The h Bar

Slereah

00:33

Oh do you mean historically or like concept-wise

Relativisticcucumber

02:29

@ACuriousMind do you have advice as to how one can learn the group theory needed for a rigorous study of quantum mechanics? is it worth it to just pick up any abstract algebra book?

Silly Goose

i mean concept wise @Slereah

e.g. like we start with a set of rays, then we say each ray represents a physical state

and so on and so on

i basically want to know where algebras of operators comes from in QM

right now my understanding is like: okay by Wigner's theorem symmetries can always be represented as linear unitary (or antilinear antiunitary) operators. so then these operators become our focus. we define these operators over sets of rays which act as physical states. then we want to use linear algebra because it's easier so we use the associated lie algebras of the lie groups of the operators and represent them as matrices and so on

but i guess like okay then is the starting point of QM to 1) assume that rays are physical states and so on, 2) assume that symmetry plays a preferred role

Relativisticcucumber

peep @SillyGoose

Silly Goose

squack @Relativisticcucumber

:D

Relativisticcucumber

how do i learn group theory @SillyGoose

whats ur favorite city @SillyGoose

Silly Goose

02:48

well idk if any of what i am saying is right XD. say we have like a connected lie group: would an example be SU(2)? Where the elements of SU(2) are the abstractions of rotation operators on the hilbert space of a spin-1/2 system. Then what does the associated lie algebra of SU(2) contain? Is it just all the generators of rotations (i.e. spin operators)? And also so each element in the lie algebra already acts on the hilbert space. but when we use the exponential map,

are we representing an element from SU(2) in terms of an element from its associated lie algebra in such a way that the exponential is now acting on the hilbert space (so in effect representing the original unitary operator in way such that it can interact linear algebraicly by acting on the hilbert space)?

and then we can further represent the exponential (and thus the unitary operator) as well as elements of the associated lie algebra as matrices, for example?

@Relativisticcucumber mayn i don't know no group theory myself D: we should learn it together hehe

@Relativisticcucumber hmm maybe tokyo hehe what about yours

Relativisticcucumber

shanghai

Silly Goose

well okay if i had to condense everything into one question. say we have SU(2) and its associated lie algebra su(2). Then, there is the exponential map $e: su(2) \rightarrow SU(2)$. Okay, so far nothing acts on a Hilbert space. Then, we define some action of su(2) over a Hilbert space? Then, do we define another map from the exponential forms of SU(2) into like their taylor expansions so that they can act on Hilbert space as elements of su(2) do?

ACuriousMind

08:12

@Relativisticcucumber you need to pick up a book specifically on Lie theory/representation theory of Lie algebras. If you do what mathematicians call "group theory" you'll usually learn about stuff like finite groups or Sylow's theorems

@SillyGoose there isn't a single way to axiomatize QM

you can choose to start with the Hilbert space and the states, or you can start with the algebra of observables, see Dirac-von Neumann axioms

Slereah

there's a big laundry list of QM axiomatizations

Slereah

08:33

Here is a question that popped up on another server : what is the spin group in zero dimension

is it the trivial group or is it not defined at all

ACuriousMind

why does it matter?

Slereah

Never said it did

ACuriousMind

then I'd say it is the trivial group

Slereah

Also my thinking but the other guy was arguing a lot that the spin group's possible definitions involved a lot of determinants and other such quantities which would not work out in zero dimensions

as a lot of usual definitions of it are based on the assumptions of Lie matrix groups

Or in the Clifford algebra definition it involves the elements being $Q = \pm 1$, which probably also doesn't work

ACuriousMind

ah no, actually I think it's $\mathbb{Z}_2$ :P

the spin group in dimension p+q is the double cover of $\mathrm{SO}(p,q)$, $\mathrm{SO}(0,0)$ is the trivial group, and $\mathbb{Z}_2$ is the double cover of the point

Slereah

08:44

Is that true in general?

ACuriousMind

I mean that's how I'd define the spin group

(and Wiki does so too)

Slereah

IIRC the Lorentz group in 1+1 is just $\mathbb{R}$ but the spin group isn't two copies of $\mathbb{R}$?

