The h Bar - 2016-10-02 (page 3 of 4)

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4:25 PM

Hi, everybody.

hi

user218912

hello

@MarkMitchison I don't know if Daniel already pinged you about this question---if so, sorry for doing it again.

I didn't already ping him because I didn't think pings work for people not in the chat already.

So, thanks for doing that.

Mark is the man for that question.

Oh, it does work if they've been in chat in the past 2 or 3 days (I forgot which one).

But I can ping anyone, anytime, as a moderator.

(did not use that power here)

4:31 PM

Ah yes, the crazy-eyes ping: @@

testing: @@ ?

@DanielSank Right...

Would any of you be willing to help with a quantum computing question?

@heather General rule of the internet: Just ask the question.

user218912

@ACuriousMind what do you think about this? "This candidate writes posts that are greatly appreciated, if he is elected he surely can devote less time to answers, and that would be a great loss, waste and pity. Let's choose a candidate who writes fewer posts and less brilliant/competent answers!" —user104372

4:38 PM

So, I'm reading about entanglement and superdense coding and I don't quite understand how it works. Basically, Bob has two qubits, both in the zero state. He sends the first through a Hadamard gate, and then has it be the control qubit in a controlled not gate that operates on the other qubit. The second qubit (the target qubit in the controlled not gate)

then gets sent to guy #3. the first qubit gets sent to guy #2 who performs one of the following operations:

1. If he wants to convey 00, then he applies a identity gate to the first qubit.

We have 00 + 11, right?

2. If he wants to convey 01, then he applies a not gate to the first qubit

(and yes, 00+11 is correct, I believe)

ok

3. If he wants to convey 10, then he applies a Z gate to the first qubit

A Z gate won't do anything.

4:40 PM

4. and finally if he wants to convey 11, then he applies first a Z gate and then a not gate

(actually, I think the Z gate does do something...googling

Actually maybe I'm wrong about that.

So anyway, after these gates, then what?

then, you send whatever this third qubit is to the third guy

Third qubit?

What third qubit?

excuse me, first qubit, I mistyped

the third guy applies a control not gate, with the first qubit as the control qubit and that second qubit from the beginning as the target qubit

then, #3 applies a hadamard to the first qubit and measures both.

any chance you could either link a diagram of the gate sequence, or write it out in math?

Hard to follow in English.

You could upload a drawing.

4:43 PM

yeah, I'll put a picture in from my notes (which are somewhat neat, don't worry), one sec'

k

user image

Sorry, it's sideways, but that's the diagram.

Jake?

What happened to Alice :o

Hey, that's how the video named them. Poor Alice, I know. =)

@heather ok

4:47 PM

Dunno why they named them differently.

@DanielSank, so what I don't understand is how guy #2 is conveying two bits of information in one bit.

guy #3 would have to know the meanings of each gate application from guy #2, right?

@heather Well, there are two detectors at the end, right?

Calculate the pair of detector outputs for each possibility.

Certainly there are four possible outcomes.

If the gate sequence is properly chosen then surely each of the four options in that box lead to the four different possible pairs of detector outcomes.

00, 01, 10, 11.

Yes, they are (after guy #2 and before guy #3) 00+11, 10+01, 00-11, and 10-01

Oh, I guess that makes sense...

but guy #3 has to have the first qubit, right?

Ok, now go through and compute what happens after that final CNOT and Hadamard.

@heather Yes.

okay, it'd be 00,01,10, and 11

@heather Well there you go then.

4:50 PM

but if they have to have the first qubit, it isn't two bits in 1 right? because they end up having 2 qubits

@heather The point is that the person sending the message only operates on one qubit.

Frankly, I have no idea why this is interesting.

oh, i see...

huh.

By the way, you may find this simulator interesting, or at least amusing.

A coworker made it.

okay, cool...i've already messed with the ibm simulator, but i'll try that one too.

sorry, one last question:

where is entanglement involved at all here? i know that bob is preparing the bell state which leads to entanglement, but how does entanglement affect the outcome? the video said that without entanglement, it would be impossible.

