The h Bar

0ßelö7

12:00

@BalarkaSen the christoffelvsymvols are the things appearing on the geodesic equation and those are nonlinear

Balarka Sen

okie dokie

Slereah

@0ßelö7 what about in this specific case, though

Since the metric is conformal Minkowski + conformal de Sitter

Emilio Pisanty

The question just made me laugh. You need to really look at my profile. Your answer is: Earth is flat. — Earth is Flat 15 hours ago

Slereah

Minkowski has a connection at 0

Emilio Pisanty

sigh

Slereah

12:02

Hm, what's the connection of conformal Minkowski

Balarka Sen

@EmilioPisanty lol

0ßelö7

@Slereah I'm worried about the inverse metric factor

Maybe you can multiply that away

Then the equation would be linear...but you'd still run into issues because you'd need to distribute if you have a sum solution

Balarka Sen

Ok, I'm totally convinced that earth is flat now

This image has changed my life officially

Emilio Pisanty

@BalarkaSen yeah, me too

Balarka Sen

My intellectual has increased 40% after seeing that footage

Emilio Pisanty

12:05

how did this make it five years without getting protected?

13

Q: How does the grid on the microwave oven window prevent microwave radiation from coming out?

If I look through the microwave window I can see through, which means visible radiation can get out. We know also that there is a mesh on the microwave window which prevents microwave from coming out. My question is how does this work? how come making stripes or mesh of metals can attenuate micr...

Slereah

The metric is like $$ds^2 = f(t) (-dt^2 + dx^2) + (1-f(t)) (-(1+x^2)dt^2 + (1+x^2)^{-1}dx^2)$$

Emilio Pisanty

five rubbish non-answers to two actual answers

Slereah

I mean I guess I could just compute it by hand

It's probably not too terribly hard

Since it's 2D and two simple metrics

though the function $f$ is rather complex so hopefully I don't need to deal with it explicitely

Emilio Pisanty

@ACuriousMind or other mods, any insight on what's going on with this one?

0

A: Why does spin arise in non-relativistic quantum mechanics?

I am sorry that my initial answer has been brief. I can expand now: the first theoretical explanation of the notion of quantum spin was in the context of special relativity (Dirac 1928 - two articles in PRSL). It was only after group theory through the work of Hermann Weyl and Eugene Wigner was s...

^ 10k tools marks it as having pending delete votes but they don't show up under the answer for me

ACuriousMind

@EmilioPisanty They show up for me

Emilio Pisanty

12:11

@ACuriousMind hmmmm

ah, I get it

only allowed to vote to delete on negative-score posts

therefore only allowed to see delete-vote counts on negative-score posts

I'm not sure that's a good thing

0ßelö7

@SirCumference ask someone in your group how they get it

I'm surprised I don't seem to have a subscription but I'm sure the library could get me whatever I needed.

@Slereah use the program you sent me

Slereah

Do you mean the SlereahSoft v. 1.0

0ßelö7

The MMA script

Then solving the equations...good luck

You could do numerical

@BalarkaSen help

@BalarkaSen So if $c_1$ and $c_2$ are homotopic curves and $\omega$ is a closed form, then $\int_{c_1}\omega=\int_{c_2}\omega$

Slereah

Well I don't need to solve the equation

I just need to show that all geodesics going through some region will end up in a specific region

0ßelö7

But I have a slightly different problem. I have a surface $\Sigma\subset\Bbb R^3$ such that the vector field $\nabla\times f$ is tangent to the surface. Thus the Stokes integral should vanish if I integrate over the surface

because you dot with the normal

So now if I have two homotopic curves in $\Sigma$, are the vector calculus line integrals the same?

Those would be $\int_{c_1}f\cdot ds=\int_{c_2}f\cdot ds$

Balarka Sen

12:26

@0ßelö7 Not sure I understand. What are you integrating over $c_i$?

$X \cdot \mathbf{n}$ where $X$ is a vector field on $\Bbb R^3$ along $\Sigma$?

0ßelö7

@BalarkaSen The vector calc line integral: $\int_{c_1}f\cdot ds=\int_a^b f\cdot c_1'\,dt$

$f$ is a vector field in all of $\Bbb R^3$

Balarka Sen

$\int_a^b f(c(t)) \cdot c'(t) dt$? That's the same as integrating the form $\omega = \sum f_i dx_i$ over $c$

$f_i = f \cdot e_i$ being the components of $f$

0ßelö7

@BalarkaSen I don't think that $\omega$ selects the tangential part tho

Balarka Sen

I don't follow you. Do we agree that $\int_a^b f(c(t)) \cdot c'(t) dt = \int_c \omega$?

