouldnt the symmtery group be $G_1 \oplus G_2$ or something

ACuriousMind

again, there's no such thing as $\oplus$ for groups

it just doesn't mean anything

DIRAC1930

Yeah, that doesn't satisfy the group axioms

ACuriousMind

I don't know what that means - that thing simply doesn't exist, there is no definition of a direct sum for groups

DIRAC1930

2:50 PM

Can't we just define it

Eveything in math seems to just be someones definition thats caught on

ACuriousMind

and as what would you define it?

DIRAC1930

I would define $G \oplus H$ as $G \times H$

how does $G_1 \times G_2$ act on a Hamiltonian

The Hamiltonian would have to be of the form $H_1 \times H_2$ right?

ACuriousMind

Sure, then it exists and is a group. But you shouldn't do that because in general $\oplus$ is only used to denote the product in categories where the product and coproduct coincident, but the coproduct of groups is different from the product of groups.

@DIRAC1930 I think you're a bit confused about how groups and representations work

bolbteppa

That notation is used for abelian groups, but it means direct product

ACuriousMind

There are essentially two main cases: Either you start with an abstract group and you try to construct representations, or you start with some space and some operators on it and you're looking for symmetry groups or whatever

I'm not sure in what sort of situation the question "How does this group act on a Hamiltonian?" arises

DIRAC1930

2:56 PM

Oh sorry, everytime I said group I meant representation

ACuriousMind

uhhhh

that, indeed, changes a lot :P

DIRAC1930

Lol

So a representation of $G\times H$ is $\rho(G) \oplus \rho(H)$ right?

ACuriousMind

although I'm not sure what a "symmetry representation" is, sooo...not everytime?

DIRAC1930

symmetry representation is just the representation of a symmetry group

ACuriousMind

Given a representation $(V_G,\rho_G)$ of $G$ and a representation $(V_H,\rho_H)$ of $H$, you get a representation of $G\times H$ as $(V_G\otimes V_H, \rho_G\otimes \rho_H)$

note that this is the tensor product, not the direct sum

DIRAC1930

3:01 PM

How exactly does this play a role in SU(3)XSU(2)XU(1)?

Actually I give up

ACuriousMind

I'm not sure what you mean

DIRAC1930

I've been trying to understand representation theory etc. for ages and I'm getting nowhere

ACuriousMind

maybe you should get comfortable with groups in basic QM before trying to understand the Standard Model :P

DIRAC1930

Whats the difference between a group and a representation in QM?

ACuriousMind

Why the "in QM"? Do you understand what the difference between a group and a representation in general is?

DIRAC1930

3:06 PM

Yes

A group satisfies some axioms

if it satisfies those properties

its a group

ACuriousMind

so then what issue do you have applying that knowledge to QM specifically?

DIRAC1930

Well explicitly written out, isn't the Hamiltonian operator a representation of the abstract H in a given basis?

ACuriousMind

I wouldn't say that

DIRAC1930

by representation I don't mean group representation, just representation in the colloqial sense

ACuriousMind

what's the "abstract H"?

DIRAC1930

3:09 PM

$H |\Psi\rangle = -\imath \partial_t |\Psi \rangle$

ACuriousMind

but that's not a thing

we often have an "abstract H" in our equations because whatever we're doing does not depend on the specifics of any particular H, but in each of these cases we still imagine this $H$ to be a concrete operator given on a single concrete space

there's a Hilbert space $\mathcal{H}$ and this $H$ is an operator on it, what does this have to do with "representation"?

(there are formulations of QM where we would talk about representations before even getting to a Hilbert space, but I don't want to bother you with that right now)

eh

if you're doing the free particle, it's $H\propto p^2$, no matrix in sight

DIRAC1930

I meant $H \propto p^2$ isn't the abstract H

ACuriousMind

but if you're on a finite-dimensional space, sure, it's given as a matrix because how else would you give someone an operator on a vector space?

