The h Bar

1:16 AM

So we start with the abstract group $SU(2)$. When we represent it, i.e. pass $U \in SU(2)$ through the map $\rho: SU(2) \rightarrow GL(\mathcal{H})$, is the map literally just making explicit that instead of an abstract matrix $M$ we have the same matrix but viewed as a linear transformation $M: \mathcal{H} \rightarrow \mathcal{H}$?

Silly Goose

4:08 AM

Hm so it sounds like what i described above is the "standard representation" of a group

so like the standard representation of $SO(3)$ is the group homomorphism into the general linear group of 3-dimensions over the real field in which the matrices of $SO(3)$ are literally taken to act on vectors $v \in \mathbb{R}^3$ via usual matrix-row multiplication

im trying to judge the accuracy of what i write here :P

uhoh

5:13 AM

0

Q: How to generate pole figures from limited amounts of area detector X-ray diffraction from thin films (explain it like I'm five years old)

Historically powder X-ray diffraction was the go-to method for reconstructing crystal structure. X-ray sources were limited in intensity and emittance and so to collect data in a reasonable amount of time one would use a (nearly) monochromatic beam and a large area detector (originally photograph...

Mr. Feynman

6:11 AM

@SillyGoose Alright, this "abstract" group thing is all about isomorphisms. Yes, you could think of defining $\mathrm{SU}(2)$ abstractly somehow and than it would turn out to be isomorphic to the group of special $2\times 2$ unitary matrices, so once you have seen they are isomorphic there is no difference.

Or you could just start defining $\mathrm{SU}(2)$ as a matrix group, namely the group of special $2\times 2$ unitary matrices and that's it. I prefer to call $\mathrm{SU}(2)$ the matrix group itself, not an "abstract" group to which it is isomorphic although that's just a personal rant. Let's now talk about standard representations.

Adopting the picture of matrix groups from the beginning, the meaning of the standard representation is much more clear in my opinion. Representations are actions of groups on vector spaces, which in principle could be of any dimension

Silly Goose

@Mr.Feynman Hm I also start by defining $SU(2)$ as a matrix group. But to my understanding a matrix is not necessarily a linear transformation. This is the point of confusion for me: does the representation simply add on this additional structure of the matrix now being a linear transformation on top of just being a matrix?

Mr. Feynman

@SillyGoose Although on a vector space there is a one to one correspondence between matrices and linear transformations (once you have chosen a basis), this is not what we are talking about. In this context matrices are just "tables" of numbers which you can multiply etc

Now, in the standard rep the group acts on the real/complex vector space of the same dimension of the group representing each matrix with the linear map corresponding to the matrix

Silly Goose

ah okay this is clear now

so the standard representation of SU(2) is not what we are interested in in Quantum

but the standard representation of SO(3) is what we are interested in Quantum

Mr. Feynman

Since we are talking about $\mathbb{C}^n$, the matrix gives you the linear map unambigously because you have a canonical basis

@SillyGoose you also deal with that

The so called Pauli spinors lie in the standard representation of $\mathrm{SU}(2)$

Silly Goose

oh

so SU(2) captures rotations which have the 720 degrees to return to original state business

Mr. Feynman

6:18 AM

@SillyGoose the standard rep of the rotation group $\mathrm{SO}(3)$ is its action on $\mathbb{R}^3$ with the rotation matrices you have known for a while

@SillyGoose I don't know why you are bringing that up now. Is that a separate question?

Silly Goose

now is there a classification scheme of representations of arbitrary groups? I am wondering what is the explicit representation used when talking about SU(2) as the group capturing non-rel quantum spin

well i associate spinors with their behavior under rotations :P so thatt is why i thought perhaps SU(2) is associated with them but idk

Mr. Feynman

It depends on the spin you are talking about. For $s=1/2$ you have Pauli spinors as above

Silly Goose

well i am wondering about the total representation not the irreducible ones

Mr. Feynman

You might have seen that the representations of $\mathrm{SU}(2)$ are labeled by the "angular momentum" $j$

@SillyGoose why?

