The h Bar

Claudio Menchinelli

00:37

guys I've begun reading about angular momentum addition. My professor's notes(taken by me during lectures) aren't clear enough, unfortunately. So I searched online and found this by Prof. Zweibach ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/…

Most of the stuff he says I did not understand completely like the thing about direct sum of representation, etc etc.

My conclusion is: we know that since the two operators $S^2,S_z$ commute with each other, we want to find a basis of simultaneous eigenkets

we then choose to work with the basis $2.7$

I have the momentum operators $\hat{S}_z = S_z^{(1)}+S_z^{(2)}$ and $\mathbf{\hat{S}}^2 = \mathbf{S}_1^2+ \mathbf{S}_2^2 + 2\mathbf{S}_1 \cdot \mathbf{S}_2$, now represented, wrt the following basis $\mathcal{B} = \{|++\rangle, |+-\rangle, |-+\rangle, |--\rangle\}$, by the following $ {4\times4}$ matrices:
$$ S^2 = \hbar^2\begin{bmatrix}
2 & 0 & 0 \\
0 & \mathbb{J}_2 & 0 \\
0 & 0 & 2 \\
\end{bmatrix};S_z = \hbar\begin{bmatrix}
1 & 0 & 0 \\
0 & [0]_2 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$$
where $J_i$ is a all-ones matrix and $[0]$ the null matrix

now I restrict myself to the subspace spanned by $|+-\rangle, |-+\rangle$ and I diagonalize $S^2$ restricted to this subspace

and I have successfully removed the degeneracy in the subspace of $S_z$ and I now have a CSCO since no two vectors have the same eigenvalues for both $S_z,S^2$

why isn't the matter presented in this way?

The question is: is what I did somehow related to this weird and obscure result: $1/2 \otimes 1/2 = 1 \oplus 0$?

Silly Goose

00:59

@ClaudioMenchinelli i would not characterize this result as being weird because it is a mathematical fact. you can prove a characterization of all projective representations of $SO(3)$ (this characterization is in math language what physics textbooks call the theory of angular momentum) in terms of irreducible representations. then taking direct sums of irreps allows you to construct any representation. in particular you can decompose a tensor product of irreps into a direct sum of irreps.

(is a mathematical fact about representations of a compact group)

Claudio Menchinelli

My professor never mentioned projective representations of $SO(3)$ nor irreducible representations. He just gave me 2.27

I forgot to say: weird and obscure result $\textbf{to me}$

so my approach is correct

?

but why do we denote the eigenvalues of $S^2$ as $1,0$?

I don't get the one unfortunately

only the zero

Ok I get it now

@SillyGoose do you have any refs about what you've just stated?

@SillyGoose maybe something very noob-friendly on the subject :P

Silly Goose

Brian C. Hall's Lie groups, lie algebras, and representations ch.4 section 6 classifies the representations of an algebra that is the relevant algebra when talking about spin in quantum mechanics. and the computation is actually mostly linear algebra. but some of the crucial facts are representation theoretic.

but it might take a bit to build up to that part of the text

are you familiar with representation theory?

the first two chapters contain the necessary definitions and relevant theories

the operational thinking is this. 1) representations of compact groups/algebras are decomposable into a direct sum of irreducible representations. 2) finite dimension projective representations of $SO(3)$ model quantum spin correctly. 3) these projective representations are equivalent to normal representations of $\mathfrak{su(2)}$ or $\mathfrak{sl}(2;\mathbb{C})$. 4) do some linear algebra and such to classify all irreducible representations of $\mathfrak{sl}(2;\mathbb{C})$

i think the subject can be a bit overwhelming though (at least it is for me) :P can try to pick it up in bits and pieces.

But if you want to learn some lie theory/representation theory, I was recommended to use Brian C. Hall's text because it deals with groups of matrices, which are very concrete. Other canonical texts deal with lie groups in the abstract, which makes things harder in the sense of increasing the pre-reqs to include some differential geometry and so on. and, applications to physics usually is for groups of matrices, not groups in the abstract.

naturallyInconsistent

01:37

@SillyGoose You say that you are following Sakurai, but you miss the wonderful result that Sakurai has; your Equations (6) do not seem to have anything to do with Sakurai and visibly seems to mix up the perturbations with the unperturbed ket, and it would not be obvious how to solve them. Your Taylor expansions in Equations (3) and (4), the middle part, is also nonsense because you already substituted zero for the parameter; you have to differentiate first before substitution.

@SillyGoose Equation (8) is just wrong. Sakurai's problem 5.15 explicitly fails if you assert the wrong thing.

Silly Goose

@naturallyInconsistent oh yes i was using some sloppy notation where $0$ in the input means evaluate after differentiating

everything in the Ignore section is to be ignored :P since it is wrong but still usable LaTex for later

naturallyInconsistent

@SillyGoose What is $O$? You should be very wrong here too; The perturbation does not have to have zero action on the unperturbed kets. It just needs to be diagonalised. That is very very different.