Semiclassical

$H$ is always concretely some operator on a vector space. it does something to vectors

DIRAC1930

What do groups act on?

Nothing

ACuriousMind

3:13 PM

what does $H$ have to do with groups?

there's no group here, it's just a single operator on some Hilbert space

DIRAC1930

Because how do find out what groups do in QM without defining the Hamiltonian first

ACuriousMind

I'm not following you

DIRAC1930

If I have a symmetry, I have $[H,O]=0$ right?

ACuriousMind

yes

DIRAC1930

$O$ is an operator in some Lie algebra

Semiclassical

3:15 PM

no

$O$ is just an operator on a vector space

DIRAC1930

if continuous

ACuriousMind

it's just an operator on the Hilbert space

Semiclassical

there's no Lie algebra here yet

ACuriousMind

I mean, the set of all operators on the Hilbert space is a Lie algebra, but I don't understand why we'd care about that at this point

Semiclassical

i mean, there is a group action here already, i.e., $e^{-i H t}|\psi(0)\rangle = |\psi(t)\rangle$

DIRAC1930

3:16 PM

Okay, O is an operator on the Hilbert space

Okay, so thats what we mean by group action

Semiclassical

i think i'm using that phrase right :P

bolbteppa

If you write a commutator you are using a Lie algebra structure or else it is not consistent

Semiclassical

what?

ACuriousMind

sure, as I said - the operators on any vector space naturally form a Lie algebra

but I'm not at the point where I can begin to understand what exactly @DIRAC1930 problem is

Semiclassical

i guess, but there's not really an "or else" in there

DIRAC1930

3:18 PM

if we were to expand psi in a particular basis and H also, $e^{\dots} $ would be a representation of the time evolution operator on the space of wavefunctions in a particular baisis right?

ACuriousMind

no, that's not a representation

at least not in the sense of group and representation theory

whatever you do in a specific basis to evaluate an expression like $H\lvert\psi\rangle$ is...just a computation, it has no larger significance

DIRAC1930

But $\hat{X}: \psi(X) \rightarrow \psi(-X)$ is one representation of the parity operator acting on wavefunctions. Meanwhile $1 \oplus -1$ is another representation of the parity operator acting on $\psi(X)_S \oplus \psi(X)_A$ or something right? where S and A are symmetric or antisymmetric respectively

Semiclassical

i'm not so sold on that being wrong. you have a 1D group of translations in time

with each possible Hamiltonian giving a different representation fo such

ACuriousMind

@Semiclassical but that has nothing to do with choosing a basis

Semiclassical

no, it doesn't

ACuriousMind

3:24 PM

@DIRAC1930 True, I'm just not sure why you used "But" there ;)

parity is, abstractly, a $\mathbb{Z}_2$ group. You've given examples for representations of it

ahhh

Semiclassical

i mean, you could define one Hamiltonian according to one basis and another according to another basis, and find they give different representations, but that's b/c they'd be different Hamiltonians

DIRAC1930

Because everyone was telling me i was wrong

lol

ACuriousMind

@DIRAC1930 well, what group do you think is being represented by the choice of $H$?

Semiclassical

the choice of basis is a distraction

DIRAC1930

Thats right, the choice of basis is a distraction lol

ACuriousMind

3:26 PM

the reason your formulae for parity define a representation are because parity squares to the identity so the set $\{1,P\}$ is a group whose representation you cn give simply by writing down a formula for $P$

DIRAC1930

But how would we find that the group was $Z_2$ is we were given a system we couldnt visualize?