Or is it?

also $\mathbb{Z}_2$ doesn't sound like a very faithful rep on a point

ACuriousMind

and also there isn't really a difference between 0 and 1 dimensions here - $\mathrm{SO}(1)$ is just the trivial group, too

Slereah

Well in one dimension things are a bit more obvious

ACuriousMind

@Slereah why would the spin group need to be faithfully represented on the point?

in 3d, SU(2) isn't faithfully represented on 3d vectors, either

Slereah

08:48

fair enough

ACuriousMind

the $\mathbb{Z}_2$ is always the kernel of that rep because we defined the spin group to be a double cover of the actual rotation group

Slereah

His argument was mostly that the rotation group isn't well defined because there's no element of det = 1

Which I guess but probably not how I'd define the rotation group geometrically

although idk how matrices work in zero dimensions

What's the determinant of a 0x0 matrix

7

A: Is the $0\times 0$ matrix (zero-times-zero matrix) a well-defined concept?

Well, one could say even more about the $0$ x $0$ matrix: Yes, it operates on the zero vector space (which contains only one single element = $0$). So, it maps $0$ to $0$ since there is no other possible image. Hence, it must be the identity mapping on the zero space, and therefore, it is its own...

ACuriousMind

hm, yeah

might be that SO(0) is actually empty

Slereah

Maybe not!

ACuriousMind

in the end I don't think it matters for anything, really

Slereah

08:59

That never stopped me before

Relativisticcucumber

@ACuriousMind okay thank you i will look for one

Debanjan Biswas

09:49

@JohnRennie What includes a PhD research on Physics? Is it theory (like GR, QFT) or experiment?

Slereah

It could be any of them

John Rennie

@DebanjanBiswas My PhD was experimental. There was some theory used to try and model the experimental data, but mainly it was experimental design and results.

Slereah

also "theory" and "experimental" can mean a variety of things

It could involve things like numeric simulations or data analysis

Debanjan Biswas

If someone research on a specific compound's nature and models it with a theory, will he get a PhD ?

Slereah

Probably ask your thesis advisor on the topic

John Rennie

10:05

@DebanjanBiswas My PhD was studying a solid state photoreaction between amorphous germanium selenide and silver. So that's an example where I was studying specific compounds. But it involved lots of different experimental techniques.

Slereah

Isn't it weird that the usual numbers are called "real numbers", because they are named in opposition to imaginary numbers

the basic properties of real numbers I wouldn't imagine being "They're not complex"

Rafa

10:36

Hello, sorry for asking the following question here! I tried to do a search on Google and physics SE but could not find anything clearly related (maybe just using the wrong search terms). Before posting it on the proper SE, I thought it would be a good idea to confirm here that it is not just a very stupid question

Basically, if I understand correctly, since the energy levels of electrons in atoms are quantised, only photons with a given energy/wavelength (corresponding to the difference between 2 energy levels) can be absorbed, which gives rise to the discrete absorption bands

But I was wondering, is there some sort of tolerance within the energy that the photons must have? I guess there must be, since otherwise, it would be impossible to get any absorption, since a specific, extremely exact wavelength would be impossible to achieve

Slereah

there's two big factors involved in this case

Rafa

From where would this tolerance of the energy of photons to be absorbed arise? Has it been measured?

Slereah

First you have the uncertainty relation, which means that even in this case, there is some spread on the energies involved

And also in actual physical cases, the atoms involved are in motions at various speeds, meaning that the atoms will typically see photons doppler shifted to various wavelengths

So in general you're gonna have a pretty wide variety of wavelengths that can be absorbed by a material

Rafa

I see, but with all those factors considered, each individual photon would have a single wavelength, and each individual atom would have a single wavelength that it would absorb. That would make it effectively impossible to ever find an exact match, wouldn't it? Unless there is some sort of intrinsic tolerance?

John Rennie

@Rafa When a photon hits a hydrogen atom the oscillating electric field of the photon perturbs the atom and it puts it into a new time dependent state that can be written as a superposition of the original orbitals.

Slereah

10:45

Also neither the atoms nor the photons really have a "single wavelength"

things aren't quite that peaked

John Rennie

Then this superposition evolves with time, and it can evolve back into the original state and a photon meaning nothing happened - the photon just carries on, or it can evolve into a new state.