Silly Goose

i think i was mixing up some of the problems and their resolutions though for sure

naturallyInconsistent

@ClaudioMenchinelli The restriction is purely for the sake of simplicity. You can choose to diagonalise the matrix in its original 4x4 grandeur; the eigenkets are clearly $$\left\{\begin{pmatrix}1\\0\\0\\0\end {pmatrix},\begin{pmatrix}0\\1/\sqrt2\\1/\sqrt2\\0\end {pmatrix},\begin{pmatrix}0\\1/\sqrt2\\-1/\sqrt2\\0\end {pmatrix},\begin{pmatrix}0\\0\\0\\1\end {pmatrix}\right\}$$

@ClaudioMenchinelli Then you note that the $S^2$ eigenvalues of this set is compartmentalised into one single one having $S^2=0$ and the other set of three have $S^2=2=1(1+1)$. i.e. simply rearranging your eigenkets to put them together, you realise that you have a spin-zero and a spin-one irrep as the result of coupling two spin-halfs together.

Silly Goose

03:20

okay i think i understand time-independent perturbation theory now lol

1) you do non-degenerate perturbation theory via inverting an operator. there are no singularities here. 2) you observe that the inverting the operator method does not work in a degenerate subspace because you get singularities. okay! then try another method. the new method is to solve the schrödinger equation proper in the degenerate subspace!!!

so there literally are no singularities. every time you encounter a singularity, you either cannot use the method you've just tried OR you have done a computational error.

oops the first equation in (12) should not have an $= 0$ on the right hand side

naturallyInconsistent

03:50

@SillyGoose what do you mean by solving the Schrödinger equation proper?

naturallyInconsistent

04:08

@SillyGoose but of course this is correct. It is physics, after all. Annoyances that are not experimentally there, must have some way to be dealt with.

Silly Goose

04:35

@naturallyInconsistent err sorry i should say solving the eigenvalue problem proper (in the degenerate subspace)

instead of doing inverse operator shenanigans

i guess i didn't understand this properly. after reading and re-reading sakurai, i was left with the impression that we are disregarding singularities. not that there are no singularities at any point in a correct procedure of perturbation theory (in this context)

naturallyInconsistent

04:52

@SillyGoose The funny part is that the presentations do explicitly state that it is about diagonalisation. Sakurai is not emphasising it, but I know that the others do.

naturallyInconsistent

05:37

Actually, Wacky Fowl, do you know what I am talking about when I say that the presentations do explicitly state this? It feels like you didn't catch it.

Silly Goose

@naturallyInconsistent I think I did not understand that the diagonalization is a method to replace our method of inverting the operator, which is invalid due to the singularities you would encounter had you inverted

I do get that sakurai and so on say to diagonalize, but I don't think I caught that we should diagonalize simply because that it is another way of approaching the problem and our first way doesn't work. the emphasis on it being a "resolution" to a "singularity problem" threw me off because it does not resolve the singularity problem. it is just another way of handling the problem more carefully

the singularity problem is not even a problem with the operator or the problem we're trying to solve. it's a problem with us as solvers of the problem being careless

but now it is a lot more clear

it is like solving $x\cdot x = x$. at first sakurai tries looking for solutions based off of $x^2/x = x/1 \iff x = 1, x \neq 0$. which works given $x \neq 0$. it so happens that $x \neq 0$ is satisfied always in non-degenerate perturbation theory. then we move to degenerate and have to look at the full statement of the problem $x^2 = x$ and solve it from there (albeit within a restricted domain and with some freedom due to degeneracy)

naturallyInconsistent

05:55

@SillyGoose But many authors should have asserted that the issue is that there is a potential for a singularity to arise, and that the place it could have been happening, is because the perturbation series involves many instances of $\frac{\left<m\right|V\left|n\right>}{H_0-E_D}$. It is plausible for singularity to arise in these expressions if the numerator is non-zero; and so the correct way to deal with them is to make the numerator always zero via diagonalisation,

Silly Goose

hm well i guess i still don't understand that logic

because one can obtain that you should choose your degenerate subspace eigenstates to be eigenstates of the perturbation $V$ without saying the numerator of something should be $0$ (at least I think so, perhasp I did something wrong)

naturallyInconsistent

so that the zero in the denominator becomes no longer destructive. Note that it is a diagonalisation: $\left<\ell\right|V\left|\ell\right>$ does not need to be zero. As long as the degenerate subspace is not having off-diagonal terms, none of the potential singularities will be an actual singularity

@SillyGoose Yes, but this is the argument that will be most clearest to physicists: do it properly or else the fire burns!

