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19:01
@ACuriousMind parity reverses the sign of helicity because even though the eigenvalue of $J_3$ does not get changed, the projection along the momentum (which is flipped) is thus reversed?
Why doesn't Mr.Rogers have anything on what you should do when you feel like your research is just an excuse to get out of the apartment?
@0celo7 I suppose you could phrase it that way. The deeper reason is that helicity classifies only representations of the identity component $\mathrm{SO}^+(1,3)$, I think, and not of $\mathrm{O}(1,3)$. Representations of $\mathrm{O}(1,3)$ are given by reps of $\mathrm{SO}^+(1,3)$ together with how the $P$ and $T$ operators act.
Because it's a beautiful day in the neighborhood?
Who is Mr.Rogers?
19:05
@0celo7 what the fuck, you don't know Mr.Rogers?
I can understand @ACuriousMind not knowing
obe
obe
@skillpatrol who are you?
Fred McFeely Rogers (March 20, 1928 – February 27, 2003) was an American television personality, educator, Presbyterian minister, composer, songwriter, author, and activist. Rogers was most famous for creating, hosting, and composing the theme music for the educational preschool television series Mister Rogers' Neighborhood (1968–2001), which featured his gentle, soft-spoken personality and directness to his audiences. Initially educated to be a minister, Rogers was displeased with the way television addressed children and made an effort to change this when he began to write for and perform on...
@obe skullpatrol
19:06
is he the pedo?
AFAIK, he is not a pedophile
never heard of him then
you have to be lying
He ran a TV show called Mr Rogers Neighborhood for like 50 years
19:07
wait did you grow up in the US? Or elsewhere?
@0celo7
@ACuriousMind digesting, I might need some exposition on that
Germany
implying the world is US and not US
@0celo7 ahhh ok
that explains it
I don't think I've met anyone who grew up in the US that didn't know Mr.Rogers
or Canada
Canada <3
obe
obe
idk who that is.
19:10
@skillpatrol Where is that? I think you mean Canadia, eh?
@ACuriousMind OK
I'm not the man Mr.Rogers wanted me to be
sniff
Waters?
hmm?
Roger Waters?
19:12
I wish I was Roger Waters
except he sings badly now
so modulo that
Too much LSD
I wish Weinberg explained why $R^{-1}_2\mathscr{P}$ commutes with a boost along the 3-axis
And I wish \mathscr would be exterminated from the face of the earth
h8ers gonna h8
Huy
Huy
19:15
h8 h8 h8 h8
@0celo7 Do you like mathscr?
I mean, look at it!
@ACuriousMind I do.
It's drunk, and not in a good way
@ACuriousMind You're drunk. (This is likely a true statement.)
who here knows the matrix for a half-rotation about the 2-axis?
Sadly not
19:18
I can't into matrix exponentials
Huy
Huy
@0celo7: What's the 2-axis?
@ACuriousMind @FenderLesPaul do you know how to prove the first statement on page 79?
@Huy y
Huy
Huy
@0celo7: So just $$\begin{pmatrix} \cos \phi & 0 & \sin \phi\\ 0 & 1 & 0 \\ -\sin \phi & 0 & \cos \phi \end{pmatrix}$$
@Huy looks right, I'll just do it explicitly then...
ugh
I'll do a back-of-the-envelope calculation on the back of an envelope
That^ brings back memories of @ACuriousMind and Sofia chatting about "shut up and calculate..." :(
19:31
yay
@skillpatrol Sofia
explicit calculation proves victorious
@KyleKanos thanks
People who have an English language barrier are at a distinct disadvantage in these kind of chat rooms.
Huy
Huy
yes
people who disagree with people like ACM are at a disadvantage
although Danu is the mean one
@Huy I agree, poor Winston
I suspect Dogeyes will die
Huy
Huy
19:37
:(
Doooog aayye
that mission was super frustrating
Huy
Huy
I sound exactly the same as those guys.
to get a gun you pretty much had to run right into the middle of a room and disarm a dude
Huy
Huy
Who needs guns when you've got martial arts.
Kung fu?
19:38
I found a broken mechanic: leg break + stun shot
grappling is very reliable
Having said that those who read too much into comments are disadvantaged too
@skillpatrol you trying to accuse me of something
Are you trying be sarcastic?
Or am I missing something?
yes
Huy
Huy
Ok YouTube stopped autoplaying the next video.
That means it's time to go to bed.
Good night.
19:42
@ACuriousMind what's the TeX command for upright sans-serif font?
later pal
night
@0celo7 \textup?
(removed)
19:45
removed
way to break immersion
$\text{ \textup{P} }$
There should be an option to star your own comments
19:46
\textsf{P}
swag
obe
obe
I like \mathfrak{}
upshape?
lol that sounds like a bra...
@0celo7 It exists :P
lift and separate
$$\textsf{P}\Psi_{p,\sigma}=\eta_\sigma \mathrm{e}^{\mp\mathrm{i}\pi\sigma}\Psi_{\mathscr{P}p,-\sigma}$$
that's pretty
obe
obe
19:48
sarcasm?
this chat makes it impossible to get work done
3
welcome to the black hole of productivity
obe
obe
What is the importance of mixed states in qm?