The probabilities for the various outcomes depend on the frequency of the original photon i.e. its energy.

Slereah

In real life you never have photon states that are entirely peaked at a single wavelength

John Rennie

The probabilities can be calculated using Fermi's golden rule

OK so far?

Rafa

I see, that makes sense! I guess from that arises the fact that absorption lines are not exact lines, but rather sort of a gaussian like curve

Then, if after the atom is perturbed by the electric field, and the system gets into a new time dependent state that will evolve with time, and might either evolve back into the original state, or could evolve into a new state, if it eventually evolves back into the original state, what happens to the photon "meanwhile"? If the system eventually goes back to the original state (i.e., "nothing happened"), and the photon just goes on

would this photon be not distinguishable in any way that another one that had followed the same path but never found any hydrogen atom along its path?

John Rennie

The atom and the photon become entangled and form a composite state that is part atom and part photon.

Rafa

10:57

Would that somehow delay the progress of the photon along its path? I'm asking because, if the evolution of this composite state is time-dependent, and therefore takes some time to either evolve back into the original state, or to evolve into a new state (photon absorbed), intuitively it makes me think that this photon would be somehow delayed from one following the same path but that never became entangled with an atom

Oh, maybe that has something to do with materials index of refraction?

John Rennie

12:23

@Rafa Yes, that is exactly how the refractive index of a material works. It's related to how much the light interacts with electrons in the material.

5

A: What causes light to refract?

The reason that light travels more slowly in a dielectric is because it interacts with the electrons in that dielectric. Light has an oscillating electric field, and if any charged particle is in the path of the light that particle will feel an oscillating force due to the oscillating electric f...

ACuriousMind

13:35

@RyderRude Please don't use comments to discuss ideas only tangentially related to the post being commented on. There are various ways to construct an analogue to "spin" in classical mechanics, which is why my answer says that "spin is not classical" really just means that the usual notion of point particles in classical mechanics doesn't have spin, not that it is somehow impossible to add spin to classical mechanics

see e.g. physics.stackexchange.com/q/450899/50583 for discussions of several approaches to "classical spinors"

but see also physics.stackexchange.com/a/284889/50583 for arguments why such classical models, while they might technically describe classical objects with non-integer spin, don't really describe anything "real" - the use of these models is as starting points for quantization perhaps, but not as describing real-world classical systems

and also physics.stackexchange.com/a/285314/50583 in that same question where the $S^2$ model of spin is explicitly discussed

Rafa

@JohnRennie Amazing, thank you so much for the explanations and pointing me to that answer! It is really enlightening to get a better understanding of what is actually going on in these processes such as refraction

bolbteppa

@DIRAC1930 finally have a serious explanation of how he introduces spinors:

Kneel before my understanding of spinors...

Absolute madness going on

Jack Rod

13:51

Hi everyone, from past one week I tried to search chatgpt but some articles on web explain that all these app are fake do any of u have real link of chatgpt app?

Any help will be very much appreciated

Slereah

This isn't a superior understanding of page setting certainly

bolbteppa

Ask chatgpt to explain spinors and you'll get a cookie cutter repetition of every other book which all say the same thing that explains none of this

Slereah

you will probably get Wikipedia on spinors

Ryder Rude

@ACuriousMind Wow I was exactly talking about the idea in that last link. I think this can be considered a classical spin, as the spins commute here. It does not strictly describe any real-world entity, but the usual classical mechanics doesn't strictly describe any real-world entity either.

Slereah

you can have non-commuting spinors classically

Ryder Rude

14:02

i mean commuting in the sense that the x, y,z components are simultaneously well- defined

the poisson brackets don't commute though

do you mean the dirac field interpreted as a classical field?

Slereah

Sure

Ryder Rude

i mean the poisson brackets aren't zero

Slereah

you can have anticommuting classical spinors

Ryder Rude

grassman variables?

Slereah

yeah

it is the usual trick to have such objects

ACuriousMind

14:04

those are fermions, not necessarily spinors

Ryder Rude

yeah

grassman variables rotate like spinors, no?

