The h Bar

Sir Cumference

17:08

It is a composite of a regular photo and a UV photo where the auroras only showed up in UV and were pasted onto the regular photo.

@0celo7 What do you mean by "applications?"

It's pretty darn cool

user54412

17:26

@JohnRennie yes

user54412

There's no matter in the simulations after all.

user54412

Also I guess you could argue some of the energy is kinetic energy too.

user54412

In any event, the Einstein equations are strictly hyperbolic linear 2nd order PDEs, so oscillating a charge generically produces waves, just like in Maxwell.

Slereah

>linear

>einstein field equations

user54412

to leading order

user54412

17:29

that's all that matters

user54412

we can say quasilinear to be pedantic

John Rennie

Thanks Chris.

For Schwarzschild or Kerr black holes presumably the issue doesn't arise since they contain no matter anyway. The energy carried away by the GWs just comes from the gravitational fields.

Slereah

yeah

John Rennie

However in the real world the objects aren't Schwarzschild or Kerr black holes and they do contain matter. If the claim were that the matter originally present is being somehow converted in GW energy that would be worrying as there is no obvious mechanism for it.

Slereah

dunno about something with matter fields, though

As said, there's no conservation of energy

and by definition, gravitational waves spacetimes aren't static

user54412

17:35

@JohnRennie By "mechanism" are you looking for something local, like how matter might emit light atom-by-atom?

Slereah

Well

user54412

It's really more the system as a whole that produces the waves.

Slereah

If you want to go all the way to quantum

It's pretty trivial

There are totally processes like $e^- \to e^- + g$

Gravitational Brehmstrallung

EVEN $e^- + e^+ \to g$

But that's pretty bloody rare

John Rennie

My point is that in a real object no matter has crossed a true horizon. All the matter is still in principle causally connected to the rest of the universe. If we're going to say that matter turns in GWs then I think we need a mechanism.

user54412

Who was saying that?

user54412

17:39

Also, waves aren't limited to black holes. Recall orbiting neutron stars.

John Rennie

@ChrisWhite that was what I was asking. Is anyone saying that, or are they saying that the radiated energy comes from the gravitational energy?

Slereah

Well

Any quadrupolar momentum, really

Waving your hand produces gravitational waves

John Rennie

And you're telling me that the energy comes from the gravitational energy, which I'm content with.

Slereah

Well

I'm not sure this is true

Since there is an exchange between potential gravitational energy and other forms of energy in non-static spacetimes

I think you could get gravitational waves coming from matter sources of energy

John Rennie

@Slereah Do we expect lepton and baryon number to be conserved? Remember, in a real object no matter has crossed a true horizon. In principle I could count the number of baryons that went in then pull them back out and count them again.

user54412

17:44

Well, set up a circuit with a time-varying quadrupole moment in stress-energy. It should produce gravitational radiation. And a whole lot EM too.

user54412

And from the far-field perspective, as the waves pass the monopole moment of the system will be seen to decrease -- it will be less massive.

John Rennie

@ChrisWhite I have no problem with that, it's essentially the same as a hydrogen atom weighing less than a separated electron and proton (by 13eV).

Slereah

My guess would be

Look into pp-wave spacetimes

I'm pretty sure there's a solution that is relevant to this

Danu

@Danu Just let me at least get a little advice before closing it. — kfnvkfrk 2 mins ago

I feel bad because the guy doesn't actually have access to the h bar. @JohnRennie can you grant him access?

user54412

Random musing: So from a QG point of view, our classical GW emitters are just systems with very complex internal graviton interactions that lead to some gravitons being sent to infinity. In which case we can neglect SM-graviton interactions. What sort of setup would have SM-graviton interactions be more important than graviton-graviton (regardless of what particles are emitted in the end)?

Slereah

17:56

Errrr

user54412

@Danu I thought we concluded only diamonds have that power?

Slereah

Isn't the basic gravitational wave emission system basically the classical equivalent of gravitational Brehmstrallung?

Danu

@ChrisWhite Ah.

Then I should be able to?

Slereah

I'd say matter-graviton interaction is more important

The basic graviton vertex for electrons is $\approx h$, while the lowest order graviton-graviton one is $\approx h^3$

user54412

@Danu Maybe? I think they need to visit chat.SE in order to have a chat account. Then try making a room for them?

