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17:03
Is the algebra determined partly by the EoM?
Maybe I should just do a SE question
But then again, I would risk it being a list of question
@Slereah I think "How does one determine the algebra of observables for a quantum system" is a perfectly fine question; I just think the answer may be the same as for "How does one determine the Hamiltonian/Lagrangian?" - you guess it.
Awesome, just found this room because I'm in need of an explain of something that has been plaguing me recently...
@ACuriousMind Yes but what are the bare minimum
What is indispensible
In Heisenberg I'd need the CCR and the Hamiltonian
@Awm0121 ask away!
Path integral I'd need the Lagrangian and some measure
What do I need to 100% determine the algebra
17:15
@Slereah Oh, I think not, at least for the free field. You'd start the Weinberg way by demanding there are creation and annihilation operators for particles. Then you simply take the algebra generated by them.
But I can find that out via the EoM
I was thinking about the fact that electrical energy can be transferred at the speed of light because of the fact that the electrons are 'tied' together, but not because they are actually travelling all that fast (electron drift is rather slow)
$i \partial_t \hat O = [\hat H, \hat O]$
That's enough to get Klein Gordon
@Slereah Why would you have such a fetish for the classical e.o.m.? ;) What does the quantum theory care?
And then you can do whatever Fourier tricks you want
Well it helps to get somewhere :p
17:17
@Awm0121 OK, though it's more like half the speed of light ...
@Slereah You can't get quantum mechanics of one particle from the classical equation of motion for a particle, either, can you? :P
Then I extended this to a simple pole (theoretical pole). What if two people held both ends of a very long pole, and one person moves their end? The other end would receive this information (that the pole was moved) after a delay that corresponds to the speed of light
Well you can
But that's cheating
@Slereah Aha!
17:18
A good quantum theory you should be able to find everything from the theory itself
@Slereah So why do you want to cheat and get the quantum mechanics of the free field from the KG equation?
Same reason I don't approve of finding out the mass of the Schwartzschild metric via Newton
Well if I don't need it fine
But I need something to do on
How do I get the propagator of the scalar field in AQFT
@Awm0121 the push would travel down the pole at the speed of sound - the speed of sound in the pole that is.
@Slereah It's just a particular expectation value, no?
Yes, but how do I calculate it
17:20
When you push the pole you generate a compression wave, and a compression wave is exactly what sound is.
Just writing that it is $\omega(f(x) \otimes f(y))$ isn't extremely helpful
@Slereah Well, what do you need it for?
@Awm0121 see:
72
Q: Is it possible for information to be transmitted faster than light by using a rigid pole?

Jonathan.Is it possible for information (like 1 and 0s) to be transmitted faster than light? For instance, take a rigid pole of several AU in length. Now say you have a person on each end, and one of them starts pulling and pushing on his/her end. The person on the opposite end should receive the push...

If all you want to do is compute Feynman diagrams, you needn't bother with AQFT at all.
I don't need to bother with AQFT for anything
But I want to
I MUST KNOW
17:21
And if you don't want to compute Feynman diagrams, I don't see why the propagator is relevant.
Also regular QFT doesn't work so well in some areas
ha, perfect
just what I was looking for
Really how do you compute ANY observables in AQFT
so to assume that it is perfectly rigid, would not be correct when using it for speed of light though experiments
I guess that was my flaw
How do you show there exists a vacuum state
How do you find expectation values for it
How do you find the time evolution of those values
etc
basic stuff
17:23
@Slereah I think that is kind of a vacuous question. You are handed the algebra of observables, and a "state" is defined as a linear functional on them giving the expectation values. So whenever you have a state, you automatically have the presciption of how to find expectation values.
Well the algebra for the scalar field is $\oplus_i S_i$
Now, time evolution, that one is an interesting question that I can't deflect
With $S$ the Schwartz space
@Awm0121 there are no perfectly rigid objects when you take relativity into account:
7
Q: Extended Rigid Bodies in Special Relativity

KitchiI was reading Landau & Lifshitz's Classical Theory of Fields and I noticed that they mention that an extended rigid body isn't "relativistically correct". For example, if you consider a rigid rod and apply a torque at one end, by definition of being rigid, the whole body must start rotating at ...