Semiclassical

by knowing what our operators are doing

ACuriousMind

but the Hamiltonian doesn't square to the identity (usually :P), so there is no obvious group whose representation given a formula for it would define

Semiclassical

if you don't have some way of mathematically describing what your operators are doing, you of course can't find a group structure

(besides just "group of Hermitian operators on a particular Hilbert space")

DIRAC1930

Oh I see I think. The Hamiltonian is the generator of a U(1) group

the group is U(1)

ACuriousMind

3:28 PM

you can indeed view all self-adjoint operators as defining a representation of $U(1)$ (or $\mathbb{R}$ in the unbounded case!) via their associated one-parameter groups

I'm just not sure why you'd want to do that

DIRAC1930

Because I don't know why we even mention groups and not just go straight to representations

$Z_2$ symmetry doesn't help you in any way

ACuriousMind

...how would you talk about representations without talking about groups?

DIRAC1930

Well don't we classify the representations and not the groups

Semiclassical

we do both

DIRAC1930

But just knowing the group tells us nothing

ACuriousMind

3:30 PM

@DIRAC1930 the definition of a representation involves groups

Semiclassical

sure it does

ACuriousMind

@DIRAC1930 depends on what you know about the group

DIRAC1930

If I just said to you, there is a $U(1)$ symmetry, that could either be time invariance, spatial invariance etc.

ACuriousMind

sure

Semiclassical

sure? but you'd still know there's an associated conserved quantity

DIRAC1930

3:31 PM

Thats true

ACuriousMind

but e.g. knowing that every non-relativistic system has to carry a representation of $\mathrm{SO}(3)$ isn't "nothing"

DIRAC1930

That's true

When people say that there is an $SO(1,3)$ symmetry, everyone immediately thinks of the Lorentz transformations on a 4 vector.

Are the Lorentz transformations in the above example a representation of $SO(1,3)$ on the space of 4 vectors?

Just so I know I'm getting my terminology correct

ACuriousMind

@DIRAC1930 yes

DIRAC1930

So $SO(1,3)$ symmetry is a completely abstract notion

How would you find, for example the Little group without knowing the representation of SO(1,3)?

ACuriousMind

ah

well, here's a thing: We often talk about "abstract" groups $G$, but how do you actually write down a group?

you usually do that by choosing some set of transformations on a space and saying "that's my group" - like with $\mathrm{SO}(1,3)$, we usually define this group as the group of Lorentz transformations

that we can mathematically treat this as an abstract group not necessarily bound to its representation on $\mathbb{R}^4$ doesn't mean you have to forget that it is the Lorentz transformations

you'll never be in the situation where you have an "abstract group" and don't know anything about it, not even a single representation

Semiclassical

3:41 PM

well. you could be given a presentation of a group

ACuriousMind

true, but I can't think of a single example in physics where that would happen

Semiclassical

yeah. physics examples have the virtue of being concrete

i wouldn't be shocked if there's some case where that can happen, but i can't think of one either

@ACuriousMind i guess there are some trivial cases, e.g., saying that "a non-relativistic system carries a representation of SO(3)" is arguably like that

you don't know which representation, so you only know the group

DIRAC1930

It is very annoying lol

Semiclassical

but eh. in practice you definitely know which rep you're dealing with

ACuriousMind

@Semiclassical what I meant was that the way you explain to someone what $\mathrm{SO}(3)$ is you'll just give them the fundamental rep, even if in their application some other rep might matter

Semiclassical

3:52 PM

yeah, true

DIRAC1930

I'm a big fan of the trivial representation

Semiclassical

a fan of infidelity, i see

2

ACuriousMind

::snort::

Semiclassical

@ACuriousMind even there, though, we do tend to get to SO(3) by looking at momentum/position operators and seeing what that implies for angular momentum operators

DIRAC1930

I don't get the joke

Semiclassical

3:54 PM

the trivial representation is not a faithful representation

DIRAC1930

Oh

googles what faithful representation means

Okay

Semiclassical

and if you're not faithful, you're unfaithful blah blah blah

DIRAC1930

I think I understand the standard model

Semiclassical

@ACuriousMind so in that sense i do think we could start from so(3) as an algebra, see that we get SO(3) as a group by exponentiation, and only after worry about how this connects to 3D rotations

DIRAC1930

Only in the trivial case however

Semiclassical

3:58 PM

(not that i'd say this is the standard way)

DIRAC1930

When people talk about group theory now with these random groups or whatever, are they not worrying about physical significance?