Danu

18:01

Yeah... Oh well

user54412

@Slereah I see what you're saying... Still it feels like we're only seeing uninteresting reactions.

Keep these mind

0

Q: (Infinite hat)-guessing problem

$2$ men are playing a game: they are wearing countably infinitely many hats on their heads. The hats are either black or white with probability $\frac 12$. They see the other's man hats but cannot see the hats on their own head. Without communicating with each other, they simultaneously point ...

Slereah

What qualifies as interesting

user54412

Clearly only kalashnikov wormholes or whatever it is you're constructing :p

Slereah

Those are p. interesting

0celo7

18:10

AK-47 is my f****king address. - Lil Wayne.

Hmm

How to censor inside of italics

Bass

@ChrisWhite But the internal graviton interactions would be mostly virtual particles, right? Similar to the case of an excited electron orbiting a proton. The electron and proton exchange virtual particles (which produce the field that lets them attract each other), and at some point in time, the electron radiates a photon and falls back to the lower energy level. Is that comparable to the case of two orbiting neutron stars?

user54412

I suppose.

user54412

Neutron stars are so far apart those gravitons must be extraordinarily close to on-shell. I think.

yuggib

18:25

my future AMA session looks quite desolate for the moment...only two questions asked in comments, and no fully-developed question T__T

Slereah

what is a quantum yuggib

are you a quantum

yuggib

that is latin...

Aaron Abraham

Alright, novice here, can SOMEONE [anyone] give me a brief idea about Noether's Theorem? [I spent quite a while framing that into a question at Physics Stack Exchange, only to get it voted down....so kinda sore about that......also thank you @JohnRennie for upvoting me here!]

Slereah

Continuous symmetry = conserved charge

Brief enough?

Aaron Abraham

Um....

That was a little TOO brief...

0celo7

18:29

given a continuous symmetry, there is a conserved charge

not sure if it works the other way around?

Slereah

It doesn't.

yuggib

continuous symmetry of the action $\Rightarrow$ there exists a conserved in time quantity (with a specific form, depending on the dynamics and relative symmetry)

Aaron Abraham

Could you explain it to the layman [ a 'layman' who possess a degree of knowledge about concepts in physics]

Slereah

Probably not

yuggib

do you know what the action is? (it is the integral of the Lagrangian with respect to time)

ACuriousMind

18:30

@0celo7 It does, at least in the Hamiltonian case where the charge simply generates the symmetry, see this question

0celo7

@AaronAbraham do you know what a Lagrangian is

freaking Qmechanic

Is he a classical mechanist?

Is Qmechanic Arnold?

Aaron Abraham

@Ocelo7 Straight and simple....NO

0celo7

@ACuriousMind how is the game

John Rennie

@Danu the best I can do is upvote the question, which I've done. Three more upvotes will give enough rep to join the chat.

Aaron Abraham

I've only ever dealt with classical so far...

yuggib

18:31

then you have to wait a fair bit

ACuriousMind

@0celo7 Nice so far

Aaron Abraham

But say i'm interested.......

Slereah

I mean

The best we can do is

0celo7

@AaronAbraham read

Slereah

Provide counterexamples

0celo7

18:32

amazon.com/…

Slereah

Show that the quantity isn't conserved without the symmetry

Aaron Abraham

I'm all ears [or is it 'eyes'???]

Slereah

well

Aaron Abraham

Would Goldstein's suffice......for Lagrangians that is....

user54412

yes

Slereah

18:34

Consider a case where we can turn off gravity

Aaron Abraham

Alright...

yuggib

@AaronAbraham take a look at landau-lifshitz 1...the best quality/length ratio ;-)

Slereah

Gravity is turned off before $t = 0$

An object at rest in that circumstance will have energy 0

Aaron Abraham

I'm with you..

Slereah

Once it turns on, it gains gravitational potential energy

Aaron Abraham

18:34

Wait...

Slereah

You have just created energy

Aaron Abraham

Ok...

Slereah

bam

0celo7

@Slereah that's illegal?

Slereah

You'll never catch me copper

Aaron Abraham

18:35

Yeah...but about 'turning off' gravity.......

yuggib

@0celo7 only in 25 states

Aaron Abraham

..............

kevin Tah N.

