The h Bar

yuggib

7:00 PM

@0celo7 wavefunction $\neq$ observable

i.e. it is not a physical quantity

Obliv

elements 1,3 and 5 are definitely in $B$ so it's just a matter of if addition between two elements is inside of $B$

0celo7

How does the Schroedinger equation work if it's not at least C2

Iszi

@JohnRennie No, if the flag was truly justified then logically action should have been taken which would be equivalent to an actually-validated flag. That is, the post should have been deleted and the user should have been kicked.

yuggib

2 mins ago, by ACuriousMind

Oh, for god's sake, stop trolling.

0celo7

Fuck, I'm not trolling

yuggib

7:02 PM

yes you are

Schrödinger equation is perfectly defined on $L^2$

0celo7

Better ban me now mods

I'm going to get banned anyway

yuggib

and you know that

0celo7

How would I possibly know that

yuggib

Aug 19 '15 at 2:10, by 0celo7

@ACuriousMind yes but he does not talk about your stone theorem

barrycarter

(sorry guys, phone)

0celo7

7:04 PM

@yuggib what?

I don't even know what that is

barrycarter

(catching up now)

yuggib

Aug 20 '15 at 2:35, by 0celo7

@ACuriousMind I want to say Stone's theorem has something to do with it.

barrycarter

@Obliv Reading what the problem says, what do you think B is and why?

yuggib

you knew 8 months ago

0celo7

What part of I CANT REMEMBER ANYTHING ANYMORE don't you get

Obliv

7:07 PM

@barrycarter I think $B$ is the set of matrices that are commutable in multiplication with $M$

barrycarter

@Obliv Right. And what matrices are in that set?

@Obliv I think the list you provided earlier was incomplete.

@Obliv If X and Y commute with M, then MX = XM and MY = YM, right?

Obliv

oh I didn't check all of them 1 sec

barrycarter

@Obliv You can actually prove it using the distributive property without necessarily enumerating the set.

Obliv

oh does $\begin{bmatrix}1&1\\1&1\end{bmatrix}$ work too?

0celo7

@yuggib Do you really not believe me

barrycarter

7:09 PM

I think so, yes.

Obliv

this doesn't help since some of these added to each other are still not in $B$ tho

0celo7

I don't see why a PDE would be defined on a discontinuous function

barrycarter

Actually, why not solve that problem analytically?

@Obliv I don't think it says anywhere that those are the only matrices in B.

yuggib

@0celo7 Hint: in the sense of distributions

barrycarter

@Obliv Multiply M by an arbitrary matrix in both directions, and see when you get equality.

0celo7

7:10 PM

I don't know what that means

barrycarter

Ocelot, PDE or PDF?

yuggib

@0celo7 do you know what a distribution is?

0celo7

No

yuggib

what is the Dirac's delta?

barrycarter

The Delta Dirac distribution has a well defined CDF, but no well-defined PDF.

0celo7

7:11 PM

@yuggib a distribution

Obliv

@barrycarter so any matrix that satisfies commutativity when multiplied with $M$ is in $B$. that doesn't change the fact that $P,Q \in B$ must mean $P+Q \in B$

yuggib

how it acts?

Obliv

Which it doesn't. for many matrices.

0celo7

Linear functional on a function space?

yuggib

and on what?

@0celo7 exactly

barrycarter

7:12 PM

@Obliv Well... (P+Q) x M = P x M + Q x M, right?

Iszi

For anyone who hasn't done so recently, this may be worth a read.

42

A: A guide to moderating chat

What tools are available to regular chat users? You, as a member of the community, have the ability help keep chat a place that is welcoming for everyone. There might be a variety of other users, including moderators and room owners, in chat with you. Users can be distinguished by the way thei...

3

> Remember: Only flag things that are truly inappropriate or offensive.

When you flag a message, you draw the attention of every mod and 10 k user currently on the chat network. This means you should only be flagging things that they’re going to see as offensive as well, which includes:

Personal attacks (characterized by you, e.g., “You’re a moron.”)
Anything that shocks the conscience (yes, this is subjective — let your conscience be your guide.)

Obliv

@barrycarter yes

barrycarter

@Obliv And P x M = M x P and Q x M = M x Q since both commute with M, right?

yuggib

and derivatives of distributions can be defined, by the corresponding action on test functions

Obliv

@barrycarter so you're saying because (P+Q)M = (Q+P)M that those quantities are in $B$?

barrycarter

7:14 PM

@Obliv If you meant Q + P, yes.

