The h Bar - 2017-04-03 (page 3 of 6)

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4:00 PM

btw

1a. For the operationalists: Do they consider the settings of the instruments part of reality?

It seem, naively speaking, pressing buttons, adjusting alignment of lasers and other operations seems to be events that really have occurred, thus physical in some way. Therefore quantum mechanics reduces to a formalism that tell us certain settings and configurations of the instruments, and the probability of getting some outcome has a linear relationship that connects them

I found the first bundle paper

mathnet.ru/links/c71ce61f0ac89c328940a951849b7025/sm5499.pdf

who's it by?

Whitney

In the МАТЕМАТИЧЕСКИЙ СБОРНИК

lol

4:01 PM

he called them spaces, though

i can see russia

@BalarkaSen Are you Sarah Palin!?

The name "bundle" seems to be from his 40's paper

who that

"On the theory of sphere bundles"

He was way into spheres

4:02 PM

@BalarkaSen seriously?

Only one of the greatest political minds of the 21st century...

i don't hope to get 21st century

1b. Is the interpretation of operationalist actually the reality?

This one is actually more philosophical and interesting. We are so used to having simple laws in physics governing phenomenon at some local scale (that is, given some basic laws of the components or foundation part of the system, you can predict everything. Similarly for realist interpretations of quanutm mechanics, we thought all the experiments and measurements are giving us more information of the underlying reality, which might be the wavefunction (many), histories (consistent histories) and etc.)

that things just happened to be related in some way, without any underlying mechanism

---
2a. For the realist: The problem of identity in many worlds. Which device and which state has the probability of moving to a given branch of the universal wavefunction. Is this subjective, or is this objective. What happens if a nearbying machine tries to record the reading of the measuring machine after the measurement, which branch will it end up?

2b. Psi ontic: If the wavefunction is real, as in it is really a property of the system, will we be able to interact with it directly, like how we manipulate electric fields?

2c. Psi ontic: Why certain interactions will project the quantum state to an eigenstate, but others don't, what is the criteria for an interaction to become projecting?

2d. Bohm: (I actually realise I knew too little to formulate a question for this camp atm)

@Secret what's another neat theorem you could learn?

what were we talking about the other day that made me prove that norm thing

@Secret Oh, do you know the Stone-Weierstrass theorem?

If you mean in real analysis and topology, then I am not sure, cause I actually has ZERO background in real analysis. I just happened to know enough and the right things given the right problem thus explains why I generally don't have much trouble following the proofs.

On that day we were talking about how two metrics with the same converging series induces the same topology. The justification lead to the norm proofs

@Secret Ah, yes. Well it's clear that when you have two norms like that, then they have the same sequences. So they induce the same topology.

4:17 PM

indeed

Thus the functional analysis of finite dimensional spaces is just analysis/linear algebra on $\Bbb R^n$, in a sense.

@AccidentalFourierTransform I thought align was the recommended standard

@Secret The whole point is that operationalists don't care about "reality". You give a description of an experimental setting, the operationalist compute the quantum mechanical predictions for the results. "Reality" does not enter into it.

@ACuriousMind Why is my "3 equations for 4 unknowns" starred?

@Secret Makes no sense, see above re operationalists not caring about reality.

4:19 PM

@ACuriousMind I think technically it is called "shut up and calculate", not operationalist

@Secret Everything ends up in all the worlds. Your question makes no sense.

@Slereah I would contest the claim that those two are the same

What would be the nuance

@Slereah The operationalist very carefully phrases their axioms to not talk about reality. The shutting-up calculator just takes the usual axioms and doesn't care about their inconsistencies at all

@Slereah SUC just goes "ontology? meh", whereas operationalism does take an explicit stance about ontology, i.e. you explicitly make no statements about what may or may not "exist" other than preparation procedures and measurement outcomes

4:23 PM

@Secret Stone-Weierstrass Theorem: Consider a compact Hausdorff space $X$. Let $C(X)$ denote the space of continuous functions $f:X\to\Bbb R$. By compactness, we know that $||f||:=\sup_{x\in X}|f(x)|$ is finite for every $f\in C(X)$. Thus $C(X)$ is a normed vector space with the norm $||\cdot ||$. Note that $C(X)$ is an algebra (in fact a unital Banach algebra) where $f\cdot g$ is given by pointwise multiplication.

or what ACM said, really

That said, I'm not sure there are actual practicing true operationalists around

lemme go dig for that survey thing

@Secret Let $\mathscr A\subset C(X)$ be a subalgebra. Suppose that $1\in \mathscr A$ and that $\mathscr A$ separates points, that is, for $x\ne y\in X$, there is an $f\in \mathscr A$ with $f(x)\ne f(y)$. Then $\mathscr A$ is dense in $C(X)$. That's the theorem.