@yuggib nice one

0celo7

@yuggib just like marrying a sister

user54412

@AaronAbraham You should be at the level of Lagrangians to understand the proof of Noether. That said, it really is just the statement that if your physics is invariant under a (continuous) change of your description of it (like shifting all your coordinates to the left, or rotating everything around the y-axis), then there is a conserved quantity.

yuggib

18:36

@0celo7 exactly

Aaron Abraham

@ChrisWhite How long do you reckon it'd take me?

0celo7

a day?

Aaron Abraham

I'm still 17

0celo7

Age is no excuse

Aaron Abraham

And I've never moved beyond classical mech...

before

user54412

18:37

And really, the only examples ever used are time translation -> energy conservation; spatial translation -> momentum conservation; spatial rotations -> angular momentum conservation.

0celo7

Noether is classical mechanics

Aaron Abraham

As in 'directly related to Newtonian mechanics'...

user54412

Yeah, usually Lagrangians (and Hamiltonians) are considered just as classical as Newton, since they are all equivalent and also all very old.

0celo7

@ACuriousMind :

Aaron Abraham

Well, when does one normally begin studying Lagrangians?

0celo7

18:40

16

user54412

@AaronAbraham sophomore or junior year of undergrad

user54412

ignore the troll

0celo7

@ChrisWhite I learned it junior year of high school, I was 16.

Aaron Abraham

Interesting....which syllabi Herr Ocelo?

0celo7

@AaronAbraham Hmm, likely Zee's Quantum Field Theory.

I don't recall if it's in his GR book.

It might also be in Shankar.

Aaron Abraham

18:41

Ocelo....I said 'normally', I doubt that would fit your description.......no offence intended...

[seriously]

user54412

I don't think a QM book would have much Lagrangians...

0celo7

Are you saying I'm abnormal?

Aaron Abraham

NO

God....

Quit jumping the gun...

0celo7

@ChrisWhite Shankar develops the Lagrangian formalism to motivate Hamiltonian stuff.

And any QFT book has Noether's theorem.

Aaron Abraham

@ChrisWhite Have you heard of the IIT-JEE ?

user54412

18:44

@Aaron Goldstein is the most famous, but there are plenty of books. Look for "classical mechanics" or "analytical mechanics." You might try Hand & Finch, or Marion & Thornton. They all define Lagrangians and prove Noether in the first couple chapters. In any event, the mathematical prereq is being comfortable with calculus.

user54412

@AaronAbraham yes, but I can't say I have any idea what level it's at.

0celo7

@AaronAbraham what is that

Arnold is the go-to for people in my field

"my field"

user218912

what is your field?

Aaron Abraham

Well, try looking up some sample papers [they don't release the actual question paper], it's a fairly tough nut to crack....

user54412

@0celo7 Funny, I thought fields couldn't have just one element.

John Duffield

18:48

@JohnRennie : for ordinary matter, gravity converts potential energy into kinetic energy. That potential energy is mass-energy, in the matter, not somewhere else. When the kinetic energy is radiated away you are left with a mass deficit. You are not left with one object that has the same mass as the original two objects, and some magical mysterious unexplained reduction in gravitational field energy.

yuggib

@ChrisWhite indeed they can't...rings maybe ;-)

ACuriousMind

@ChrisWhite They can if you allow for $F_\text{un}$

0celo7

@ChrisWhite what

user218912

fun?

yuggib

@ACuriousMind allow? you can allow for anything, it does not mean it exists

Aaron Abraham

18:50

@ChrisWhite Try looking for a pdf version of the book 'I.E.Irodov's problems in general physics' it has questions of the level the normally appears for the IIT-JEE. I intend on writing that particular test, and I wanted to know if whether knowledge of Noether's theorem would in anyway prove useful [ it isn't actually in the syllabus though]....Hence I'm trying to figure out what i'm getting myself into.....

user218912

that's like high school level dude.

Aaron Abraham

O.O

ACuriousMind

@AaronAbraham Noether's theorem has nothing to do with solving typical physics exercises.

Aaron Abraham

But would it help knowing it in any way?

user218912

no.

user54412

18:52

Lagrangians on the other hand are often very useful in solving exercises.