Silly Goose

i am wondering if there exists a unique (up to some condition) representation of SU(2) over a Hilbert space

Mr. Feynman

6:21 AM

That's not what you need for spin though

You only need to consider the irreducible representation of the right dimension

Silly Goose

my impression is that you start with the rep of SU(2) and then by a theorem you are guaranteed to decompose the rep of SU(2) into a direct sum of irreps (corresponding to each 1/2 integer of spin)

but is it that you just start with looking for irreps in the first place?

If so I suppose that solves that question because it seems like each irrep is unique up to isomorphism

Mr. Feynman

@SillyGoose Are you talking about the Clebsch Gordan decomposition? $\frac{1}{2}\otimes\frac{1}{2}\cong 0\oplus 1$

@SillyGoose irreducible very loosely means that you have a single spin content

Silly Goose

I think so--but i have only seen the CG decomp in the context of simplifying multi- spin particle systems into a single spin system-- but i mean to consider SU(2) in a more abstract context

because the way i understand it now, you have some hilbert space of all possible spin states (for every spin). then a rep of SU(2) should partition the space into a direct sum of invariant subspaces (each corresponding to each spin)

Mr. Feynman

@SillyGoose That's what happened in the previous case

It is understood we are acting on $\mathcal{H}=\mathbb{C}^2\otimes\mathbb{C}^2$ there

Silly Goose

I guess my question is: do we not care about the rep $\rho: SU(2) \rightarrow GL(\mathcal{H})$? Are the reps of SU(2) like determined by the irreps or something?

Okay I think this is the core question: When is it sufficient to know the irreducible representations of a group to construct every representation of said group

(in the context of representing SU(2) in $GL(\mathcal{H})$ ) because if it turns out that SU(2) is special in that all of its reps can be constructed from its irreps, then it makes sense to only look for its irreps because its irreps are unique up to isomorphism

Mr. Feynman

6:33 AM

@SillyGoose For single particles we only care about irreps. For more particles you have the Hilbert space is a tensor product of the single particle Hilbert space and you care about the tensor product of irreps, which is reducible and can be decomposed in a direct sum of irreps

@SillyGoose In general I don't know, you'll need someone more grounded in Lie theory like @ACuriousMind to answer that question

Silly Goose

oksy so in that sense the irreps are the atoms of any rep we care about for SU(2) (in quantum mechanics)

okay now for the meta question: what points us towards using SU(2) as the group representing non-rel quantum spin?

i couldn't find anything online answering this q :P

well i found one thing on the wikipedia: that SU(2) is the double cover of SO(3) but this means nothing to me :P

Hm perhaps this is a statement of the more general case

so SU(2) is a compact matrix Lie group. then for every $n$-dimensional Hilbert space $\mathcal{H}$, the representation $\rho: SU(2) \rightarrow GL(\mathcal{H})$ is completely reducible; that is $\rho \cong \bigoplus_i \rho_i$

where $\rho_i$ are irreps; thus, proving the importance of characterizing the irreps of compact matrix Lie groups (including SU(2))

Mr. Feynman

7:14 AM

Another question about the connection 1-form. In the way I define it is as a matrix of one forms. As we said yesterday, when we have a $G$-structure we also require that the connection 1-form is $\mathfrak{g}$-valued. How does that make sense in general? Not all Lie groups are matrix Lie groups, so I have no reason to expect that the Lie algebra is made up of matrices...

Slereah

7:31 AM

Connections in general aren't gonna be in matrix form, they're just projection maps on the tangent bundle of the principal bundle

They are like $TE \to VE$ maps

In most cases of our interest the vertical space has Lie algebra values

which are matrices

But unless you get into really weird groups that's not usually an issue

most Lie groups in physics are matrix Lie groups

ACuriousMind

@SillyGoose every semisimple rep is the sum of irreps, and every rep of a semisimple algebra is semisimple.