The survey was basically half Copenhagen half many world

And a smattering of Bohm

@EmilioPisanty I think it's a bit difficult to care enough about all of this to not be in the "shut up an calculate" camp but then be satisfied with the operationalist approach.

@ACuriousMind but it's a tenable position though

I would say that it's synonymous with being a bit weird in the head

4:26 PM

is it

but you could argue that the only thing that "exists" is measurement outcomes

(I have tried my best to not enter dangerous territory, but this is what I find very hard to wrap my head around when using many worlds)Ok one philosophical question that arises when I use the many world interpretation is the following:

I measure the spin of an electron right now. One second later, where am I, am I everywhere. But how is this consistent with I only remember seeing only one outcome of the electron spins and not all possible outcomes all at once. Hence the personal identity problem?

Copenhagen is a bit weird

user image

(d) is not quite operationalist but it's close

@EmilioPisanty It is, no question about that. But I think it's really more of a position that the "shut up and calculate" people would choose if forced to choose one than one many people pick of their free volition

4:27 PM

arxiv.org/abs/1301.1069

AccidentalFourierTransform

@EmilioPisanty recommended standards are not always correct

@EmilioPisanty those do not add up to 100%...

AccidentalFourierTransform

here, the correctest way to do it is with equation+aligned

because I say so

@Secret Would you like to see the proof? I had planned to write it up anyway one of these days.

@0celouvsky don't blame me

4:30 PM

@0celouvsky don't worry I am reading

@0celouvsky Multiple answers allowed

presumably a multiple-answers-ok poll

anyways

user image

oic

AccidentalFourierTransform

@0celouvsky that... that could be a new loophole to Bell's theorem...

^ probably the closest we can get to the modern distribution of interpretation popularity

4:30 PM

where is psi ontic (no hidden variables) on that list?

"other"

@Secret I don't see the problem. In each world, there's a different version of "you". Each version only remembers being in a single world.

Apparently Zeilinger doesn't care much for operationalism

@ACuriousMind Ah that explains it

@Secret psi ontic is not an interpretation by itself

it contains no resolution to the measurement problem

MWI is one such

but e.g. objective collapse is also psi ontic

AccidentalFourierTransform

4:32 PM

user image

@Secret The proof is based on a Lemma, which is actually one of the most important theorems in analysis. :P Bernstein's Lemma. Let $f\in C[0,1]$ and define the polynomials $$P_n(x)=\sum_{k=0}^n {}_nC_kf(n/k)x^k(1-x)^{n-k}.$$ Then $P_n\to f$ uniformly in $[0,1]$. This is a "quantitative" version of the classical polynomial Weierstrass approximation theorem.

0celo: Hmm, that kinda reminds of how you can approach any reals by a sequence of rationals

indeed

@EmilioPisanty What psi ontic interpretation I am using if I say the following?

"Every physical state has a property known as the wavefunction. Besides wavefunctions, there are also interactions that can occur when two or more physical states approach each other or evolve in time. Among these interactions, some of these project the wavefunction to an eigenstate, while others only unitarily evolves it. It is not clear however what is the criteria that makes some interactions projecting while others don't"

@Secret So basically, what Bernstein is saying is that given $\epsilon>0$, there is an $N$ large enough so that $|P_N(x)-f(x)|<\epsilon$ for every $x\in [0,1]$. Sadly I do not know of a formula for $N$.

I think the proof might give a hint

AccidentalFourierTransform

4:43 PM

$N=10000$

But it involves the uniform continuity of $f$, so you'd need to know how "wavy" $f$ is

$N = 2^{1000}$

AccidentalFourierTransform

$N=\frac{1}{0^+}$

I won

$N = \infty + \aleph_0 + \omega^\omega$

jesus

AccidentalFourierTransform

4:45 PM

$N=N_\mathrm{Slereah}+1$

Nooooo

doesn't +1 do nothing at the level of infinite cardnials?

it does for ordinals, though

$\omega + 1 > \omega$

I don't want to know the difference between those two

AccidentalFourierTransform

1

4:46 PM

what?