That's definitional, I mean

Pullback $\omega$ along $c : [a, b] \to \Sigma$

0ßelö7

ah yeah there's a pullback

contintue

Balarka Sen

12:34

Eh, actually, $\Sigma$ is not very relevant here, because $\omega$ is a form on $T\Bbb R^3|_{\Sigma}$ (not $T\Sigma$)

Just extend everything (vector fields/forms) to a small tubular neighborhood of $\Sigma$ so you don't have to worry about it at all

0ßelö7

everything is defined everywhere

Balarka Sen

Alright, great, then dump $\Sigma$

ACuriousMind

@EmilioPisanty Yeah, that interaction is weird - in general, the vote threshholds for deletion can lead to unintended (or at least non-obvious) consequences

0ßelö7

@BalarkaSen Well the kicker is that $\nabla \times f\bot \hat n$

if the curves are disjoint, then there's a $\Sigma'\subset\Sigma$ smooth such that $\partial\Sigma'=c_1-c_2$

but I'm concerned with the case when they intersect

ACuriousMind

Also, @EmilioPisanty, it might interest you that you voted "Looks OK" here 3 years ago

Balarka Sen

12:40

@0ßelö7 Oh, hm

So we're claiming $\int_{c_1} \omega = \int_{c_2} \omega$

0ßelö7

@BalarkaSen Context: solid torus in a homogenous flow such that the vortex lines lie on the boundary

Balarka Sen

Does this really have anything to do with the surface? Seems like it suffices to understand that $c_1$ and $c_2$ are homotopic on $\Bbb R^3$

Which they are

I am confused

0ßelö7

@BalarkaSen It would be pretty hard to prove anything then. It's clearly not true if they are merely homotopic in $\Bbb R^3$

One could be along the meridian and the other longitudinal

Balarka Sen

But $\omega$ is a 1-form on $\Bbb R^3$; integrating it along two different loops shouldn't give different results, right?

Where is it that I am messing up?

ACuriousMind

@BalarkaSen Why shouldn't it give different results?

If $\omega$ is closed, the result is the same, namely zero. But if it isn't closed, there's no reason to expect the integrals to be the same that I can see

Balarka Sen

12:45

Ah, thanks.

I automatically assumed $\omega$ is closed.

I haven't had my daily dose of refreshing memes

Slereah

12:57

Here is your daily meme

0ßelö7

@Slereah what happened to you?

Slereah

0ßelö7

@ACuriousMind So do you know how to resolve the doubt?

Slereah

what u mean

Secret

slereah had turned into a geometric entity

ACuriousMind

12:58

@0ßelö7 Which one?

0ßelö7

Slereah

Aren't we all geometric entities

in the end

0ßelö7

@ACuriousMind If the two curves are homotopic, are their integrals equal?

Slereah

I linked my fancy professional website to my StachExchange account

ACuriousMind

@0ßelö7 No

Slereah

12:59

so I'm putting on a less whacky avatar for now

0ßelö7

@ACuriousMind No it's not true?

ACuriousMind

@0ßelö7 Yes

0ßelö7

@ACuriousMind Why not

If they are disjoint it is true

I would hate to think them intersecting makes it untrue

If they intersect in a corner it's also true

ACuriousMind

@0ßelö7 Why?

0ßelö7

@ACuriousMind Use Stokes theorem.

If they are disjoint and homotopic then the homotopy sweeps out a smooth manifold with boundary

And since $(\nabla\times f)\cdot n=0$, Stokes theorem gives the result

ACuriousMind

13:02

@0ßelö7 Yes, but the boundary is not the original curves

Emilio Pisanty

@ACuriousMind ¯\ _(ツ)_/¯

ACuriousMind

If you consider two circles in $\mathbb{R}^2$, the homotopy sweeps out a shape whose boundary is not the disjoint union of the two circles

0ßelö7

ACuriousMind

@0ßelö7 Well, you just drew a case where that happens to be true!

0ßelö7

proof by picture

what's a case when it's not true?