Aaron Abraham

[also that exam doesn't really ask 'typical' physics questions...]

user218912

uhhh

user218912

it obviously isn't as difficult as the physics olympiad.

user218912

it's just a lousy exam that's high school level.

user218912

probably equivalent to first year physics.

Aaron Abraham

18:53

You might SERIOUSLY want to reconsider that statement...

At least after seeing a few sample papaers...

*papers

user218912

when I was in grade 11 I did work through some of them.

user218912

and I did do problems in irodov.

Aaron Abraham

Mains or Advanced? I found Mains fairly easy....

user218912

I have no idea.

0celo7

jesus

user54412

18:55

Just looked at Irodov. All the problems I saw can be done just as quickly with Newton as any other method.

0celo7

I couldn't do these

Aaron Abraham

Irodov's problems are graded....you'll find tougher question towards the end of each chapter...

user218912

I'm bad at low level physics because I am unable to visualize the problem.

user218912

[I'm bad at advanced physics too, but for other reasons]

0celo7

at least you're not bad at all physics

user218912

18:59

I'm only lazy and have low experience with physics because I skim when I read.

user218912

I'm sure when I do my undergrad I'll be fine.

0celo7

I can't develop an intuition for physics

Aaron Abraham

Have a look at this year's paper.... admission.aglasem.com/…

0celo7

I couldn't answer Q.3 after a semester of thermodynamics

I don't even know what Q.4 is

user218912

@0celo7 didn't you do first year physics?

0celo7

19:02

yes

I'm bad at physics

user218912

I disagree but whatever.

user218912

you're good at real physics.

user218912

this kind of physics is lame.

Aaron Abraham

You're expected to answer the paper in 3 hrs or so [includes chemistry and math of a similar level], also you aren't allowed ANY sort of reference material in the examination hall and NO calculators..

0celo7

What physics am I actually good at?

I despise most physics

user218912

19:03

ok dude why are you so obsessed with this exam?

user218912

@0celo7 gr duh

0celo7

I'm pretty sure I don't know any GR

I'm not even sure what's up with crossing the event horizon

Aaron Abraham

To prepare me for another exam :P

0celo7

I just understand the math

Secret

@yuggib @acuriousmind The Hamiltonian of 2 spins is described by $H=\frac{\omega}{2}\sigma \cdot \tau$. Suppose I have the following partially entangled state $\lvert @\rangle=\sqrt{0.7}\lvert ud\rangle-\sqrt{0.3}\lvert du\rangle$.

We know that $H\lvert @\rangle=\sqrt{0.3}\lvert ud\rangle-\sqrt{0.7}\lvert du\rangle$

The density matrices of the system before and after time evolution by H are

$$\rho_{AB}=\begin{pmatrix}0 & 0 & 0 & 0\\0 & 0.7 & -\sqrt{0.21} & 0 \\ 0 & -\sqrt{0.21} & 0.3 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$$

Aaron Abraham

19:05

The physics questions asked in that exam are pretty much as tough as they can get for a high-schooler.....so if I reach that bar........i'll feel gratified...

[Plus I wanna write that exam too...]

Normally anyone who's made it to the first thousand ranks or so are revered as gods....

0celo7

I would probably be a street urchin if I were indian

I don't get competition

Aaron Abraham

Those who top the exam normally score in an excess of 325

marks

It's out of 360 marks...

And there's negative marking for wrong answers...

ACuriousMind

@Secret 1. I have no idea what that "general form" is supposed to mean. 2. That Hamiltonian doesn't tend towards anything (it doesn't depend on distance), but at some point the system won't be described by it anymore. 3. I don't understand the question. When you do a measurement on the state, you change the entire state, not just "Bob's part". The no-communication theorem however says that you cannot actually use that to transmit information.

user54412

@3750 I've never met an Indian who wasn't.

user54412

I suppose that means there are a billion people at least who share this obsession.

user218912

19:12

@0celo7 I figured out how to answer my question by defining a rank-2 pseudo-tensor and then using the 4 dimensional $\epsilon$.

user218912

Is there any other way to do it?

user218912

@ACuriousMind do you know how to answer my question without working backward because 0celo7 said you need to work backwards in order to do it.

Aaron Abraham

This obsession....looking at how many people wrote it......would be found in the 1.2 million people who wrote the test THIS year..0.

user218912

@DavidZ wait why was my question put on hold? I asked about a specific physics concept and showed my work.

