5:13 AM
@RonaldVilliers You're a Brit too?
We generally refrain from political comment as it tends to start arguments that can get noisy. However I will say I have considerable sympathy with your views :-)

3 hours later…
7:45 AM
Just searched the meaning of wank in my browser and now i need to clean the historic XD

8:16 AM
@Semiclassical that's a good way to put it I'd say, yes. Which definition you use will probably depend on context or what one is trying to do. I'd say, from a purely information pov, the individual labels "0" and "1" hold little importance, which is why |00>+|11> and |01>+|10> would be regarded as equally correlated. It is also often the case that local operations are "free", so flipping the second qubit's labels is easy to do anyway
@Semiclassical it's more than that though. Any maximally entangled (two-qubit) state does that. In other words, any state of the form $(U\otimes V)(|00\rangle+|11\rangle)$ for any pair of unitaries $U$ and $V$. This includes, but is certainly not limited to, the four Bell states. Mathematically, this corresponds to the set of 2x2 matrices with degenerate singular values summing to 1, the Bell states corresponding to the four special cases $I,Z,X$ and $iY$

8:39 AM
What's going on here

@bolbteppa looks like nlab shenanigans
multisymplectic stuff
Hamiltonian mechanics before picking a specific time

Seems to say ch.'s 2,3,4,5 are different approaches, sec. 2 seems to look familiar-ish

Section 2 is the Legendre bundle stuff I think
the classic Hamiltonian mechanics for fields stuff
I don't know much about Peierl brackets
seems like a fun paper
I may give it a read

"Since $C^{\infty}(M)$ represents a Fréchet space, the Bastiani differentiability of functionals is equivalent to the convenient differentiability mentioned above."

of course

8:59 AM
Ah, Gieres. I remember that name from the neat review paper that points out cases where e.g. the domain of definition of the operators in the uncertainty principle matters

French fellow, too
mb he will hire me
Gieres please take me out of this hellhole

9:21 AM
Seems to have a few interesting ones

9:40 AM
@Slereah this Jet bundle stuff is confusing. I thought would could think of Lagrangians as $L:TQ\to\mathbb R$. Is the 1-jet bundle the same as the tangent bundle, $J^1(TQ)=TQ$?
the wikipedia page (en.wikipedia.org/wiki/Jet_bundle) says 1-jets are equivalence classes of sections of a bundle, defined by sections having the same first order derivatives.... so in the case of $TQ$, 1-jets are just vector fields, right?

Well, for point particles, the jet bundle is roughly equivalent to the cotangent bundle, yes
But this isn't true for all physical objects
for fields that isn't true
You have to use what is called the jet prolongation
if you have a section $\phi \in \Gamma(E)$ of your vector bundle $E$ (like a scalar field), the jet prolongation is a section of the jet bundle, so that roughly $$j^\infty \phi = (\phi, \partial \phi, \partial^2 \phi, \ldots)$$

@Slereah what's "point particle" in this language? Is it just saying $Q=\mathbb R^3$?

Well if we're going full bundle here, a point particle is defined by a sigma model
Your "spacetime" is time $\mathbb{R}$, the bundle of the field has fiber $\mathbb{R}^n$, and the momentum is the tangent bundle of that bundle
the jet bundle is roughly gonna be $\mathbb{R^n} \times \mathbb{R}^n$

@Slereah never go full bundle...
3

sometimes you gotta!
It's trivial for point particles but if you're using weird gauge theories it becomes more complicated
the Lagrangian in this case is defined as a map from the jet bundle to $n$-forms on the base space
So that you can integrate it
then you just have $$S[\phi] = \int L(j^\infty \phi)$$
for point particles this is just $$S[x] = \int L(x(t), \dot{x}(t)) dt$$

9:56 AM
@Slereah mh.. wait, I was thinking in terms of purely classical non-field mechanics here. Does that change? For me $Q$ is the manifold of possible configurations (positions). Possible "motions of a particle" would be paths $\mathbb R\to Q$ satisfying some EOM. I suppose these amount to sections of a trivial bundle $\mathbb R\times Q\to\mathbb R$, but does thinking it this way give any advantage?