Wouldn't that be an unanswerable question if you just started off from the group?

Because SU(2) for example could lead you to isospin or non-rel spin

Semiclassical

well, you might need to start from the group to figure out what the physical implications are

there's some pretty weird groups though

DIRAC1930

Does knowing $S_z$ commutes with $X,Y,Z$ help you in terms of physical significance?

ACuriousMind

@DIRAC1930 historically, we almost never start with a group and then wonder about physical significance

DIRAC1930

Do similar things happen with isospin?

ACuriousMind

4:02 PM

we develop the physics, and then in retrospect recognize that group theory can explain a lot of the structure we see

Semiclassical

@DIRAC1930 yes, very much. it tells you that you can measure $S_z$ and $X$ independently

DIRAC1930

Does that happen with isospin or something similar?

Well there will always be one element of the Lie algebra that's diagonal so I assume it does happen

ACuriousMind

@Semiclassical yes, you could do that, but it's not a very physics-y way of thinking

Semiclassical

yeah, fair

DIRAC1930

Do many string theorists care about physical significance anymore?

Except it looks like the people right at the top of the field

ACuriousMind

4:04 PM

depends on the string theorist

Semiclassical

ask a string theorist :P

ACuriousMind

there are certainly some mathematical string theorists who care more about the mathematically interesting aspects than physical implications

DIRAC1930

Witten seems to get so much criticism but he at least seems to be trying to address the real world

ACuriousMind

but there's also plenty of string theory that tries hard to do model-building that results in realistic low-energy theories

DIRAC1930

So in QFT, do we postulate a symmetry, enforce that the Lagrangian transforms as a scalar under that group and see what happens?

ACuriousMind

4:08 PM

usually not

why would we postulate random symmetries?

Semiclassical

maybe for effective field theory stuff

ACuriousMind

I rant about that conception of doing theory a bit here

Semiclassical

there's not a lot of symmetries we usually worry about

DIRAC1930

I'm not sure why. Isn't SU(3) postualted?

Semiclassical

that might be how it's taught now, but

ACuriousMind

4:09 PM

@DIRAC1930 the people who developed QCD had a lot of real-world collider data they were trying to explain

Semiclassical

the reason we went to SU(3) was b/c of the evidence

(don't ask me to explain said evidence, mind)

ACuriousMind

and they figured out that a gauge theory with SU(3) explains the results

there's nothing about SU(3) that would make us postulate it, it just happens to be the symmetry group of the theory that turned out to actually model reality

of course, the success of such theories led theorists to more generally consider the properties of gauge theories with some symmetry group $G$

DIRAC1930

Interesting

ACuriousMind

but that's more than just postulating a symmetry - it's considering a specific class of physical theories, namely gauge theories, and there's an explicit action there

it's the action that contains the physics, not the mere notion of a symmetry group

DIRAC1930

You can derive the Maxwell Lagrangian just from symmetry principles right?

ACuriousMind

4:14 PM

depends on what you mean by that

it's pretty much the only invariant Lagrangian you can build for a U(1) gauge field in 4D

but who told you you needed a gauge field? Symmetry can't tell you that, especially not because gauge symmetries are unphysical in the sense that you can't see the local symmetry in any measureable quantity

DIRAC1930

Yes I read your post you linked here

Do we usually start with a classical field theory and then quantize it?

It seems like finding symmetries of classical fields is easier

ACuriousMind

100% yes - we essentially don't know how to build a QFT except by quantizing a classical field theory

DIRAC1930

Okay, so i'm thinking about this in the completely wrong way

Semiclassical

i do think we sometimes 'postulate' symmetries in order to go from an unsolvable problem to an approachable one

but that's b/c we want to get some useful toy model

ACuriousMind

the closest to "building a QFT without building a classical theory" I know is the approach Weinberg takes in his QFT books

because he constructs the fields out of c/a operators only after doing a lot of heavy lifting

Semiclassical

4:22 PM

that's one reason why QFT feels more mysterious to me than QM

ACuriousMind

but still, he ends up with something that's clearly also an action for a classical field theory, so...

