The h Bar

John Rennie

4:17 AM

@EnthusiastiC These days many are available on the Internet but you'll need to Google for them as there is no single repository for theses. For those of us who graduated in pre-history you need to contact the university where they graduated as the university will keep a printed copy.

Arjun

7:30 AM

@LuckyChouhan are you from India?

Mr. Feynman

8:17 AM

@LuckyChouhan Hi (?)

ACuriousMind

9:36 AM

@imbAF Just to make sure: Ryder linked exactly the post I would have linked, too, and you really should not think about "virtual particles being produced from the vacuum". It's not a helpful picture to actually start to understand quantum field theory, though once one understands it, one can see how people arrived at that picture. The "energy-time uncertainty" for virtual particles we hear so often about does not actually exist in that form.

Slereah

10:28 AM

> What is a Comonad? – Comath and Mputer Science

heh

Nairit Sahoo

10:48 AM

When doing Wigner classification why do we look for stabilizer subgroup corresponding to momentum and not angular momentum or boosts?

ACuriousMind

@NairitSahoo now that's the method of induced representations :P

Ryder Rude

hi

Nairit Sahoo

@ACuriousMind Can you please tell why they fix momenta and not something else? Is it because we know the representations of Lorentz group, and from that we build up representation of Poincare... so we have to look at the little group of the extra element i.e. translations?

ACuriousMind

11:05 AM

@NairitSahoo Well...for an elementary answer: It works, doesn't it?

Nairit Sahoo

@ACuriousMind That's how the book said about it. I just went with the flow when I read it for the first time a few months ago

They were like "let's do this and see what happens..."

ACuriousMind

I don't think when Wigner first did this there was any big theory behind it, he just happened to notice that a momentum vector + a representation of its stabilizer actually suffices to construct a representation of the whole group

I don't remember if Weinberg has a physical motivation for this procedure (it might be worth checking), but mathematically this is Mackey's theory of induced representations via systems of imprimitivity developed later, where in particular you can prove that groups of the form $G = L\mathbb{R}^n$ (which is the form of the Poincaré group) generally can have their representations constructed from a vector from that $\mathbb{R}^n$ and its stabilizer in $L$.

Nairit Sahoo

What is $L$?

ACuriousMind

in the Poincaré case the Lorentz group

the Poincaré group is the (semi-direct) product of the Lorentz group and translations

Nairit Sahoo

yes. But I never saw that notation before.

Mackey's theory of induced reps. ok

I don't know if this is an equivalent question, but the other thing that I have been thinking is when we have the transformation law of fields at x=0 and from that we need to have transformation law of fields at arbitrary $x$. From $0 \to x$ do we really have to go via translations i.e. do we really have to take conjugate w.r.t. translations, or is it a choice?

ACuriousMind

11:14 AM

I don't know what you mean; the alternative way to get the field reps I told you yesterday doesn't do this "$0\to x$" step at all

Ryder Rude

It might b relevant that we want unitary reps in QM. if we were doing non unitary faithful reps, we could do the tensor product of the infinite dimensional rep of Poincaire and the finite dimensional rep of Lorentz

ACuriousMind

what

Ryder Rude

i mean that the finite dimensional reps r non unitary wrt boost

Nairit Sahoo

@ACuriousMind Yes. I was referring to the other method.

Ryder Rude

so if we simply did the above tensor product, we would get non unitary reps

Nairit Sahoo

11:17 AM

But okay... I think you don't like it that much :p

ACuriousMind

@NairitSahoo sure, but if one method does not involve choice, how could the other?

they have to yield the same result because they're doing the same thing

Ryder Rude

what i mean is that we take the spin-0 infinite dimensional rep of Poincaire, and tensor product it with a finite dimensional rep of Lorentz. This gives us a non unitary rep of Poincaire, which we don't want

Nairit Sahoo

Hmm true

Ryder Rude

am I making sense

ACuriousMind

@RyderRude Barely :P I understand what you're saying but I don't understand how it's relevant to the discussion since no one asked about this weird idea of constructing reps

Ryder Rude

11:20 AM

@ACuriousMind it's not that weird... the construction of unitary reps is just a patch on this idea, right?

EnthusiastiC

@JohnRennie Thanks for the reply. Actually, that's what I'm doing right now. Also, researchgate.net alongside journals and articles, hosts theses from many authors

Ryder Rude

@ACuriousMind i mean that for unitary reps, we remove some degrees of freedom

ACuriousMind

@RyderRude now you've lost me again :P

have you looked at the construction via the little groups/stabilizers?