@Slereah With what norm and what multiplication and what is the sum over?
17:25
Of $R^{id}$
Lemme see
Because an algebra of functions on any space is commutative with the standard operations, and hence not the algebra of any quantum system
@Awm0121 for a more technical discussion see Born rigidity
great, really appreciate the information to help me understand it better
@Slereah You might be mistaking the algebra of test functions (for the operator valued distributions that are the fields in the rigorous phrasing of standard QFT) for the algebra of observables.
The multiplication is $(f_n g_n) (x_1,...,x_n) = \sum_{\nu = 0}^n f_\nu(x_1, ..., x_\nu) g_{n - \nu}(x_{\nu+1},...,x_n)$
Yeah I'm having trouble finding out which is which here
But they do write that a state is $\omega(f)$
With f a member of that algebra
17:28
@Slereah Although I have no idea what the subscripts denote, that does look pretty commutative to me.
Your algebra of observables should not be commutative.
Hm
Other papers do seem to mention that the algebra is from $\varphi(f)$
So it's a bit hard to determine
for instance
"We construct $\mathcal{A} (\mathcal{M}, g)$ by starting with the free star-algebra generated by a unit 1 and elements $\varphi(f)$, with $f\in C^\infty_0(\mathcal{M})$"
They do seem to drop the KG equation straight into the definition of the algebra
@Slereah Just from skimming, it seems pretty clear to me. The field $\phi$ is an algebra-valued distribution on the test functions, and the algebra of observables is the algebra generated by the image of $\phi$
I guess it is better than this one then :p lqp.uni-goettingen.de/events/aqft50/slides/1-4-Yngvason.pdf
The algebra that $\phi$ takes values in and hence its multiplication and norm seems "god given" though
yeah
would be nice if there was a little step to derive at least the field equation
from something or other
Although then again
Are field equations any more arbitrary than lagrangians
17:37
Oh!
@JohnRennie, one more thing. Your answer to the transfer of electrical energy being more like half the speed of light, can you explain? This is just because of the medium that the energy is travelling through?
@Slereah The multiplication on that algebra is not commutative
Hm
Then what is what!
Are they somewhat equivalent?
They are equivalent if the $\phi$ constructed in the first paper is injective.
Also I find the propagator being right in the axiom list to be a bit crass :p
17:40
At least, then you get that $\phi$ also defines a representation of the BU algebra from which you can recover that algebra
Oh right, I suppose that both the algebra and the operators you get from the mapping have the same algebra?
@Slereah Considering that the propagator is just the Green's function of the KG equation, and the KG equation is essentially an axiom since the Lagrangian from which it is postulated, is that really all that different from the standard procedure?
@Slereah for the very basics, see "foundations of the theory of dynamical systems of infinitely many degrees of freedom I and II" by I. Segal
It just dumps all those unnecessary things like "action" or "lagrangian" :P
@ACuriousMind A bit inelegant, though
17:42
best starting point
To put such a giant ugly function
or his book with baez and zhou on constructive qft
@Slereah Why? Postulating directly that which actually determines the dynamics instead of postulating something on which you have to do a lot of work to get the same dynamics seems...efficient.
I like axioms I can remember :p
Also how do you get the momentum from this?
Derivative maybe?
@BernardMeurer whats the name of that octopus?
17:53
@HariPrasad Can't tell you
@BernardMeurer Am I close to guess it?
Not at all
@BernardMeurer Is it the real name or something that you made up?
@HariPrasad Can't tell you
It' supposed to be a password
@BernardMeurer ha ha ok But what is there in that page you encrypted?
17:56
@HariPrasad The photo of a girl I met that I wanted to show DS
@BernardMeurer DS?