Obliv

yes

again, very tired q.q

barrycarter

@Obliv Yes, if P and Q are in B, then P + Q must also be in B.

Iszi

It's also worth noting that spam/offensive is not your only flag option. You can also "flag for moderator". The process for that includes a short form where you can provide some amount of context and/or describe what action is really required for the message. And that only goes to moderators - not mods+10k.

yuggib

so if an equation is satisfied in a distributional sense, it means that it is satisfied by a distribution (when acting on test functions)

barrycarter

@Obliv By the distributive property of multiplication/addition on matrices.

Obliv

7:15 PM

@barrycarter that just means that matrices P and Q are in B. That doesn't mean the sum of P and Q are in $B$.

@barrycarter wait no.. that does mean the sum exists.

barrycarter

@Obliv OK, (P + Q) x M = P x M + Q x M, right?

Obliv

that makes no sense WTF.

barrycarter

@Obliv Dude, get some sleep.

0celo7

@yuggib I know you want me to say Schwarz space but I can't prove it.

So why would I know any of this?

I'm an engineer

Obliv

@barrycarter are you so sure that the distributive property of (P+Q)M is commutative?

I must prove this.

Iszi

7:18 PM

@James Hm. I think that answers a question I just had too.

Thanks.

barrycarter

@Obliv I'm saying you can write it out step by step, yes.

yuggib

@0celo7 you always try very hard to convince yourself and the others that you know less than what you actually know

and knowing that a result holds is not equivalent to knowing all the proofs involved

0celo7

Because people always overestimate me

amazon.com/b/…

WHAT

I must move to one of these cities

Slereah

Lifting of the day is done B)

My arms

I've got a whole lot of fitness to do to take care of that dishful of chicken grease I ate yesterday

0celo7

@Slereah amazon.com/Foundations-Differential-Geometry-Vol-1/dp/…

Slereah

7:23 PM

not too expensive

0celo7

@Slereah amazon.com/Comparison-Theorems-Riemannian-Geometry-Publishing/…

what was the other one

Slereah

O'neil

0celo7

Too expensive

Will wait until it gets cheaper again

Obliv

@barrycarter how rigorous does a proof have to be? If $P,Q \in B$ then $PQ \in B$. It follows that $PQ(M) = QP(M)$. Therefore, $PQ \in B$. Is that not rigorous enough?

barrycarter

@Obliv Wait, how do you know that P, Q in B means that PQ is in B... or is that what you're trying to prove?

Obliv

7:29 PM

that's what I'm proving, yes.

barrycarter

@Obliv Why does PQ(M) = QP(M) ... P and Q are only guaranteed to commute with M, not each other.

Obliv

but they're all matrices

barrycarter

@Obliv And?

Obliv

isn't matrix multiplication commutative?

yuggib

@Obliv o.O

Obliv

7:30 PM

oh it's not always. DAMN.

barrycarter

@Obliv If it were, the problem would be trivial. B would be all 2 x 2 matrices.

Obliv

can I prove the statement with a constraint?

or does it have to be true for all $P,Q \in B$ do you think/

barrycarter

@Obliv All P, Q in B, obviously.

Obliv

ok I have an idea i'll brb.

barrycarter

@Obliv You're trying to show the B is closed with respect to addition and multiplication.

Obliv

7:41 PM

hey @barrycarter in $P(QM)$ QM is commutative so it is the same as if it were $P(MQ)$ right? And $(PM)$ is commutative so it would be the same as $M(PQ)$ so that proves commutativity for $(PQ)$ on $M$ doesn't it?

barrycarter

@Obliv Very good, yes. You're using associativity to prove commutativity.... wait.

Obliv

i'm using the commutativity of the parts to prove the commutativity for the quantity.

barrycarter

(PQ)M = M(PQ)... yes, taht wrks

Well, yes, but you need P(MQ) = (PM)Q at one point

Obliv

yeah matrix multiplication is associative right?...please god just say yes

barrycarter

yes :)

I was just pointing out that you were using it implicitly

Obliv

7:47 PM

psh yeah i knew that

0celo7

@yuggib I never use a result that I don't know how to prove.

Jimmy Kudo

Hey guys, in magnetism i encounter phrase like "Time varying magnetic field" , "increasing or decreasing magnetic field". Can some one please explain their meaning and significance? A small help can change my life :P

ACuriousMind

They mean exactly what they say. A "time varying magnetic field" is a magnetic field that changes with time. What about that is unclear to you?