Aaron Abraham

Well, auf wiedersehen.....

ACuriousMind

19:14

@3750 What question?

user218912

@ACuriousMind Maxwell's equations in covariant form.

ACuriousMind

Oh, that question

@3750 "I don't know how to proceed" is not a specific physics question.

user218912

hmm

0celo7

@3750 I was going to suggest that, I think

ACuriousMind

@3750 Why would you forbid "working backwards"? If you only use equivalences, then it doesn't matter from which part you start.

0celo7

19:17

My solution would have been to use the canonical isomorphism $C^\infty(\Bbb R^3)\cong \Omega^3(\Bbb R^3)$.

Because you get something like $\epsilon^{ijk}\partial_{[i}F_{jk]}=0$

user218912

@0celo7 0.o

0celo7

Which as equivalent to $\partial_{[i}F_{jk]}=0$

Slereah

What is omega three

0celo7

3-forms

Slereah

o

So 0 forms to 3 forms

0celo7

19:18

Yes

In particular, $\partial_{[i}F_{jk]}=f(x)\epsilon_{ijk}$ for some function $f$

user218912

@ACuriousMind well are my indices correct? I remember 0celo7 saying the $i$ was wrong in the combined equation.

0celo7

Then use $\epsilon^{ijk}\epsilon_{ijk}=2$ (or something)

@ACuriousMind ACM!

We've been over this

Sometimes people want to work a problem as the author intended

Secret

1 and 3. I am not sure how to describe that without referring to Susskind (and I think his proof in p.223-226 might be a specific case of non communication theorem, but he is trying to show that any time evolution that happens at bob's will not change alice's density matrix (which I am not sure how measurement will be described here, because as you mention it will change the entire state as the apparatus became entangled with the composite system, thus obviously will not leave alice's state alone))

0celo7

@3750 it's given by the Hodge star.

So, $\star \partial_{[i}F_{jk]}=f(x)$

Up to a factor.

user218912

I know what the hodge star is but I have no idea what you're doing.

0celo7

19:21

@ACuriousMind can you explain

user218912

@ACuriousMind also can you please answer my question (updated) please?

ACuriousMind

@0celo7 What?

@3750 I didn't check them

0celo7

@ACuriousMind :(

user218912

@ACuriousMind please check them?

Secret

1 and 3. I am not sure how to describe that without referring to Susskind (and I think his proof in p.223-226 might be a specific case of non communication theorem, but he is trying to show that any time evolution that happens at bob's will not change alice's density matrix (which I am not sure how measurement will be described here, because as you mention it will change the entire state as the apparatus became entangled with the composite system, thus obviously will not leave alice's state alone))

ACuriousMind

19:24

...why did you post that again?

Secret

I have an internet hiccup thus the msg duplicated

sorry

just ignore that duplicate

ACuriousMind

@3750 In your second-to-last equation, the k from before became an i.

0celo7

@ACuriousMind I don't think he understands why $\epsilon^{ijk}f_{ijk}=0$ where $f\in\Omega^3(\Bbb R^3)$ implies $f_{ijk}\equiv 0$.

$\Bbb R^3$ is being thought of as a smooth manifold ofc

user218912

@ACuriousMind $\epsilon^{ijk}\partial_j E_k + \partial_0B^i \neq \epsilon^{ijk}\partial_k F_{0i} + \epsilon^{ijk}\partial_0 F_{jk}$

user218912

?

ACuriousMind

19:29

@3750 Yeah...why would the k suddenly shift to the derivative?

user218912

so it should be

user218912

$\epsilon^{ijk}\partial_j E_k + \partial_0B^i = \epsilon^{ijk}\partial_j F_{0i} + \epsilon^{ijk}\partial_0 F_{jk}$?

user218912

or is $E_k$ not equal to $F_{0i}$

user218912

should it be

ACuriousMind

@3750 Why would the index change?

user218912

19:31

$E_k = F_{0k}$?

ACuriousMind

You have $E_k = F_{0k}$.

user218912

ok

user218912

wow

user218912

it works now I think.

user218912

@DavidZ can you remove the hold from my question so I can answer it?

user54412

19:36

This is a national holiday, so of course my campus visualization machine is oversubscribed by 50%, almost entirely thanks to my department.