The configuration space for point particle corresponds to fibers of your bundle, yes
For classical mechanics your fiber bundle is roughly just $\pi : \mathbb{R}^3 \times \mathbb{R} \to \mathbb{R}$
Your jet bundle here is a subbundle of the tangent space of this bundle

@Slereah and how do you define "momentum" in this context? Velocities would be just elements of the tangent bundle I guess, $v\in T Q$. Is momentum defined wrt some Lagrangian/Hamiltonian, or is it something else here

Tangent space is gonna just be $T(\mathbb{R}^3 \times \mathbb{R}) \approx \mathbb{R}^8$, but we're only interested in the... vertical bundle, IIRC?
So we're only interested in the subspace that's tangent to the fiber here
Which also has fiber $\mathbb{R}^3$
ie the velocities
that's how you can do basic classical mechanics in this formalism
Your first jet bundle is gonna be the space $\mathbb{R}^3 \times \mathbb{R}^3 \times \mathbb{R}$
And you can project it on either time $\mathbb{R}$, or you can use the projection on the configuration space $\mathbb{R}^3 \times\mathbb{R}$
ie there's a projection from $(x, v, t)$ to $t$ and one from $(x, v, t)$ to $(x, t)$
This is what this image means:

2 hours later…
12:22 PM
@Slereah I still don't understand why you are interested in the tangent bundle of $\mathbb R\times Q$ with $Q$ configuration space, rather than just the tangent bundle of $Q$. What's the advantage?

Well you are basically only considering the tangent bundle of $Q$ really
That's why you're looking at the vertical space

@Slereah well, ok, but then why consider the bundle $\mathbb R\times Q$ in the first place?

Well remember that in the EL equation, you do also have to consider time

also, the vertical bundle of $E\to B$ would be the one whose fibers are the tangent bundles of the fibers $E_x$, right?

yes basically

12:26 PM
@Slereah sure, but I thought one would just consider the time/parameter when asking for the paths $\mathbb R\to Q$ satisfying a given EOM
or maybe the point is that you can consider such paths $\gamma$ as elements of the bundle $\mathbb R\times Q$
(or rather, sections of it)

The thing being that when you're working Lagrangian mechanics, you are using both bundles
that's what the variational bicomplex thing means, as it is called
the "real" derivative of the bundle is $d$, which you split into the horizontal and vertical differential $d_H$ and $d_V$
You need both of those to work in Lagrangian mechanics
also I was wrong, the configuration space is horizontal, not vertical
I think the basic idea is that when you're using the jet bundle, you're not quite considering what the coordinates are

@glS the problem is that, in general, the Lagrangian might depend on time explicitly

ie you're using $(x, v)$, but neither of those are related
they're just two coordinates

so if you want a general formulation of mechanics the Lagrangian is not a function on $TQ$, but you need to involve time prior to the paths
if your Lagrangian is time-independent, you can forget about the $\mathbb{R}$-factor

But in Lagrangian mechanics, they are not unrelated, they correspond to a specific section, and the time derivative is involved
Like there's a lot of relations between the two derivatives in the context
The horizontal derivative is just what gives you the EL equation, technically doesn't involve time at all, but the way it is usually written, it does involve the time derivative
and the action is invariant up to any term $d_V \alpha$

12:39 PM
@ACuriousMind ah, that makes sense, yes. So you need to consider $L:TQ\times\mathbb R\to\mathbb R$.. and then you want to still think of this as a map from some bundle, and you consider the bundle $Q\times\mathbb R\to\mathbb R$ for this purpose? Is that it?
@Slereah I'm losing you here. What are you referring to exactly? You are saying that the horizontal bundle of configuration space (is this $Q$ or $Q\times \mathbb R$ here?) is what? The domain of the Lagrangian? The space of possible physical paths?