ACuriousMind

13:04

But the circles in the plane show it doesn't always happen

1 min ago, by ACuriousMind

If you consider two circles in $\mathbb{R}^2$, the homotopy sweeps out a shape whose boundary is not the disjoint union of the two circles

Don't pick concentric circles

0ßelö7

I don't know what that means

@ACuriousMind I want these curves to go around the torus hole

ACuriousMind

Pick the unit circle and then translate it by 10 units.

0ßelö7

I'm pretty sure that forces them to be concentric in a sense

ACuriousMind

These two circles are homotopic, but the homotopy sweeps out something whose boundary is homotopoic to a single circle.

0ßelö7

@ACuriousMind I want both curves to go around the large hole

How does your counterexample work then?

ACuriousMind

13:06

Well, you didn't tell me that!

0ßelö7

I think by Jordan curve theorem there's always an "interpolating" annulus

Emilio Pisanty

@0ßelö7 If ACM's example vaguely applies, then the boundary of the area swept out by the homotopy doesn't even need to be a manifold

0ßelö7

I'm arguing in the case when both go around the large hole and they are disjoint, it's definitely a manifold.

ACuriousMind

You just asked me whether integrals over homotopic curves are equal, but the correct statement is that integrals over homologous curves/cycles are equal.

In your torus example, it may well be that all homotopic curves are also homologous

0ßelö7

@ACuriousMind I hate using homology because I want smooth stuff

Emilio Pisanty

13:09

so e.g. take the ellipse homotopy $$\frac{x^2}{(1-\lambda)^2}+\frac{y^2}{(1+\lambda)^2}=1$$ for $\lambda\in[-1/2,1/2]$

0ßelö7

We've had this argument before, and the answer was that it's hard to prove that things are actualy smooth simplices

@EmilioPisanty I am talking about a very specific case

Emilio Pisanty

@0ßelö7 yeah, but doesn't your specific case fit a copy of that?

0ßelö7

@EmilioPisanty are the endmembers of that homotopy disjoint?

is one contained in the other?

Emilio Pisanty

no, they intersect

0ßelö7

right, my question is about the equality of the integrals when they intersect

Emilio Pisanty

13:11

at least if I did it right

0ßelö7

I can prove it in the case when they don't, which @ACuriousMind didn't believe

he now does

so moving on to the case when they do intersect...

ACuriousMind

@0ßelö7 Ah, I just realized the two circles are homologous, since the simplices can be singularly embedded. In any case, you need to make the statement much more carefully than "homotopic loops have the same integral".

0ßelö7

@ACuriousMind I think one can apply Stokes theorem over smoothly triangulable things. But I have no idea how to prove something is smoothly triangulable

ACuriousMind

@0ßelö7 The problem is not with applying Stokes'. The problem is that you cannot in general show that the two loops are the boundary of the surface swept out by the homotopy.

I realized that homology doesn't help because its notion of "boundary" is not the one you use when applying Stokes' to a submanifold.

0ßelö7

@ACuriousMind If the form were closed, that would not be the issue. Via a completely different proof one can show if that if $\omega$ is a closed form and $c_1$ and $c_2$ are homotopic, then $\int_{c_1}\omega=\int_{c_2}\omega$

ACuriousMind

13:15

Well, sure.

Your point being?

0ßelö7

But here I don't have that $\omega$ is closed, but that the projection of (the Hodge star of) its exterior derivative along the normal direction is zero

So I'm looking for something along those lines

Emilio Pisanty

this is what I meant

0ßelö7

@EmilioPisanty I'm not disagreeing with you

Emilio Pisanty

@0ßelö7 ok

0ßelö7

I'm just saying that I can prove what I want for a large class of curves, and I find it a little unbelievable that slighly altering how the curves intersect ruins things

Because even if they intersect, then there will be pieces that have interpolating manifolds

And those things can have corners, and Stokes theorem with corners will apply

But if you have a cusp, game over. Seems dumb

@Slereah arxiv.org/abs/1709.06608

@ACuriousMind The real issue is that the homework needs to be more clear

samjoe

13:28

Hello friends I have a doubt regarding angular momentum.

If a point is accelerated and we have to find angular momentum of a system about that point, do we have to take frame of reference as that point? I mean do we have to apply fictitious forces on the system? Is same true for torque about that system?