0celo7

@ChrisWhite Hmm?

Just run it on your laptop

user54412

::goes off to buy extra hard drive and LN2::

Secret

@acuriousmind I have the following state $\lvert @\rangle=\sqrt{\frac{1}{3}}\lvert ud\rangle - \sqrt{\frac{2}{3}}\lvert du\rangle$

The density matrix of this state is calculated as follows

$$\rho_{AB}=\begin{pmatrix}0 & 0 & 0 & 0\\0 & \frac{1}{3} & -\sqrt{\frac{2}{9}} & 0 \\ 0 & -\sqrt{\frac{2}{9}} & \frac{2}{3} & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$$

I have calculated the probabilities for the outcome $P(uu)=0$, $P(ud)=\frac{1}{3}$ and $P(du)=\frac{2}{3}$, which is consistent to the description of the state vector.

ACuriousMind

@Secret Your calculation of $P(rl)$ makes no sense as written. Why would $P(rl) = \langle rr\vert rr\rangle$, and how is the inner product of the state rr with itself not 1? Regardless, I don't understand the question - if you compute a probability, the "physical interpretation" of such a probability is that it's a probability! I don't get what you mean by "physical interpretation", I guess.

David Z

19:53

@3750 the first part of your question is basically asking what you did wrong, which is not okay, but the second part is fine so I guess I can reopen it. I would advise you to edit out that first part of the question where you just ask whether your math is correct.

user218912

@DavidZ thanks

user218912

fixed

Balarka Sen

@0celo7 Meh, IMO, solving complicated/laborious/made-up physics/math question is not physics/math.

I can write down a complicated diophantine equation and ask for it's number of solutions over Z[2^1/3+2^2/3] or something. That's not number theory.

Secret

@acuriousmind (fixing some careless mistakes). I am suspecting that the probabilities $P(uu)$, $P(ud)$, $P(du)$ is sort of telling me that suppose there are 99 replicates of the state $\lvert @\rangle$ is being prepared and the measured by bob and/or alice (and then they later check their results together),

then there will be 0 times bob and alice will both get their spins pointing up, but there will be 33 of the time Alice will measure spin up and bob will measure spin down, and 66 times Alice will measure spin down and bob will measure spin up

Balarka Sen

I do not find thinking about such problems contributing to my intuition, or giving new insight.

(which is precisely why I do the math I do: to gain new insight!)

ACuriousMind

20:02

@Secret They do not give 1, you're just computing them wrong.

Balarka Sen

Problem solving skills are good but should not be pushed to the extreme so that it becomes more important than appreciating the beauty of math (I think the case for physics is analogous).

0celo7

I would comment on the beauty of physics but I think I'm enough of a pariah around these parts.

0celo7

20:16

@ACuriousMind How's the game?

Danu

@JohnRennie I don't think that's a good reason to upvote a post.

user218912

20:34

@0celo7 I answered my question is it right?

0celo7

I would not be happy with it.

user218912

why CAN you divide out $\epsilon^{ijk}$?

user218912

because I know it is legal

user218912

how do I otherwise get rid of it

user218912

oh man I got rek'd

Secret

20:46

@Acuriousmind
Ok using the density matrix
$$\rho_{AB}=\begin{pmatrix}0 & 0 & 0 & 0\\0 & \frac{1}{3} & -\sqrt{\frac{2}{9}} & 0 \\ 0 & -\sqrt{\frac{2}{9}} & \frac{2}{3} & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$$
After recalculating (I overlooked that the basis elements such as ud of Alice and ud of bob are not necessary orthogonal as the density matrix shown) I got the following:
$$P(rr)=\frac{1}{2}(0+\langle uu|uu\rangle+\langle ud|ud\rangle+\langle ud|du\rangle+\langle du|ud\rangle)=\frac{1}{2}(0+\frac{1}{3}+\frac{2}{3}-\frac{\sqrt{2}}{3}-\frac{\sqrt{2}}{3})=\frac{1}{2}-\frac{\sqrt{2}}{3}$$

ACuriousMind

@Secret Yes, that's what a probability is. Your way of computing them still doesn't make any sense to me, though.