@Secret what does "there are also interactions" mean? Are you saying interactions "exist"?

@0celouvsky he's technically correct

@ACuriousMind The difference between ordinals and cardinals is 1?

What does that even mean?

No, the difference between $\omega + 1$ and $\omega$ is 1.

That's not what I asked.

what the heck is the "topological image" of a map

4:48 PM

That is a technically correct interpretation of your question since you didn't use the reply feature :P

@Secret that said, it mostly looks like what folks would call "objective collapse" without committing to the reasons for that collapse.

oh please

@EmilioPisanty yes, as real as the physical states themselves. They will corresponds to the unitary evolutions and the projection operators resepctively. Thus measurment can occur if say a photon strikes a cloud of electrons in a certain way that result in the electorn state to end up in an eigenstate

It probably might sound too classical, since it basically generalise the idea of "electrons get deflected by electric fields" to all possible types of changes to the system induced by another system

@EmilioPisanty I see

@Secret I think you're being way too liberal with your use of "exists"

but whatevs

For me, something 'exists' as long you can "interact" with it some way, whether via the sense, via insturment readings, or via applying the concept to some settings or applications

This is why I have this weird philosophical view that the concept of nothingness exist

(oops too philosophical, better get back to real analysis)

But anyway thanks guys for clarifying about quantum interpretations

4:53 PM

@0celouvsky in Lie theory and applications

infinite ordinals are basically like

As you may recall, the construction of integers from sets is basically $n = \{n-1, n-2, ..., 0\}$

an infinite ordinal is like $\omega = \Bbb N$

So that it's larger than any integer

@Slereah Typo, you have $0 = \{0\}$ there.

Since the order relation is

$a < b$ is the same as $a \in b$

The ordinals also form a proper class

it's a simple way to construct infinite numbers

5:01 PM

There are also 2 types of ordinals. successor ordinals and limit ordinals

Limit ordinals is the supremum of all ordinals before it. For example, $\omega$ is the supremum of the set of all finite ordinals

Successor ordinals= (I forgot how to phrase this formally, except they generalise the concept of successors in natural numbers)

> By dyon we shall mean a Dirac monopole carrying electric charge.

wtf is that?

So if I have an uncountable subset $A\subset\Bbb R$, I should expect to have accumulation points in lots of places, right?

In physics, a dyon is a hypothetical particle in 4-dimensional theories with both electric and magnetic charges. A dyon with a zero electric charge is usually referred to as a magnetic monopole. Many grand unified theories predict the existence of both magnetic monopoles and dyons. Dyons were first proposed by Julian Schwinger in 1969 as a phenomenological alternative to quarks. He extended the Dirac quantization condition to the dyon and used the model to predict the existence of a particle with the properties of the J/ψ meson prior to its discovery in 1974. The allowed charges of dyons ar...

Othrwise I knew nothing about it

Ah, yeah

If it doesn't have accumulation points, all points are isolated

@Secret d'ough, of course

I just assumed it was some arcane term for dipole

5:05 PM

So cover by small intervals, and by separability of $\Bbb R$, you can only have countably many intervals

does that require C?

@ACuriousMind yo what's that strange thing with choice (or without choice) with covering R with lots of sets or something

even simpler, you can only have countable intervals covering reals as each interval contains a rational and there are only countably many of them

thus if you have $\mathfrak{c}$ many intervals covering the reals, it will implies the rationals are uncountable, a contradiction

@Secret I know, that holds in any separable topological space

But nothing should require C

I was just remembering that strange result you get without choice

How do I do the little circle on the A?