@FenderLesPaul Not sure what you guys will do when I stop showing up :/
19:50
:(
@KyleKanos finally overtake you in rep
lol
@KyleKanos why would you stop showing up?
@Jimself I'll answer questions in the evenings, but not during the day
19:51
Don't leave us Kail
@FenderLesPaul Doing that "job" thing
@KyleKanos but Kyle :(
@obe They represent statistical states, where we only know the probabilities that the system is in a certain state, as opposed to knowing its state and calculating probabilities for measuring it in a different one.
@vzn well, I have progressed a lot.. I am right now modelling open quantum system of the photosystem...It's fun but a lot of work and research...
@FenderLesPaul I can't stay unemployed forever you know
19:52
he can't leave us. We know where his frame of reference is
@KyleKanos yes you can
@KyleKanos You can! I believe in you!
@KyleKanos what's going on? you are leaving Physics.SE?
I'll send you pizzas to eat
obe
obe
@ACuriousMind Ok.
19:53
@FenderLesPaul I have a wife & 4 kids, no I really can't
hmm that complicates things
@ACuriousMind I have a wife & 4 kids, no I really can't
@TanMath Sorta. I'll be getting a job soon & won't be as active
So I'm not leaving per se, just my availability will be reduced
@KyleKanos aw.. that's sad..thanks for all the help you've done for all of us...
something > nothing
^ that
19:54
@KyleKanos So you got the Pascal job?
I'll try to be doing the 20 reviews/day
But I think Brandon's going to get that Late Answers badge before me :(
This reflects my feelings atm: youtube.com/watch?v=PFO28_CQQf4
@alarge The offer was extended over the phone, we just have to finalize the details
@KyleKanos
he got all the other ones before you too, might as well be consistent
19:55
@Jimself Very true
@KyleKanos Congratulations. What kind of a job is it?
A quant job
quantum?
In a bank?
you're a heretic now
19:56
@0celo7 Yes, I'm in the process of being quantized
@alarge Yes
@KyleKanos it happens to the best of us D:
@ACuriousMind was quantized when he took a shitty GR class
@0celo7 quant = quantitative finance, I believe. Heresy, but of a different sort :P
lol
we should replace puberty with quantization
ACM thought GR is index manipulations
now he knows GR is index manipulations and a cauchy sequence here and there
19:58
GR is pretty much index shit
And once in a while, PDEs and integrals
And topology if you are mad
@skillpatrol I've seen that. Considered a few times of being a member
The best Stack Exchange
I want to post dumb questions
"How do I get money"
@alarge the book list is finally online, but the grad stochastics class is not being offered this semester
obe
obe
@ACuriousMind $A = \sum_{n} \ket{a} a \bra{a}$ is the same as $A = \int \ket{a} \braket{a|a} da$ ?
20:00
"Is Mayonnaise a currency?"
obe
obe
fail latex.
@FenderLesPaul yes
anything can be a currency
@0celo7 upvoted
@obe Those commands only exist in the physics package, which, for whatever reason, is not active in MathJax
in the US, we use "in god government we trust" as a currency
20:00
Sometimes I want to go to the meta SE board and ask
obe
obe
@ACuriousMind Ok well, is it right? If A is a hermitian operator.
@0celo7 Well, advanced ug/grad analysis will get you started with measures and integrals anyway.
"I have several stacks and I would like to exchange them for other stacks"
@alarge I'm definitely taking the grad Real/Complex analysis modules
then probably functional analysis
haha it's funny because probability theory
20:02
maybe stochastics, although I'd have to find a time for the prereq, grad probability
excuses self
@obe Well, I'm not exactly sure how the expressions are supposed to look, but no, a sum is rarely the same as an integral
it is only in the limit
@KyleKanos So what's your role? Say, looking at the wiki article, which description fits the position the best? Also, why Pascal?
They may be "analogous", though, in the sense that at some points, you might need to "replace" a sum with an integral, which is caused by the physicist sweeping infinite-dimensional subtleties under a very large rug.
20:03
@ACuriousMind sums are integrals -.- did you even take calculus
yeah
you can't spell integral without sum
@ACuriousMind a comfy rig
they don't even share a single letter
yes they do
egral is synonymous with sum
20:04
@Slereah major derp on your part
do you even SAT bro?
the integral sign is literally an S
But I meant
The word "integral"
means integer
20:06
@alarge Pascal was mentioned over the recent phone call, though I was told C++ elsewhen by these guys. More on the Risk Management side of things.
@KyleKanos do you have to move?
I will, but I'll still be East coast
up? down?
closer to Jim?
or self?
:D
20:10
closer to self
wow
deep
Moving north
eh land?
Boston?
Cuba?
20:11
Bawston?
This has suddenly turned into an AMA...
Cuba is south of me, last I checked
lies
Is it though?
CONSPIRACY
Moving north of the north pole?