@heather Well, if the states weren't entangled, you wouldn't be able to get the action on the top qubit to influence the measurement results on both qubits in the way that they do.

I'm not sure exactly how to explain. Let me think.

If I hand you a qubit, there are only two measurement possibilities, 0 and 1.

4:55 PM

Right.

Now suppose I pick four different operations on this qubit.

Okay...

The ones in that box in your diagram.

If I hand you 0, you get 0.

Right

If I do a NOT gate to make 1, and give you that, you measure 1.

4:56 PM

Okay.

If I do something to make 0+1 and hand you that, I have no idea what you'll measure.

It's random.

Sure, yeah.

There are only two states I can give you that unambiguously lead to specific measurement results.

Yeah.

Ok, now suppose you have a qubit in 0, and I hand you my qubit, which can be in any of the four states prepared in the box in your diagram.

Nothing you do with those two qubits can possibly distinguish all four states I might have handed you.

However, if our two qubits were entangled before I do my operation, then you can do stuff to unambiguously figure out which of the four operations I did.

@MAFIA36790 o/

user116211

4:58 PM

amazon.com/Sets-Groups-First-Course-Algebra/dp/0710205570/…

user116211

This is the most insane price of a hardbook I have ever seen.

Why does entanglement make the difference? What specifically does entanglement do?

user116211

@DanielSank \o

@heather, in essence, with an entangled system it's nonsense to think of the individual constituents as independent entities. The entangled system is one full physical thing.

@MAFIA36790 Dude

4:59 PM

@MAFIA36790, wow, I'm impressed. Would anyone even buy that?

I once tried to buy a Gutenberg Bible

The entangled system is a two-bit system.

Those things are CRAZY expensive.

user116211

$780.73 !!

@DanielSank, so, you're saying that because the entangled system connects the two qubits,

4:59 PM

So, when I do my four operations, I'm doing them on a two-qubit system. This allows me to encode more unambiguous information than on a single qubit.

user116211

Price of Samsung Galaxy s7 ;/

get an iphone

user116211

@0celo7 then? You bought it?

But, I mean, you have two qubits either way, just in one case they aren't entangled...

@MAFIA36790 No, it was over a million...

5:00 PM

@heather If they're not entangled, then when I do my operation on the top qubit, I'm operating on a one-qubit system.

user116211

@0celo7 7 series is insanely expensive for me.

If they're entangled then I'm operating on a two-qubit system.

oh, I got you now! that makes sense.

user116211

@0celo7 Was it put in auction?

So you really are putting in two qubits and getting out two qubits, because the second qubit is connected too the first, but it has the appearance of getting two qubits out of one.

user116211

5:01 PM

Some crazy history buff which has lots of bucks must have bought it.

Wow, thanks!

user116211

@heather No one; not made for us common students ._.

@MAFIA36790, no kidding. I'll stick with a used bookstore near me. Got a linear algebra book for only $8 there...and it was probably just as useful.

And didn't wipe out my bank account.

Which, y'know, that's always a good thing.

@heather The "superdense" part is that one person can encode N bits of information but with only operating on less than N qubits.

It's cute.

user116211

@heather Got the best Linear Algebra book in $4! New piece!!

5:04 PM

I don't know if/why it's interesting or important.

@heather I'm more into actually building qubits. I actually don't know all that much about the quantum information aspect.

@MAFIA36790, got you one better: I found Gilbert Strang's textbook online for free.

@DanielSank, but isn't important to know the theory to build more efficient qubits?

(Though I'm certainly interested in the practical aspect as well.)

@heather What does "more efficient" mean to you?

Also, I'm not making a value judgement. I'm just telling you what my strengths are.

user116211

@heather Free O.o

Quicker. Uses less memory. Stuff along those lines.