ACuriousMind

no, not necessarily

you can package them into spinors if you want something like the classicla Dirac field

Ryder Rude

i mean they're defined the rotate like spinors in the case of a dirac field

yeah

Slereah

The commuting quality of a field and its representation under the rotation group are two different things

ACuriousMind

but a priori it's perfectly possible to extend classical Hamiltonian mechanics to arbitrary super-Poisson manifolds, there's no spin-statistics for classical models

Slereah

14:05

You can have commuting or anticommuting scalars

and commuting and anticommuting spinors

Ryder Rude

i read that there's this theorem in which starts by considering a fock space of particles, assumes some nice properties and then proves that the fock space must be a representation of a super lie algebra

Slereah

In QFT they are related due to the spin statistics theorem

but otherwise that is not the case

Ryder Rude

i think this is a very deep result

Slereah

You actually do use anticommuting scalars for some applications in fact

Ryder Rude

it's called... i'll look up the theorem name

13

A: What is the spin-statistics theorem in higher dimensions?

When formulating a physical theory, one usually begins with a set of axioms. The theory itself will be just as useful as its axioms are accurate. In particular, when dealing with a supposedly fundamental theory, the following set of axioms seems natural and well-motivated by experiments: There ...

do you think this proves why we only care about commuting and anti-commuting variables

ACuriousMind

14:11

@RyderRude that's...not what the HLS theorem says

the theorem says that the (super)Lie algebra that is the symmetry algebra of a physical theory must be a direct sum of the super-Poincaré algebra and another algebra, i.e. "spacetime and internal indices don't mix"

Ryder Rude

but that answer says a different thing

it says they proved why the super lie algebra arises in the first place

ACuriousMind

oh, you mean Deligne's theorem

Ryder Rude

yeah

Slereah

@ACuriousMind Walked a mile and my EM gauge shot up 😨

ACuriousMind

then you've still not correctly paraphrased it, it's not about starting with a Fock space

Ryder Rude

14:15

what i understood is that they started with a fock space and assumed some nice properties under particle exchanges

i havent seen the proof

only an exposition article

do you think this theorem proves why our classical theories are either commuting or anti commuting and why our evolution laws are according to a super lie algebra?

ACuriousMind

it doesn't prove anything about classical theories

Ryder Rude

i thought it might be eventually connected to that..like a line of dominos

because the classical theories are also super lie algebras

it is immediately only connected to super poincaire algebra, right?

@ACuriousMind does this really have nothing to do with the fact that the classical theories employ super lie algebras?

Slereah

Super Lie algebras are a pretty generic thing

What else would you include

Ryder Rude

variables that neither commute nor anti commute?

ACuriousMind

the actual statement is more like this: The assumptions about particles and combining them that AFT talks about there turns the set of all possible state spaces (not just the Fock spaces, the spaces of all the possible different finitely many particles) into a tensor category

Slereah

14:22

There is also that

ACuriousMind

Deligne's theorem states that every such tensor category is the category of representations of a super-Lie algebra

Slereah

If you're interested in that issue you can look up braiding

That's the theory involved in the commuting of variables

Ryder Rude

idk category theory lol

Slereah

It is not directly related to category theory

Ryder Rude

i googled braiding

is there any natural explanation for why the classical theories are super lie algebras?

Slereah

14:25

In two dimensions you are allowed to have fields that neither commute nor anticommute

Ryder Rude

nothing related to physics came up

oh

is there any natural explanation for why the classical theories are super lie algebras?

ACuriousMind

I don't know what you mean by that

Ryder Rude

there's something called the braid group that showed up

ACuriousMind

as in, I don't know what it is supposed to mean that a classical theory "is" a super-Lie algebra

Ryder Rude

i mean, why are super lie algebras the natural things to consider

ACuriousMind

14:26

who says they are?

who is doing this considering?

Ryder Rude

every classical and quantum theory has them

why

ACuriousMind

well, yes, because continuous transformations naturally form (super-)Lie groups and their infinitesimal versions are (super-)Lie algebras

Ryder Rude

i had that i mind..

is there all there is to it

ACuriousMind

Lie groups and algebras are just the mathematical "packaging" for what we mean by a continuous symmetry/transformation

Ryder Rude

but why the super stuff?