20:12
@KyleKanos give us a hint
hi @JohnDuffield
Cuba is the pink blob that is beneath the Florida wing of the US
Flo-rida
right round
Like a record?
get the job finalised first and good luck pal
20:13
^ agreed
@0celo7 : hi, but drat, I'm afraid I have to go.
later pal
13 ads blocked on this^
Ghostery found 15
They will not let me star it in the homotopy theory room :(
ABP found 27
20:27
"As he talked, Tao carved shapes in the air with his hands, like a magician. "
@ACuriousMind Let $U(R_2)=\exp(\mathrm{i}\pi J_2)$. Why does $U(R_2^{-1})$ reverse the sign of $J_3$?
who the hell writes like that?
ny times
@Jimself : the speed of light in vacuo is not constant, see my answer here for more references. The locally-measured speed of light is only constant because we use the local motion of light to define the second and the metre, and then use them to measure the local motion of light. It's a cargo-cult tautology I'm afraid. Define the second and the metre the old way, and the locally-measured speed of light isn't constant at all. It varies with elevation, just like Einstein said. — John Duffield 16 mins ago
::eye twitch::
20:29
I'm not sure how many times I can hear exactly the same argument and say "no that, again, is wrong for these reasons"
@Jimself I remind you, for no particular reason, of this meta post of yours.
2
@0celo7 ...by the commutation relations of $J_3$ and $J_2$?
An accurate post if ever I've written one
@ACuriousMind I figured but I'm not smart enough :( halp
what does $J_1$ do here
I'm assuming $J_3 U(R_2^{-1})=-U(R_2^{-1})J_3$, but the commutation relations do not help...I don't think
No need to bring up Duffield, we're all having a good time!
$[J_2,J_3]=\mathrm{i}J_1$
20:33
The misconception about spatial variation of speed of light is just one of those situations where you have to know the mathematical foundations of the theory
I don't see it...
or you're doomed to never understand why it's a wrong viewpoint
@0celo7 Isn't the statement "$U$ reverses the sign of $J_3$" that $U J_3 U^{-1} = -J_3$?
doooomed
he's backed himself into a corner at this point, he can't take his word back
@ACuriousMind that's exactly what I wrote, with a U moved over
20:34
So I guess you get it by applying the BCH formula
oh my god that thing
@0celo7 Ah, yes, it is
@Jimself Yes, yes it is
Baker-Campbell-Hausdorff?
@0celo7 No, all he has to do is say "Huh, wow, I never really saw it that way. I always thought that he meant this." And then change almost every one of his answers
I need the one I can't find
the one with the exponents and the things
argh
20:37
we would then drop the issue and not make a big deal of it. Face saved
where the heck is the one that I need??
@0celo7 I am the one that you need
All you need is love
@ACuriousMind uh, halp?
20:41
isn't that like
The Lie Trotter Katto formula?
Or related to it
BCH, Hadamard lemma, Lie-Trotter, they're all interrelated
Lie Trotter Katto sounds like a magic spell
@0celo7 Well, tbh, my idea wasn't more than "some variant of BCH should give this" :/
@FenderLesPaul halp
@FenderLesPaul He's a witch!
20:42
Witch!
@0celo7 with?
@FenderLesPaul the claim right above eq. 2.6.20 on page 78 of Weinberg 1
Since $U(R_2^-1)$ reverses the sign of..."
that part?
yes
oh
I drew a picture
it's like 4th grade :(
$U(R_2)$ rotates the $z$-axis by pi lol
I love how ACM told me to use BCH
@0celo7 The last time I gave you an intuitive answer you didn't want it!
Diagrams are your friends
20:47
@ACuriousMind got me there
obe
obe
Is it possible to take QM and QFT at the same time?
That's about as unadvisable as it gets, I'd say
really? try taking GR without knowing vector calculus
"Professor, what's a partial derivative?"
@0celo7 do you want a verification in terms of operators?
because to convince myself I'd just say
@FenderLesPaul if you have one
my picture was fine for my purposes
obe
obe
20:52
@0celo7 Though you can learn that in like 5 minutes.
@obe you can learn Shankar is 5 minutes
that $U(R_2^{-1})$ acting on a J_3 eigenstate, say with the spin aligned with the z-axis, rotates the x-z plane by pi so it brings the spin to -z
obe
obe
@0celo7 If he would talk less.
@FenderLesPaul that's exactly what I did
but I'm not entirely sure how to verify it in terms of operators
20:53
exactly
it has to be possible, right?
because my go to method is to use infinitesimals and just work with generators
obe
obe
Anyhow I'm going to be the first to do this, let's see how it turns out in 5 months.
but you can't do that here since it's pi degree rotation
You can't understand it in 5 minutes
@0celo7 it should be possible yeah
so something like $e^{i\pi J_2}J_3 e^{-i\pi J_2} = -J_3$?
20:55
@obe why are you trying this?
or is what Weinberg says only true for J_3 eigenstates and not as a statement about $J_3$ itself?
Without the basics you're wasting your time pal.
$J_3$
above 2.6.21 he says the same thing about a $J_3$ rotation changing the sign of $J_2$
yeah he does
hmm
I'm not sure how to use BCH here because the J_3 isn't exponentiated

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