The h Bar - 2016-01-09 (page 3 of 3)

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10:01 PM

@TanMath How are you sure? Are you trolling?

@ACuriousMind So in my picture above, if you boost along the $x$ axis, the green vector changes its direction and aligns more closely to the positive $x$ axis because of relativistic length contraction. Similarly, the blue vector rotates to align more closely to the negative $x$ axis. Colloquially, in the $v\sim c$ limit, those vectors have to be parallel to the momentum, because for $v=c$, they are parallel (like for photons).

Does that make sense?

Yes

@ACuriousMind My question is, what happens to the red vector, the one that's orthogonal (in all $x$-boosted frames).

@DanielSank why would I troll????

@Bass It stays orthogonal.

But what should "happen" to it, anyway?

You shouldn't expect to get meaningful results about the massless spin from the limit $v\to c$.

10:09 PM

@ACuriousMind So this would be a helicity 0 eigenstate?

@ACuriousMind Dunno. As I said, I found this "there are no spin eigenstates orthogonal to momentum" in my notes, but apparently that's complete rubbish.

@Bass No, why would it? Just examine the helicity operator as a 4x4 matrix, it has no zero eigenvalues.

Your argument is a completely classical thinking about two vectors. It doesn't dictate the behaviour of the quantum theory at all.

@Bass Yes, it rubbish because that's not an invariant notion, not even non-relativistically. For instance, it's a completely non-sensical statement in the rest frame of the particle.

hi @ChrisWhite

user54412

hi

Don't know if people spend more time learning classical mechanics or trying to "forget" it to be able to understand quantum physics.

@ACuriousMind as always, muchas gracias!

@Bass Since I grew up with quantum mechanics, it was never so difficult for me to grasp.

10:19 PM

@Danu you grew up with it? meaning?

55

A: Help us identify micro-privileges for top users

See a list of inbound links to your questions and answers On occasion, I'll run in to someone referring to my questions or answers out "in the wild" -- kind of neat! It isn't easy to get at that kind of information without manually searching Google; as a privilege it would be awesome to see a li...

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@TanMath Meaning nobody ever assured me something like that could not exist

@Danu Well, the basics of QM are indeed not difficult to grasp, but I find it very hard to avoid classical arguments when I'm thinking about quantum stuff. Classical thinking is what we learn as babies. The thing is here or it isn't. It goes through that hole or not.

Or did you grow up observing superpositions? :P

3

@Bass Exactly.

The concept never really bothered me.

(this is not meant as some kind of bragging)

I don't have any problems accepting quantum "weirdness"

user54412

10:27 PM

Glad I'm not the only one.

user54412

Actually, I don't buy it when everyone says we're intuitively used to the classical world somehow.

^^

Or for instance the question "what is REALLY happening" between measurements?!?!?!

I find it very unnatural and have strong doubts that it's something we should even care about

user54412

We've just been fed classical models since they day we were born. But take your stereotypical tribe outside modern civilization and see if they intuit the ontology of forces and inertia.

@ChrisWhite Exactly

All the discomfort that most people seem to have with quantum stuff is ontological

But why are the Newtonian terms accepted as "real"?

@Danu I'm not talking about accepting the weirdness or wondering about what's REALLY there (though I admittedly do that sometimes). I'm talking about making QM your intuition. In the quantum world, a spinor is an everyday object, nothing to wonder about. In our everyday lives, have you ever met a spinor object?

user54412

10:30 PM

@Bass Have you ever met a Newtonian scalar? ;)

@Bass No, but have you ever met a fucking Atwood thing?!

now that is counter-intuitive to me

The Atwood machine (or Atwood's machine) was invented in 1784 by the English mathematician George Atwood as a laboratory experiment to verify the mechanical laws of motion with constant acceleration. Atwood's machine is a common classroom demonstration used to illustrate principles of classical mechanics. The ideal Atwood Machine consists of two objects of mass m1 and m2, connected by an inextensible massless string over an ideal massless pulley. When m1 = m2, the machine is in neutral equilibrium regardless of the position of the weights. When m1 ≠ m2 both masses experience uniform acceleration...

These fuckers :P

@Bass Actually, Roger Penrose gives an interesting spinor-like demonstration in The Road to Reality. All you need is a book and a strip of paper. I don't quite remember it, though.

What, Möbius strip stuff?

What?