Jimmy Kudo

whats is need?

ACuriousMind

What?

Jimmy Kudo

7:58 PM

why time varying? does it gives the direction?

i mean why do we have to vary the magnetic field?

ACuriousMind

We don't "have to" vary the field - it varies all by itself (in most situations because the current that generates it varies).

Jimmy Kudo

You mean to say Current varies so Magnetic field varies?

So will it do any change if its a constant Magnetic field and not a time varying?

yuggib

@0celo7 yes you do

ACuriousMind

@JimmyKudo Any change to what?

Jimmy Kudo

I am particularly concerned about numerical not experiments

ACuriousMind

8:02 PM

I don't really get what your issue is, what situation are you looking at, and what troubles you about the field varying in time?

Jimmy Kudo

To every aspect related to the magnetic field

ACuriousMind

Of course a magnetic field that doesn't change with time is different from one that does change in time!

Jimmy Kudo

I am just asking the significance of the "time varying" magnetic field

Can u say an example which produce time varying magnetic field please

ACuriousMind

@JimmyKudo Alternating current in a wire.

Jimmy Kudo

ok now I understood, thanx

For DC its constant right?

ACuriousMind

8:04 PM

Yep

Thanx Acuriousman :)

@yuggib Proof?

nya nya nya

@Acuriousman thank you for all of your help too :)

ACuriousMind

I'm a mind, not a man ;)

3

Obliv

8:05 PM

PROOF

you're an a.i

kevin Tah N.

i second that

Jimmy Kudo

U are bot? I see you always online :P

kevin Tah N.

he knows too much

ACuriousMind

Damn, you've finally unmasked me :O

Jimmy Kudo

:)

Bernard Meurer

8:09 PM

@ACuriousMind Are you watching the eagles?

ACuriousMind

@BernardMeurer Should I?

Bernard Meurer

ustream.tv/channel/Ws8PbVWq5Xs

ACuriousMind

Do they something except being eagles?

Bernard Meurer

Both eagles and both babies on the nest

Idk

Obliv

I keep thinking you're talking about football then I realize you're both not from the U.S. lol @BernardMeurer

Bernard Meurer

8:10 PM

It's weirdly entertaining

THEY ARE FEEDING

ACuriousMind

...they're named "Liberty" and "Justice"?

Bernard Meurer

@ACuriousMind Hell yeah

Jimmy Kudo

any one watches The Flash?

arent those eagles scared of the cameras and all

Obliv

How can I determine that $f: \mathbb{Q} \to \mathbb{Z}$ defined by $f(a/b) = a$ is well-defined? I feel like $f(.5)$ and $f(1/2)$ will both return $1$, no?

ACuriousMind

Meh, they're just being eagles

Also I keep getting an ad every minute

Jimmy Kudo

8:13 PM

use adblock

Bernard Meurer

Use adblock

ACuriousMind

NoScript perfectly kills all other ads, no need to install adblock

barrycarter

@Obliv Hmmm, can you write the same rational number in more than one way?

ACuriousMind

@Obliv It's not well-defined.

Obliv

great. I'll just write : because ACM told me it is not well-defined.

yuggib

8:14 PM

@0celo7 what is the proof of Stokes theorem?

ACuriousMind

@Obliv Oh, did you mean "whether" and not "that"?

barrycarter

@ACuriousMind Good catch

@Obliv Are we getting a little snippy with ACM?

Obliv

lol

just foolin, m8s

how did you come to that conclusion, though. @acuriousmind

ACuriousMind

To show it's not well-defined, do what @barrycarter said.

Obliv

wow that was obvious.

can't believe I didn't catch that >:(

barrycarter

8:16 PM

@Obliv I can believe it.

Obliv

not helping >_>

barrycarter

@Obliv Have you looked at the chat history? You should get some sleep if you're tired.

@Obliv At one point, you claimed the elements of B were in B because I had confirmed it for you.

Obliv

Yeah I kind of assumed $B$ was a closed set because it said choose the elements that belong to $B$. VERY VERY dumb assumption.

so I thought maybe the elements I chose to be in $B$ were not actually in $B$ that's what I meant

barrycarter

@Obliv Oh, OK, that makes more sense.

kevin Tah N.

