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John Rennie
John Rennie
5:04 AM
@cOnnectOrTR12 Yes.
The object travels from a starting point s₀ at an initial speed u to an ending point s₁ at a final speed v with a constant acceleration a, and the distance s is the distance between s₀ and s₁.
 
 
2 hours later…
Nihar Karve
Nihar Karve
6:53 AM
@Slereah Urs himself has something to say about that: physics.stackexchange.com/a/360010
 
Physics Meta
Physics Meta
7:38 AM
0
Q: Closed Question Sample

LordlyAmigoMany new users who are quite unaware of format of asking question i.e. sometimes they take data information to be obvious should be assisted with some sample closed questions and how to improve them. One time but detailed analysis on kinds of closed question could be assistive for most of new use...

discussion closed-questions
 
 
1 hour later…
Slereah
Slereah
8:59 AM
I wonder if there's a rigorous version of Einstein unified field theory
It's a bit of a hand wave in some aspects
The metric tensor is just a generic matrix, but I'm not sure if there is a condition on the determinant
Can the matrix just have determinant zero?
 
Nihar Karve
Nihar Karve
9:18 AM
It has to be non-degenerate, i.e. non-zero determinant, no?
 
Slereah
Slereah
I'm not sure
Most of it was written in like 1917 to the 1930's
Before mathematics was invented
It's the sum of the metric tensor, which is any symmetric matrix, and the EM tensor, which is any antisymmetric matrix
Theoretically it could be zero I guess, unless that gives unfortunate results?
Although the metric tensor part still needs the appropriate signature, so idk
I'm not sure what space that would be
 
Slereah
Slereah
9:40 AM
Is Arxiv down?
yup
time to panick
 
ACuriousMind
ACuriousMind
the journals strike back
 
Slereah
Slereah
it's not at all competing with them, it's for "preprint"
A legally distinct entity from articles
Not at all the same thing
 
glS
glS
@Semiclassical I ended up asking a closely related question to the Werner thing on qc.SE btw, in case you figured out the answer in the meantime: quantumcomputing.stackexchange.com/q/21523/55
@ACuriousMind you can compute that explicitly btw, assuming the integral is over the uniform measure on unitaries. There are formulas for integrals of products of coefficients of unitaries. A reference is mit.edu/~18.338/2013s/projects/iw_report.pdf, check out section 3. The problem is decoding the notation to get to the actual formulas one needs. The wikipedia is useful here en.wikipedia.org/wiki/Weingarten_function
uuh pinged the wrong person there, sorry
 
glS
glS
10:14 AM
@Semiclassical finally found a reference with the explicit relevant formulas: see iitis.pl/~miszczak/files/papers/puchala17symbolic.pdf, eq (10). Modulo the high probability of me messing up some index in doing the calculation, you get $$\rho_{\rm avg}=\frac{1}{d^2-1}\left({\rm tr}(\rho)-\frac{{\rm tr}(W\rho)}{d}\right)I + \frac{1}{d^2-1}\left({\rm tr}(W\rho)- \frac{{\rm tr}(\rho)}{d}\right)W $$
where $W$ is the swap. This result actually makes perfect sense: it's a Werner state.. an unseemly similar one to the state in eq (3a) in Werner's original 1989 paper
and where by "unseemly close" I mean it's the identical form they get there, though through a completely different reasoning. That's pretty nice
 
glS
glS
10:47 AM
ok, it's not *exactly* the same formula: Werner's one (https://journals.aps.org/pra/pdf/10.1103/PhysRevA.40.4277) writes a Werner state (let's call it $\rho_W$ here) in terms of its expval with the swap: ${\rm tr}(\rho_W W)$. We instead have here a Werner state $\rho_W$ written in terms of some other state $\rho$.