@heather Uh, well... I'm not sure what you mean.

Qubits don't "use memory".

5:06 PM

@MAFIA36790, yep. Dunno why it was out there for free, but it was. Newest edition, too.

Maybe you're saying that for a certain quantum computation we'd like to use as few qubits as possible?

user116211

@heather Kindle? pdf?

If so, yes, that's an important consideration.

There is a member of my group, Austin Fowler, who works on that.

@DanielSank, yes, I do mean as few qubits as possible

...and we have a whole team of theorists in LA working on algorithms etc.

5:07 PM

You work on quantum computers in your daily job!?

Yeah.

I work at the Google Quantum AI Lab.

Okay, that's so awesome...

WHAT!?

Okay, insert just pure awe here.

@MAFIA36790, pdf. I can give you the link if you want...

plus.google.com/u/1/+QuantumAILab

No pirating books in chat.

2

user116211

@heather Yeh, he works for Google and finds extraterrestrial and makes AI.

user116211

5:08 PM

@heather Noo!

@heather if you're even in southern California stop by for a lab tour.

user116211

Sep 25 at 6:53, by MAFIA36790

@JohnRennie Man, you can't get satisfied without turning a page by your hand and enjoy the smell of the old pages of the book (most Dover books do have their signature oldish smell); that's why I prefer actual books to e-books unless the price of the book is beyond my reach.

@DanielSank, no, not in southern California, sadly. More like Iowa.

user116211

Pdfs are bleh...

@MAFIA36790, @Ocelo7, okay, sorry...

5:09 PM

Where even is Iowa?

user116211

They are the only option when the book is out of print or too expensive.

@0celo7 here

@Ocelo7: look at a map of the U.S. jab your finger in the middle at the squarish state below Minnesota and above Missouri. You've got IOwa.

user116211

Iowa(/ˈaɪ.əwə/) is a U.S. state in the Midwestern United States, bordered by the Mississippi River on the east and the Missouri River and the Big Sioux River on the west. Surrounding states include Wisconsin and Illinois to the east, Missouri to the south, Nebraska and South Dakota to the west, and Minnesota to the north. In colonial times, Iowa was a part of French Louisiana and Spanish Louisiana; its state flag is patterned after the flag of France. After the Louisiana Purchase, people laid the foundation for an agriculture-based economy in the heart of the Corn Belt. In the latter half of the...

Oh, and we do not grow potatoes, that's Idaho.

5:10 PM

@heather Indeed.

@heather I could get the general area.

@Ocelo7, =)

But I'm not sure which flyover state is Iowa.

:P

@DanielSank, you'd be surprised how many people make that mistake.

Ok, I would have been too far down.

5:11 PM

@0celo7 Right, 'cuz Tennessee isn't a flyover state...

user116211

Theorem of the day: Power Set with Intersection is Commutative Monoid.

@DanielSank (1) I'm from Virginia (2) TN has ORNL

got a picture that shows this perfectly:

Also Nashville and Memphis.

user image

5:12 PM

So yeah, not a flyover state.

user116211

I'm proving it @0celo7...

@0celo7 Virginia is the state you go to accidentally when you screw up on the Beltway trying to get from Maryland to DC.

::drops the mic::

I'm out.

@DanielSank You're drunk.

@DanielSank, thanks for all your help!

@MAFIA36790 WHY

@MAFIA36790 I will give you some theorems to prove, ok

user116211

5:15 PM

@0celo7 \o/

interesting, legitimately useful results

user116211

okay, would try.

user116211

Lumo's latest post: Flooded Czech cave found to be the deepest one in the world

user116211

@0celo7 Because I'm proving it now.

user116211

@ManishEarth o/

5:18 PM

Let $E$ be a Banach space, $X$ a metric space, and $C^b_E(X)\subset E^X$ the set of bounded, functions. We give $C^b_E(X)$ the uniform norm. Let $\mathscr{F}\subset C^b_E(X)$ be a family of continuous functions. Define $\mathscr{F}(x)=\{f(x)\mid f\in\mathscr F\}\subset E$.

user116211

Also, checking the proof of Trivial ring is Commutative ring...

user116211

@0celo7 I don't know functional analysis, if you want to mean that ._.