Well, usually what people call the configuration space is $\mathbb{R}^3$
So the configuration space is $Q = \mathbb{R}^3$, and the bundle you're considering is $E = Q \times \mathbb{R}$
A path is then a section of $E$, so that $x(t) \in \Gamma(E)$

ok. So if I'm getting this right, the vertical bundle of $E\equiv Q\times\mathbb R$ is $V(Q\times \mathbb R)\equiv\operatorname{ker}(\mathrm d\pi)$ with $\pi:Q\times\mathbb R\to\mathbb R$. Now, $\mathrm d\pi: T(Q\times\mathbb R)\to T\mathbb R\simeq\mathbb R$. So isn't this vertical bundle just the bundle $TQ\times\mathbb R\to\mathbb R$?
because we are seemingly looking for the smooth paths $I\to Q\times\mathbb R$ which project to $0\in\mathbb R$

I think so yes?
Since your configuration space is usually $\mathbb{R}^3$, this means that $TQ$ is trivial, too
So it is really just the bundle $(x, v, t)$

@Slereah so the thing about this, is that I interpreted one of the advantages of this type of formalism is that you can consider more complex manifolds of configurations, which might be non-flat etc, so that's why I tend to try and think in terms of a generic configuration manifold $Q$ rather than just some $\mathbb R^n$

Yes
They may fail to be flat, the fiber may not be a vector space, etc

12:52 PM
the general statement is that local coordinates on the bundle are $(x,v,t)$

so this is pretty cool. So the Lagrangian can be written as a bundle map with domain $V(Q\times\mathbb R)$... which then I guess also generalises more nicely when time loses its special role with relativity and stuff

hi
I have a question regarding history of science

then you might want to visit History of Science and Mathematics ;)

that's why when you do field theory, you don't use $d/dt$ for Lagrangian stuff
You use $\partial_\mu$

its about electricity and benjamin franklin

12:54 PM
Because the base space is different

@Slereah field theory works a bit differently - you already start with some bundle $E\to M$ for the fields over your space(time) $M$

Benjamin Franklin gave the sign convention for charges and I know about the famous kite experiment. But why did he give positive sign to the glass rod rubbed.

why not

because there are 50-50 chances so why positive to glass

Maybe he had a specific reason but I think it's not worth worrying much about

12:57 PM
perhaps he flipped a coin

uhh that's not how conventions are made

I mean sometimes it is!
you have to pick one

It has to do something about more deep than I could think of

that's why sometimes there are different conventions

@SamyakMarathe on the contrary, that's how a lot of conventions are made!
physically, it does not matter whether you assign electrons negative charge and protons positive charge or vice versa
so there cannot be a physical reason for the assignment

12:58 PM
Well, some people have Opinions on such matters
There are good odds that people might have had ideas on the topic
the electric fluid comes from the rod or something like that
but they would be reasons related to obsolete theories on electricity

ok then what about this, when he explained about the kite experiment, he gave positive sign to the ground and the clouds being negative and for the glass experiment, he gave positive sign to glass and negative to silk cloth. Why both the sign matched

Historically interesting but not much more

but really, if you are interested in an actual historical reason, History of Science and Mathematics is the right place for that

@SamyakMarathe are you certain Franklin did it like that, or is that a modern re-interpretation of what he did
much of physics "history" that we learn is filtered through a modern lens when the people at the time talked about it in very different terms

1:01 PM
@ACuriousMind yes he indeed did that. He clearly mentioned that the glass has excess of electric fluid and he gave it a positive sign

ah, there you have it - the historical answer probably lies in the understanding of the fluid theory of electricity at the time

oh yeah i think i got it
btw it was interesting to know a lot about franklin from assassins creed 3 and youtube

so you'll need to find someone who knows about that, which is really the sort of stuff History of Science and Mathematics specializes in

thanks to ubisoft and google, lol

hello there

1:03 PM
hi

hello

did anyone play games?
like assassins creed or so

I stopped playing AC after the second game, I think

I think string theory in spacetime would have a pretty weird bundle
I'm not 100% sure what it is

woh, thats a high level science

1:08 PM
@Slereah I have no idea what you mean by that - are you talking about string field theory?

I m just a high school student who has not entered quantum realm yet, so better for me to leave.