I take the bloody physics gre again on the 16th. . . me so scared :O

Obliv

8:27 PM

good luck man :P @kevinTahN.

yuggib

:28843115 first, you need to know how to integrate (differential forms) ;-P

0celo7

@yuggib I do actually know how to prove that

I wrote the full proof once in my own words, it's in my private notes.

kevin Tah N.

@Obliv thanks man!! :D

barrycarter

Ocelot, and Fermat's Last Theorem too?

0celo7

@barrycarter Nah, margin was too small.

yuggib

8:31 PM

@0celo7 so you know how to define integration of differential forms as well

with all the proofs of the relevant properties

0celo7

Of course, pullback onto Rn using charts

Partition of unity

All that good stuff.

I don't know why are trying to knock me

IF there is something I can do, it's geometry

kevin Tah N.

@0celo7 just accidentally checked out your blog. cool beans!!

yuggib

I am not trying to knock you, you asked me for a proof that you use results of which you do not know the proof

and I am not sure that you know all the relevant proofs about integrals (in $\mathbb{R}^n$ if you used charts) that you need

it is not something bad, it is just that there are a lot of things that people use without knowing the proof

0celo7

@yuggib The only thing I can't do is the Fubini theorem.

You need that to prove the Stokes theorem.

But I do know """most""" of the proof.

yuggib

as I told you, it is not a bad thing to use a result without knowing all the underlying proofs

but it is unfair to say "I do not use results of which I do not know the proof"

maybe unfair is not the correct word...exaggerated is better perhaps

Bernard Meurer

8:43 PM

@DanielSank I summon you!

kevin Tah N.

I strongly believe you just need to know what a set is and how to draw commutative diagrams and you will understand 30% of math, which is enough to pretend :D

ACuriousMind

Diagram chasing proofs are really those proofs you do once and pray never to have to do again

Snake lemma, horseshoe lemma, etc. They're not really difficult, but they're so lengthy.

And most people really do prove them by diagram chasing without having seen the proof of that embedding theorem that shows that that suffices for categories other than abelian groups/modules.

kevin Tah N.

interesting, I have only chased diagrams for very basic proofs recently ~ a few weeks now. Mostly about little things in hopf algebra. It was amazing. All I had to do was draw a diagram and make statements concerning where the arrow were pointing and composing them. It was amazing!!

Also I realized if I drew a venn diagram things became very clear! I have seen this in algebra, probability theory and I even used it to see what some of the statements in rosenlicht's into to real analysis meant

If someone had told me this a few years ago, I would have been kinda smart by now

I mean, I have no formal training as a mathematician or anything lol

Also i noticed from the intro to analysis text that you can proof anything by using supremum and infinimum concepts

. . . . . .most of the time

DanielSank

8:59 PM

@BernardMeurer What is thy bidding, my master?

kevin Tah N.

lol

Physics Meta

0

Q: Is it okay to bump questions?

Most of my questions are about quantum field theory. They tend to get extremely few views and a handful of positive votes before sinking off the bottom of the active list. My non-QFT questions are pretty well-received, so I'd like to think I'm not asking totally trivial questions, but almost all ...

discussion

0celo7

Well I just sat in the most boring talk of all time

What I thought it was going to be on: power series

Obliv

is the reflexivity of a relation (say $\sim$) guaranteed in sets? Like {$a,b,c\}$ and $a\sim a$ is guaranteed?

0celo7

What it was on: combinatorics and graph theory

kevin Tah N.

9:01 PM

sounds like the later should be interesting

0celo7

@Obliv reflexivity is by definition required of all eq. rels

Obliv

let me reword

0celo7

@Slereah Are you around

Obliv

@0celo7 how about now? If it's just a relation and not equivalence relation

0celo7

@Obliv no, is $1>1$?

kevin Tah N.

9:03 PM

hmm sets. . .

what is "reflexivity" as used above with regards to sets

0celo7

@ACuriousMind Do you have a "cooler" definition of $TM$ than $TM=\sqcup T_pM$?

kevin Tah N.

Reflexive relation

In mathematics, a reflexive relation is a binary relation on a set for which every element is related to itself. In other words, a relation ~ on a set S is reflexive when x ~ x holds true for every x in S, formally: when ∀x∈S: x~x holds. An example of a reflexive relation is the relation "is equal to" on the set of real numbers, since every real number is equal to itself. A reflexive relation is said to have the reflexive property or is said to possess reflexivity. == Related terms == A relation that is irreflexive, or anti-reflexive, is a binary relation on a set where no element is related to...