Show that $\mathscr{F}$ is precompact provided (i) $X$ is compact, (ii) $\mathscr{F}$ is equicontinuous everywhere, (iii) $\mathscr{F}(x)$ is precompact for all $x\in X$.

@MAFIA36790 You should learn

I can help

user116211

@0celo7 Oh, sure; let me complete first real analysis, abstract algebra, linear algebra, general topology (multi-variable calculus is implied).

user116211

@0celo7 Very well.

user116211

5:21 PM

@0celo7 I'm reading Kreyszig in of-times, though.

@MAFIA36790 To be fair, we learned this the other week in my (not functional) analysis course, which has just single-variable analysis as a prerequisite.

user116211

@0celo7 ohh. So, what should I have to know to, at-least, tackle the theorem above?

@0celo7 @MAFIA36790 Guys I have a rather basic set theory question: If I have two infinite sets A and B that has the same cardinalty (that is, they can be mapped to each other with a bijective function) how to check which one is a proper subset of another if say the elements of A and B are listed like A={10,90,a,0,u,r,1,789,ff,&,d,89,...}, and B is also somethign similar that has no explicitly formula to enumerate the elements in the sets?

First 9 sections of Bredon's Topology and Geometry is more than enough.

user116211

@0celo7 noted.

5:24 PM

28 pages.

You can probably skip the stuff on nets.

user116211

But I have only Kelley in Topology, lemme check it online.

This is a very, very general form of the Arzela-Ascoli lemma.

user116211

Price is not that bad ;)) Good.

user116211

> This title is not currently available for purchase.

user116211

._.

5:28 PM

@IceLord Look at my rate of answering in the last months. I don't think moderatorship will significantly impact that rate.

What is the rate?

How does one tell?

user116211

@0celo7 data explorer?

I can't code SQL

Just go to the answers tab in my user profile, sort by newest and look at it

@ACuriousMind Yeah, I did that. Is this a lot or not?

@MAFIA36790 you should get it

5:31 PM

@ManishEarth @DavidZ @dmckee I apologize, but could I speak with one of you privately?

Thank you.

@0celo7 That's irrelevant - what's relevant is that I write maybe 1-2 answers/day, and much more of my time is spent in review queues and commenting around the site. I expect moderator duties to take approximately the time the latter activities currently take me, so I don't expect my rate of answering to change at all

Very useful for studying topology, it's much more condensed than Munkres

@ACuriousMind Ok, I see.

Gotta go again now

@ACuriousMind Remind me to tell you an amusing story later.

@heather Private chat room: chat.stackexchange.com/rooms/46211/chat-for-dmckee-and-heather

@Ocelo7 Why?

5:35 PM

Curiosity killed the cat.

And you are a cat. So beware.

:(

@MAFIA36790 I don't think you should commit to buying books just on the recommendation of a single person---but that's just me.

@Danu h8r

Also Hatcher's material is probably better, plus for free.

5:42 PM

Hatcher does not cover general topology

@BernardMeurer I prefer to keep communications about the site through on-site channels. But I can set up a private chat room: chat.stackexchange.com/rooms/46212/…

The sections I was recommending are on basic topics like compactness and metric spaces.

@0celo7 He does, in separate notes.

It's a complete waste to buy Bredon just for the basic general topology :P

I agree

I'm saying he should get it in general, and I find it a very useful (condensed) reference for general topology.

I don't think Hatcher talks about de Rham at all.

I'm saying he shouldn't buy it. Getting it is something else (e.g. libraries, internet)

5:45 PM

Fair enough.

He should buy Lee.

Speaking of, I should pencil in the errata one of these days.