Silly Goose

i have been stuck on this for so long...

naturallyInconsistent

Also, Sakurai problem 5.15 is a direct example where V is zero in the subspace, so that the degeneracy is only lifted in 2nd order perturbation. i.e. it can be arbitrarily complicated how this thing is actually done.

Silly Goose

idk i guess to me the loose analogy i wrote above makes it clear that this "inverting operator" method in general destroys information

in the simple case, you lose the solution $x = 0$.

then one can immediately observe that if a destruction of information is happening due to the method, you must use another method

@naturallyInconsistent i am def going to do this problem soon hehe

naturallyInconsistent

06:02

Ah, but here we are explicitly saying that x=0 is unphysical and so it needs to be gone...

Silly Goose

right i get that and in non-degenerate p-theory that is what is done and I am cool with that

naturallyInconsistent

In degen p-th it is also what is done

basically, it is just generally the case that you are always diagonalising the perturbation in your degenerate subspaces. It just so happens that in non-degen, the subspace is dim=1 and trivially diagonalised

Silly Goose

hm but I don't see how you can justify throwing away the singular points

naturallyInconsistent

there is thus no singularity anywhere

Silly Goose

here I justify it by never inverting the operator and so never encountering singular points because the equation proper is mathematically equivalent to an eigenequation in the degenerate subspace

naturallyInconsistent

06:06

Eigenequation = diagonalising...

Silly Goose

in non-degenerate, for example, you can justify never encountering singular points by orthogonality of $\langle n^{(0)} \lvert n^{(j)}\rangle$ for $j = 1, 2, ...$

in particular, any time you act with the inverse of the operator will be on higher order corrections to the eigenstate, each of which have no contribution from the singular point.

naturallyInconsistent

@SillyGoose In degen, $\left<\ell,\alpha|\ell,\beta\right>=\delta_{\alpha\beta}\propto\left<\ell,\alpha|V|\ell,\beta\right>$ after the process of diagonalisation, i.e. eigen-equation

Silly Goose

well i am not arguing against a diagonalization

naturallyInconsistent

And from then on it is free for us to invert the operator because it is never singular

Silly Goose

hm well in my books $\frac{0}{0}$ is singular

naturallyInconsistent

06:11

@SillyGoose If the numerator is a stronger zero than the denominator, this is always zero. That's what the authors are trying to say

Silly Goose

hm but i just don't get why even hand-wave in that way here

naturallyInconsistent

Because it is natural for physicists to reason in such a way?

Silly Goose

well I think it is confusing :P

naturallyInconsistent

If you can find a way that is demonstrably better without causing other students to be confused, then we might consider using that treatment in the future

Silly Goose

i personally like the way i have written it up (though it is probably more idiosyncratic than pedagogical)

naturallyInconsistent

06:16

I can tell you that it is not going to be readable for other students. You will have a much better time if you write it out as a matrix, because then the diagonalness is extremely visually striking.

Silly Goose

oh yeah i want to draw some images...since that helps when dealing with the degenerate subspace and what not business

naturallyInconsistent

I think part of the thing that got lost in translation is that the fundamental problem of perturbation theory is about diagonalisation in the first place. The statement that $H_0+\varepsilon V$ is a difficult problem to solve is exactly the statement that it is difficult to diagonalise that. Essentially, what is happening in degen p-th is that we have no choice but to manually diagonalise in the degen subspace, leaving the p-th to diagonalise over the non-degen bits.

Silly Goose

yes!! that is what was not getting across to me as i read

but now it is clear

naturallyInconsistent

yay

Silly Goose

i think i will try 5.15 now

naturallyInconsistent

06:33

good

I havent tried the p-th expansions on it, but the diagonalisation is fun.

Relativisticcucumber

06:57

what is the point in debating what the "best" way to write something up is? i dont feel that this concept is useful at all.

naturallyInconsistent

@Relativisticcucumber treatments can be arbitrarily ugly! Notation can be so bad that it hinders scientific progress of nations for centuries on end. Such things do matter...

Relativisticcucumber

i think there are for sure bad ways

but i think as someone who is learning whats the point of debating what is the best way or why something is wrong? seems its more productive to just learn and move on?

naturallyInconsistent

But a debate on the best way to present something can dramatically change the student's understanding of stuff. For example, I have been trying to write up the Thomas precession answer that I wanted to write, but it is still mysterious even to myself. The treatment of "consider the proton orbiting the electron" is so bad that I am annoyed that people even tried it. The working I was typing up was trying to use some Lie algebra trickery, and then I couldn't get the correct rotation.

And then I found a totally different argumentation that is about putting a cone on top of a sphere and deriving the shrinking-ness of the rotation angle that way, i.e. a global rather than a local derivatives way to get the answer, and it is a fundamentally different kind of understanding.