Slereah

14:28

Sometimes things do not commute

Ryder Rude

right, lie groups basically correspond to a manifold

this is the story i told myself

but then the super stuff came up

Slereah

In addition to the gauge groups, you may also have to consider the symmetries of interchange in your theory

ACuriousMind

well, I would argue the "super" stuff isn't even known to be natural :P

we have no evidence for supersymmetry

Ryder Rude

and now selereah says 2d theory need neither commute nor anti commute

bolbteppa

There are no spinors in classical mechanics, so if you want to describe quantum objects as the quantization of a classical field theory, you better use anti-commuting variables in your classical theory so that they are non-physical on a classical level

Slereah

14:30

@ACuriousMind I mean you can use supersymmetry for things that aren't SSM

ACuriousMind

oh, sure

but then it doesn't classify particle irreps :P

Ryder Rude

so i guess the super stuff isn't the end of story either

ACuriousMind

which is what Deligne's theorem is about

Ryder Rude

@ACuriousMind but the classical dirac theory uses the super bracket

ACuriousMind

so what?

Ryder Rude

14:31

idk.. why are anti commuting objects natural

i thought deligne's thoerem was about this lol

tough luck

Slereah

They're a possible braid group representation

ACuriousMind

@RyderRude I mean that's just spin-statistics

Ryder Rude

i will look up braid groups then

Slereah

You fool

ACuriousMind

unless you're in a dimension with anyons, spin-statistics is the reason you need anti-commutators

Slereah

14:33

These braid groups are too strong for you traveller

Ryder Rude

yeah, you can motivate it using experiments

wdym, it's just a bunch of threads

i will save it for things to study later

bolbteppa

You have to remember that one can't trust a single theorem in physics

Ryder Rude

friendship ended with deligne's theorem

now braid group is my best friend

bolbteppa

I'd bet internet points that one can apply the old proof of spin statistics to the higher dimensional case in some form without any of these big words and end up with the same thing, one doesn't need to study the classification of super lie algebras before beginning their study of classical mechanics just so that years down the line they are equipped to say they are working in a sort-of-maybe-maybe-not well-defined maybe/maybe-not really appropriate vector space

that allows us to use fermions and bosons in the same action where we can do things like brst quantizatione etc...

Ryder Rude

i wasnt interested in a proof of spin statistics theorem as much as a proof of why super lie algebras arise in physics

but i get your point, yeah. we shouldnt rely on a single theorem to justify the foundations of physics

Slereah

14:42

the basic issue is that a lot of those weird behaviours only really occur in QFT and trying to extrapolate why they are there in a classical context will probably not give you much satisfaction

bolbteppa

The first time it comes up is when you try to introduce supersymmetry, years later after the fact people realized this stuff on a super pedantic formal level is the 'right' language to describe a bunch of things people easily do in physics without even thinking or caring about the formal underpinning

Slereah

You can try to fit in a lot of different mathematical structures in classical mechanics, if you are loose enough with what you consider classical mechanics

We consider super lie algebras because that's the sort of objects we find in modern theories

maybe there are even more general objects involved that will pop up later

Ryder Rude

i agree bolbteppa, math can really be a big red herring sometimes

Debanjan Biswas

@JohnRennie

Ryder Rude

all of our models are approximate anyway. so it can waste time to study the minute details of the mathematical objects

bolbteppa

14:45

It's part of a broader study of trying to extend the Poincare group to allow non-trivial extra symmetries, e.g. including internal groups in non-trivial ways

Coleman–Mandula theorem

In theoretical physics, the Coleman–Mandula theorem is a no-go theorem stating that spacetime and internal symmetries can only combine in a trivial way. This means that the charges associated with internal symmetries must always transform as Lorentz scalars. Some notable exceptions to the no-go theorem are conformal symmetry and supersymmetry. It is named after Sidney Coleman and Jeffrey Mandula who proved it in 1967 as the culmination of a series of increasingly generalized no-go theorems investigating how internal symmetries can be combined with spacetime symmetries. The supersymmetric ge...