For Angstrom

$\r{A}$

hm

It should be \AA apparently but it doesn't seem to work in mathjax

I get slightly mad when I see people doing the dihydrogen monoxide joke

Because that's not the systematic name of water

5:23 PM

IUPAC name: water, oxidane

user image

Remember to kiss your earphones

I'd rather not

5:46 PM

To that guy you wanna end up swallowing the magnets, right, that is a fate worse than death

Lol

@0celouvsky If you don't accept it, there's a non-empty partition of the reals with greater cardinality than the reals itself, I think

yeah

You need choice to prove that the cardinality of the disjoint cover of a set is $\leq$ the cardinality of the set

That's actually kind of insane

@ACuriousMind how does the partition then not have more points than real numbers?

that's why you need finitism

5:52 PM

I guess. 5 million is my max

@0celouvsky That's the paradox

@0celouvsky It's just a theorem, that's a different thing than being able to construct it

I wonder what finitist set theory looks like

To show that the partition can have at most as many sets as the reals have points, you'd need a choice function that selects an element from every choice of the partition. So in the absence of choice, such partitions are not forbidden

Yeah that's pretty strange

But C is stranger

Although, does the theorem say that such a partition exist, or just doesn't say that it doesn't?

5:54 PM

Well ordering is a greater sin

@Slereah I'm not sure, probably the latter

But I think you can construct some model of ZF + not choice where it does actually exist

Was Hilbert a Jew?

"V-omega, the set of hereditarily finite sets is a model of the theory you get by taking ZFC and replacing the axiom of infinity with its negation, and it is bi-interpretable with Peano Arithmetic"

neat

"Hilbert was baptized and raised in the Reformed Protestant Church"

Even worse, a heretic

Although i'm guessing you don't need the axiom of choice in finitist ZFC

Since it's always true for finite sets

So basically yeah

If you want math that makes sense just use the Peano axioms?

6:03 PM

I don't need axioms

I just need what I can see in the night sky

There aren't more numbers than stars

there are no numbers in the night sky

Get a stronger microscope.

Macroscope*

No, you want to make things larger

The frame bundle is a bundle of frames for a vector bundle

But

it is a vector bundle

6:09 PM

No

Is there a frame bundle frame bundle

You could take the double frame bundle

The double tangent bundle is allegedly useful as a technical artifact

O'neill does use the double tangent bundle a lot

@Slereah Well, is GL(n) a vector space?

Hm

I guess not

$\mathrm{diag}(1) + \mathrm{diag}(-1)$ wouldn't be in it

6:10 PM

exactly

@Slereah he does?

well maybe not a lot

But I have seen the double whammy

p. 71 has it

$$d \exp_o : T_0 (T_o M) \to T_o (M)$$

That's not the double tangent bundle

That's the double tangent space

Double tanngent bundle is the tangent functor applied to the tangent bundle as a whole

plz no category theory

This is a family channel

Would that be like

The tangent space of phase space

For instance

@Slereah phase space is a cotangent bundle

6:17 PM

I don't judge a bundle by the duality of its space

You should, these are not isomorphic as bundles

In what case are they not isomorphic?

When they are not trivial

Given a bundle $L$, you have that $c_i(L^\ast) = (-1)^i c_i(L)$, so their Chern classes differ (if not all the odd ones vanish), so they are not isomorphic

What is $c_i$

i-th chern class

6:33 PM

Does a trivial bundle basically always mean that the base space is $\Bbb R^n$?

Well, a reduction to the trivial bundle

I mean

@Slereah No, it means that the bundle is just the base space times the fiber, i.e. the bundle is a direct product

I am aware

But the reduction of say $GL(n)$ to $I$

Is it possible with more than one coordinate patch

Or is the fact that there's just one the reason why

I'm not sure I understand the question

The trivial bundle rather trivially (:P) has a trivialization that's just one set - the entire base space

that is what I wondered, yes

But the trivial bundle has nothing to do with coordinate patches, since you can form the trivial bundle $M\times F$ over any base $M$ for any fiber $F$.

6:39 PM

Well my question is, for a bundle that has a non-trivial structure group, in what circumstances does it have a reduction to the trivial group

If and only if the bundle is trivial

Didn't we do this already yesterday? I feel I had this conversation already

But the tangent space isn't trivial, no?

@ACuriousMind something related, certainly

@Slereah The tangent space is not a bundle

And the tangent bundle may or may not be trivial, it depends on the manifold

well, tangent bundle, yes

So, what's the question, exactly?

6:46 PM

My question is why @Slereah doesn't read a book about this...