@ACuriousMind I mean the classical theory
1 dimensional submanifold of $M$

@SamyakMarathe It's fine if there are multiple conversations here simultaneously

My friends say string theory is uncertain... how far it is logical ?

1:15 PM
@Slereah that's just a non-linear $\sigma$-model - the "field bundle" for $\sigma$-model fields $\Sigma\to X$ is just $\Sigma\times X\to \Sigma$
@Ishwaran what do you mean by "logical"?
(Most) people are not uncertain about whether or not there's logic involved in string theory, but they are uncertain whether or not it actually describes our reality

The fact that it is so internally logically consistent is why people work on and develop it further, obviously it's not complete and there are going to be some issues

also bc it's cool
who cares for physics

@Slereah yes

1:30 PM
I think people are a bit too worried about string theory being useful or not
It's not like it's costing a lot of tax payer money
people may have the wrong notion about what theoretical physicists are paid
"I would estimate there are between 1000 and 2000 string theorists in the world. The main annual meeting on string theory is called, sensibly enough, strings. It typically has around 500 participants. But, not every string theorist goes to strings every year. If we assume that a third go, we get 1500 as our estimate. That seems about right."
I think society can withstand paying 2000 string theorists

I wonder why we couldnt take a colorized pic of the blackhole back in 2019

well we could, but then we couldn't see any relativistic effect
most stellar black holes are covered in schmutz
here's one
can't see much going on with it

isnt the accretion disk supposed to be blue ?... because the energy level should make it blue

I mean this isn't like a photo
this isn't a raw photo either and even then you can't see what's going on
most astronomy photos you see aren't what a person would see looking at it
they have weird exposition times and wavelengths

False color (or pseudo color) refers to a group of color rendering methods used to display images in color which were recorded in the visible or non-visible parts of the electromagnetic spectrum. A false-color image is an image that depicts an object in colors that differ from those a photograph (a true-color image) would show. In this image, colors have been assigned to three different wavelengths that our eyes cannot normally see. In addition, variants of false color such as pseudocolor, density slicing, and choropleths are used for information visualization of either data gathered by a single...

1:44 PM
A classic example

@glS sure. what's nice about the four Bell states is that their positive/negative correlations in the X,Y,Z bases are simple
the reason i like the singlet state in particular is that I can absorb $U\otimes U$ factors into it. Huzzah for rotational invariance
which in particular should mean maximally-entangled states can be generated by starting with the singlet state and doing some unitary to just one of the qubits

@Slereah looks like black hole will explain and answer every physics theory

probably not

why not ?
in what ways it will fail to ?

well, not everything is a black hole
A rock is very much not a black hole but geology is still a field

1:49 PM
black holes do seem like they'd be good settings to resolve questions about quantum gravity
but doing experiments on black holes is not exactly within reach right now :)

the closest one is 1500 light years away

Even sending a laser to it will take a while

what's the opposite of hyperbole

Hopefully the last roman emperor shined a laser onto it back then
euphemism

1:51 PM
"understatement" seems more to my intention

@Semiclassical literally that would be 'hypobole'

unfortunately there's this rule of thumb where people will not usually make experiments that last longer than about 40 years
unless it's a pretty cheap experiment

but I don't think that's an English word yet, so you have to pull a Shakespeare and just pretend it is one

well, the experiment can last a long time, but they won't bother if you won't get results for 40 years

well, it's in 1913 Webster's apparently: en.wiktionary.org/wiki/hypobole
though the definition there isn't quite what one would want

1:53 PM
40 years is about the lifespan of a physics career

@Semiclassical huh, but with a different meaning, interesting

Also 40 years is about the time it takes in the realm of propulsion technology to make something obsolete
If you send a rocket to space for 40 years, odds are good that by that point, you can send a much faster rocket that will arrive before that one

> When Walther Nernst learned of Einstein's 1906 paper on specific heat, he was so excited that he traveled all the way from Berlin to Zürich to meet with him.
... I'm ... unimpressed ...
aaaaaaaaaaall the way from Berlin to Zürich. mmmm hmmmm.