5:59 PM

user image

Caption: Hmm I guess that might explain why ACM saw in one of my presentation of alleged division by zero structures some months ago, additive associativity breaks down

@dmckee Okay, great, I just got home & requested access

I think we now have it. I will prove this later:

[Scifi maths]
Division by zero theorem 2:
Division by zero cannot exist unless the algebraic structure have no additive identity

(We cannot break the distributive law because it will make $\times$ and + loses meaning)

(Breaking additive associativity, only ensure the additive identity is not zero, but it still prevent 0x form somehow absorbing (only in that case it maps any elements into some additive identity e)

That might explain why absorbers tend to go hand in hand with additve identities...

Will try to construct a ringoid to test this later..

I am guessing the reason why simialr restriction does not exist due to the existence of the multiplicative identity might have something to do with the fact that + does not distribute over $\times$ thus that breaks the symmetry somehow

6:14 PM

@Secret What's a ringoid?

That sounds like an STD

mathworld.wolfram.com/Ringoid.html

@0celo7 has a ringoid infection

@Secret Ah, that's a very simple definition though

Basically a $(S,+,\times)$ structure where the only axiom is the distributive law

@Secret Where multiplication is distributive in regards to addition

@BernardMeurer Yes, my butt hurts so much

6:16 PM

They are even simpler than rg, cause even rg have to obey associative laws

@Secret Are quaternions a ringoid?

quaternions are meme math

because they're not a ring, right? Because it's not associative

where did you hear about quaternions?

@0celo7 A homeless guy told me about quaternions

6:19 PM

What?

@BernardMeurer Sorry I forgot to set your access before posting the link. Should be fixed now.

His name was JOHN CENA

Request access?

What is going on?

@dmckee Yeah, it works now :)

Well, I tend to think of algebraic structures as a hierarchy. E.g. you start with magma, then you have semigroups, groupoid as subobjects, then semirings, near rings, rg, rig, rng as subobjects of these, then you have rings as subobjects of these, and then finally you reach fields. You can also start with ringoids and go through from rg, to rig or rng, then ring, then fields

6:21 PM

Jesus!

I highly doubt one needs all of those

user image

> If you want to divide by zero ,you must be famialr with everything in abstract algebra

Ringo is a Beatles, not an algebraic structure

2

> If you want to formulate back to the future time travel, you must have the knowledge level to be a historican, to formulate the theory of everything, to make good art and to be a skillful politician

> NB one need to be near omniscient in order to crack type 3 time travel dynamics

I am not, obviously

@Slereah Do you know what a "monodromy argument" is in the context of principal bundles?

No

I assume it involves some kind of drome

But only one

user116211

6:32 PM

@Secret First ringoids, then semirings, then rings.

yes sorry, (have not sat down and compare the rules in details)

user116211

@BernardMeurer A ringoid is an algebraic structure with a set and two binary operations; one being distributive over other.

@MAFIA36790 I googled it :D

But thanks!

user116211

@Danu okay, noted.

user116211

@BernardMeurer good.

6:34 PM

Actually if + ever distributes over $\times$ I predict we cannot have addition by 1, in an analogous symmetry to division by zero

@Slereah Apparently a "standard monodromy argument" shows that a flat connection on a simply connected manifold implies the bundle is trivial.

Also a deeper thought lead me to here:

The key to whether you can divide by zero, lies on this proof:

user image

The moment zero absorb itself, you are screwed, as any mutlpicative inverse of zero will give 1=0

So to divide by zero, we need to break this proof and prevent zero from absorbing itself

user116211

@Secret Generally it is advisable to talk about ring addition and ring product. When the set is real, you can go with the conventional addition operation and product operation.

user116211

@0celo7 Got the book; I mean pdf.

@MAFIA36790 Only terrorists pirate books.

user116211

6:37 PM

@0celo7 I just googled; and it was in the topmost result.

user116211

I would check the library though; there would be a copy; I'm sure.