Ryder Rude

i looked at it long ago. It did mention little groups

ACuriousMind

if you don't know how the construction works in detail this is a pointless discussion :P

Ryder Rude

11:22 AM

yeah..maybe

ACuriousMind

it doesn't just "mention little groups", they are the crucial ingredient, and it was the nature of those little groups that Nairit asked about

Ryder Rude

yeah... i mentioned my point because i think the little group trick is for constructing unitary reps, but i don't remember the details. For non unitary reps, we can use the trivial method i gave

by removing degrees of freedom, i meant something like "choosing two basis vectors for the internal degrees of freedom ar each $p ^ {\mu}$, in case of the Maxwell field" @ACuriousMind

The above trick is related to little group and it gives us unitarity, right? @ACuriousMind

ACuriousMind

nope, I still don't know what you mean by that

what "internal degrees of freedom" do you mean? Who's picking "basis vectors" for them in the context of Wigner's classification?

I think you're free-associating random techniques one does in the setup for QFT with each other that are not really related :P

Ryder Rude

yeah... maybe

@ACuriousMind i meant, like, the expansion of Maxwell field as $\int \sum _r $\epsilon ^{\mu} (p^{\mu}) a_r (p) + \text {the other term}$. The $\epsilon (p)$ is what I'm calling the basis vector chosen at $p$

Sorry i will have to fix it

This $\int \sum _r \epsilon _r^{\mu} (p) a_r (p) + \text {the other term}$. The $\epsilon _r ^{\mu} (p)$ is what I'm calling the basis vectors at $p$

ACuriousMind

yeah, that has nothing to do with the construction of the representations

that's do to with the gauge symmetry

Ryder Rude

11:34 AM

thanks... I will have to read about it

ACuriousMind

note that, if anything, the $a_r(p)$ is the operator here that transforms in the unitary representation, not the $\epsilon$

Ryder Rude

yeah...the epsilon r just fixed basis vectors

ACuriousMind

so why would you think they have something to do with Wigner's classification :P

Ryder Rude

I read this really long ago. The book started with this technique to get unitarity... and later introduced the little group idea

It's in Schwartz's book

ACuriousMind

...I'm willing to bet that that wasn't about unitarity of the representation, but to get "unitarity" in the sense of getting rid of the negative norm states during Gupta-Bleuler quantization

those are entirely unrelated

Ryder Rude

11:38 AM

yes, it was exactly that, but it's related to unitarity because the transformation of creation/annihilation operators induces a transformation of the states

I mean states creates by those operators

ACuriousMind

yes, but that has nothing to do with unitarity of the representation of the Poincaré group

you're not looking at operators from the Poincaré group at all here

Ryder Rude

yeah.. the method doesn't mention generators at all

ACuriousMind

There are many places in quantum physics where we care about something being "unitary"; they are not related just by virtue of using the same word

@RyderRude so now you can understand my confusion as to why you brought this up in the context of someone asking about Poincaré group representations, right? :P

Ryder Rude

@ACuriousMind to be clear, this is the procedure i have in mind for the Poincaire group (does it make sense): we do a tensor product of the infinite dimensional spin-0 rep, and a finite dim Lorentz grp rep.. this is non unitary rn. And later, we remove degrees of freedom to get unitarity

ACuriousMind

no, that's not at all what happens

Ryder Rude

11:42 AM

yeah.. but i don't mean Wigner's method. But can this be a method too?

ACuriousMind

no

Ryder Rude

okay :P. Then i am wrong. Sorry for the confusion

i will have to re-read it

ACuriousMind

this would imply the unitary representations are quotients of this tensor product, which is as far as I know simply not the case

Ryder Rude

oh

ACuriousMind

Again, you are probably confusing this with Gupta-Bleuler quantization where we quotient out the unphysical states to end up with a unitary Hilbert space for the quantization of the EM field

your confusion with the $\epsilon$ points in that direction, but these two things are completely different procedures

Ryder Rude

11:45 AM

With the Dirac field, i had the $\mu _r (p)$ in mind... i wasn't specifically thinking of Maxwell

But again, thanks for clarifying this..

@ACuriousMind one last thing, would u say that the little group trick is for unitarity, and that non unitarity is easier to get by the other method?

ACuriousMind

I don't know what that means

a unitary representation in the infinite-dimensional case is fundamentally different from non-unitary ones

it's not like in the finite-dimensional case where you might first construct random representations and then think about how to unitarize them

Ryder Rude

okay...thanks for clarifying this

Lucky Chouhan

12:49 PM

@Arjun yess! I think you're also :)

@Mr.Feynman Nothing, just wanted to say Hi to Prof. Feynman :) btw how are you doing?

@RyderRude yeah, I saw a video. In mirror image everything gets flipped

Ryder Rude

1:11 PM

@LuckyChouhan great

Nairit Sahoo

1:36 PM

Why do certain people prefer to vary the action to get EOM when they can just plug and chug in EL equations?

