The h Bar

12:33 AM

Oh do you mean historically or like concept-wise

2:29 AM

@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

2:48 AM

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

8:12 AM

@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

8:33 AM

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

8:44 AM

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

8:48 AM

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

8:59 AM

That never stopped me before

Relativisticcucumber

@ACuriousMind okay thank you i will look for one

Debanjan Biswas

9:49 AM

@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 AM

@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 AM

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 AM

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 AM

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 PM

@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

1:35 PM

@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

1:51 PM

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

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