Silly Goose

7:57 AM

ah okay i shall take a look

muahahaha beautiful map diagrams strike again

oh oops i need to add something about what invertibility means for the diagram

Mr. Feynman

8:16 AM

@Slereah is this statement general though? I understand that it's fine for practical cases

Maybe it's something about Lie's third theorem

Slereah

I mean like the metaplectic group isn't a matrix Lie group I guess

but if you end up with that group you've got bad luck

Mr. Feynman

The point is that I don't understand in which part of the definition I know for connection as a map taking $E$-valued 0-forms (sections) into $E$- valued 1-forms satisfying [...] we lose generality and the connection form is just a matrix of forms

Is that definition involving the vertical bundle more general?

Slereah

8:32 AM

yeah it doesn't have to be in matrix form

it can just be a tangent vector to the bundle

It's just that for some Lie groups, you can represent the basis vectors by matrices

Silly Goose

Hmm okay I think I have hopefully summarized this all accurately

Mr. Feynman

@Slereah I see

Silly Goose

ive changed any to "a given"

Silly Goose

8:57 AM

hm from Qmech's answer here: physics.stackexchange.com/questions/407116/…

Firstly, we care about angular momentum. Sure that is okay. Define angular momentum to be the generator of rotations. Can't use SO(3) as per usual, need to use the double cover because we need to allow projective representations. Then we discover spin?

So it's not that SU(2) is right for spin, but SU(2) is right for angular momentum?

Silly Goose

9:21 AM

i have found perhaps the only set of notes in the universe (i say this half-jokingly) that properly address the topic of angular momentum in quantum: scholar.harvard.edu/files/noahmiller/files/…

Mr. Feynman

9:42 AM

@SillyGoose as soon as you are concerned with non-integers angular momenta you need $\mathrm{SU}(2)$

Silly Goose

10:59 AM

Hm okay so we furnish a projective representation of $SO(3)$ via a genuine representation of its double cover $SU(2)$

we consider the irreps of $SU(2)$ then. the odd ones turn out to be legit representations of $SO(3)$. so we identify these with both orbital angular momentum as well as integer spin?

the even dimension irreps are just identififed with half integer spin?

I don't think i understand this overlap in irreps.

Silly Goose

11:37 AM

okay so 3D rotations are a different projective representation of $SO(3)$

lol the theory is angular momentum is quite baffling

I suppsoe this provides some motivation for dividing the spin and other degrees of freedom

orbital angular momentum and spin are both derived from representations of $SU(2)$, but on these different degrees of freedom

Mr. Feynman

11:53 AM

@SillyGoose Yes. The reason is that the Lie algebras are the "same" $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ (this is what being locally isomorphic means), so they have the "same" (up to composing with the isomorphism) representations, labeled by "angular momenta". Now, $\mathrm{SU}(2)$ is simply connected (otherwise it couldn't even be the universal cover of $\mathrm{SO}(3)) and all representations of the algebra induce a representation of the group.

That is not true for $\mathrm{SO}(3)$, which is not simply connected and it turns out that only integer angular momenta representation can be lifted to a genuine (linear) representation of the group

The remaining ones, namely the even dimensional ones can only be regarded as projective reps

Silly Goose

and all of these irreps are spin; namely, even are half integer and odd are integer

Mr. Feynman

The fact that you have either orbital angular momentum or spin depends on the situation. The algebra is the same

Silly Goose

but now my question is how do we know that there should be two projective representations of $SO(3)$? one for orbital and one for spin

I am trying to answer "what is angular momentuM" by talking about rotations. but my exposition falls apart if someone asks okay you assumed that the projective irreps will be spin. what about orbital angular momentum

Mr. Feynman

@SillyGoose The mathematical fact is that you now have these projective representations, right? After that, calling it spin or orbital angular momentum is just Physics

@SillyGoose For a system having non-zero spin and non-zero angular momentum, then what you must regard as the generator of rotations is the total angular momentum $J=L+S$

Which more properly should be written as $J=L\otimes I+I\otimes S$

Ryder Rude

Hello

Silly Goose

12:02 PM

hm so you're saying the algebraic structure is always the same: even and odd irreps

but when we command "represent orbital angular momentum!" the weights change (or whatever the correct terminology is) relative to what the weights would be if we were representing spin?

because I say that the even irreps correspond to half integer spins. but there shouldn't be half-integer anythings for orbital momentum

Mr. Feynman

@SillyGoose Yeah, of course half-integers can only be spins. Integers may be either

Let's make an example. Suppose you have a spin $s=1$ particle with orbital angular momentum $\ell=1$, then your space of states is $1\otimes 1\cong 0\oplus 1\oplus 2$, ok?