Slereah

Coleman-Mandula is a QFT-specific theorem, btw

it will not tell you in a classical context why you can't consider more general object

bolbteppa

Supersymmetry is a way to bypass this - by adding non-Lie algebra generators you can get non-trivial additional symmetries. Why would you even think to do this? A good place to look is the early papers and their motivations

Slereah

you totally could make some unholy classical theory that mix gauge and diffeomorphisms

bolbteppa

The proof in section 4 here is enough on CM for the average bear

Ryder Rude

@bolbteppa so this has everything to do with supersymmetry and nothing to do with why the classical dirac theory employs a superalgebra

then my interest dies lol

@Slereah i agree. there's no strict definition of a classical theory

bolbteppa

14:49

Yes, you will not gain any intuition about spinors from this nonsense

Slereah

You will never get intuition about spinors

bolbteppa

But you can do things like talk about classical brst quantization using this stuff, where you need to add anti-commuting variables, but even when you get to that stuff you can avoid the dark black rabbit hole of formalism

Ryder Rude

thanks for saving my time. i'm not pursuing any understanding of supersymmetry

Slereah

bolbteppa

Well my little picture above will give this month's answer to what a spinor is

It doesn't fully explain the link to the description of a spinor in the quote I gave yesterday, but it's apparently/potentially the most fundamental way to think about them as crazy as it is...

Ryder Rude

14:53

i never felt any need to get a pictorial intuition for them

the math is all there is imo

Debanjan Biswas

@JohnRennie Thank you for it. Now I understand about experiment part. But will someone get PhD if he only provides a mathematically correct theory without any experimental evidence?

bolbteppa

You can see one link in that the above description is talking about a world of flat (higher dimensional) planes living in a (higher dimensional) quadric and the intersections of these planes, all associated to a single point, so that rotations transform this configuration of planes into another configuration, where the spinor is a description of the configuration of planes... But I don't get the ${\nu + 1 \choose 2} E_{\nu}$'s thing...

Ryder Rude

who will go see ant man quantumania

ACuriousMind

I don't like superhero movies

Ryder Rude

15:08

it's about a guy who rules the multiverse and he's in the quantum world for some reason. so there's some overlap with the physics lover crowd

if you can tolerate shitty physics

Slereah

15:24

Everything in the universe is about physics

Many movies involve rocks

which are very physical objects

Slereah

15:49

People have a lot to say about Euclid's life but what I want to know is what are the portraits of Euclid based on

Why does he have a sleeping cap on

Ryder Rude

i thought he was 16 when he died

Slereah

Basically nothing is known of his life, so idk where you'd get that

Ryder Rude

maybe i read about someone else. a very influential mathematician. died at 16 in a sword or gun duel

Slereah

You're thinking of Galois, for reasons that are unknown

they have nothing in common outside of being mathematicians

Ryder Rude

sorry. it is indeed galois.

Slereah

15:59

I don't know why people are always so shocked that people would change books of famous people in ancient times

Wouldn't you make a better version of a textbook if it was legal

Ryder Rude

maybe people published the corrections with their own name?

ACuriousMind

If Euclid died in a gun duel that would raise a lot more questions :P

Slereah

@ACuriousMind He was ahead of his time

Ryder Rude

in the ancient times, there used to be something called math dueling

people would reserve their greatest math formulas for the duel

winner took the lead mathematician spot in the official committe or something

ACuriousMind

@Slereah shot by a time traveller trying to stop Euclidean geometry?

Cyber-Pythagoreans from the future who try to kill Euclid for math sins

great B movie plot

Ryder Rude

16:04

euclid thought the bullet would go straight. he forgot that earth's gravity curved the space

is that a good joke? :P

Slereah

@RyderRude Do you mean like this

Albert Einstein vs Stephen Hawking. Epic Rap Battles of History

►

bolbteppa

Sleeping caps were all the rage back in the days of barley and rye

Ryder Rude

@Slereah LMAO

Slereah

> I'm the giant whose shoulders you'd have stood on, if you could stand

Brutal

The general vibe I'm getting of ancient greece is that this is what basically happened rly

Ryder Rude

i remember there was a duel about finding the solution to the cubic equation

it is in a veritasium video

the winner dude had found the solution and had reserved it to himself

Slereah

16:08

Not so much dueling but certainly debating on the topic

greek mathematicians were wont to slander each other for their dumb ideas

bolbteppa

Weyl:

> "Only with the spinors do we strike that level in the theory of its representations on which Euclid himself, flourishing ruler and compass, so deftly moves in the realm of geometric figures. In some way Euclid's geometry must be deeply connected with the existence of the spin representation."