Avantgarde

@samjoe First, angular momentum is about an axis, not a point. The same point can lie on very different lines giving different angular momenta, so you need to specify the specific axis. And no, why would you need fictitious forces for angular momentum/torque?

samjoe

@Avantgarde sorry I meant axis. If axis is accelerating perpendicular to itself. So we are just finding the angular momentum of system about that axis but not sitting in the axis frame?

mathvc_

14:02

Does anyone know why in QM one should symmetrize all x and p? I mean instead of X P, one should consider 1/sqrt(2)[XP+PX]?

0ßelö7

@mathvc_ because it's hermitian

Slereah

One should not symmetrize it

But it is convenient in some cases

In the case of the harmonic oscillator it simplifies everything greatly

mathvc_

@0ßelö7 what about X^2 P?

0ßelö7

does $P$ commute with $X^2$?

Slereah

it dudnt

0ßelö7

14:06

this process is called Weyl quantization btw

it's actually not consistent

Slereah

There's a variety of weyl quantizations you can do IIRC

From $xp$ to $px$ and every weight in between

(this corresponds for path integrals to different stochastic measures)

0ßelö7

implying path integrals make sense

Slereah

In non-relativistic QM it's fairly well defined

0ßelö7

can you define it off the top of your head?

Slereah

Something something Gaussian measure

check Jaffe and Glimm I guess???

0ßelö7

14:10

it's done in Hall, which is not quite so obscure

@JohnRennie my god there's an unholy garlic stench coming from my drain

Slereah

Which one is Hall

Is that the one you got for QM done right

0ßelö7

yeah, it's like Reed and Simon lite

Slereah

might be worth getting

0ßelö7

It's a very good book

Slereah

I asked for some physics book for my birthday btw

0ßelö7

14:13

good reference for spectral theory stuff and general unbounded operator stuff

@Slereah I asked for Reed and Simon

@Slereah give me steenrod for my birhdayv

I will send you something

Slereah

I asked for Henneaux and for the big Cauchy GR book

ACuriousMind

@mathvc_ You "should" not. But symmetrization (Weyl ordering) is one way to define a map from classical to quantum observables that's as good as any quantization map is going to get. Where did you get the impression you "should" consider $xp+px$ instead of $xp$? It's fine to consider both in QM - you just have to say to what end you are considering them.

GPhys

@0ßelö7 GR prof "next homework is less calculational I sort of got some complaints"

(wasn't me)

0ßelö7

@Slereah QoGS?

@ACuriousMind ok why do you repeat everything I say

makes me feel bad

@GPhys who the fuck complains about homework difficulty?

John Rennie

@0ßelö7 on the bright side your drain will be clinically free of vampires

ACuriousMind

14:21

@0ßelö7 I didn't say what you said. If anything, what I said is closer to what Slereah said.

@0ßelö7 Also, this is wrong. It's perfectly consistent, just doesn't fulfill all the properties the (non-existent) ideal quantization map should have.

Secret

in Mathematics, 2 mins ago, by Semiclassical

(at least, I think that's true. Is there a Bell-type experiment involving two distinguishable particles?)

ACuriousMind

@0ßelö7 People who honestly thought the homework was a bit too difficult? :P

GPhys

I didn't think it was difficult, but I did think it was a bit on the long side for 4 days to due date

0ßelö7

@ACuriousMind so it's a non-consistent quantization scheme

Semiclassical

oh hey

GPhys

14:24

I imagine that's what they meant

Semiclassical

you stole my question :P

But, yeah.

ACuriousMind

@Semiclassical Nothing in the notion of an entangled state requires the particles be indistinguishable.

Secret

I did not stole it, I just redirect it to h bar

Semiclassical

I didn't think so. But I'm not sure what the setup would be in that case.

ACuriousMind

I think your main issue will be a practical source of such entangled states

Semiclassical

14:26

So in principle you could have entanglement between distinct particles, but in practice it's hard to do that?

ACuriousMind

Absolutely yes to the first part, cautiously yes to the second since I'm not an experimentalist

GPhys

@0ßelö7 we're talking about covariant derivatives

Semiclassical

hmmmm

Secret

I don't even know what reasonable kind of observables can be correlated if the particles are distinguishable...

ACuriousMind

@Secret The same ones as in the indistinguishable case.

0ßelö7

14:27

@GPhys horizontal/vertical bundle?

ACuriousMind

Why do you think the distinguishability changes anything?

GPhys

have not heard those words yet

Secret

Well, how can you make two electrons or two photons distinguishable as you only have spin and polarisation states?