Rewriting this in a form I'm personally more familiar with, in terms of projections over eigenspaces of the swap, we get $$\rho_{\rm avg}=\frac14({\rm tr}(\rho)+{\rm tr}(W\rho))\frac{\Pi_+}{\binom{d+1}{2}} + \frac14({\rm tr}(\rho)-{\rm tr}(W\rho))\frac{\Pi_-}{\binom{d}{2}}.$$
 
bolbteppa
bolbteppa
11:29 AM
@NiharKarve If that really is what's going on with LQG, yikes... These notes seem to indicate the 'generalized connection' point (in Sec. 4.1.2) is right
 
Slereah
Slereah
I suspect that a lot of people like LQG because it's the plucky underdog but not obscure enough to support
Nobody will die on the hill of Euclidian gravity
I'm not sure LQG is even the best candidate against string theory
 
bolbteppa
bolbteppa
What else would be
 
Slereah
Slereah
idk, there's a laundry list of theories
None of which I know enough to really judge
 
bolbteppa
bolbteppa
That's one of the big ones on the list
 
Slereah
Slereah
11:46 AM
Science fans tend to have strong opinions on QG but I doubt there are many people in the world that know enough about any two QG theories to compare them
Although the people that do still have strong opinions
 
bolbteppa
bolbteppa
The early chapters of those lqg books look good though
 
Slereah
Slereah
I'm sure Rovelli, Schreiber and Motl have that knowledge and yet they will still scream at each other
@bolbteppa I mean you know, the points that Schreiber bring up aren't trivial
I wouldn't be able to guess that it was the problem of LQG if I read it
It's a bit handwavey, but no more than a lot of quantum "proof"
 
bolbteppa
bolbteppa
Yeah
 
Slereah
Slereah
Apparently it's basically just "Let's consider a larger class of functions"
It could be a huge problem, but it could also work!
 
bolbteppa
bolbteppa
Going through his post briefly
> To recall, the starting point of LQG is to encode the Riemannian metric in terms of the parallel transport of the affine connection that it induces... The idea of LQG is ... to equivalently regard the configuration space of gravity as a space of parallell transport/holonomy assignments to paths (in particular loops, whence the name "LQG").
i.e. you replace working with a metric by working with loops in the space, already sounds like a bad idea and potentially already denying QM, but okay let go with it
> But now in the next step in LQG, the smoothness condition on these parallel transport assignments is dropped. Instead, what is considered are general functions from paths to group elements, which are not required to be smooth or even to be continuous, hence plain set-theoretic functions. In the LQG literature these assignments are then called "generalized connections". It is the space of these "generalized connections" which is then being quantized.
 
Slereah
Slereah
11:54 AM
Isn't working with holonomies pretty standard gauge stuff
 
bolbteppa
bolbteppa
That next step is the key ingredient, turning off the continuity/differentiability
 
ACuriousMind
ACuriousMind
@Slereah that's what he's saying - it's standard to think about holonomies, but it's not standard to drop all continuity requirement on the holonomies and pretend they still have geometric meaning
 
bolbteppa
bolbteppa
> The trouble is that there is no relation left between "generalized connections" and the actual (smooth) affine connections of Riemanniann geometry. The passage from smooth to "generalized connections" is an ad hoc step that is not justified by any established rule of quantization. ... The passage to "generalized connections" amounts to regarding spacetime as just a dust of disconnected points.
It makes sense that the continuity etc... has to go, but just switching off continuity etc... is just turning a structure into a set of points (dust), anybody could do that without dressing it up in the language of holonomies
> Much of the apparent discretization that is subsequently found in the LQG quantization is but an artifact of this dustification.
 
ACuriousMind
ACuriousMind
If I had to come up with a motivation for dropping the continuity, it would probably be the path integral quantization of other theories, where you also have to drop smoothness requirements on the paths when you integrate over them (the integral measure is typically on something like the dual space of Schwartz functions, which aren't even functions but tempered distributions)
 
bolbteppa
bolbteppa
Dustification is a pretty brutal phrasing of it, it looks like that's what's going on in those notes above from skimming at least
 
ACuriousMind
ACuriousMind
12:00 PM
alas, I don't know anything about LQG so I can't tell whether this analogy has anything to do with what they're doing
 
bolbteppa
bolbteppa
Where are the nlab string criticisms
 
Nihar Karve
Nihar Karve
there are none
any "criticisms" can be rightfully dismissed through a combination of vague gesturing at GSW and ad hominem attacks
 
fqq
fqq
@Slereah apparently it was planned maintenance
 
bolbteppa
bolbteppa
> (under construction)
 
fqq
fqq
12:17 PM
See also

Amelino Camelia.
 