I mean it was 1906
it probably took 3 days
in one of those old school steam locomotives
chugging at 30km/h

Europe is physically smaller than what most people think

2:01 PM
Not me
i live there

@EmilioPisanty it's a long journey for an European even today!

Berlin to Zurich feels like a short trip for me though

@ACuriousMind is it, though?

@Ishwaran It's only 670km

2:04 PM
@VincentThacker much closer than what I expected it to be

@EmilioPisanty I'd say so

I don't think I'd go 9h by train to meet uuuuh

I think everyone I know that does not travel a lot professionally would consider it a pretty long trip just to meet a single person

who's a current german physicist
also it was 1906, Einstein wasn't that big yet I think?
up and coming but not quite a superstar yet

(my parents live half that distance away from me and I consider that an annoyingly long journey :P)

2:08 PM
he must have spent about a day to reach from berlin to Zurich back in 1906
nominally it took 20 hours to reach Zurich from berlin back in 1900
by train

The same distance won't even get you from Sydney to Melbourne

@ACuriousMind language is wacky

In India, one can travel 750 kilometers in 9-10 hours
But it takes 4 hours to cross Banglore limit

Yeah but you know
People just died travelling that long back then

@glS out of curiousity, why $iY$ rather than $Y$? just so that it's a real matrix in the Z-basis?

2:15 PM
That's just what it was like going through the US back then

i think dysentery is mostly a matter of not having clean water?

I remember my grandfather saying he travelled all the way from Chennai to Delhi in 44 hours by train back in 1970's

which would make sense for the Oregon trail

absolutely
Can anyone suggest a material ideal enough for aperture blades for a cam ?

Domain walls?
it is very ideal
I don't think it has been proven to exist yet but it would probably make a good aperture

2:25 PM
That sounds good

right
or just think in terms of density matrices instead of state vectors

sorry, got the question wrong actually. I mean that was true, but also yes, $iY$ was to get the equivalent to $|01>-|10>$. But given that multiplying by scalars doesn't matter anyway, I could have used $Y$ just the same
@Semiclassical but if you do that things have to be treated a bit differently. In particular, two-qubit states are 4x4 density matrices etc. As long as you deal with pure states, there's no need for it

right. i just meant as far as overall complex phases
since $(e^{-i\phi}|\psi\rangle)(\langle \psi|e^{i\phi})=|\psi\rangle\langle \psi|$

@Semiclassical I understand that, but also at the end of the day the singlet is not that special even from that point of view. Any state of the form $\sum_i |u_i\rangle\otimes|v_i\rangle$ for any pair of orthonormal bases is stable under a local operation $U\otimes V$, upon appropriate choice of $U$ and $V$. For the singlet you get $V=U^*$ (I think). More in general, $U$ and $V$ are matrices with the same action wrt the bases $|u_i\rangle$ and $|v_i\rangle$ (up to a complex conjugate somewhere)

I think it's $V=U$ directly, but that's a small point

2:34 PM
@Semiclassical if you mean "think in terms of density matrices" in order to avoid working with complex projective spaces, then yes, sure. But you also lose something in that dimensions increase etc. And you still have to manually fix the trace of the matrices, as the mapping $|\psi\rangle\to|\psi\rangle\otimes\langle\psi|$ only eliminated the $U(1)$ "extra freedom" of vector states

since $e^{-i(\theta/2)S_1^x}e^{-i(\theta/2)S_2^x}=e^{-i(\theta/2)S^x}=1$ when acting on the singlet state

@Semiclassical is it? You have $(U\otimes U)(|ii\rangle)=|jk\rangle u_{ji} u_{ki}$ (summing over stuff). To get the identity you need a complex conjugate on one of the copies