0celo7

> For the proof, see Steenrod

AHHHHHHH

@yuggib can you prove Stokes' theorem?

yuggib

@0celo7 no

0celo7

@yuggib why not

kevin Tah N.

9:15 PM

I thought people accidentally learned to do this when they learned about maxwell's equations

differential and integral forms

ACuriousMind

@0celo7 No. Sadly, it's not the étalé space of the sheaf of vector fields.

Bernard Meurer

@DanielSank I need your help

Regarding UCSD

Are you aware of any QC over there?

I recall reading something about a Fermionic QC

but I don't even know what that is

0celo7

@ACuriousMind ...What?

0celo7

9:29 PM

@yuggib ever heard of amazon.com/Mathematical-Physics-Chicago-Lectures/dp/0226288625/… ?

@ACuriousMind Are you going to get the F3 DLC?

yuggib

@0celo7 never looked into it

Bernard Meurer

@ACuriousMind Are you there?

yuggib

@0celo7 no...

don't know it

DanielSank

@BernardMeurer I honestly don't know.

Bernard Meurer

@DanielSank What's a fermionic QC though?

yuggib

9:37 PM

0

Q: Are physical functions always differentiable

I know that physicist usually don't really think too much about differentiabillity of functions. Usually there are at most finite many points where functions aren't differentiable and if there are infinite many of them they are all isolated. On the other hand you sometimes see examples (especiall...

classical-mechanics mathematical-physics differentiation

related to before...

Bernard Meurer

9:49 PM

@DanielSank csqc.ucsb.edu

These guys are over there

ACuriousMind

@BernardMeurer Somewhat, what's up?

@0celo7 Nope

Bernard Meurer

@ACuriousMind What are Bell inequalities?

ACuriousMind

@BernardMeurer Inequalities discovered by Bell that many people make a large fuzz about

0celo7

@BernardMeurer Did you try googling

@ACuriousMind wtf if I asked that you would tell me to google

ACuriousMind

Also, that^^

Bernard Meurer

9:58 PM

@0celo7 Yes, it didn't yield me a good answer

@ACuriousMind Got it

ACuriousMind

@0celo7 Note that I didn't actually tell him anything of substance :P Also, you're not my children, I'm not required to treat you all equally

0celo7

oO

You openly admit you treat me like less than @BernardMeurer

Bernard Meurer

@0celo7 The fact that I don't ask him a 100 things a day might be why

ACuriousMind

@0celo7 I treat you differently. I'll refrain from value judgements.

0celo7

I haven't asked him anything lately

ACuriousMind

10:00 PM

53 mins ago, by 0celo7

@ACuriousMind Do you have a "cooler" definition of $TM$ than $TM=\sqcup T_pM$?

ArtOfCode

cough

Bernard Meurer

@0celo7 I'm clearly the favorite child

ACuriousMind

@BernardMeurer Honestly, I never cared much about those inequalities, but then, I never had issues with accepting quantum mechanics.

Bernard Meurer

@ACuriousMind Meh, I trust you people way too much to not accept QM

0celo7

@BernardMeurer apparently :(

ACuriousMind

10:02 PM

::hands @ArtOfCode cough drop::

ArtOfCode

cheers

Obliv

Trying to prove $\sim$ is an equivalence relation. $a\sim b$ if and only if $f(a) = b$. There exists an inverse such that $g(b) = a$. does that mean I can say $b \sim a$ and that $\sim$ is symmetric?

0celo7

@ACuriousMind that's not even a question

@ACuriousMind Have you heard anything about Finsler geometry?

ACuriousMind

Something about manifolds with path length metrics but no actual metric tensor or somesuch, but I know nothing specific.

0celo7

@ACuriousMind you're so smart

@Slereah Got the Asian edition of KN coming, dat classic tho

It's gonna take a month

0celo7

10:27 PM

@Slereah hopefully it's in English

DanielSank

@BernardMeurer Not a clue.

Bernard Meurer

Ged demmit dan

@DanielSank Why does SB have to be far from LA and SF

DanielSank

@BernardMeurer You mean these groups at UCSD, right?

@BernardMeurer You know... because.

Bernard Meurer

I found this quantum.ucsd.edu

And this ccom.ucsd.edu/~qc

But the latter seems dead

and the former is the fermionic QC I was talking about

Can't wait to hear back from IQC

Holy

ACuriousMind

Chill

Bernard Meurer

10:42 PM

I need to answer UCSD before the results for the IQC come out

What do I do

ACuriousMind

And don't eat any coffee beans

Bernard Meurer

@ACuriousMind I already finished my package of those, they're so good man, @ChrisWhite was right

0celo7

dammit I was doing something

then I was interrupted

and have no clue what I was doing

Bernard Meurer

@0celo7 What do I do?