DanielSank
@BernardMeurer Alright
@HariPrasad :)
I like such challenges :D
18:06
Welcome anyone to make a physics version of both genders
@Secret lol. Money reducing agent?
well, that guy is basically saying how most of them drain the wallets of their boyfriends

That's a common stereotype but it is not true in general
18:37
@BernardMeurer this name guessing game is kind of funny.
It's particularly funny that someone would keep asking you.
Even if they guess right, you could say "no". Unless they just try it then they'll never know.
@HariPrasad She is very pretty.
@Secret Can we...not post sexist stereotypes with an implication that they're funny?
5
^ That
How much of a jerk am I if I answer my own question after several other people have offered good answers that don't quite get at what I wanted?
@DanielSank 0% percent jerk, unless that was your intent all along.
@ACuriousMind Was not intent.
Is there an error bar on that 0%? :-)
I'm a theorist. What's an error bar? ;)
18:43
You can do theory with error bars!
@ACuriousMind ::blood temperature rises 0.3 degrees::
Crap, I mean
:: blood temperature rises 0.3 +/- 0.1 degrees ::
Fahrenheit or Celsius?
@ACuriousMind Kelvin, duh.
@DanielSank I never know with you Americans!
@ACuriousMind I know! We're super inconsistent about that.
18:45
Any Mathematica geeks around that fancy a go at this one?
0
Q: Can a package append its context to $DistributedContexts?

Emilio PisantyI am writing a package which I use extensively in parallel computations, and to lighten the coding load on the 'run-time' notebook, I want my package to sit on the $DistributedContexts list of sessions in which it is called. The way I'm doing it now is to simply call $DistributedContexts := {$C...

@DanielSank Nice edit, I was already looking up "ornery" :)
@ACuriousMind It was a misuse of the word anyway.
@DanielSank It is pretty funny yeah :p
And Regarding beauty I agree
@BernardMeurer Did he see the picture?
@HariPrasad Yes, the thing was meant to him
19:06
@BernardMeurer Wow she is beautiful
@HariPrasad :)
What does $C^\infty_0$ mean, again?
Is it compact support
@BernardMeurer well all the best
@Slereah I'm used to $C_c^\infty$ for that, but could be.
Ah yes apparently momentum is just a derivative
$\pi = n^\mu \nabla_\mu \varphi$
"A state which cannot be written as a non-trivial convex linear combination of others is called pure."
What is a non-trivial combination
19:15
@Slereah One that has more than one summand.
Errr
What would be an example
Mixed states are an example - their density matrices are convex linear combinations of the density matrices of pure states.
I meant more in the context of AQFT here :p
The combination example they give is $\lambda \omega + (1 - \lambda) \omega'$
That's just the definition of what a convex linear combination is.
The "trivial" convex linear combination is the one with $\omega = \omega'$.
Most certainly
Oh
So just $\vert 0 \rangle + \vert 1 \rangle$ is not pure
19:23
Sure it is
How would you write it as a convex linear combination?
Errrr $(\langle 1 \vert + \langle 0 \vert) A (\vert 0 \rangle + \vert 1\rangle) = \langle 1 \vert A \vert 1\rangle + \langle 0 \vert A\vert 0 \rangle = \omega_0 (A) + \omega_1 (A)$?
And how is that convex linear? What is $\lambda$ in $\lambda\omega_0 + (1-\lambda)\omega_1$?
Half, if I don't forget the normalization factor
Wait
Where did your cross terms go? :P
"It follows from the classical “Hamburger moment problem” "
Hm, hamburgers
19:29
$\langle 1 \vert A\vert 0 \rangle$ is not zero!
Oh right, they may not be orthogonal with $A$ in between
I guess it's gonna be the usual $\vert 0\rangle \vert 1 \rangle + \vert 1 \rangle \vert 0 \rangle$
Not quite sure how to write it as a linear thingamagig though
What is gonna be that "usual" thing?