ACuriousMind

08:11

@ClaudioMenchinelli Your subspace is not an invariant subspace - the result of applying $S_\pm$ to $\lvert +-\rangle + c\lvert -+\rangle$ is not again a vector of this form (but contains $\lvert ++\rangle$ and $\lvert --\rangle$, too), so this doesn't mean anything.

The "obscure" result you're talking about is physically just the statement that $\lvert ++\rangle, \lvert +-\rangle + \lvert -+\rangle, \lvert --\rangle$ span an invariant subspace isomorphic to the usual spin-1 space, and $\lvert +-\rangle - \lvert -+\rangle has spin-0.

Slereah

08:22

I have a Hunch

That question I had a while back about the necessity for the compactness of the kinematic group's timelike component

What if it relates to ordered topological vector spaces

Since a spacetime is basically that

ACuriousMind

what exactly do you mean by "kinematic group"?

and how are you identifying the "timelike component"?

Slereah

It is a somewhat vague notion of a group acting on a space in such a way that it is "like a spacetime"

Basically having to obey a few axioms

It comprises the Poincaré group, Galilean group, De Sitter group, Carrolian group, etc

Basically it admits the rotation group as a subgroup and has something akin to a time direction

one condition of which being that the subgroups generated by the "boosts" are compact

errr *non compact

3

Q: Why must boosts be non-compact?

It is a common argument in the theory of kinematic groups (the groups of motions for a spacetime) that the subgroups generated by boosts must be non-compact[1][2][3]. This is true of all commonly used kinematic groups such as the Poincaré group, Galilean group, De Sitter group, Anti De Sitter gro...

^cf this

ACuriousMind

I'm very confused

Slereah

Mayhaps using the theory of ordered topological vector spaces might be a good way to define them basis-free

ACuriousMind

here in chat you say that the timelike component should be compact, but in your question you say the subgroups generated by boosts should be non-compact

Slereah

08:29

What has you confused

ACuriousMind

which is it?

Slereah

Non-compact

I misspoke :p

ACuriousMind

I mean if these subgroups were compact, what that would mean physically would be that you can get so fast that you're at rest again (with respect to your starting velocity), right?

ekardnam_

After a wick rotation it becomes compact no?

ACuriousMind

@ekardnam_ no, Galilean boosts are non-compact, too (no speed limit)

naturallyInconsistent

08:32

@Slereah my view of this is with sezik. This is due to the highly non-trivial fact that we have inertial frames. For every boosted frame, there exists a frame boosted even faster, because either you can just do the same boost twice, or you can find an inverse boost to bring the frame to rest, and find out that there are faster frames than it. Thus, boost subgroup has to be non-compact

Slereah

@ACuriousMind I think the fundamental reason behind it is that if the group is compact, there's no notion of causality

You don't have a split of your spacetime into past and future

ACuriousMind

yes, because you cannot distinguish between being at rest and moving forward; I think this would mess up all sorts of assumptions about physics

in order to make a technical argument I'd need to see a technical definition of "kinematic group" :P

naturallyInconsistent

@ACuriousMind Now im confused. After a wick rotation the metric become Euclidean and the boosts should become rotations.

ACuriousMind

@naturallyInconsistent The isometries of the Euclidean metric are rotations SO(4), but the Galilean group is not that

Slereah

ACuriousMind

08:37

I guess which of the two applies here depends on what exactly you mean by "Wick rotation"

naturallyInconsistent

@ACuriousMind Yes, but SO(3,1) turns into SO(4) after a Wick rotation, isn't it? Galilean group isn't relevant because it is Lorentz group that converts to rotation group

Slereah

SO(4) is very much not a kinematic group

ACuriousMind

@naturallyInconsistent On a formal level, there really isn't a process where Wick rotation "turns" one group into another

naturallyInconsistent

Formally presents shocked pikachu face

ACuriousMind

Wick rotation, as we use it in QFT, is just analytic continuation of scattering functions; there is no fully consistent "Euclidean physics" underlying it

Slereah

08:41

the kinematic algebra basically follows these rules with the usual number of generators for a group acting on a spacetime :

\begin{eqnarray}
\left[J, H\right] &=& 0\\
\left[J, J\right] &=& J\\
\left[J, P\right] &=& P\\
\left[J, K\right] &=& K
\end{eqnarray}

Plus $J$ has the SO(3) algebra

also there's some condition on discrete symmetries but that is less interesting

ACuriousMind

@Slereah I still don't really know what a kinematic group is

Slereah

It's a Lie group with a Lie algebra obeying those conditions

ACuriousMind

saying it has to be vaguely of the form of the Poincaré algebra isn't really a definition :P

Slereah

Why not

The kinematic groups are well classified

ACuriousMind

Because all of these algebras also have compact forms, and apparently you want to say they're forbidden from being compact