Semiclassical

Take a proton and an electron. They're both spin-1/2, so you can ask about Sz for each one along a certain axis.

Secret

right, that would work

Semiclassical

14:28

But I still would want a practical example to be satisfied.

ACuriousMind

@Secret I don't know what you're asking. We're not talking about "making" two indistinguishable particles distinguishable.

GPhys

@0ßelö7 he accepted my bullshit variational argument (it wasn't wrong it just wasn't quite what he as asking for)

ACuriousMind

@Semiclassical Schrödinger's cat is entangled with the poison vial ;)

Very practical.

Semiclassical

eh, that's kinda cheating in that both of those are macroscopic objects

Secret

Actually, take a condensate of hydrogen atoms, and then ionise them, perhaps some of the observables of the proton and electron can remain entangled?

0ßelö7

14:30

I have a quantum mechanics doubt

ACuriousMind

@Semiclassical Find a very small cat, then.

0ßelö7

what happens if we hook the cat up to an atomic bomb

Secret

I wish I am more knowledgeable at experimental quantum mechanics

Semiclassical

:/

@0ßelö7 then it's not an isolated system anymore

0ßelö7

can we measure the bomb and kill the cat?

GPhys

14:31

I have QFT homework due tomorrow Q___Q

how did I pass the graduate QM I and II I'm so stupid T_T

0ßelö7

@Semiclassical so?

Semiclassical

if having the atom decay produces a signal you can observe from the outside, then I'm dubious that you can consider the box as closed in the first place

0ßelö7

how is anything an isolated system

Secret

arxiv.org/abs/quant-ph/9709052

0ßelö7

there are always neutrinos

Secret

14:31

This paper about hydrogen atom is VERY old

0ßelö7

can a neutrino observe the cat?

or do we have to observe the neutrino first?

GPhys

@0ßelö7 oh god

ACuriousMind

@0ßelö7 No, neutrinos are dog persons

Semiclassical

if i can't observe the neutrino, then it's irrelevant

Secret

cat is made of stuff that can interact with the weak force, thus there should be nonzero probability for a neutrino to interact with a cat

Semiclassical

14:34

interacts != interacts in an observable manner

0ßelö7

@GPhys I took graduate QM too and I don't understand this

what if there's a camera in the box

does the cat die when we look at the screen or when we turn on the camera

or does the food in its belly observe it?

what if you lleave the cat in there for a week

can then you open it

can't you tell that the cat died from thirst 4 days ago?

or is it alive the whole time

Slereah

@0ßelö7 The very same

Semiclassical

@0ßelö7 Here's a variation on that. Suppose you don't open the box, but your friend does.

0ßelö7

@ACuriousMind please

I mean, what if you hook up an EKG to the cat

Semiclassical

Is your friend then entangled with the cat until you ask them about it?

Secret

14:36

in Mathematics, 17 mins ago, by Phase

Is it valid to say that in Hyperreals and surreals, $1 - 0.9999.... = \epsilon = \frac{1}{\omega}$ ?

hyperreal expert needed

0ßelö7

then the EKG measures when the cat dies and can have a timer on it

Secret

slereah=hyeprrreal expert

ACuriousMind

@0ßelö7 I don't discuss interpretations :P

John Rennie

@ACuriousMind That's the (first) law

Semiclassical

@0ßelö7 you could boil it down even further and just say "suppose I leave the box open and I have a pair of binoculars"

0ßelö7

14:37

@ACuriousMind You're such a troll. You can't bring up Schroedinger's cat in a physics chatroom and not expect it to cause an epistemological crisis or two.

I need answers

Semiclassical

well, to be fair, I'm the one who brought up Bell-type experiments

0ßelö7

@Semiclassical yeah, how does this stuff work?

it's not even an interpretation, it's literally asking what would happen

Semiclassical

well, I think the point is no different than the following. Suppose you set up a double slit experiment with electrons and you observe an interference pattern

Suppose you then have the idea to put some kind of device at one of the slits which would register whether or not the electron passed through that slit

The thing is, if you do that, then you won't see the interference pattern anymore.

0ßelö7

@Semiclassical Yeah, but "device" is too vague

And it doesn't really explain what "measuring" means

Anonymous

@Semiclassical What do you classify as a device?

0ßelö7

14:40

Can an errant neutrino measure the cat?