What you wrote there is a Bell state, which is still pure, just entangled.
I don't believe in density matrices, though :p
I believe we should ALWAYS KNOW HOW TO WRITE OUR QUANTUM STATE
How do you write a not pure state as a linear combination of Hilbert vectors?
@Slereah You do realize that the partial states of the subsystems of a system in an entangled state necessarily are mixed?
No :(
19:34
@Slereah What?
I do not
I am never quite sure what a mixed state is
@Slereah A density matrix that is not just a projector on one state in the Hilbert space
Why are they always expressed in terms of density matrices, though
And if you take an entangled state on $\mathcal{H}_1\otimes\mathcal{H}_2$, you'll find taht the partial traces give necessarily mixed states on $\mathcal{H}_1$ and $\mathcal{H}_2$
Can't they be expressed with the usual Hilbert vectors
19:36
@Slereah No, that is the whole point!
They are a statistical mixture of the Hilbert state vectors
Oh so they only happen if you don't use the broadest Hilbert space
You can "purify" a mixed state, but that requires "doubling" the Hilbert space
Is it doubling or is it just using the proper Hilbert space :p
And it doesn't becomes a "pure state" in the algebraic sense then, just a vector state.
If you have two particles you use the Hilbert space for two particles!
19:37
So? Mixed states of two particles are not vectors in the Hilbert space for two particles
Aren't they?
Maybe you should go back to density matrices in QM as a more general notion of state before using the algebraic notion of state.
Yeah mb
What's a good source on them?
Uhhhhhh
Landau hopefully
cuz it's my non-relativistic QM book :p
I mean it's probably not totally indispensible, because all things considered I doubt I'll encounter a lot of mixed states :p
But it would be nice to know
I have odd gaps in my physics knowledge
Also some things I just kinda forgot
I don't think I'm still capable of doing thermodynamics by now
Or... what do you call in english classical mechanics for extended objects
Like THE WHEEL
Solid mechanics
19:47
0
Q: Where does the mexican head logo of this stack stands for?

Marijn Beside the word physics is drawn a figure which looks like a mexican head/ sombrero. Where does it stands for and why is this used. Does it has to do something with condensations??

" This criterion can be applied straightforwardly to the free KG field, but not to non-linear observables such as Wick powers discussed in sec. 3.1. Thus, for general, nonlinear observebles, the determination of the probability distribution from the state remains open."
Dang it :V
First time a hamburger has betrayed me
$WF(W_2) \subset \{ (x_1,k_2;x_2,k_2) \in T^* \mathcal{M}^2 \setminus \{0\} \vert k_1 \in \dot V^+, k_2 \in \dot V^-, k_1 \tilde{} k_2 \}$
Argh
20:40
Apparently the vacuum state in AQFT is a gaussian state
$\frac{\partial^n}{\partial t_1 ...\partial t_n} \ln [\omega(e^{t_1 a_1} ... e^{t_n a_n})]_{t_i = 0} = 0$ for n > 2
Oh wait, everything I learned was a lie!
You need to expand the algebra to even get non-linear observables!
@DanielSank Isn't list comprehension the most beautiful thing?
21:09
Argh
The nonlinear observable part is starting to be a bit much for my brain
I think I need some shut eye
@BernardMeurer No.
@DanielSank Forgot you sold your soul to Scala, traitor...
@BernardMeurer Python is nice. I just don't think list comprehension is the most beautiful thing.
@BernardMeurer Have fun with memory efficiency.
:)
21:18
@DanielSank She doesn't need to be smart if she's that hot
The comprehension that is
Thought you might be talking about your annex for a minute there.
Hahahaha, she's going for geology
that's so cool
So smart and good looking?
As it seems, it must be a trap
@BernardMeurer Not necessarily. I found one like that.
21:23
@DanielSank Yeah but you work at Google :p
you're hacking the game
Found her way before that.