Slereah

08:45

Well yes, that is an extra condition on the group

And one which I would like to investigate

ACuriousMind

or, well, if you really only want this to be a real Lie algebra and you don't implicitly complexify as physicists often do, then this actually suffices as a definition but then there's just nothing to show - as a real Lie algebra this isn't compact (compute the Killing form), so it will have non-compact subgroups

Slereah

The existence isn't the issue, it is more the motivation

Silly Goose

@naturallyInconsistent i just did it :D

Slereah

Like the boosts being non-compact is a pretty reasonable sounding condition, but otoh it still allows for some pretty weird group, so obviously there's more to it than just being like the Poincaré group

Which I suspect is just causality

Like you wouldn't have a partial ordering from causality that's invariant under the group I think

naturallyInconsistent

@SillyGoose 🥳

ekardnam_

08:57

@ACuriousMind what i meant is what @naturallyInconsistent is saying

SO(4) is often called the Lorentz group of "Euclidean Minkowski spacetime" which is Euclidean 4-space

in the physics litterature

as much as hyperbolic space is often called "Euclidean AdS"

@ACuriousMind I always felt it was the opposite, stuff is more consistent in the Euclidean side of things and then we continue analitically to Minkowski

ACuriousMind

@ekardnam_ have you ever looked into how Wick rotation works for spinors? it's a mess

also, the Euclidean world isn't really physics as we usually understand it - since there is no distinguished time direction, the equations of motion aren't well-posed initial value problems

Slereah

Can't just turn around to go back in time

ekardnam_

@ACuriousMind not really, but I can see how spinors would be more "sensitive" to the change of the signature

Slereah

Gets even worse when spacetime is curved

ekardnam_

@ACuriousMind yeah i can see that

Slereah

09:10

Doesn't seem to be a whole lot on group represetations for ordered topological vector spaces alas

Might have to figure it out myself

ekardnam_

@Slereah what you mean by kinematic group is the isometry group of spacetime? (In a general spacetime not just Poincare for Minkowski)

Slereah

Roughly speaking it maps allowed trajectories to allowed trajectories

ekardnam_

@Slereah oh so isometries seem to be only a subgroup maybe

but maybe it is enough to prove it is not compact

Slereah

I mean the two are connected

If you have your kinematic group, you can find some structures which are preserved by it

That's the whole theory of Klein spaces thing

ekardnam_

is the isometry group of a lorentzian manifold always non-compact?

Slereah

09:22

You can define the Lorentz group as the group that leaves the metric tensor invariant, but in some sense the metric tensor is also defined by the Lorentz group

ekardnam_

because if yes then since it is a subgroup of the kinematic group also the kinematic group has to be not compact

Slereah

@ekardnam_ Kinematic groups are local notions

As you may know manifolds may not have any isometry group at all

(outside of the identity)

The kinematic group wrt manifolds is what people mean when they say that GR is "locally Minkowski"

ekardnam_

local in what sense? defined on the tangent space like lorentz group

@Slereah oh so isnt it always the lorentz group once you fix the signature to be (3, 1)

Slereah

GR's kinematic group is the Poincaré group yes

Different kinematic groups are for other theories

@ekardnam_ yes

ekardnam_

@Slereah oh okay I see what you mean

I was thinking less general than you

Slereah

09:28

The simple example being that classical mechanics has the Galilean group as its kinematic group

naturallyInconsistent

@ACuriousMind Do I have the courage to get a glimpse of this monstrosity? Not today, not today.

ekardnam_

10:02

sick way from the Russians to draw Young diagrams

imgur.com/a/cdYw1Zi

Ryder Rude

also note that particles are hard to define in the Euclidean QFT

Slereah

If I want the notion to be applicable to all kinematic group I'm still gonna have to deal with manifolds because the De Sitter group doesn't act on a vector space

ugh

Ryder Rude

let's say i expand $T(e^{iH(t) t})$ perturbatively, H(t) is the interaction picture hamiltonian. and then i expand $<0| T(\Pi \phi (x_i)|0>$ using the path intgeral perturbatively

how r these two expansions related

both sort of give the S-matrix. the former more explicitly

the latter needs to be fed to lsz before sort of giving a portion of the S matrix

sorry i think i meant $T(e^{\int H(t) dt})$ for the former

ekardnam_

10:34

i think I have been dealing with the $\Omega$-background for more about an year now

I still don't know what it is

Ryder Rude

10:56

there is this great result that particles can only have 0 momentum in a massless euclidean qft

so the Fock space shud b like : vacuum, particle with zero momentum, two particles with zero momentum.....

this is because E^2+p^2=0 for massless, and energy cant b imaginary

wait does this also hold in the classical.theory then

so in a classical euclidean universe, massless particles stay at rest

Mr. Feynman

11:32

Is a congruence of null geodesics always hypersurface orthogonal?