Silly Goose

okay

Mr. Feynman

Suppose that now you don't care about orbital angular momenta anymore and consider a completely different situation, namely two particles having spin $s=1$

Then the situation is completely analogous mathematically, you just have to compose the angular momenta in the same way and get the same result with a Clebsch-Gordan decomposition

$1\otimes 1\cong0\oplus1\oplus2$

Ryder Rude

The general analysis about finding representations shows that the eigenspace of any three operators, satisying the angular momentum commutation relations, would be a subspace of the full Hilbert space that the analysis gives u

Silly Goose

Hm but I think I am confused about why there cannot be $l = 1/2$ in the first place

Ryder Rude

12:10 PM

The orbital angular momentum operators satisfy the same com relations, so their eigenspace is a subspace which excludes the 1/2 integer stuff

Similarly, the Pauli matrices satisfy the same com relations. Their eigenspace only has the 1/2 eigenvalues and it excludes the other eigenvalues

Mr. Feynman

@SillyGoose That comes from how the orbital angular momentum is defined. It turns out that its eigenfunctions are the spherical harmonics and there cannot be half-integer angular momenta spherical harmonics for a matter of single valuedness iirc

Silly Goose

hm so then the irreps are different for orbital ang than for spin

but isn't this like getting a projective representation of $SO(3)$ only to keep the genuine representation parts of it?

Ryder Rude

The eigenspaces for particular sets of operators satisfying the so3 algebra r allowed 2 b be different. e.g. The Hilbert space of Pauli matrices is different frm the Hilbert space of orbital angular momentum, and is different from Hilbert space of 2+1+0 rep, etc

But, in general, all of these r subspaces of the Hilbert space that the general analysis gives u

Silly Goose

Hm okay so what is the diff between the hilbert space on which we discover the $SO(3)$ proj rep modeling orbital ang and the one modeling spin

Ryder Rude

They r hilbert spaces of different operators. E.g.the Pauli matrices are 2x2 matrices. The orbital operators r $ix\frac{d}{dy}-iy\frac{d}{dx}$, etc

So they hav different hilbert spaces

Mr. Feynman

12:17 PM

@SillyGoose Is there a difference between the $\mathbb{R}$ representing time and the $\mathbb{R}$ representing the $x$ coordinate?

Silly Goose

@RyderRude im trying to work this out without referring to the underlying lie algebras

Mr. Feynman

And again, orbital angular momentum is integer - not half integer -so you don't need projective reps for it

Silly Goose

@Mr.Feynman sorry where is this $\mathbb{R}$ coming from?

Hm but following this argument

Mr. Feynman

It is an analogy to explain that the same mathematical concept can represent different physical situations

Silly Goose

then orbital angular momentum could only be considered for odd dimension systems. unless the problem does not exist for infinite dimension systems

from this comment it seems like perhaps the problem doesnt exist in infinite dimension systems

Mr. Feynman

12:21 PM

@SillyGoose What is the meaning of dimension in this case for you?

Silly Goose

ah dimension of the representation space

or here Hilbert space

Ryder Rude

The orbital angular momentum eigenspace is just the eigenspace of the particular operators $ix\frac{d}{dy}-iy\frac{d}{dx}$, etc. In this case, we have an independent definition of the operators. For spin, our definition of the operators comes directly from the representation analysis

For example, Pauli matrices hav been lifted directly from over there

Mr. Feynman

@SillyGoose Well, so as soon as I tell you that you have angular momentum $\ell$ (and we don't consider other degrees of freedom) it follows that the dimension of the representation is $2\ell+1$

Ryder Rude

Same for spin 1 matrices

So this is y there is no restriction on representations of spin

Mr. Feynman

And since as we said $\ell$ is integer, this implies odd dimension

Ryder Rude

12:23 PM

But orbital operators hav an independent definition. We hav to do their eigenvalue analysis

And 1/2 values r not found in that analysis

Mr. Feynman

@SillyGoose The comment is only about the Clebsch-Gordan decomposition. What the book above is about to tell you is precisely that there are no linear even dimensional $\mathrm{SO}(3)$ irreps

and that's where projective reps come into play

Silly Goose

Hm I think it is to do with ACM's comment. We split an arbitrary system into finite and infinite degrees of freedom $\mathcal{H}_s \otimes \mathcal{L}^2$. Then, we develop the projective representations of $SO(3)$ on both, yielding spin for finite dimensional case and orbital angular momentum for the infinite dimensional case.