Tsc, forgot about that
Bet she'll tell you she's actually a string theoretician the second after you marry. The biggest treason of all
@DanielSank It's a peculiar sound I do with my mouth
@BernardMeurer Got it. I know the sound.
21:25
Did you buy a new phone after all?
@BernardMeurer No way. A fate worse than death.
@BernardMeurer Hahaha I do have a smart phone now.
Thanks god, holy mother finally
Google gave us Nexus 5X's as holiday gifts.
Yeah it's sweet.
Last year they gave smart watches, but since I didn't have a phone to use it with I gave it to my friend's wife.
21:27
Linkedin keeps suggesting that I work at google, I'm always like "Soon, soon"
Do it.
I'm pretty sure I gotta graduate first
and they pick their interns from fancy ass schools :/
@BernardMeurer That's not how it works.
I don't think.
Demonstrate that you can do stuff.
There are lots of people out there who get reasonable grades but don't know how to actually build a project, etc.
Show what you've done.
I get unreasonable grades and build too many projects :p
@BernardMeurer Balance that.
It's important.
21:30
Yeah I've made a promise to try and study harder during College, but I think that'll be easier since at least some of what I'll be studying I'll like
Yeah that makes a huge difference.
High school was rough because I didn't care about half of the work.
In college you care about 90% of it.
But regarding internships, I just said that because all the people who I know that interned at google went/go to Harvard. That kind of made it feel miles away
I had the same thing, I had 14 different, simultaneous subjects at HS, out of which 2 I liked
@BernardMeurer Do you have a Google account? I want to share an image with you.
@BernardMeurer You know... most of the people I know who went to Harvard for grad school didn't really like it that much.
@DanielSank Google account? My main email is one I think, [email protected]
@BernardMeurer excellent
21:36
@DanielSank Who's that?
@BernardMeurer ?
The picture you shared
@BernardMeurer One moment, I'm trying to understand how Google sharing options work.
I think it's different than I expected, and I'm kind of annoyed.
Wait a second... @BernardMeurer do you see the image I shared?
I got a b/w picture of someone with a hair similar to yours a little
Yeah, I was watching my GDrive's shared folder for it
@BernardMeurer Ah. I see.
That's my fiance. Pardon the potato quality image.
21:40
@DanielSank Is her first name Anna?
@BernardMeurer ?
Just want to know if my stalking ability is still on spot :p
And you were right regarding having found someone with both the qualities we were talking about
Poof, gone
Being internet stalked once before is enough to make me a little cautious.
@DanielSank I literally just googled the filename which I guessed was a name and "astrophysicist" and it was the first result :p
Yeah, I'm not claiming CIA level security here.
21:45
Oh, regarding what you said about harvard, Idk, I'm having a hard time kicking out the inferiority complex that comes with not getting into any good schools
"Even if the interacting fields were elements of the same algebra W for all $\lambda$ and had a common set of states, as Dyson noted more than 60 years ago, for a theory
such as the one with interaction Lagrangian $\lambda \varphi^4$, a ground state $\omega_0(\lambda)$ cannot be expected to be analytic in $\lambda$ at $\lambda = 0$, since no ground state can exist when $\lambda<0$. Thus, perturbative expressions for quantities such as the S-matrix should not converge—even if it were the case that notions of “in” and “out” particle states could be defined."
@BernardMeurer Don't worry about it.
My colleagues are not all from high name schools.
One is from Binghamton, which I hadn't heard of before.
Sounds like a fake british town
21:49
@DanielSank Yeah it's just that most of the people I admire and think "that person is da bomb" when I go look into it are from those kinds of schools
@BernardMeurer Perhaps.
Like you being from Yale for example
@Slereah Not a feature or insight of AQFT - that the perturbative expansion around 0 is asymptotic but not convergent because convergence would imply a well-defined theory for negative coupling is a bit of folklore you find in many places.
Ok @BernardMeurer but the school doesn't make the person.