I would say no, but then why a congruence of null generators is?

Man, I can't get this stuff about null hypersurfaces from Physics books

Is there any diff geo book discussing these things?

Slereah

12:43

In ordered topological spaces, is what the math people calling a cone-saturated set the causal diamond of a set

The domain of dependence of the set I guess

Slereah

13:05

lol

Slereah

13:20

apparently Pythagoras got a bit of the runaround from the priests in Egypt

They were not enthused by having to teach some random greek kid because his important dad asked the pharaoh to do it

^this guy apparently

Ryder Rude

13:49

what do u gain from learning physics and math?

Ryder Rude

15:04

Pharaohs mustve been really powerful

Slereah

God on Earth

Ryder Rude

yeah. both religious and political power at its max

how come they got god status without even doing magic

sounds like North Korea

to be god on earth, you should be expected to do miracles everyday

Pharaohs died in wars

Slereah

He made the nile rise every year

What more do you want

Ryder Rude

lol

ACuriousMind

@Slereah ~~on Earth~~ in da big house

Ryder Rude

15:08

did Pharaos consciously scam others

or they believed they were gods

Slereah

I don't think they left any writings of it if they did

Although the greek pharaohs kind of phoned it in

I think the Ptolemy's weren't super convinced of their divine stature

Ryder Rude

oh

Slereah

Also like the first few Roman emperors that humored the egyptians by pretending to be pharaohs

But IIRC they gave it up

ACuriousMind

15:21

it's generally safe to assume that people in the past largely believed in what they professed to believe in, just like people today do

Slereah

Considering the egyptians considered the greeks to be basically children as far as civilizations go it must have been pretty humiliating for them to be conquered like that

Like if the Vatican was conquered by the US

Gonna replace the eucharist with hamburgers

ACuriousMind

the interplay of the different polytheistic religions of the Mediterranean is interesting because both Romans and Greeks were very willing to identify foreign gods as their own; I'm fairly certain the Greeks did not consider the Egyptian gods to be different gods in the sense we would understand the term

ekardnam_

@Slereah a free hamburger would make going to church suddenly more appealing

ACuriousMind

generally - before they became Christians - the Romans were pretty content with leaving local religious practices as they were as long as they paid tribute and accepted some form of the interpretatio Romana that identified their gods with the Roman gods

Slereah

@ACuriousMind I know that they certainly found the animal heads amusing

Also this is one of those things that people say but it's not entirely true

Pre-Christian religions placed less emphasis on orthodoxy and were more syncretic but it's not entirely true that they didn't care about it at all

IIRC at some point the Ptolemies basically stopped doing their pharaoh duties because they couldn't be arsed

ACuriousMind

15:31

@Slereah oh, I don't mean to imply they somehow cared less about their religion, they just found other aspects like proper ritual practice more important than the specific names or myths the others told about the gods

Ryder Rude

sounds like they were into equivalence classes

Slereah

I mean they were fine with letting it go on, but from what I've read they weren't as interested in participating

ACuriousMind

and the Ptolemies were the worst anyway :P

they spent 300 years without learning the language of the empire they ruled

Slereah

> If you consider the effects of Alexander's instruction, you will see that he educated the Hyrcanians to contract marriages, taught the Arachosians to till the soil, and persuaded the Sogdians to support their parents, not to kill them,and the Persians to respect their mothers, not to marry them. Most admirable philosophy, which induced the Indians to worship Greek gods, and the Scythians to bury their dead and not to eat them!

Plutarch was certainly not that respectful of foreign traditions :p

> And Socrates was condemned by the sycophants in Athens for introducing new deities, while thanks to Alexander Bactria and the Caucasus worshipped the gods of the Greeks. Plato drew up in writing one ideal constitution but could not persuade anyone to adopt it because of its severity, while Alexander founded over seventy cities among barbarian tribes, sprinkled Greek institutions all over Asia, and so overcame its wild and savage manner of living.
Few of us read Plato's Laws, but the laws of Alexander have been and are still used by millions of men.

He had an axe to grind

Ryder Rude

15:47

genghis was secular

you would almost think he was a good guy

he was good post-war according to historians

WaveInPlace

In a question on gravitational time dilation John Rennie gives a calculation for gravitational potential outside a mass and says it's, "the same result we get from the Schwarzschild metric, though you should note that the Schwarzschild r co-ordinate is subtly different from the r co-ordinate used in Newton's law."

How do the coordinates differ?
https://physics.stackexchange.com/questions/69043/the-higher-you-go-the-slower-is-ageing/69048#69048

Slereah

Didn't Genghis Khan have muslim killed for following some rituals that he disagreed with on tengrist grounds

Ryder Rude

havent heard of that. maybe historians whitewashed it

Slereah

@WaveInPlace $r$ isn't the radial distance

The radial distance you have to integrate

WaveInPlace

I'm not 100% sure I follow. Is this a variable-length-close-to-mass issue?