3

Q: Infinite dimensional representations of $\text{SO}(3)$

In the theory of angular momentum, we wish to study the projective representations of the rotation group $\text{SO}(3)$, for which we turn to the representation theory of the double cover $\text{SU}(2)$. I understand the finite dimensional representation theory of the Lie algebra $\mathfrak{su}(...

quantum-mechanics angular-momentum group-theory representation-theory lie-algebra

Ryder Rude

Yes. This is correct. For the finite Hilbert space part, we can lift whatever matrices we want from the general representations of SO3. e.g. we lift Pauli matrices to model electrons, spin 1 matrices to model photons, spin 2 matrices for gravitions

We can lift 1+0 to model two electrons

But orbital operators hav an independent definition. $xp_y-yp_x$, etc

So u r no longer lifting the operators according to ur modeling choices

We r instead doing the eigenvalue analysis of these operatoes $xp_y-yp_x$, etc

Mr. Feynman

@SillyGoose I'm leaving now. Anyways, if your question is about ACM's comment, who could answer better than ACM himself? :P

Silly Goose

cya!

ACuriousMind

2:17 PM

@Mr.Feynman I'm on vacation and only look in here briefly, if at all, so don't expect elaborate or quick answers

Silly Goose

i am down the angular momentum rabbit hole :P

Ryder Rude

Happy vacation :)

Slereah

2:51 PM

@ACuriousMind Is it a pentecost vacation

or is that not a thing in Germany

Loong

Yes, Whit Monday is a holiday in Germany.

I should remember not to go to work on Monday.

However, I am on call.

Mr. Feynman

3:15 PM

@ACuriousMind Oh, I didn't ping you. :P Actually I have no questions for you. Enjoy your vacation!

Ryder Rude

Do u guys subscribe to the materialism philosophy or something close or something very different

Ryder Rude

3:30 PM

I am close to materialism but i also attribute some un-describable aspects to materials themselves

These aspects shud be how consciousness emerges out of materials

So i dont believe consciousness emerges out of if-else instructions. I think qualia r real

naturallyInconsistent

3:45 PM

@SillyGoose completely drowning.

ACuriousMind

4:18 PM

@Slereah well, it is Pentecost, but the vacation consists of me spending the weekend playing MtG with a bunch of other nerds

Slereah

as the lord would want

2

Ryder Rude

4:46 PM

@Slereah lolol

user 858770

Amen.

Amit

@RyderRude Materialism assumes to know what is matter that's a builtin blunder

Slereah

Homer Simpson's First Appearance

►

Silly Goose

5:06 PM

@naturallyInconsistent Lol

I keep telling myself that after one more answer i will get down to the bottom of angular momentum

Slereah

it's how little things spin

user 858770

or how strongly they spin

Ryder Rude

5:27 PM

@Amit yes. I can think of two well-defined deifinitions of materliasm : 1. Mind is made of the same stuff that everything else is. But this school is just monism. So it can encompass idealism too, which materialists wouldnt agree with.. 2. The universe is information. This second school is the closest thing to materialism that is well defined

The second school is also synonymous with a mathematical universe

I think materialists mostly just dont bother making their school well-defined

Like u say, claiming that everything is matter is making no progress. Becuz the nature of matter itself is unknown.

But if materialists say that universe is information, then it amounts to making a positive claim. Becuz this wud be claiming that the nature of matter is exhaustively information

Amit

6:11 PM

Then you have another problem, because if it can be described accutately, it is information.. so it becomes unfalsifiable, because you can't produce an accurate account that describes why you can't describe, hence convert into information, something

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