@Bass I guess the main point is that I do not, at all, rely on visualization etc when doing physics

@HDE226868 Spinors

10:32 PM

@ChrisWhite Temperature is an everyday scalar function. (if I understand you correctly).

@Danu Ah. No, not Mobius strips.

@HDE226868 Isn't that just the Dirac belt trick?

@EmilioPisanty Possibly. Let me dig it up.

user54412

@Bass A common example. But how do you know what temperature is, besides reading a thermometer and noting that it only returns scalar values? If you design an apparatus to measure a system involving spinors, haven't you done the same?

@EmilioPisanty NO

That fucking belt is the bane of my existence

user54412

10:34 PM

A prof in undergrad taught an intro QM lesson on spin by taking off his belt and doing something. I don't know what, since everyone I asked about it just remembered the prof taking off his belt in class...

@HDE226868 I think you can do that with your arm. Take some small object, note its orientation, then move your hand that holds the object under the shoulder of the same arm and continue the rotation until it has the same orientation. Your arm is twisted now. Continue the rotation, moving your hand above your shoulder, until it has the same orientation again, which is the start position.

@0celo7 Just because you don't like it it doesn't mean it isn't what Penrose talks about

that's Dirac's belt trick, isn't it?

@Bass Yes, pretty much

@EmilioPisanty Yeah, it's the example I was thinking of.

10:36 PM

@Bass Example: youtube.com/watch?v=JDJKfs3HqRg

@ChrisWhite The important (and counter-intuitive IMO) difference is that with a classical scalar, you can measure everything there is. With a spinor, you can't, since if you rotate a spinor about $2\pi$, you get it's negative, which is ill-defined if you could measure it.

I mean the function $\psi$ is ill-defined if it has both a value $\psi(x)$ and $-\psi(x)$ at the same place, using the same coordinates. The only thing that saves quantum spinors from being ill-defined, is that we cannot measure their absolute value.

Correct me if that doesn't make sense.

What framework are you even in?

user54412

I see what you're saying, and I bet most of the world agrees with you. But I still don't buy it.

QFT or relativistic (non-existent) QM?

user54412

What about a vector? What really is the x-direction? Is that somehow less abstract than a spinor?

10:42 PM

I also think that spinors are in principle also classically useful, not solely "quantum hocus pocus"

user54412

It's like we've elevated $\mathbb{R}^n$ and functions $\mathbb{R}^n \to \mathbb{R}$ to privileged ontological status, and any other mathematical constructs are somehow mystical and tenuous.

After all, spinors are understood mathematically through representations of some (orthogonal, I think) group

@Danu Just talking about spinors (not restricted to QM or classical physics), that they are more counter-intuitive than ChrisWhite's Newtonian scalar IMO.

@Bass Which spinors, though

In which model exactly

@Danu: Not sure what the above statement means. Spin(n) is the double cover of SO(n). This is related to the belt trick precisely by the fact that if we follow the nontrivial loop in SO(n), we want its endpoint to be different than where we started in Spin.

10:45 PM

@ChrisWhite I don't mean abstract vs. non-abstract. I'm talking about intuitive or counter-intuitive, which is a subjective notion, but I suppose it is quite similar for many people. A 3-vector can be explained to a 5 year old, just tell him it's the direction of some object.

@MikeMiller Which above statement?

@Danu Just some universal cover of $SO(n)$. In other words, something that transforms to its negative under a $2\pi$ rotation.

"Spinors are understood mathematically by..."

@Bass No, which physical model?

Or do you claim that the existence of these objects as mathematical constructs is already too strange?

@Danu Not specific, really. I'm just saying that spinors are something that's not in our intuition. We have to learn it when we learn quantum physics.

10:47 PM

@MikeMiller I'm not sure either, that's why I kept it vague; Remember, I haven't gotten to spin geometry yet :(

:)

But do you know what spinors are, @MikeMiller?

Nah.

I thought they were elements of a certain module associated to some representation of $SO(n)$

some "different" representation somehow

very different from the fundamental rep

@Bass Well, this is too vague for me to really say anything about; I think that in practice one really doesn't encounter too many problems dealing with spinors.

@ChrisWhite I think I understand what you mean. If you spend enough time working with complex numbers, you see that they are something very natural, you're starting to think they are not more abstract than $\mathbb R$. But this is something you learned by taking lectures or by reading books. It's not something you learned by playing as a child. Many basic facts about $\mathbb R$ we learn as children. IMHO :)

10:51 PM

wait, of course I do. They're elements of a certain representation of Spin(n). It's reasonably large.