Slereah
Slereah
@ACuriousMind Path integrals usually don't drop continuity that hard
 
ACuriousMind
ACuriousMind
@Slereah I said smoothness, not continuity ;)
 
Slereah
Slereah
"dustification" doesn't seem to be a mathematical word
is that a slur against LQG among string theorists
Or is that some super specific term
 
ACuriousMind
ACuriousMind
@Slereah huh? It's just a word Urs made up because he described LQG as viewing spacetime as a "dust of disconnected points"
 
Slereah
Slereah
Well it could be an existing term, idk
It's just a forgetful functor or something
 
ACuriousMind
ACuriousMind
12:22 PM
It's also a neat rhetorical trick because the proper mathematical term for forgetting about the usual topology would really just be discretization/discrete topology, but that would give the people he's criticizing more math credit than he's willing to give :P
 
bolbteppa
bolbteppa
LQG is just the forgetful Dustification functor with no canonical kernel
 
BannedUser
BannedUser
I think I understand the pressure problem
Just tell me if I am right
when hot air balloon reach maximum height it's pressure is same as pressure outside right?
since pressure decrease as height increase
 
Slereah
Slereah
Approximatively, probably?
I'm not 100% sure of the approximation for balloon pressure
Is it assumed that the pressure in the balloon is constant
Air pressure around the balloon isn't going to be, since air pressure varies with altitude
But it's a small difference usually
Let's talk about stuff
 
BannedUser
BannedUser
it's true about balloon but hot air balloon with constant heat won't have same pressure I guess no I feel so stupid now
the reason is it is being constantly heated thus it is have same air pressure as outside
i feel so stupid to ask this to my professor ;_;
 
Slereah
Slereah
 
BannedUser
BannedUser
12:35 PM
i didn't thought about atmospheric pressure change too that time
 
Slereah
Slereah
Stuffology
"F forgets at most stuff regardless."
 
BannedUser
BannedUser
btw am i right again lol
 
Semiclassical
Semiclassical
1:10 PM
@glS “ Werner's paper shows this by explicitly building such a model, but the construction is not the easiest to follow.” this, so much this
My stab at Werner’s construction for generic d goes like this
For each run, let $U$ be some random unitary in U(d). Also let $|\omega\rangle=U|0\rangle$ where $|0\rangle$ is some reference qudit state.
Werner does his integration over $\omega$ not $U$, but that’s the same thing (?)
 
Semiclassical
Semiclassical
1:33 PM
For Bob, we suppose he!s measuring the qudit state $\omega$ using a complete set of projectors $\{Q_\nu\}_{\nu=1}^d=\{|\nu_B\rangle \langle \nu_B|\}_{\nu=1}^d$. The probability to measure one of these is the standard $$|\langle \omega|\nu_B\rangle|^2=\langle \omega |Q_\nu|\omega\rangle$$
This gives us some distribution over the random state $\omega$
 
Slereah
Slereah
what's a qudit, base 10 quantum state?
 
Semiclassical
Semiclassical
d-level system
I don’t make the terminology I just use it
Anyways. Bob’s distribution is derived from QM but can be replicated classically since there’s no entanglement etc going on
The distribution for Alice is trickier
She’ll have some complete set of projectors $\{P_\mu\}_{\mu=1^d}$
And Werner chooses her probability to obtain state $\mu$ as:
 
ACuriousMind
ACuriousMind
so for d=2 this is a qu2it???
 