@Slereah $(P,M,G)$ is a principal bundle, $\phi$ a flat connection on it, so that $P$ admits local horizontal sections

then assume $M$ is simply connected

user116211

I should skip Net (Moore-Smith Convergence) then as you said.

I think we need to use parallel translation to extend a local section globally, then use the theorem that a single section suffices to trivialize the bundle

But Kobayashi-Nomizu's proof seems quite long and hard

so maybe I'm overlooking something.

Therefore in order to show whether any system can have an inverse of zero, we only need to check whether $0\cdot 0=0$

Kill this equation (and all possible proofs that lead to it) and you open the door to divison by zero

Actually, I might just have everything I need to formally proof the Division by Zero no go theorem:

will post thsi in my blog shortly...

6:44 PM

you have a blog?

@BernardMeurer How does the analysis?

My Wiki is my blog

You have a Wiki?

@0celo7 Reading now, mostly studied chem today

Yeah, started that to store some of my ideas and scribbles

@Secret You should get a proper blog man, I'd be happy to help with that if you want

1 hour later…

8:02 PM

Maybe

Maybe the simplest embedding of distributions in hyperreal functions is just

Take the sequence definition of the distribution

And then make that the hyperreal function

$f_n(x) \to (f_0(x), f_1(x), f_2(x), ...)$

@Slereah What's a hyperreal?

$\Bbb R^\Bbb N / \sim $

Where $\sim$ is the equivalence relation of a non-principal free ultrafilter on real sequences

@BernardMeurer ^ See that's the kind of shit I don't do any more.

Wat da fook

It's sequences of real such that two sequences are equivalent if most of their terms are equivalent

8:07 PM

He probably doesn't know what an eq. rel. is.

You might want to start there.

It's a relation that follows the properties of an equivalence

as the name states

amazon.co.uk/gp/product/B00ELHD0XW/…

there are some technical properties @Slereah

Also what the hell is the difference between hyperreals and robinson's field

Apparently the answer lies in "function theory on some nonarchimedean fields"

The real question is what the hell is the utility of hyperreals :-)

I just said

You can embed distributions into them to form an algebra

8:18 PM

oh really?

why didn't you say that 10 months ago

Were you looking for an algebra of distributions

no

but now you seem less crazy.

you can always trust my judgement

It is always firmly rooted in deep physical principles

Also apples

Apparently it's yet another extension of the reals

Why

Aren't they all isomorphic anyway

Why bother to make ten different non-archimedean extensions of the real

8:34 PM

Apparently $^\rho \Bbb R$ is a subset of $^*\Bbb R$

not quite sure what makes it better for function analysis

explain the notation.

The first one has a tiny rho next to it

While the second has a tiny asterisk

no ban pls

Ban @0celo7

I'm flagging that hate speech

goodbye

don't make me do this...

8:41 PM

That paper is a bit annoying because he doesn't really explain the motivation behind the set

@Slereah If you apologize I won't flag

Go to hell

:(

why did you say that

Anyway

I think it's supposed to be a set of hyperreals with a cap?

only numbers smaller than a certain scale are kept

and it is more granular since infinitesimals only go down to a certain scale

What is an infinitesimal @Slereah

8:55 PM

x such that $\forall n \in \Bbb N. nx <1$

Doesn't that violate the Archimedean property?

yes, that's why it's a non-archimedean field

How does one construct such a field

A variety of ways

Usually it's done by making each member a sequence of reals

Then the order relation might be such that some sequences would be infinitesimals

For instance in hyperreals the canonical infinitesimal is $(n^{-1})_{n \in \Bbb N}$

9:49 PM

I'll miss JD.

@JohnDuffield Despite what everyone else thinks about you, I admire you.

You have conviction to stand up for what you believe in.

PSA: JOHN DUFFIELD HAS BEEN CHAT BANNED FOR A YEAR.

The Weather Girls - It's Raining Men

I meant chat banned.

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