@EmilioPisanty it's a terrible analogy

because then you have people like Duffield thinking a vector going around a Mobius band is a spinor

Should be a complex vector space, dimension 2^{something like n/2}

user54412

@EmilioPisanty I just went through that thread. I had no idea I could find my own deleted posts!

@0celo7 I'm not saying it's good or bad

@ChrisWhite Yeah, it's really new

@Danu I think it's more clear if you compare them to normal vectors. If you rotate a vector by $2\pi$, you get the same vector again. This is something that many people grasp, without having to learn geometry or vector analysis.

10:53 PM

Like, a month old

user54412

@Bass Perhaps we should rethink how we teach children ;)

@0celo7 I'm just saying it is in fact probably what Penrose was talking about

However, if you rotate a spinor by $2\pi$, you get it's negative.

@EmilioPisanty ok, well I am

@EmilioPisanty ok...not disputing that

@Bass I'm not even sure what the precise statement would be here.

Rotate where? In what space?

10:54 PM

@0celo7 I think people with the sort of mindset I think you're referring to will find stuff to misinterpret no matter what's floating around

I think you have to commit to some explicit realization

@Danu: Think of this as a statement about Spin(n) instead of the space of spinors.

@MikeMiller Think of what?

10

A: How do you rotate spin of an electron?

You have fallen prey to a popular simplification of spinors. The statement "you have to turn electron by 720 degrees in order to get the same spin state" does not refer to an actual rotation of an actual electron. In quantum mechanics, we describe the states of objects as elements of a Hilbert s...

K

10:56 PM

@Danu This is not restricted to a certain space. It's like, what's a tensor? It's something that transforms like yaddayaddayadda. The same holds for a spinor. A spinor is something that transforms under the fundamental Spin(n) representation.

(caveat this is just what I learned, don't count too much on it.. corrections appreciated)

@Bass Welp, I still don't see what is so surprising

@Danu The surprising thing is that if you rotate it by $2\pi$, it points in the opposite direction.

I don't really understand why that's surprising if it's by design.

@Bass ...in some abstract, auxiliary space where things that point in the opposite direction are identified anyways

@Bass But...that's only counterintuitive if you insist describing the application of the $-1$ element of $\mathrm{Spin}(3)$ as "rotating by $2\pi$"!

10:59 PM

I mean, if I walk a circle around an electron it'll be the same one after one circle.

@ChrisWhite If you manage to write a book that teaches children spinors, I withdraw everything I said.

While, well, "rotating by $2\pi$" rather by definition means the identity operation.

I think we all have the same objection but am amused that we all ohrased it in ways that seem legitimately different.

I think this way of wording things is just confusing because of the words, not because of the physics

@MikeMiller Who is "we" and objection against what? :P

I am going full precision mode :P

I'm just opposing those who say spinors are not less counterintuitive than vectors. For vectors, we have many real-world examples. For spinors, we have to resort to some "belt trick" analogies.

11:01 PM

I don't think so.

user54412

@Danu everyone better look out!

what's a vector?

@ChrisWhite Hide yo' kids, hide yo' wives

@MikeMiller Oh SHUT UP :P

ok

elements of the tangent space*

@MikeMiller I didn't mean it. Plz u no be offended

11:04 PM

@Danu Okay, let's take rotations by $\pi$ instead of $2\pi$. If you rotate a vector by $\pi$, you get the opposite direction. You don't need any math course to understand that. If you rotate a spinor by $\pi$, you get something.. uhh.. it might be $i$ times the original vector. Or $-i$. Very intuitive, I must say.

@Bass but not in physical space

@ACuriousMind If we had classical spinors (that could be observed classically) then I could slowly rotate around that spinor, and as soon as I reach $2\pi$, I see that it's pointing in the negative direction. Wrong?

@Bass What do you even mean by "pointing" somewhere?

@Bass Yes.

@Danu Why not? If we had a spinor we could observe?

11:07 PM

"Rotate a spinor by $\pi$" does not sound to me like a well-defined operator.

I think you've read too much "oh quantum physics is so strange, spinors are so crazy" things :P

It doesn't exist in 3D space. You can't "look at a spinor".