Lucky Chouhan

@RyderRude where are you from?

Slereah

2:09 PM

> The monadic curse is that once someone learns what monads are and how to use them, they lose the ability to explain them to other people

Ryder Rude

2:34 PM

@LuckyChouhan i wanna stay private :)

Mr. Feynman

2:56 PM

@LuckyChouhan character ai would probably play a much better Feynman

I'm drowning in Ising models

Ryder Rude

ok now i re-read what Schwartz is doing... he starts with tensor fields which which r obviously reps of Poincaire group (this is the "trivial aproach" i talked about above, but Schwartz only uses tensor fields instead of tensor products of the spin-0 rep with arbitrary finite dimensional reps of Lorentz group)

And then to get a unitary theory, he comes up with Lagrangian to remove degrees of freedom from the tensor fields to match the degrees of freedom from Wigner's classification

Also, he does not do Wigner's classification.. he just states its results and mentions the little group once, which is y i didn't remember much :P

Ryder Rude

3:14 PM

he uses this stuff to derive gauge invariance and stuff, because he is guessing the right lagrangians

also, the tensor fields he starts with r classical fields in position space, so they're not a Hilbert space rep of Poincaire group. what i was thinking is that this was equivalent to starting with a non unitary infinite dimensional rep of Poincaire group, but that's incorrect

imbAF

4:55 PM

I have a question about the W.E theorem because I am confused with the notation of the general expression. The fact that C.G coef. are present, it means that we are dealing with a basis shift, from an uncoupled tensor product basis to a coupled one. From what is explained in the wiki:

$|j,m\rangle$ is an eigenstate in the coupled basis while $|j',k;m',q\rangle$ is an element of the uncoupled tensor product basis. And the fact that the relation between the quantum nr. j,j' and k is as follows $j'+k=j$, makes my assumption correct. Now the problem is with the left side of the equation. There…

Wigner–Eckart theorem

The Wigner–Eckart theorem is a theorem of representation theory and quantum mechanics. It states that matrix elements of spherical tensor operators in the basis of angular momentum eigenstates can be expressed as the product of two factors, one of which is independent of angular momentum orientation, and the other a Clebsch–Gordan coefficient. The name derives from physicists Eugene Wigner and Carl Eckart, who developed the formalism as a link between the symmetry transformation groups of space (applied to the Schrödinger equations) and the laws of conservation of energy, momentum, and angular...

Additionally the following is said :
"The Wigner–Eckart theorem states indeed that operating with a spherical tensor operator of rank k on an angular momentum eigenstate is like adding a state with angular momentum k to the state"

What it means to add a quantum state to another? How is that mathematically translated? I know addition of angular momenta, but of states?

ACuriousMind

@imbAF Oddly enough, you asked about this theorem already 2,5 years ago :P I have nothing to add to my answer from then

imbAF

I know I did and was waiting for you to link me the old convo ahhaha

@ACuriousMind I don't understand you answer to my first question

Tensor operator of first order is a vector operator, which is what the angular momentum operator is.

But if the order is diffeent than 1, the operator shouldn't be classified as a vector operator, hence it cannot be an angular momentum operator

is this wrong?

angular momentum generator?

ACuriousMind

5:14 PM

hm, I don't know what I was thinking with that example, either, I'll edit

imbAF

Is it correct to say that in my thread I am mixing Hilbert spaces with different representations of the basis of the Hilbert space?

ACuriousMind

no, I don't even know what that means :P

what is a "representation of a basis"

imbAF

I mean

coupled basis and the tensor product basis

this alone:
"We have a space of states H
upon which we have a reducible representation of the rotation group SO(3)"

I have no clue

I don't understand the answer

I have 0 knowledge of representaiton theory

SO(3) and how one can show that angular momentum and it's components represent geometrically, or algebraically or in whatever other way, a rotation

Could you try and answer my question here? I feel that here I have written a better question and my understanding is better then 2 years ago

ACuriousMind

5:38 PM

I genuinely do believe that my abstract explanation is, once understood, much clearer than the standard presentation. But anyway:

@imbAF I can answer your specific questions, but the answers won't help you much: a) There are no two different bases here, the thing in the WE theorem is a CG coefficient but it is not actually the inner product between specific coupled and uncoupled states in the Hilbert space (you can see this if you look at the normal proof) b) they just mean addition of angular momenta, i.e. imagine tensoring with an eigenstate of momentum k

imbAF

"i.e. imagine tensoring with an eigenstate of momentum k" ? what?

tensoring? what does that mean ?