0celo7

About?

I don't remember our conversations

Bernard Meurer

10:48 PM

The application decisions for the IQC are released after my deadline for accepting UCSD's admission offer

0celo7

What's the penalty for accepting and then not going?

ACuriousMind

Eternal shame, probably

Bernard Meurer

Dunno, how do I google for that?

Does that penalty have a name?

0celo7

@ACuriousMind Don't you talk to me about eternal shame

Mr. "let's shame 0celo7 for asking questions"

@BernardMeurer Call and ask?

Bernard Meurer

HA

NO PENALTY

grad.ucsd.edu/admissions/faq/admitted-students.html

"Yes, you may decline after accepting admission. The application site will not allow you to change your decision from “Accept” to “Decline”, so you must contact your department directly to inform them of the change in your decision."

ACuriousMind

10:54 PM

@BernardMeurer Um...I may be wrong, but that's for graduate students, isn't it?

you're applying for undergrad, right?

0celo7

@ACuriousMind You're right.

Bernard Meurer

@ACuriousMind Crap

0celo7

@BernardMeurer Call them tomorrow.

Bernard Meurer

Shit I don't want to call them, they'll get angry that I'm considering other schools

What if I get a bitchy admissions officer and they're like "Oh, you're considering not taking it? Let me make your life easier and deny you"

HDE 226868

@BernardMeurer I'm going to go out on a limb and say they won't. Not if they want to stop you from going to those other schools.

Bernard Meurer

10:57 PM

I'll call with a fake name

I'll be Björn Merkwürdigliebe

HDE 226868

They'll be real happy if they find out you pretended to be someone else, yeah.

Bernard Meurer

Shit dude I'm going to have a stroke

0celo7

@BernardMeurer My love for Bajoran is pretty strange.

Bernard Meurer

That's a cool name right?

0celo7

No.

Bernard Meurer

11:01 PM

@ACuriousMind Do you think I could be a Björn?

Will you welcome me into the clan?

0celo7

I like Björn Kleinschwanz.

Bernard Meurer

lel

David Z

11:39 PM

Well this is odd... recent answer posted to a 2-year-old question and accepted? Also, it sounds like nonsense

-6

A: What is the significance of Planck force?

The maximal possible force in universe is something that scientists missed to consider, and if they would consider it, then everything what they have done so far would not be worth a dime, ranging from "point-like particles" and up to the "Big Bang". Point-like particle (i.e. photon), or any othe...

ACuriousMind

@DavidZ It probably got accepted because it has nice TeXed formulae :P Every answer on that page except yours is nonsense.

@BernardMeurer You're either born a Björn or not. Sorry, I don't make the rules!

Bernard Meurer

@ACuriousMind Dang it

ACuriousMind

@DavidZ: I won't do it because we don't delete wrong answers, but I see a delete button there. Can a negatively scored accepted answer be deleted by 20k delete votes?

David Z

@ACuriousMind I'm not sure

If you see the button there, you should use it as appropriate

While we don't delete answers for being wrong, we do delete answers that don't represent serious attempts to answer the question using relevant knowledge.

(apparently)

ACuriousMind

If we're ever tired of discussing homework, we could do that policy next ;)

David Z

11:52 PM

That's on you ;-) I have my hands full with homework and stuff

skill patrol

How's the bot coming along?

ACuriousMind

Actually, I wouldn't care much about leaving wrong downvoted answers on the site if they didn't act as net rep generators

That user derived 21 rep points from posting that nonsense, although the post sits at -7!

Which isn't a concern if they are drive-by posters, but it is if they're persistent.

David Z

It's a little bit of a concern

Bernard Meurer

@ACuriousMind Can you have more than two particles in an entanglement?

ACuriousMind

3

Q: three-particle quantum entanglement

So I know that two particles can be entangled in a quantum way, but is it possible that more than two particles be entangled in a quantum way? Most descriptions provide with two-particles cases, so I wonder. (It's hard to think of three particles entangled in spin, or so.)

quantum-mechanics quantum-field-theory entanglement

skill patrol

11:59 PM

Use your Google-fu :P

Bernard Meurer

Holy

Imagine a 10 qubit entanglement

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