@Danu Same thing I mean by a vector pointing somewhere?

@Bass But it's not like that

@ACuriousMind Why not if we had a classical spinor?

user54412

11:08 PM

@Bass and what does that mean intuitively?

@Bass What does that even mean?

user54412

try pointing out something to a cat: they'll look at your finger, not where it's pointing

@ChrisWhite lolwat :P

"A spinor" is an abstract element of a vector space with the fundamental rep of SU(2) on it. Nothing about it is "quantum" or "classical" as such. What you do mean, "if we had a classical spinor"?

Note, the above ^ vector space is not the spacetime

11:09 PM

IIRC this discussion grew from talking about how intuitive vectors and spinors are. I said, vectors are something that's very intuitive to us, since a "direction" is something we get acquainted with as babies. If a baby rotates an arrow by $\pi$, it points in the other direction. So when we learn vectors in school, we just have to think about the arrow we played with, and we're done.

user54412

@Danu my point is our "intuition" for Euclidean 3-vectors is merely a set of learned manipulations ("extend the line passing through the endpoints", etc.)

Also, I'm pretty confident spinor is pronounced "spine-or", regardless of what anyone else says. It has one n for christ's sake.

@ChrisWhite Sure, and I agree, but the cat example didn't really convey that in any way IMO haha

@MikeMiller wrong

It's SPIN-or

it's about its spin, not its spine

user54412

@Danu Now I know what it's like to be a raving nonmainstream madman.

Yes, but it has one n, so no.

11:11 PM

and yes, this is a big deal for me :P

Fucking Germans all get it wrong

I think this is the important discussion to have about spine-ors.

With a classical spinor, I mean something we could play with, that acts like a spinor. Like an arrow toy for a baby. (of course that doens't exist). This arrow would point in the other direction as soon as the baby rotates it by $2\pi$.

@MikeMiller Shaddap

SPIN-or

@MikeMiller Physicists write down the exponential map from the su(2) algebra to SO(3), and then they get take the fundamental rep of su(2) and are surprised that then the same thing that exponentiated before to 1 in SO(3) is now the -1 in SU(2). They're really just confused by double covering :P

@Bass You seem to be thinking only in terms of spacetime

11:11 PM

My entire point is that this does not exist, and so spinors are less intuitive to us than vectors.

user54412

@MikeMiller that actually makes a lot of sense

@Bass That's not a "spinor". That's a toy you engineered to somehow act like your preconceived notion of a spinor behaving weirdly under rotation.

@Danu Yes, I'm talking about intuitiveness, the one we learn as children. No projective Hilbert spaces here.

You cannot "look" at a spinor anymore than you can "look" at a 4-vector.

@ACuriousMind But don't you see my point that a vector has very concrete real-life correspondences, which a spinor hasn't?

11:13 PM

The only things you can look at are 3-vectors and 2-vectors, and that's it.

Look, but don't touch!

user54412

Did particle physics really invent a suffix, -or, that has the unique ability to override English pronunciation rules?

Yes.

@Bass: This is a losing argument for one simple reason. Your argument was that "Spine-ors are not intuitive". You have in front of you three people who disagree. Therefore you are not uniformly correct: some people find the notion of a spine-or very intuitive.

@ChrisWhite Obviously

@MikeMiller Argument by sociology

Very clever

user54412

11:14 PM

Take that, English majors!

If it was supposed to be pronounced spinner, it would be spelled that way.

That's not how I pronounce it

spinor

Wow, nobody thinks you're correct! :)

user54412

Interestingly, my mind's voice would pronounce spinner and spinnor differently, even though I would vocalize them the same.

@MikeMiller The question is not whether some people find spinors intuitive. As I said, it doesn't need much time to get acquanted with complex numbers, so they seem at least as natural as real numbers. But that's only after you spent hours on lectures and exercises. With real numbers, everyone has some intuition.

11:16 PM

I suppose I could be content with spinnors. But spine-ors they are, and spine-it's they will remain.

@Bass: Sorry, can you state what you're arguing for, then? I seem to have misunderstood.

@ChrisWhite are there English pronunciation rules?

user54412

@Bass More than most people realize.

@Bass Yes. But most of your "counter-intuitiveness" comes not from the spinor being "weird", but from you applying pre-conceived notions to it like "rotation by $2\pi$" or "pointing in the other direction" that simply don't make sense. Just stop imagining it as a 3-vector.