Ryder Rude

hmm.. i guess the connection is euclid geometry --> pythagoras theorem --> rotation group --> spin

bolbteppa

$x^2 + y^2 = z^2$ is $x^2 + y^2 - z^2 = 0$ which leads you to the spinors I've been talking about above

parz

‘lo!

bolbteppa

Lorentzian spinors are associated to the Pythagorean theorem...

Silly Goose

16:13

What’s up with spinors being so topical xD

Ryder Rude

say u have a spinor floating in space. now you just turn around 360 and then face it again. it's flipped now. this really bothered me when i first learned them.

Slereah

Not with me

I think spinors are lamd

I think spinors are lame

Ryder Rude

lol

Silly Goose

Are spinors always states in qm

Slereah

I don't know what you mean by that

Ryder Rude

16:18

they're also used to describe the dirac field (which is a classical field)

bolbteppa

This 360 degrees direction thing is more complicated in higher dimensions

Ryder Rude

@bolbteppa are you defining them in higher dimensions as a 1/2 integer representation of the higher dimension rotation algebra?

ACuriousMind

1/2 for which number?

higher dimensions have more than one spin number - much like 4d already has reps labeled by a pair of numbers $(s_1,s_2)$

Ryder Rude

oh

yeah, like the lorentz group

in that case, one of them being half classifies as a spinor

bolbteppa

The $n = 2 \nu + 1$ dimensional analogue is: "Any pure spinor can be defined in a definite manner as a polarised isotropic $\nu$-vector"

ACuriousMind

16:21

now, for a 4d field theory, you still have that the particles have little group SO(3), so you can still reduce this to a single spin number and figure out the "non-relativistic spin number" is $s_1 + s_2$

@RyderRude but that would mean you get more than just Dirac and Weyl "spinors"

Ryder Rude

yeah

ACuriousMind

the algebraic unambiguous definition is to say that Dirac spinors transform in the unique irreducible rep of the Clifford algebra of that signature

Ryder Rude

yeah, they're well defined

ACuriousMind

Weyl spinors happen in even dimensions because this Clifford irrep decomposes into two SO(p,q) irreps

Ryder Rude

maybe we can call any representation that isn't part of the tensor group a "spinor"

by tensors, i mean objects built out of tangent vectors

spinors cannot be built like that. they're like the square root of a vector

bolbteppa

16:24

The spin rep is just a big bag of tensors all mixing into one another

ACuriousMind

you can and this ambiguity in what exactly "spinor" means makes discussions in arbitrary dimensions very annoying

Ryder Rude

yeah, it's very annoying. agreed

Slereah

There's tons of objects transforming under the rotation group that are neither spinors or tensors

Ryder Rude

like what?

fields?

bolbteppa

Right, the question is why this big bag of tensors is fundamental, i.e. a 'fundamental representation of the orthogonal group' vs being some irreducible thing you just construct by hand

Ryder Rude

16:26

scaler fields

ACuriousMind

I'd always use "spinor" without further qualification to mean a Dirac spinor, but of course people also call things like a Rarita-Schwinger field "spinorial"

bolbteppa

and what is the geometry behind it, again we're led back to this collection of isotropic planes

ACuriousMind

the broad algebraic definition is to just say that spinor representations are the ones that are representations of Spin(p,q) that do not descend to representations of SO(p,q)

Ryder Rude

i call any representation a spinor which you can't get by outer producting tangent vectors

they're like the missing ninja representation

you only see them after you solve the representation problem exhaustively

ACuriousMind

that's equivalent to what I just said

Ryder Rude

16:28

agreed.

any freaky representation is a spinor

ACuriousMind

you can't produce non-SO(p,q) reps of Spin(p,q) if you start out by tensoring SO(p,q) representations

bolbteppa

None of the old boys in all these links I've been giving were satisfied by that picture

ACuriousMind

I'm not beholden to any boys, old or not :P

Ryder Rude

i was very satisfied by this personally

it's the perfect motivation to spinor

first figure out tensors, then ask if any ninja representations exist

and if you take outer products of these ninjas, you may end up with one of the normal tensors

so they're like the roots of the non-freaky representations

right?