Semiclassical
Semiclassical
Yup
I mean not really but that is the joke
Qutrit for d=3 etc
 
ACuriousMind
ACuriousMind
what if I use a variable that's not d for the number of independent states :P
 
Semiclassical
Semiclassical
1:46 PM
Then yer a heretic
 
ACuriousMind
ACuriousMind
I feel like scientific naming has steadily gotten more and more silly over the decades :P
 
Semiclassical
Semiclassical
Someone get an annihilation operator for this blasphemer
 
Slereah
Slereah
I have a deep traumatism over using rm -rf
One time I used it wrong like 8 years ago
Now I am forever terrified of using it
 
Semiclassical
Semiclassical
$$\Pi’_{\mu’}\Theta(\langle \omega|P_{\mu’}-P_{\mu}|\omega\rangle)$$
 
glS
glS
@ACuriousMind love it. We should definitely go with that notation
 
Semiclassical
Semiclassical
1:53 PM
Where Theta is the Heaviside theta, and the primed product is over all $\mu’$ other than $\mu$
 
glS
glS
@Semiclassical gotta keep my skills in the art of British understatement well-honed
 
Slereah
Slereah
@ACuriousMind Sometimes it gets more boring
 
glS
glS
@Semiclassical I lost you there. What's this? And what's $\Pi'_{\mu'}$?
 
Slereah
Slereah
Now there's standard names for mesons and they're super dry
 
Semiclassical
Semiclassical
@glS probability for Alice to measure using some set of projectors and obtain outcome $\mu$
Sum is over all other possible outcomes
This is a step I’m lifting more or less directly from Werner, to be clear
Aside from change in notation
It’s a pretty mysterious step to me as well
 
glS
glS
2:03 PM
@Semiclassical I'm actually confused here. Are you saying that the hidden variable is itself a state?
 
Semiclassical
Semiclassical
I’m saying that you calculate as though it was
Ultimately it’s just a label
 
glS
glS
@Semiclassical mh, right. I suppose you can do that without problems.
 
Semiclassical
Semiclassical
But what you compute is in terms of one-qudit inner products
It is, of course, incredibly asymmetric between Alive and Bob
Their functions are entirely different
 
glS
glS
so, to summarise: hidden variable is uniformly distributed over $S^2$, and denoted with $\omega$. We assume A&B do projective measurements, $\{Q_\nu\}_\nu$ and $\{P_\mu\}_\mu$, respectively. Now, the probabilities prescribed by the LHV are, for Alice, $p_\omega(\nu|\{Q_\nu\}_\nu)=|\langle\omega|\nu\rangle|^2$, denoting the probability of getting outcome $\nu$, conditioned to hidden variable being $\omega$ and measurement choice being $\{Q_\nu\}_\nu$.

On Bob's side... I'm still not clear
 
Semiclassical
Semiclassical
it's actually the other way around. Bob is the one whose measurements are easy
Alice is the wacky one
i did mess up a bit, though. Werner's paper uses $\mu$ for Bob and $\nu$ for Alice.
 
glS
glS
2:10 PM
right. Don't think it matters much anyway
 
Semiclassical
Semiclassical
...mostly
yeah
 
glS
glS
but so, are you saying $\Pi'_{\mu'}$ is something like $|\langle\omega|\mu\rangle|^2$?
what's $\mu'$ there?
 
Semiclassical
Semiclassical
not really. here's Werner's original equation, for context:
so to unpack that...
suppose you want the probability to obtain the $P_\mu$ state. Then what you do is look at all the other states, and ask if $|\omega\rangle$ is closer to all of them than to $|\mu\rangle$
the farthest i could get on the logic here was to think of the two-qubit case:
 
glS
glS
@Semiclassical so, return (deterministically) the outcome $\nu$ that has/corresponds to the smallest overlap with $|\omega\rangle$?
 