3:12 PM
hmm
in general Werner states have symmetry $(U\otimes U)\rho (U\otimes U)^\dagger = \rho$
with the singlet state as the antisymmetric case for two qubits
also the claim for singlet state is $(U\otimes U)(|01\rangle - |01\rangle)=|01\rangle - |01\rangle$
it's acting as the identity on the singlet state, not on product states
correction: it's only equal up to a phase. (don't have to worry about this in the density matrix statement)
blah, meant to say |01>-|10> above
if one restricts to $U\in SU(2)$, then $(U\otimes U)(|01\rangle-|10\rangle)=|10\rangle - |01\rangle$ full stop
generic unitary matrices have an overall phase factor which complicates that slightly

3:50 PM
I agree, but that's more of a coincidence of the 2x2 case with the singlet defined that way. In particular, you use $\det U=1$ in the calculation. The generalisation of that is asking the local unitaries fixing $\sum_i |u_i\rangle\otimes|v_i\rangle$, not $\sum_i |ii\rangle$ as I was. In this more general case you should have $(U\otimes V)\sum_i |u_i\rangle|v_i\rangle=\sum_i|u_i\rangle|v_i\rangle$ iff $U=c_{jk}|u_j\rangle\!\langle u_k|$ and $V=\bar c_{jk}|v_j\rangle\!\langle v_k|$, for any unitary matrix $(c_{ij})_{ij}$.
this should work for arbirary isometries and dimensions
notice that $\mathbf{SU}(2)$ matrices have the form $\begin{pmatrix}x& y\\ -\bar y&\bar x\end{pmatrix}$, and thus the coefficients $c_{ij}$ in the above will be related by $c_{11}=\bar c_{22}$ and $c_{12}=-\bar c_{21}$. Using these you get that, in the specific case of the singlet, $U=V$

4:48 PM
What I’ll admit I don’t grok right now is in what sense the other Bell states are maximally-correlated. For instance, if I start from the two-qubit Bell state I don’t think the following is true: If Alice measures the spin along some direction, then she can predict what Bob will obtain for that same state. (It’s true for the Cartesian directions but should fail, say, in the (0,1,1) direction.)
By contrast I think it’s true without qualification for the singlet state.
I’m so used to spin being intrinsically “rotational” that I’m probably missing the more general mindset
(I know you had a comment on that definition so let me go find that)
Ahh, so making the change in von Neumann entropy as large as possible. I’ll assent to that, yeah

5:06 PM
I'd say that idea is that the "local labels" don't really matter from the point of view of trying to understand the way the states are correlated. You often care, in this context, about the "quantumness" of the correlations.

In the scenario you mention, Alice will be able to fully predict Bob's outcome from hers, provided they coordinate beforehand. If the shared state is $\sum_i|u_i\rangle\otimes|v_i\rangle$ (which any maximally entangled state is), then Alice measuring $|u_i\rangle$ means she know Bob will/did measure $|v_i\rangle$. Or in other words, if she measures the $i$-th outcome,
and applying arbitrary local unitary operations to such a state will give you another state of the same form. So the $|u_i\rangle$ and $|v_i\rangle$ will rotate, but the correlations between the parties remains untouched. In other words, maybe you started with |00>+|11>, and after local operations you now have |01>+|10>. Now what "first outcome" and "second outcome" mean for Bob changed, but the correlation is the same in the sense that A measuring I outcome implies B measures I outcome

Ohh, that’s interesting. So what makes singlet special is that they can each focus on the same observable. But there’s no particular reason to obsess over this: it’s enough that any measurement by Alice can predict some measurement of Bob
I’m so used to the original Bell inequality case where anti-correlation makes things simple

@Semiclassical pretty much. The point of view is more or less that local operations are "free" anyway, so you don't care much about what "$|0\rangle$" and "$|1\rangle$" mean physically, as that is easily changed via local operations. What matters is the "quantum link" between the parties, as that's what usually hard to create, interesting, etc

Right
Local operations are free
I’ve seen a bit about that in the context of LOCC vs LOSR resource theories (not that I really understand either)
But the LO part is simple enough
One merit I will still reserve for the quint singlet state: it should make it easy to deduce the pairing of observables for another Bell state
Like, if Alive measures in some weird direction , then writing the Bell state in terms of the singlet state should make it apparent which observable Bob should use