ACuriousMind

...in the usual setting where you have "addition of angular momentum", you have a tensor product between two states with definite angular momentum, no?

that's where you first encounter CG coefficients

imbAF

yes

ACuriousMind

So the point is that applying an operator of momentum $k$ to a state is an analogous situation, and you can decompose the result via CG coefficients exactly as you would if you had formed the tensor product of the state with another state of angular momentum $k$

that's what the WE theorem is

imbAF

And the description in wiki, does imply that you have two angular momenta with main angular quantum nr. j' and r (order of the tensor op.) and one that corresponds to the total angular momentum

ACuriousMind

5:43 PM

I'm not really sure where the difficulty here is: With states you have two state with momenta $j_1,j_2$ and the resultant total momentum $j$, here you have one state with momentum $j_1$, an operator with momentum $j_2$ and the resultant total momentum $j$

imbAF

I understand this part

ACuriousMind

it's addition of angular momenta, just one of the two momenta being added comes from the tensor operator, not another state

imbAF

I have no difficulty here

I know

ACuriousMind

@imbAF then I don't understand what the question is

imbAF

Well

ACuriousMind

5:44 PM

because your question sounds very much as if you don't understand this :P

imbAF

Not really

Let's start with this:
$|j,m\rangle$ is an eigenstate in the coupled basis while $|j',k;m',q\rangle$ is an element of the uncoupled tensor product basis.

Do you agree?

Notation, being the main point of my question

ACuriousMind

no, cf. my answer a)

> There are no two different bases here, the thing in the WE theorem is a CG coefficient but it is not actually the inner product between specific coupled and uncoupled states in the Hilbert space (you can see this if you look at the normal proof)

imbAF

Wow

ACuriousMind

you have to look at the $\langle j,m\vert j', k; m' q\rangle$ not as the inner product of specific states, it is the reason my answer only uses the $C_{\dots}$ notation there

imbAF

so ....... is the Clebsch–Gordan coefficient for coupling j′ with k to get j,

is wrong then?

from wiki

ACuriousMind

5:46 PM

no, that's right, it is a CG coefficient

imbAF

yes but a CG of coupling of j' with k to get j

so j is the result of coupling

ACuriousMind

yes, it's the CG coefficient for $j'$ and $k$ coupling into $j$

imbAF

But this statement somehow does not imply what I wrote above

Which, when I encountered the cg coef. this was the textbook definition when they are present

and the def. in wikipedia

ACuriousMind

you'll have to be more specific what you're talking about, you've lost me :P

imbAF

Ok

Well, first of all I gotta say

why tf would you teach stuff like this without even TOUCHING representation theory

clown system

ACuriousMind

5:49 PM

...I don't disagree

imbAF

I will sound repetitive, but I will start from the beginning and perhaps we can find a way for me to understand what I struggle with

ACuriousMind

which is why I would heavily recommend just leaving this be until you understand more about it; the WE theorem isn't so central that you'd really need it anywhere else (or rather it's completely fine to just use its results)

I'm not really interested in trying to "fix" the poor standard presentation

imbAF

it's like the 3rd lecture of a previews semester

I want to try one last time and understand it. Or make you understand what I don't understand

So, hopefully you have the patience, once again to help me with that

ACuriousMind

well, not today, I'll be gone in a few minutes for the rest of the evening, I'm afraid :P

imbAF

ah

Just one other question

$R_{\vec u}(\alpha)|\psi\rangle=e^{-i\alpha \frac{\vec L \vec u}{\hbar}}|\psi\rangle$

How does one mathematically execute/perform the RHS? Meaning, what do I write as result when an exponential expression containing an operator acts on a state

Sir Cumference

6:04 PM

You know, when we're first taught physics we learn a bunch of terms like "impulse" and stuff

But I don't think I've ever used the word "impulse" since freshman year of college

Isn't that kind of weird?

imbAF

that's the german word for momentum, isn't it?

Sir Cumference

Well it's what we call a change in momentum

Don't know about German tho

I feel like there's probably other first semester physics terms that I've strangely never encountered since

imbAF

wouldn't momentum change be force?

Sir Cumference

Impulse is the integral of force over time

So it's the change of momentum

Force is the derivative of momentum

imbAF

and if you were to change only direction and no magnitude, how would the integral give you that change?

Sir Cumference

6:11 PM

Well it'd be the difference between two momentum vectors

imbAF

7:00 PM

I was able to understand the W.E theorem a bit more. And this does make sense now.
If I understand it correctly, we consider a quantum mechanical system to which we can associate an angular vector operator. The W.E theorem considers two eigenstates of it, one initial one namely $|j_2,m_2\rangle$ and another one to which the system finds itself after the "interaction" (whatever we mean by that, what should I mean actually?) between the tensor operator and the angular momentum.

Dannyu NDos

10:53 PM

My mom is overly concerned about lithium-ion batteries exploding. What should I do?

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