@ChrisWhite thought tough though through thorough :)

user54412

Non-native speakers have asked me if we English-speakers just had to memorize all pronunciations growing up, and my response is always "definitely not." I certainly could pronounce most words correctly the first time I saw them.

11:19 PM

@ChrisWhite Heh, same here

user54412

@Bass Everyone likes to bring up that example. But I think native speakers just don't realize how many actually-obeyed rules there are. It's a selection bias.

@Bass I don't agree that you have any real idea about what real numbers are from just playing around :P

@ACuriousMind Makes sense, point taken. Still, you have to learn spinors. With vectors, you can just start with the intuitiion you got as a baby.

Actually, I think most people would disagree that "The metric completion of $[0,1] \cap \Bbb Q$" captures their notion of line segment.

@Bass I don't believe that even one second!

11:21 PM

@ChrisWhite Yes, I realized that when I heard some text-to-speech engine pronouncing german text in english

Remember struggling with free body diagrams for 3 years in high school physics?

I don't think very much beyond natural numbers is "intuitive"

No.

Maybe positive rationals

user54412

@Danu And only finitely many natural numbers at a time!

@Danu I don't mean that you learn everything about vectors as a baby. Neither about real numbers. But you got a head start.

11:22 PM

@Bass Did you? I'm not sure about vectors.

I think you just forgot what it feels like not to know them already

@Danu Everyone learns some basic properties of their addition, scale and rotation operators.

It's unclear to me what anybody is arguing anymore.

We just don't learn the mathematical notation or names of these operators, but we have some feeling about them even if we don't know we have it, and even if we have no name for it.

user54412

@MikeMiller Welcome to the h-bar

@Bass I doubt that

For instance, would you ever have intuitively accepted that rotating a cube is noncommutative?

(very simple object, simplest possible rotations...)

11:25 PM

Why don't we all calm down and read my favorite post from yesterday?

In fact, my father didn't believe me when I told him last year.

He had to try it out and see for himself.

@Danu only in three dimensions. I think our intuition that rotations are commutative stems from our intuition from 2D rotations, since that's what most babies do.

...because we live in a plane, right? :)

I also disagree that an orientation-preserving orthogonal transformation should be called a rotation. I think they're generated by rotations.

I think you're just adapting your story for convenience

@MikeMiller Ehh?

Why would you not call all of them rotations?

11:28 PM

@Danu And for you, it's surprising that it's surprising for him. That's what I mean. If you spend enough time with spinors, they seem as intuitive to you than the things that were intuitive to you long before them.

@Bass No, don't turn this into an argument in favor of you :P

@Danu Why not, with such an assist :P

The point is that your so-called "intuitive" notions are not intuitive at all

Well, nice discussion guys, gotta go now, cu!

And you just forgot what it feels like to not know them

11:30 PM

Because rotation already has a fixed meaning even outside of 2 domensions: when you say "rotation of 3-space", the layman who doesn't look at you funny probably thinks you mean rotation around some line. In any case, it also means I can state thing like "Every element can be obtained by two rotations" simply, and there are frwuently times where it's easier to check something for actual rotations and then see it's true for SO(n) as. corollary.

and then you're surprised that some of us don't mind spinors?

:P

Perhaps I should say that, intuitively,

@Danu I could turn that into an argument in favor of me too, but let's call it a day.

@Danu Who said I minded spinors?

@Bass You say they're not "simple" or "intuitive" compared to other things

That's fine, but saying that that's an inherent thing is what's wrong

Yep. Makes them certainly more interesting than vectors. But I couldn't explain them to my grandma.

11:31 PM

I'm convinced vectors are super difficult when you first learn about them.

It's just that we do it early on

@Danu Because you learn them much earlier. If you learned both vectors and spinors at the same time, uhh...

Bye! good nihgt

@Danu Really?

@ACuriousMind Thx

Is the $N = n$ thing of SUSY algebras the number of generators $Q$

user54412

11:55 PM

Off to see a movie. Since I see about 1/year, I guess this means the next 11 months will be pretty bleak.

@ChrisWhite Awesome, have fun!

I'm watching a ton of movies

...but not in cinemas

user54412

well I do that

user54412

I'll probably get around to Episode VII in another... year or so

....screw Star Wars anyways

Bad acting, bad story, apparently good special effects, who cares?

I saw an amazing movie last night

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