5:36 PM
@Semiclassical the way I see these is that at the end of the day it's really just akin to fixing a coordinate basis. It doesn't matter which state you use, singlet or any other maximally entangled one. They are all equivalent up to changing the physical meaning the two parties decide to attach to their "|0>" and "|1>". What matters is the relation/overlap between the different states involved
so if you are studying a specific experimental implementation, where there are constraints or other special roles played by the different states, then sure, it matters which specific maximally entangled (or whatever) state you consider. But if you are discussing more in the abstract without thinking of a specific experimental scenario, then it doesn't

5:50 PM
@glS right. the kind of statements you make for a historical context in particular
part of the reason i have hung on to thinking like that is that it makes the Bloch sphere, for instance, seem less odd to me. the Bloch sphere to me parametrizes observables rather than states. (obviously they're equivalent, but it prevents silly things like thinking that 90-degrees on the Bloch sphere means orthogonality in Hilbert space.)

physicsforums.com/attachments/image-jpg.288929 When we try to find field due to a shell it comes same as a point charge kept at its center. But doing so we take E outside of integral when applying gauss law. How is E constant. Does it behave like a charged conductor which I guess has constant field perpendicular to surface.

i think the bell state |00> + |11> allows the following statement. Suppose Alice makes a measurement of the spin in some direction $(a_x,a_y,a_z)$. (Or equivalently the observable $A=a_x \sigma^x+a_y\sigma^y+a_z\sigma^z$.) Then Alice can predict Bob will get the same result if he measures in the $(a_x,-a_y,a_z)$ direction
For the +X and +Z directions in particular, this means Bob should measure the same observable as Alice, i.e., X1=X2 and Z1=Z2. For the Y direction, they should measure in opposite directions (Y1=-Y2).
That's a nice enough statement for me

@Semiclassical !

@Semiclassical sure, but all of that remains true for any other max entangled state. You just rotate every basis states in the statements accordingly
@Semiclassical a way I like to think about it is that $\mathbb{CP}^1\simeq S^2$

6:06 PM
Can somebody hear me ?

i.e. the complex projective line is isomorphic, up to a point, to $\mathbb C$, and $\mathbb C$ (plus point at infinity) can be thought of as a sphere

6:58 PM
@glS yeah, i was just specifying to the original comparison
it's a pity that there's nothing so nice as the Bloch sphere for two qubits
one sorta gets spoiled by how pretty that case is

7:13 PM
To state the above for a generic maximally-entangled state: Any such state can be generated from the singlet state by applying a unitary transformation to the second qubit. Such a unitary is equivalent to a rotation of the Bloch sphere. So if you know Alice's outcome for choice of observable, you just apply the rotation to Alice's observable to determine what Bob's observable should be
easy-peasy

7:27 PM
(the smart way of putting this is probably: take a maximally-entangled two-qubit state, make a projective measurement on Alice, then trace her out. what you're left with is always some pure state for Bob.)

7:45 PM
@Semiclassical I agree. My current naive understanding of this is that it's a topological issue: $\mathbb{CP}^n$ doesn't embed "nicely" in $\mathbb R^m$ except for $n=1$. We know there is a circle bundle $S^{2n+1}\to\mathbb{CP}^n$, and thus, I think, $\mathbb{CP}^n\simeq S^{2n+1}/S^1$. It just so happens that for $n=1$ we have $S^3/S^1\simeq S^2$.
I asked a related question on this stuff on math.SE some time ago, math.stackexchange.com/a/4182341/173147

neat

however, this isn't true for higher dimensions, and $S^{2n+1}/S^1$ isn't as nice as a hypersphere

if it's not one qubit, then QM laughs at puny mortal attempts at visualization :P
i am curious about stuff like the Majorana stellar representation but that has at best a very restricted use

I worked out a few sections of the space for a qutrit here quantumcomputing.stackexchange.com/a/8439/55. You get some nice surfaces, but the complexity is quite obvious even in such simple cases

reminds me of some of the pictures from here: arxiv.org/abs/1710.05892
though the set of quantum correlations has the decency to at least be convex :P