Semiclassical
Semiclassical
right
i think
Suppose the unknown state that Bob measures on is really |0>, so that he's guaranteed to get +1 if he measures in the Z direction. In the singlet state, this means that Alice is guaranteed to get -1 if she measures in the Z direction.
getting notation for this is a pain
best i could come up was a product of Theta functions
so that if any of Theta functions is zero, then the product vanishes
 
glS
glS
2:18 PM
maybe $p_\omega(\mu|\{P_\mu\}_\mu)=\delta_{\mu,\mu^\star}$ with $\mu^\star\equiv \operatorname*{argmin}_\mu |\langle\omega|\mu\rangle|^2$ (I'm changing notation wrt Werner, to have an expression compatible with the one in my message above)
 
Semiclassical
Semiclassical
i think you've got the right idea, yeah
that does seem simpler than what i had
in the d=2 case, of course, none of this is so terrible
especially since $|\langle \omega|\mu\rangle|^2$ has a simple expression if you express everything in terms of the Bloch sphere
probability for Alice to get $+1$ measuring in the $\hat{a}$ direction is $\frac12(1-\alpha\cdot a)$ where $\alpha$ is the Bloch vector of $|\omega\rangle$ and $a$ is the measurement direction for alice
 
glS
glS
also probably a bit misleading, in that the min is only over the directions specified in the choice of projective measurement, but anyway. So the overall correlations should read:
$$p(\nu,\mu|\{Q_\nu\},\{P_\mu\}) = \int_{S^2} \mathrm d\omega\, p_\omega(\nu|\{Q_\nu\}) p_\omega(\mu|\{P_\mu\})=\int_{S^2} \mathrm d\omega\, |\langle\omega|\nu\rangle|^2\, \delta_{\mu,\mu^\star}$$
 
Semiclassical
Semiclassical
right
yeah, that looks good
somehow Werner actually manages to compute those integrals
 
glS
glS
the tricky thing is how to handle the $\delta$ thing, which is defined via a minimisation..
 
Semiclassical
Semiclassical
yeah
it looks like he goes to a basis in which the measurement on $P$ is diagonal, and argues from there
 
glS
glS
2:23 PM
but also, what's this LHV model for? I mean, which Werner state?
 
Semiclassical
Semiclassical
the 'minimally-separable' one
i.e., if you mixed in any more of the antisymmetric state, you'd get an entangled state which can't be generated with a LHV model
 
glS
glS
so... projection onto the $-1$ eigenspace of the swap?
 
Semiclassical
Semiclassical
right
 
glS
glS
i.e. $\frac{1}{\binom{d}{2}}\frac12(I-\mathrm{Swap})$ with $d$ dimension
 
Semiclassical
Semiclassical
in the case of $d=2$, it'd be the 50:50 mixture of the singlet state and the maximally-mixed state
 
glS
glS
2:26 PM
so what's the idea, once you have an LHV model for this one, the other ones follow because they come from mixing with separable states?
 
Semiclassical
Semiclassical
think so. i think you could also replace Bob's distribution with something like $\lambda \cdot 1/d+(1-\lambda) |\langle \omega |\nu\rangle|^2$
so adding white noise to Bob's measurement so that it'll give weaker correlations
(i'm not totally sure i've got stuff normalized right, so don't take that as an exact statement)
 
glS
glS
@Semiclassical no, wait. How's that the "minimally separable" state? The minimally separable would just be the singlet, no? Which is what you get with $I-\mathrm{Swap}$ for $d=2$
 
Semiclassical
Semiclassical
no. singlet state is entangled not separable
think separability criterion. singlet state violates it, maximally-mixed state doesn't
 
glS
glS
? "minimally separable" = "maximally entangled" = "singlet"
unless you mean something else with "minimally separable"?
 
Semiclassical
Semiclassical
oh. tbh i made up that terminology on the spot
 
glS
glS
2:31 PM
@Semiclassical yea it's not standard terminology, I just assumed that's what you meant. So what did you mean?
 
Semiclassical
Semiclassical
i meant in the sense of: a werner state is a convex combination of a maximally-entangled state and a maximally-mixed state, with weights $p$ and $1-p$ respectively
if $p>1/2$, it's entangled. if $p\leq 1/2$, it's separable
Werner's LHV construction gives the same correlations as the $p=1/2$ Werner state
 
glS
glS
@Semiclassical but $p=1/2$ is separable, so we already know there must be an LHV model.. isn't the point here to present an LHV for the cases in which the state is entangled?
 