If I measure the neutrino afterwards and notice that it has a dead cat smell, can I conclude the cat has died?

Semiclassical

It's as clear as it needs to be, really. If you have a device which interacts sufficiently with the system as to be able to measure which slit it goes through, then the interference pattern won't be there.

Slereah

@Secret It is not valid to say.

0ßelö7

"interacts sufficiently"

@Secret what is $0.99999$ if not $1$

Anonymous

^

Semiclassical

I suspect Wheeler's delayed choice experiment is relevant here

Anonymous

14:43

Even I was confused with so many books mentioning stuff like "interacts sufficiently" and "device" and some other vague terms...

Slereah

A real number $x$ is represented by the hyperreal sequence $(x,x,x,x,x,...)$

John Rennie

@Semiclassical the electron starts out as (approximately) a momentum eigenstate. Your measurement converts it to (approximately) a position eigenstate.

Semiclassical

Right.

the joys of complementarity :P

0ßelö7

I want to talk about cats, not electrons

Slereah

So $$0.999... = \sum 9 \times 10^{-n} = \sum (9 \times 10^{-n}, 9 \times 10^{-n}, 9 \times 10^{-n}, ...)$$

John Rennie

14:44

@Semiclassical So Fourier transform the position eigenstate and you'll find it doesn't have a well defined de Broglie wavelength.

Slereah

Which will itself be 1

Although of course it depends what you mean by "0.9999"

If it's the hyperreal sequence $(0.9, 0.99, 0.999, ...)$... I still think it's 1

$$1 - 0.999... = (0.1, 0.01, 0.001, ...)$$

Now to check whether it's 0 or not...

Semiclassical

@JohnRennie sure. My point is really that there's no need to be too detailed about how the choice of slit is measured.

Slereah

Wait actually I think it's > 0

Since every element is

Semiclassical

Once you commit to measuring the choice of slit, you're no longer able to observe the interference pattern.

Slereah

but of course it's not really a standard definition of the decimal expansion

Secret

14:51

hmmm... if that hyperreal sequence is nonzero, what value would it be...

Slereah

An infinitesimal number.

John Rennie

@Semiclassical I'm not sure that's true.

Secret

Interesting, so $1-0.\dot{9}$ has an infintesimal part

John Rennie

You can't convert the momentum eigenstate to a position eigenstate because that would take infinite energy. You convert it to a more localised state.

0ßelö7

@JohnRennie infinite energy?

John Rennie

14:53

Depending on the measurement you will get different degrees of localisation.

Semiclassical

sure, but even in the double slit you don't insist that the two slits are infinitely narrow

you just insist that the slits are small compared to the slit separation

Slereah

@JohnRennie Also because momentum eigenstates aren't states

John Rennie

@Slereah details, details :-)

Anonymous

In page 7 of this document I don't understand their explanation for why 100% of the electrons come out to be "white" at the end. They seem to say something along the lines of "once the hardness is measured it is in a superposition of being white and black". That seems very vague.

Anonymous

ACuriousMind

15:06

@Blue Are they using color and hardness as stupid nicknames for spin along different directions?

Anonymous

"To build a better definition of superposition than “we have no clue what is going on” requires
a new language. That is the language of quantum mechanics, whose underpinnings will be
the topic of 8.04. No matter how we describe it though, superposition is really weird, but
true."

Anonymous

I'm not very sure. They mention this at the beginning: "Let us focus on 2 properties of electrons. For the sake of this discussion, let us call them color
and hardness. An empirical fact is that the only observable colors are individually “black”
and “white”, while the only observable hardnesses are individually “hard” and “soft”. There
are no other observable values that these properties can take, as no one has ever seen any
other such value. What I mean is that it is possible to build a box that measures color or

Anonymous

"2 properties"

ACuriousMind

Ah, yes, they are.

Anonymous

Aha. :P

Anonymous

15:09

Could you explain what's happening in that Figure 10 ?

Anonymous

Perhaps I should not follow MIT OCW for QM...lol

ACuriousMind

I personally think it is very silly to use these analogies, but whatever

In any case, what they're saying is that the "combine" operation does something magical - the output of the experiment in figure 10 is purely white, where a reasoning that follows the hard and soft electrons along individual paths would predict only 50% to be white

0ßelö7

@Blue learn QM from Reed and Simon

ACuriousMind

The quantum mechanical explanation is that quantum objects do not follow paths, so if you split by hardness and then combine again without ever actually measuring the hardness in-between, this is a do-nothing operation.