7:53 PM
I'm sure there's some Zyczkowski's paper which fully explains all of this stuff though. I long for the day I'll actually understand half of it

there was at least one of those pictures I could make for myself

@Semiclassical well, you certanily get convexity if you consider all states (and if you don't, something's wrong). I was only drawing pure ones here

ahhh
fair
so really need to take the convex hull of that figure. but good luck doing so in mathematica
(there's probably a way but i don't know it)

@Semiclassical ConvexHullMesh works decently well =)

yes
i'd forgotten that

7:55 PM
actually, I dealt a fair amount with working out convex hulls of sets of pure states. It gets pretty nasty in many situations, both analytically and numerically

the closest thing to the above story i know goes like this
suppose, as with a standard CHSH test, we take Alice and Bob to measure A,A', B, B' respectively
if Alice measures the singlet state in direction $\vec{a}$ and Bob measures in direction $\vec{b}$, then their observables are $A=\vec{S}_1\cdot \vec{a}$ and $B=\vec{S}_2\cdot \vec{b}$ respectively. as such the correlation can be computed as $\langle A B\rangle \propto \vec{a}\cdot \vec{b}$
(if you're taking the outcomes as $\pm 1$ then it's $=-\vec{a}\cdot \vec{b}$ but the minus sign is irrelevant)
So now suppose Alice and Bob's directions are all in the plane, and (in the order A, B, A', B') they each have relative angles of $\theta$ between them
then the correlations are $\langle AB\rangle =\langle A'B\rangle=\langle A'B'\rangle \propto \cos\theta$ and $\langle AB'\rangle \propto \cos3\theta$
and we can recognize (ignoring any annoying coefficients) that $\langle AB'\rangle = 4\langle AB\rangle^3-3\langle AB\rangle$
if we plot that in the $(\langle AB\rangle, \langle AB'\rangle)$ plane, and take the convex hull, we get:

8:21 PM
which, beyond being a neat looking picture
the points at (1/2,-1) and (-1/2,1) are examples of so-called extreme non-exposed points
on the one hand, these points can't be expressed as the convex combination of any two others in the set
but there's also no linear function i can pick which would 'expose' them uniquely
so they're a bit funky :)

2 hours later…
10:03 PM
@Semiclassical interesting. I don't quite get the justification for taking the convex hull here, though. You're considering different pairs of local observables, presumably on a fixed state. $\theta$ is a parameter quantifying a relation between two pairs of such observables. Now you plot one expectation value against the other, varying the relative "angle" between the pairs of observables
I understand convex hulls in this context when the things you take the convex hull on are points representing expectation values of states. But here the base points represent different measurements on the same state?
@Semiclassical regardless, I had fun some time ago going through a couple texts on convex euclidean geometry. It's fun stuff really, exposed/extreme points at all. The best idea I got from that was the support function though, which allows you to get analytical descriptions of convex hulls in a relatively easy way. Things like exposed and extreme points emerge in the support functions as points of non-differentiability and non-continuity

@glS The way I remember is that, since all the correlations are dot products of the measurement directions, we can form a symmetric 4-by-4 matrix of dot products between the four measurement directions
Such a matrix is positive semidefinite, and PSD matrices form a convex set
This guarantees that, when you look at the four matrix elements for correlations specifically, you’ll still have a convex set
What I don’t remember is a way to do this constructively: given two different quantum correlations, generated by different measurement combos, how do you find a measurement combo which gives the midpoint of these correlations
(I’m not particularly confident there’s a nice way in general)
What you’d need, basically, is a way of smoothly changing one set of measurement directions into another, such that the resulting correlations are convex combos of the edge cases
I’ll see if I can dredge up anything

1 hour later…
11:39 PM
@JohnRennie oh that's probably for the best. Actually American but I've been getting into UK TV and consequently politics :P
Do you have any recommendations for comedies ? I've been watching Chewin' the fat with Ford Kiernan and Greg Hemphill. So bloody funny, Still Game isn't as funny in my opinion but it is more family friendly :P
I haven't really watched British TV besides the BBC..
I'll work my way through the north I suppose