I do :(

@EmilioPisanty Is that, or something similar, available in text form? I'm kinda reluctant to say that a 3-hour lecture is "required watching" but if there was some document I could link I'd start giving that as required reading to all the people going on about hidden variables, Bell's theorem etc.

transcribe it you lazy AI

I will do it in exchange for mod powers @ACuriousMind

That sounds like a Faustian bargain :P

@ACuriousMind I would be a good mod!

2

Friiiick

Of course there's a subtlety in the definition of $\mathscr D$ vs. $\mathscr C$

functional analysis can't be easy, can it

no, that would be too nice!

7:17 PM

@ACuriousMind ooooh, mate, I dunno

not that I know of

frankly, the best I can think of is writing to Spekkens and seeing if he can recommend something

Paradox: While I found Bohm interesting especially its notion of nonlocaity, for some reason to be figured out I don't really buy any hidden variable models

One suspicion might be I found intrinsic randomness/indeterminism natural

AccidentalFourierTransform

7:36 PM

> Grace period ends in 52 minutes

so... I guess Im not gonna get that bounty :-P

oh well

Mar 18 at 20:44, by AccidentalFourierTransform

is it unethical/discouraged to post an empty answer so as to get the bounty, and offer the bounty back?

Mar 18 at 20:49, by ACuriousMind

@AccidentalFourierTransform Yes, that is discouraged - you'd be posting not an answer and it would be deleted as such.

:-P

but thanks anyway

@Fawad You have been warned before about discussing circumventing the system in chat in this chat room.

AccidentalFourierTransform

also, it was about collecting the bounty, not awarding it :-P

@AccidentalFourierTransform you are welcome!

@ACuriousMind that 5 second hurt me

@0celouvsky I will vote for you when you stand for election. I know you will make physics.se great again!

2

Yep. We'll build a wall to keep out math homework problems and make math stack exchange pay for it!

Sounds fair.

7:46 PM

Oh and I will prosecute @Slereah to the fullest extent of the law

AccidentalFourierTransform

and play golf

every day

and hit on your daughter

I'd reveal Qmecanic's true identity

AccidentalFourierTransform

I am Qmechanic

I have multiple accounts

Wrong

You're not smart enough

AccidentalFourierTransform

I am also Chris White

and anonymous

7:50 PM

We are legion

We are one

AccidentalFourierTransform

@0celouvsky I dumb myself down when talking to you

Feb 8 at 16:48, by AccidentalFourierTransform

23 hours ago, by AccidentalFourierTransform

im hella smart

0celo7 being a mod would be like the ultimate prank

best troll ever

I want to see that happen

when are we having elections again?

can we fire ACM?

3

nobody likes him anyway

3

oh man we just missed April Fool's Day

@AccidentalFourierTransform please don't make such wild accusations

AccidentalFourierTransform

"just"?

did you sleep for three days?

@DavidZ SE wasn't that fun this year, anyway. "Direct dance identification" or whatever it was called was pretty lame compared to shooting unicorns :P

7:54 PM

@ACuriousMind yeah, true. Well, I guess we can't have unicorns every year.

I was kind of tuned out of the whole AFD thing this year

I've been busy

AccidentalFourierTransform

so we hate ACM and David is busy. We need elections now!!

0celo7 for president

Hi, everybody.

AccidentalFourierTransform

hi there

@AccidentalFourierTransform I liked your communist rebel side better :P

AccidentalFourierTransform

ah yeah, sarcastic communism is fun

7:56 PM

Wtf @ACuriousMind you don't want me to be a mod?

AccidentalFourierTransform

I wonder how much of the historic communism was actually sarcastic

stalin_irl

@0celouvsky Dude you can't even usually follow the chat guidelines.

@DanielSank uh, yeah I can?

AccidentalFourierTransform

we need a mod that speaks his mind

I can't help it that the flag system is broken

7:57 PM

@0celouvsky Perhaps, but I doubt it.

@0celouvsky -_-

I haven't even been suspended in a week

At least!

So there goes your argument.

@0celouvsky That's...not true.

Lol

@0celouvsky I'm not really in the mood for pointless teasing.

AccidentalFourierTransform

its an alternative fact

7:59 PM

What was I suspended for?

Oh, the Clinton rapist thing?

Yeah I can't help that there are salty Clinton voters here

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