Semiclassical

it's easy to come up with simple QM models which lack a classical analog

ACuriousMind

there are some CFTs that lack classical actions

DIRAC1930

Is Landaus book good for classical field theory

Or does it it not go into the stuff I want to learn?

ACuriousMind

or only have them "locally" or whatever, I don't understand what's going on there fully - the thing with 2d CFT is that having the Virasoro algebra gives you so much information you don't need an action to build a quantum theory

Semiclassical

4:24 PM

QFT at present is much more tied to classical field theory than QM is tied to classical mechanics

oh, speaking of things involving 2D cft

there was a talk at my uni (which was recorded, i couldn't see it in person) talking about getting 'simple' models of quantum gravity via averages of CFTs :)

which...bizarre, but also neat

e.g. this paper by the presenter + Ed Witten: arxiv.org/abs/2006.04855

DIRAC1930

What's the best book on classical field theory that's short?

Do these gauge groups pop up there too?

ACuriousMind

you can have classical gauge theories, sure - EM is one, after all

but non-Abelian classical gauge theories are rare

Semiclassical

classical CD, oof

DIRAC1930

So non-abelian gauge theories are really only used in QFT?

Semiclassical

i imagine there's some weird examples you could come up with in statistical mechanics

something something biaxial liquid crystal

(though i don't think those have ever been observed)

DIRAC1930

4:31 PM

I had a course on classical field theory that had a whole chapter on non-abelian gauge theories but I imagine that part was really quantum but didn't really make a difference

ACuriousMind

well, there is stuff like the gauge theory of the falling cat

that's SO(3), I think

Slereah

@ACuriousMind You can even make it $E(8)$, but then the cat complains

ACuriousMind

if the cat complains, you can't make it that :P

Slereah

@DIRAC1930 GR is a non-Abelian gauge theory

ACuriousMind

just like you just can't stand up if there's a cat lying on you

it's natural law

Slereah

4:40 PM

@ACuriousMind I've done it

I'm not letting a cat boss me around

ACuriousMind

y-y-y-y-you monster!

DIRAC1930

@ACuriousMind What did you do your PhD in?

Nihar Karve

funny you should say that

ACuriousMind

@DIRAC1930 I don't have a PhD

Slereah

I don't know what gauge groups arise in normal people physics but I know condensed matter physics has a lot of weird shit in there

Semiclassical

4:47 PM

biaxial nematics would have non-abelian topological defects

too bad i don't think they've ever been seen

DIRAC1930

You should write a small monograph summarizing everything

Semiclassical

"small"

DIRAC1930

As in just the bare essentials

Starting from classical field theory and ending with gauge theory

Slereah

@DIRAC1930 is it gonna be one of those monograph that starts with "Consider a $\infty$-topos"

@ACuriousMind What about the non-parity term

ACuriousMind

@Slereah which one?

Slereah

4:52 PM

The one that's like uuuuh

$\varepsilon_{abcd} F^{ab} F^{cd}$

Or some garbage like that

ACuriousMind

do you mean $F\wedge F$ that's just topological?

Slereah

Maybe?

ACuriousMind

you can add it to the Lagrangian but it doesn't do anything classically

Slereah

Peskin talks about it

DIRAC1930

@ACuriousMind When did you start learning everything you know?

Was it in your spare time after university?