Anonymous

@ACuriousMind Could you clarify the meaning of "quantum objects do not follow paths" ?

Anonymous

15:15

I don't think they teleport from one point to another :P

ACuriousMind

@Blue Not really, I'm afraid. They just do not "move" in the sense a classical particle does. In the double-slit you can see this as "spreading as waves", and for such beam splitters, the electron that reaches the hardness box then moves as the superposition of two states - 50% hard electron on the left path, 50% soft electron on the right path. It "takes both paths" until it is measured, only that that doesn't really mean anything

The "location" of a quantum object simply need not be well-defined when its location is not being measured

And I see that the pdf you linked puts it better - the quantum object has a mode of motion that is unlike anything we can visualize.

0ßelö7

@ACuriousMind we can visualize diffusion

ACuriousMind

@0ßelö7 And how is a photon passing a beam splitter "diffusing"?

0ßelö7

@ACuriousMind it's a wave packet

Anonymous

@ACuriousMind That doesn't make intuitive sense. But yes, assuming that "it takes both paths until measured" does explain the phenomenon. That clears it a bit. Thanks!

0ßelö7

15:22

It's a spread out blob, we can visualize that

ACuriousMind

@Blue It's not supposed to make intuitive sense. Our intution is gained in the classical regime.

2

Anonymous

@ACuriousMind One more thing. Is "spin of an electron along two different directions", "spin" in the classical sense...or just eigenvalues we get from solving Schrodinger's equation ?

ACuriousMind

@Blue It is not spin in the sense anything is spinning classically, but I do not understand what you mean by the alternative.

Anonymous

Spin (physics)

In quantum mechanics and particle physics, spin is an intrinsic form of angular momentum carried by elementary particles, composite particles (hadrons), and atomic nuclei. Spin is one of two types of angular momentum in quantum mechanics, the other being orbital angular momentum. The orbital angular momentum operator is the quantum-mechanical counterpart to the classical angular momentum of orbital revolution: it arises when a particle executes a rotating or twisting trajectory (such as when an electron orbits a nucleus). The existence of spin angular momentum is inferred from experiments, such...

ACuriousMind

@Blue What is that supposed to tell me?

Anonymous

15:34

I'm not sure if spin of electron is calculated from Schrodinger's equation...

Anonymous

How is spin measured or calculated ?

ACuriousMind

Spin has nothing to do with the Schrödinger equation.

vzn

@Blue fyi that formulation looks like directly from a book by DZ Albert. hup.harvard.edu/catalog.php?isbn=9780674741133 ... hint: hes a professor of philosophy, lol

ACuriousMind

Or, well, you can of course have SEs in which the spin operator plays a role

But you don't need to talk about time evolution at all to talk about spin.

@Blue The classic spin measurement is a Stern-Gerlach experiment.

Anonymous

@ACuriousMind Ooo...thanks. I'll read it

Anonymous

15:37

@vzn I found that document on MIT OCW though. The lecturer Allan Adams uses that formulation in the first lecture of the series

vzn

@Blue think he should give DZ Albert some credit, suspect it originates with him

Anonymous

@vzn Oh, he does mention that book's name in the video. I remember...

Anonymous

"Quantum Mechanics and Experience"

vzn

@Blue cool. yes Albert has formulated a particularly counterintuitive pov. the bohrian "complementarity" rears its head. reminds me also of "duality".

Anonymous

@vzn Well, he seems to have a PhD in theoretical physics (en.wikipedia.org/wiki/David_Albert)

Anonymous

15:40

@vzn I see. Interesting

vzn

@Blue a rare scientist comfortable with bridging the gap between physics + philosophy, to some degree, in contrast to some others around here :| ... (have/ read his book yrs ago)

John Rennie

@vzn are you trying to be provocative?

0ßelö7

@Blue spin is from the Dirac equation

vzn

@JohnRennie lol, science itself is provocative sometimes wouldnt you say? :)

John Rennie

Quoting other peoples posts in an attempt to show they have contradicted themselves is just provocative not science.

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Anonymous

15:46

@0ßelö7 Thanks. I'll see

John Rennie

→ 1 message moved to trash

0ßelö7

Can you please stop deleting messages?

Anonymous

@0ßelö7 He just "transported" them ;)

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