Semiclassical
Semiclassical
no. the point is to give such an LHV model for p=1/2
that's all Werner was doing in that paper
i mean, here's the abstract: "A state of a composite quantum system is called classically correlated if it can be approximated by convex combinations of product states, and Einstein-Podolsky-Rosen correlated otherwise. Any classically correlated state can be modeled by a hidden-variable theory and hence satisfies all generalized Bell's inequalities. It is shown by an explicit example that the converse of this statement is false."
he was giving a state which has a LHV model and thus can't violate bell inequalities, but nevertheless is not classically correlated
i think classically correlated amounts to the stronger constraint $p\leq 1/3$
 
glS
glS
@Semiclassical "not classically correlated" = "entangled" here, yes?
 
Semiclassical
Semiclassical
b/c in that case you can just pick $p_\omega(\nu|\{Q_\nu\})=\langle\omega| Q_\nu |\omega\rangle$, $p_\omega(\mu|\{P_\mu\})=\langle\omega| P_\mu |\omega\rangle$
@glS hmm. no, i don't think so.
hmmmmmm
 
glS
glS
2:39 PM
@Semiclassical then what do you mean with it?
 
Semiclassical
Semiclassical
just what i said, I think. the boundary case is that the probabilities are obtained by using some two-qudit product state
if you do convex combinations of that, i don't think the correlations get any stronger
my reading, at least when $d=2$, would be:
classically correlated -> $p\leq 1/3$,
EPR correlated -> $p>1/3$,
separable -> $p\leq 1/2$,
entangled -> $p>1/2$
(i specify $d=2$ b/c i'm not sure if the cutoff for 'classically correlated' is the same for all $d$)
 
glS
glS
@Semiclassical I don't really understand what you mean with "classically correlated" here
 
Semiclassical
Semiclassical
"A state of a composite quantum system is called classically correlated if it can be approximated by convex combinations of product states, and Einstein-Podolsky-Rosen correlated otherwise."
 
glS
glS
@Semiclassical that's the definition of a separable state
 
Semiclassical
Semiclassical
hrm
then i'm confused
 
glS
glS
2:43 PM
@Semiclassical well, it's the definition replacing "approximated" with "can be written as"
 
Semiclassical
Semiclassical
yeah
 
glS
glS
but that's probably what they meant anyway
 
Semiclassical
Semiclassical
the claim in the paper is that classically-correlated does not imply "expectations factorize"
so maybe it's separable vs. simply separable?
 
glS
glS
for reference, you might have a look at Wiseman et al. followup: arxiv.org/abs/quant-ph/0612147:
[11] is Werner's paper
 
Semiclassical
Semiclassical
right
tbh i'm not sure how one directly shows that the p=1/2 case is separable. i can check it for d=2 via the PPT criterion
but that's necessary not sufficient
and i definitely don't know how one goes from "separable" to "LHV model"
in general
so Werner's construction might not be the simplest in that regard
 
glS
glS
2:51 PM
@Semiclassical yea it's not trivial for larger dimensions. Watrous does it, but I didn't go through the derivation. See page 319 of his book, and then Example 7.25
 
Semiclassical
Semiclassical
probably for d=2 there's some brute force way to deduce the decomposition
Werner's paper is sorta sidestepping that by going directly from the desired state to the LHV model
 
glS
glS
@Semiclassical that's not hard actually. Separable means $\rho=\sum_k p_k \rho^A_k\otimes\rho^B_k$ for some probability distibution $p$ and states $\rho^A_k$ and $\rho^B_k$. Then, the hidden variable is the $k$ itself: doing POVMs $\mu^x$ and $\mu^y$, the correlations are $p(ab|xy)=\sum_k p_k \langle \rho^A_k,\mu^x_a\rangle \langle \rho^B_k,\mu^y_b\rangle$
 