ACuriousMind

yes

that's at least one of the possible interpretations of what the "square root" phrase means

Ryder Rude

16:34

yeah, the square root phrase usually refers to something else entirely

it felt too complicated to me

it was something to do with norms and determinants and stuff

forgot about it

bolbteppa

> "But in almost all these works, spinors are introduced in a purely formal manner, without any intuitive geometrical significance; and it is this absence of geometrical meaning which has made the attempts to extend Dirac's equations to general relativity so complicated."

That's how he motivated this geometric way of introducing them

Ryder Rude

i also thought there might be problems in extending this idea to GR. because lorentz transforms no longer hold a special place there. one still has tangent vectors and tensor transformations. but idk how to define their roots

ACuriousMind

@RyderRude the answer is spinor bundles

Ryder Rude

one approach i read is to apply the clifford algebra technique with a non minkowski metric

ACuriousMind

you indeed need additionally a spin structure on the manifold, i.e. not all in principle possible spacetimes possess spinors

Ryder Rude

16:39

so we have already figured this out too

is this related to the clifford algebra technique with a general metric?

ACuriousMind

you can do this in terms of the Clifford algebra since effectively you need to consistently define what the $\gamma^\mu$ mean at every point, yes

Ryder Rude

thats very pretty. i love it when something generalizes nicely

Obliv

would the acceleration of a two object system due to gravity be additive

also objects on earth experience acceleration due to earth's gravitational field but is it more accurate to say every object experiences a different acceleration if you couple it with the earth together

since the earth would be accelerating to the object however negligible it may be

Silly Goose

17:42

I am having trouble understanding how the definition provided is equivalent to the qualitative description of equivalence above. I get the condition of permuting the indices of the algebras of operators. however, I am uncertain of how to connect the idea of conjugation by local unitaries to the conjugation of what seems like not necessarily local unitaries in the definition

Additionally, i am unsure how to interpret $UHU^{-1}$ since $U: \mathcal{H} \rightarrow \mathcal{H}'$, but $H: \mathcal{H} \rightarrow \mathcal{H}$?

bolbteppa

@SillyGoose what is the motivation for reading a paper this formal

Silly Goose

it is enjoyable and also for thesis

Mr. Feynman

@SillyGoose Oh, what's your thesis about if I may ask?

ACuriousMind

@SillyGoose the $U$ in the definition is unitary and the condition that the $A_i$ are mapped to $A_{j_i}$ by it is the condition that it's local

Silly Goose

i think very roughly it is going to be about: given a hilbert space and hamiltonian create an algorithm to factorize the hilbert space based on some decoherence related criteria

this paper is proving the existence and uniqueness (or attempting to) of a similar idea except using hamiltonian locality as the criteria for factorization

Mr. Feynman

17:52

That sounds cool :P

Silly Goose

yeah im really really excited :DD

Oh i see @ACuriousMind !

but still if the Hamiltonian is a linear transformation on one hilbert space, how do you compose a unitary map which is between two different hilbert spaces and the hamiltonian?

ACuriousMind

@SillyGoose you can compose a map $f : X \to X$ with an invertible map $g : X\to Y$ to get $g \circ f \circ g^{-1} : Y\to Y$, that's just how function composition works

oh, they put the $-1$ on the wrong $U$ in the definition you posted :P

pretty common typo

Silly Goose

ahhh okay

i was also wondering like they use the convention THT^{-1} prior to this definition

and they just switch it up XD

okay good to know it is a typo

@Relativisticcucumber omg hi

John Rennie

18:31

@DebanjanBiswas Yes, you could be working in a theoretical area where experimental data isn't available. String theory is an obvious example. Then you're basically doing an (applied) maths PhD.

Mr. Feynman

18:45

@ACuriousMind I think that's the most terrifying $-1$ in QM along with the minus in $\sigma_2$ matrix

Really, it's one of those things you think you remember but forget all the time and have to check it up or think about it