Slereah

15:56

Coordinates do not have to be connected to distance in general relativity, they're just numbers that give you the location of a point

The actual distance in the Schwarzschild metric will be the calculated distance by checking the length of a spacelike radial geodesic, and this turns out not to be equal to r

WaveInPlace

It's bigger?

Claudio Menchinelli

@ACuriousMind I read your messages

You said that the action of the ladder operators on the vectors that span the subspace returns vectors not belonging to that subspace

but I'm interested in the action of $\mathbf{S}^2$

If I let the latter act on the subspace-spanning vectors, I obtain a linear combination of them, which is totally fine. Why is my reasoning a failure then?

ACuriousMind

@ClaudioMenchinelli No, not really. What you're really interested in is irreducible representations of the rotation algebra. It just turns out those are labeled by values of $S^2$

or, at least that's what the texts you're reading are interested in even if they often fail to present this in a coherent manner :P

Claudio Menchinelli

Is there a reason we work with the basis $|++\rangle, |+-\rangle, ....$ spanning $V_1 \otimes V_2$ to begin with?

is there a definition of irreducible?

ACuriousMind

I'm not really sure what kind of reason you're looking for here

and of course there's a definition of irreducible :P

Claudio Menchinelli

16:08

I meant a definition I can understand, even intuitevely

ACuriousMind

@ClaudioMenchinelli it's a space on which the rotation operators act that doesn't have a smaller subspace that you can restrict to and they still all act on it

Slereah

There's no definition of irreducible

It's more of a feeling

ACuriousMind

your $V_1\otimes V_2$ has two such subspaces, one spanned by $\lvert ++\rangle , \lvert +-\rangle + \lvert -+\rangle, \lvert --\rangle$ and one spanned by $\lvert +-\rangle - \lvert +-\rangle$

these spaces are interesting, for instance, precisely because $S^2$ is a constant on them (by Schur's lemma), i.e. they are spaces of constant total angular momentum

Claudio Menchinelli

1 and 0 respectively

I see

Ryder Rude

oh so Schur's lemma applies to S^2.

becuz it commutes with others

i hadnt seen an application of this lemma yet. thanks

Silly Goose

16:13

A representation sends group elements to linear maps—it lets us make contact with groups and our Hilbert space. as linear maps (with an emphasis on where they send vectors to), we have something like $O: V \rightarrow V$. A representation is irreducible when the action of every group element represented on $V$ sends it to $V$ and the only other subspace for which this happens is the trivial subspace.

@ACuriousMind is $S^2$ an intertwiner between the tensor product and the direct sum representation?

Ryder Rude

ok so this is how we know that diagonalising S^2 would produce irreducible reps

becuz S^2 must b diagonal in irreducible reps

ACuriousMind

@SillyGoose sure, because both are the same space, just written differently - and then the intertwiner condition is just that it's an operator that commutes with all algebra elements

Claudio Menchinelli

wait, what would be the correlation between this sentence: "it's a space on which the rotation operators act that doesn't have a smaller subspace that you can restrict to and they still all act on it" and the subspace $$|+-\rangle, |-+\rangle $$

ACuriousMind

well, technically the $\mathrm{e}^{\mathrm{i}S^2}$ generated by it is the intertwiner (since $S^2$ itself can have 0 eigenvalues) but same difference

Claudio Menchinelli

this subspace is not irreducible because....

ACuriousMind

16:21

@ClaudioMenchinelli the relation is that that subspace is not such a space - the ladder operators $S_\pm$ and hence also $S_x$ and $S_y$ act on it such that they produce states that are not inside of it

Ryder Rude

@SillyGoose is trivial subspace the 0 vector?

Claudio Menchinelli

@ACuriousMind that's what I thought as well, ok I get it

ACuriousMind

it fails the condition of actually being a representation of the rotation algebra, not the condition about non-trivial subspaces

Silly Goose

@RyderRude yes

Oh oops i said group when i should say algebra

Ryder Rude

i think u said correct?

Claudio Menchinelli

16:23

So $S_x,S_y$ are two of the $O$ elements in @SillyGoose's answer

which do not satisfy the relation

Silly Goose

@ACuriousMind hm wait i think i misunderstood. so we are actually taking $1/2 \otimes 1/2$ to $0\oplus 1$ and then applying shur’s lemma to $S^2: 0\oplus 1$? This would make more sense to me for the statement of shur’s lemma that i have from Hall (which is defined in terms of irreps)

ACuriousMind

@ClaudioMenchinelli I'm not really sure what's going on with the $O$ tbh :P

Silly Goose

@RyderRude it is true for groups, but i think Claudio is dealing with algebra elements (S_z and so on)

Ryder Rude

oh

Claudio Menchinelli

"a group element sent to a linear map..."