0celo7

@3750 lol

You should have listened to me

Secret

@Acuriousmind We know that for the singlet entangled state, the probability that when Alice measured spin up and bob will also measured spin up is 1, meaning that whenever Alice get spin up, she knew that Bob will also get spin up, likewise, when Alice measure spin down,she also knew that when she compared results with bob later when they meet again, he will also get spin down 100% of the time, despite that for them individually they have equal probability of getting spin up or down each

(hence why when they only look at their own results without comparing with their partners, they will on…

For the state $\lvert @\rangle$ looking at the probabilities of the spins that can be measured for both parties in the x and z directions, we knew form our previous discussion that for the z component Alice and Bob will get both spins measured in the same direction 0% of the time (since P(uu)=P(dd)=0)

but they will always measure their spins being opposite always (Since the remaining two possibilities for spin measured in the z direction that are eigenstates of both spin operators $\sigma_z$ and $\tau_z$ have probabilities described by P(ud) and P(du)), and they sum to one). In particular,…

@Acuriousmind
Now for the spins in the x direction, based on what we saw from the probabilities $P(ll)$, $P(rl)$, $P(lr)$ and $P(rl)$, we should expect that the result that alice and bob will measure both spin left, both spin right, alice spin left and bob spin right, and alice spin right and bob spin left to be given by the aforementioend probabilities.

So in their paper with 100 records, shouldn't all these correlated results of the spins in the left and right direction show this pattern as indicated by the aforementioned probabilities? (e.g. if alice and bob got spin left 28.6% of the t…

In short, based on the probabilities calculated, do we expect the correlated (whether spin measured to be opposite or same direction) results for each spin component to display probabilities that corresponds to what is calculated from the density matrix?

Or in other words, do we expect the results of alice and bob, when considered in pairs, give a probablity distribution as descriebd by the calculations, thus showing that entanglement result in the measurement results to be correlated or anticorrelated in a probabilitistic fasion?

ACuriousMind

21:13

@Secret I'm sorry, but what exactly about the notion of a probability confuses you? A probability of p% for some event means that in p of hundred trials said event will occur.

I don't get why you seem to think there is anything special about this Alice/Bob situation - a probability is just that, a probability.

Secret

I am wondering whether those probabilities are reflecting the statistics on how often alice and bob's results will be correlated (spin parallel) or anticorrelated (spin opposite), or that we cannot interpret it that way and simply treat them as probabilities of them getting the states uu, ud du etc.

because if we cannot say that the probabilities are reflecting how oftne their results correlated or anticorrelated, then I have no idea how to interpret e.g. what's the probability P(uu)=1/3 will be for

ACuriousMind

??? The correlated spins correspond to uu and dd, the anti-correlated ones to ud and du. So the probability of correlated spins is P(uu)+P(dd) and that of anti-correlated ones is P(du) + P(ud).

I'm still not sure what the issue is.

@Secret P(uu)=1/3 means that there's a 33.33..% probability of getting the result uu upon measuring z-spin.

Secret

Ok I see, in that case I seemed to have understood entanglement correctly.
You also mentioned that my calculation on the probabilities P(ll) P(rl) P(lr) P(rr) don't make sense to you. I am actually not very confident myself. How does one calculate e.g. the probability of alice measuring spin left and bob measuring spin right, is P(lr) the corretc probability to descibe it and thus the calculation will be what's the probability of the state $\lvert @\rangle$ being projected to the product state $\lvert lr\rangle$?

ACuriousMind

There are so many nice symbols in the alphabets. Why would you choose @ to denote a state?

And you calculate the probability of measuring $\lvert \psi \rangle$ to be in the state $\langle \phi \rvert$ as $\lvert\langle \psi\vert\phi\rangle\rvert^2$

Secret

21:31

I have a habit of notating the key focus of my questions (and sometimes a bunch of constants in an equation that is not important) with spirals (@ is the closest thing to a spiral symbol, as mathjax don't do spirals), I tend to do that when letters like A,B,C $\alpha, \beta\, \gamma$ might lead to confusion

Ok I see, I will try again later (correcting that mistake), but it seems that I am on the right track in computing and understanding entangled systems

Sanya

22:19

Hey ... is someone still here? :D I wanted to ask whether it was considered ok to post a question about possible improvement in the formulation of a non-closed question on meta?

Physics Meta

22:33

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Physics Meta

22:52

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