The horror
21:50
@DanielSank Agreed; I'm not sure it's just a hard conception to break I guess
Apparently there is Operator Product Expansion for non-perturbative QFT
It's a bit skimmed over though
How do you even find the propagator for a nonlinear theory
Is the Green function even defined in the nonlinear case
@DanielSank How old is your linkedin picture?
@BernardMeurer hey buddy.. did you ever implement the euler function lol?
@Obliv It's on my TODO, which right now is a list catching on bloody fire :p
I have to compute all inputs of $n \leq 30 \in \mathbb{Z^+}$
22:00
I had to work on a MIDI visualizer today
Alrighty, let me build one now
Gimme the euler function
What we really need is a QFT that doesn't require a bloody renormalization at every turn, I say
yeah
it just gives the amount of relatively prime numbers to the input
so all you really need is to find the in-between numbers since you have a sieve for the prime numbers
"In addition to physical applications, vertex operator algebras have proven useful in purely mathematical contexts such as monstrous moonshine"
I want some of that monstrous moonshine
oh @BernardMeurer maybe it's not that simple lol. for ex: $\phi(12)$ returns 3 (1,5,7,11) I think
22:05
@Obliv It'll be having arbitrary precision
because I gotta get some experience with that
since 2,3,4,6,8,9,10,12 all have either 2 or 3 in common with 12
uhh @BernardMeurer I'm not sure if that's a good idea (depending on the method you use) I was thinking of decomposing the numbers into their prime factors somehow and then comparing each decomposition to the input to see if there were common prime factors. I'm not so sure that there's an algorithm out there that can do that.
since we don't even know all the primes in $\mathbb{Z}$ anyway
oh wait it would be outputting numbers in real time. Okay, no problem with that I think. I thought it would try to compute something indefinitely until it got a finite value and then output it.
No no, I'm just saying it will support MASSIVE numbers
let me get me a sieve now
okey
I'll be using the primesieve.org implementation which is built on the sieve of Atkin I think
22:21
@BernardMeurer Old. Why?
@DanielSank It has a gangsta look to it
like you're from the lab-hood
@BernardMeurer I hard as f--k.
What? Hahahaha
Seriously, someone flagged that?
What rule did that break?
3
Could someone explain to me why that was flagged?
I have no idea what is offensive about that message. I voted for "invalid".
22:25
I suspect this was a non English speaker who doesn't know what that phrase means.
@DanielSank Because your words! :-P -- I'm not the flagger.
@PaulVargas What does "because your words" mean?
That doesn't go against the policy though..
@DanielSank Sorry! I have to go!
only people that are inexperienced with slang/swear words actually take offense to phrases like that. Maybe you should write a translation in parentheses next time ;) @DanielSank
22:28
@Obliv I'm not going to tailor my speech in an English chat room for people who don't know English.
lol i'm being sarcastic m8
He even censored the swear word. It's hardly more offensive than "I hard as %&§!", which doesn't seem to be offensive at all to me.
I appreciate the need to be aware that not everyone has a common vocabularly, but being flagged for that is ridiculous.
@ACuriousMind I think someone must have thought it was sexual.
@DanielSank Oh
Um
Which is funny.
22:31
I Still don't get what you meant there Dan ;-;
@Obliv Is it okay if I absolutely OCD over efficiency?
@BernardMeurer "Hard" means "tough", "street wise", "able to handle yourself in dangerous situations".
You clearly don't listen to enough rap.
@DanielSank Ah! Yeah I saw how you were gagnsta when that gas container spraied all over
@BernardMeurer Container spray?
What was it again, Liquid Nitrogen or Helium
Helium I think
Was there a mishap while you were here?
22:39
Yeah Ben dun'goofed on the valve
then you rushed to open the lab doors
@BernardMeurer Hah. ok
and said "Yo dood ez on dem valves"
Yes, that's an exact quote
@BernardMeurer Sounds about right.
I'll get you one of these as an engagement gift
made of aluminium tho, don't want you to hurt your neck
22:54
hello
@ChrisWhite Halp
It's catching fire, halp

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