ACuriousMind

16:26

@SillyGoose it's still Schur and yes, the application of Schur's lemma is to the individual summands, i.e. $S^2$ restricted to the irreps is an intertwiner of those irreps with themselves and hence constant on them

Silly Goose

Oh god yes schur

So this schurs lemma business is mostly useful for compact groups? Or it is generally still useful

ACuriousMind

what does any of this have to do with compactness?

Silly Goose

Because finite dimensional representations of compact groups are decomposable into direct sums of irreps (at the least).

ACuriousMind

not sure why that would imply that Schur's lemma is only useful for compact groups

naturallyInconsistent

@Slereah To lead, yearly, to ... rise ... and donate via onanism ... barely contains amusement

Silly Goose

16:30

Perhaps you want to deal with a representation that cannot be decomposed into irreps. Which is conceivable as happening for non compact groups

ACuriousMind

So? Schur's lemma still gives you that any intertwiner of reducible representations is either zero or an isomorphism between them

naturallyInconsistent

@WaveInPlace The usual flat space spherical coördinates we are very used to r being both the radial distance of the origin and the area of the sphere that is centred on the origin. In the Schwarzchild metric, because of the curvature of spacetime, these two things differ, and we are using the area of the sphere centred on the origin to define the radial coördinate r. This causes the length from the singularity to any point on the sphere (or from the Event Horizon to the point on the sphere)

to differ from what $r$ would have led you to believe it was.

Ryder Rude

how is r area of sphere?

naturallyInconsistent

@ClaudioMenchinelli Remember the thesis of the thing you are trying to do. In general, under reductionism, one studies complicated stuff by composing simple stuff together. It does not matter if you are trying to add the contributions of many different spins together, or you are trying to add orbital angular momentum to spin angular momentum, there are just some angular momenta you have to add together. They couple together, and so the natural beginning point is to use the basis of each initial

WaveInPlace

@naturallyInconsistent, I get that they're different, it's more a matter of how they differ. If I'm understanding this correctly, is it true to say that the radial coordinate r is always >= the radial distance r?

Ryder Rude

16:35

theres too many interesting fields to learn

u can never learn all of them

naturallyInconsistent

angular momentum as the basis for the coupled space. Then you can obtain the result that adding (irreps of) angular momenta together leads to new irreps, which is inherently an interesting result.

@WaveInPlace I do not care if it is always larger or always smaller. They just differ. Yes, under Schwarzchild metric, they differ only in one direction. I am not sure if there are metrics whereby they can differ in the other way, or vary in the way that they differ, but as long as you are aware that they differ, that is more than good enough for meow

Claudio Menchinelli

@naturallyInconsistent I see

adding apples to apples hahaha

WaveInPlace

Good enough for a test, but not good for an intuitive understanding. :)

Thanks!

naturallyInconsistent

@WaveInPlace What are you talking about? I am literally giving you the intuitive understanding...

Silly Goose

@ACuriousMind hm i didn’t know this. I guess i am confused because the statement of Schur’s lemma i have seen applies only to irreps. But if my representation cannot be decomposed as a direct sum of irreps, how can i apply Schur’s lemma?

Ryder Rude

16:39

physics.stackexchange.com/a/18582 this says it's circumference

ACuriousMind

@SillyGoose Oh, sorry, I was briefly confused; you are correct, Schur's lemma only applies to irreps

WaveInPlace

@naturallyInconsistent Knowing how they differ let's me visualize how increased curvature relates to the size of radial coordinate r. Only knowing that they differ isn't enough for that.

ACuriousMind

that doesn't mean it's "mostly" useful for compact groups, though, you'll plenty of times still look at irreps for non-compact groups (e.g. Poincaré group!)

WaveInPlace

I tend to be pretty visual with these things. Probably a hold over from my chemistry training.

Silly Goose

Oh okay i see makes sense

So for the lorentz algebra case, we get the complete reducibility result because it is semi-simple?

naturallyInconsistent

16:47

@WaveInPlace It is kinda the opposite? The important intuitive takeaway is that one realises to question (and know what the actually useful convention is) that it is even a possibility for these two things that have always been the same prior to curvature, to differ. How precisely it differs is a grunt work brute force integration. It is the opposite of intuitive.

ACuriousMind

@SillyGoose yes

Silly Goose

Hm okay i see. So at least the import of Schur’s lemma is clear for semi-simple lie algebras and compact lie groups

WaveInPlace

Fair enough.

naturallyInconsistent

The weather is too cold to be fair.

And it is too rainy.

Silly Goose

same here

is the import of casimir elements that they label irreps?

ACuriousMind

16:59

yes

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