ACuriousMind

4:55 PM

@DIRAC1930 no - I studied physics at uni

DIRAC1930

Yes but you know way more than a standard undergraduate course

I'm assuming you learnt all of gauge theory youself

ACuriousMind

well, for one thing, I didn't just do an undergraduate course - I did a Bachelor's and Master's course (Master and PhD are two entirely separate stages here)

for the other, Heidelberg is pretty atypical in that it is feasible (and recommended for people interested in theoretical physics!) to take QFT in the third year of undergraduate/Bachelor

DIRAC1930

In my 2nd course on QFT, I studied non-abelian gauge theories etc., however as you can tell, it was way to advanced for me so I learnt nothing

ACuriousMind

I essentially just took every math and theoretical physics course I could, I don't have anything special I did to learn what I know

DIRAC1930

I think I'm just dumb

More Anonymous

5:03 PM

@DIRAC1930 welcome to the club

:P

DIRAC1930

haha lol

More Anonymous

Everyone is part of it except Acuriousmind :P

ACuriousMind

well, you seem to lack a lot of prerequisites- we did proper theoretical mechanics (so also Lagrangian and Hamiltonian mechanics and basic classical field theory) in our first year and proper quantum mechanics in our second year

I don't think a lot of places do that to the extent we did

More Anonymous

@ACuriousMind I never learned classical field theory and was thrown straight into the pool of QFT and drowned

DIRAC1930

Me too

More Anonymous

5:05 PM

@ACuriousMind what book would u recommend for self study?

ACuriousMind

I can't recommend any books because I didn't learn this stuff from books :P

More Anonymous

@ACuriousMind Any free lecture notes?

:P

ACuriousMind

the only textbooks I have are more specialized stuff

More Anonymous

@ACuriousMind QFT for the gifted curious mind I suppose :P

ACuriousMind

@MoreAnonymous Here's the notes of the course I took

5

More Anonymous

5:08 PM

@ACuriousMind Thank u so much ... I feel the genius flowing in me

ACuriousMind

the Weinberg book is probably the most "elementary" book on QFT I've really read, and I wouldn't recommend that as a starting point to anyone

DIRAC1930

I think my problem is that I hate doing long calculations

ACuriousMind

not because it's bad but because it's very different in approach from what the rest of the QFT world does :P

DIRAC1930

That looks like the way two of my courses were done

Slereah

I would probably recommend Schwarz for QFT

More Anonymous

5:11 PM

I recommend giving up on physics and living a worthwhile remaining life

:P

DIRAC1930

I think the issue is that most QM books can be self contained because the math is a lot easier

QFT assumes so much background knowledge

More Anonymous

Honestly there are too many ways to think of QFTs ... Personally I like the many body approach

Its similar to QM

Slereah

Every physical theory has a bunch of different formalisms

GR and QFT are particularly horrendous

I'm not sure what's the worst GR formalism, but special mention to that paper I found once that was "GR from quaternions"

ZeroTheHero

6:08 PM

would anyone know of situations where the Wigner-Brillouin perturbation series is clearly advantageous to the usual Rayleigh-Schrodinger series? Methinks in degenerate (or nearly degenerate) perturbation theory for large Hilbert spaces but that just reflects my ignorance of applications of the WB scheme.

John Rennie

7:01 PM

An entertaining little experiment:

I Rented A Helicopter To Settle A Physics Debate

►

fqq

7:59 PM

@DIRAC1930 most QM books and courses just skip all the hard maths. Functional analysis, $C^*$ algebras etc are definitely not easier than representation theory

Semiclassical

most intro books stick to being 'formally correct' without attempting to be rigorous

e.g. expressions like $\psi(x)=\langle x|\psi\rangle$ and $\langle x|x'\rangle=\delta(x-x')$

within the standard construction, these expressions don't really mean anything: mathematically there's no such thing as an $|x\rangle$ ket

you can make that rigorous by including a rigging for your Hilbert space, but i doubt most physics people care about that

myself i tend to treat them as mnemonics

DIRAC1930

10:28 PM

What do people think of The Road to Reality by R. Penrose?

'Any reader who is not too concerned with the full details of the formalisms
that I shall be presenting is recommended simply to ignore this issue
completely. (Most experts would do the same—until the moment comes
when they have to write articles or books on the topic!)' lol

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