Semiclassical
Semiclassical
hmm
might be worth working out what the decomposition for $d=2$ is then
there's probably some 'smart' way to do it using SVD
in terms of brute force the annoyance is the number of parameters
 
glS
glS
@Semiclassical finding separable decompositions can be nastier than that. The derivation Watrous does in the book doesn't seem straightforward. Not sure if it becomes so for $d=2$
 
Semiclassical
Semiclassical
yeah
in terms of parameters in d=2, there should be three parameters (Bloch vectors) for each of the density matrices, so 6 parameters per k
if i assume k=1,2 (which may be optimistic) that's 6+6+1=13 parameters
yaaaaay
in that respect Werner's paper does have the following advantage: rather than showing a state is separable and then deduce the LHV, determine the LHV directly and deduce that it can't be entangled
i wonder if they're complementary
like, could one deduce a particular $\rho=\sum_k p_k \rho^A_k\otimes\rho^B_k$ from knowing Werner's LHV
 
Nihar Karve
Nihar Karve
3:04 PM
@bolbteppa Even worse: Lubos claims that the kinematical Hilbert space in LQG
- essentially the $L^2$ space over connections with a background-independent Haar measure defined using holonomies on graphs (and then projecting down using the Gauss, diff and scalar constraints) isn't even separable lol
 
53Demonslayer
53Demonslayer
didn't realize that you couldn't make two non-interacting accounts on physics stack exchange like you can on math stack exchange
oops
 
glS
glS
@Semiclassical well, no, in general you cannot go from LHV to separable decomposition. That's kinda what Werner is showing
@Semiclassical they show that a state can admit LHV and be entangled/nonseparable
 
Semiclassical
Semiclassical
that latter one doesn't sound right. he's showing that the p=1/2 state has an LHV, but the p=1/2 state isn't entangled
hmm. or should i just be saying HV
 
glS
glS
@Semiclassical if the state is not entangled, it having an LHV is trivial (given the separable decomposition of course). But they show more than that:
 
bolbteppa
bolbteppa
@NiharKarve not sure about that point if I understand it right
 
Semiclassical
Semiclassical
3:15 PM
sure
but i don't think EPR-correlated / classically-correlated is equivalent to entangled/separable
hmm
 
glS
glS
@Semiclassical There are two different types of properties here: entangled/separable, and nonlocal/local. Werner's paper uses old notation, and they call separable states "classically correlated" and nonseparable states "EPR-correlated". Nobody uses that terminology anymore, but there's that
 
bolbteppa
bolbteppa
5
A: Nonseparable Hilbert space

user91126The main thing that goes bad in nonseparable Hibert space is the loss of Stone-von Neumann theorem. Loosely speaking, the Stove-von Neumann theorem assures us that Schroedinger representation of the canonical commutations rules is irreducibile, and it is unique modulo unitary equivalence. Hence S...

 
Semiclassical
Semiclassical
notational drift, yaaaay
 
bolbteppa
bolbteppa
Is that really right, axiomatically they (artificially) impose separability even though something as trivial as a free particle isn't separable (until you start artificially also denying that trivial solutions are physical etc)
 
glS
glS
from this more recent review (https://arxiv.org/abs/1303.2849):
> For mixed states, it turns out that the relation between entanglement and nonlocality is much more subtle, and in fact not fully understood yet. First, (Werner, 1989) discovered a class of mixed entangled states which admit a local model (i.e. of the form (11)) for any possible local measurements. Hence the resulting correlations cannot violate any Bell inequality. While Werner considered only projective measurements, his results where later extend to the case of general measurements (POVMs) in (Barrett, 2002) ...
 
bolbteppa
bolbteppa
3:20 PM
Wait it was this one too
18
A: Is there any physical system with a non-separable Hilbert space?

Valter MorettiThe standard formulations of QM and QFT are such that the resulting Hilbert space is always separable, namely there exist a finite or infinite countable Hilbert basis (and thus every Hilbert bases are of the same type correspondingly). Separability is required as an axiom from scratch or it ari...

 
Semiclassical
Semiclassical
well, to be precise, let's go to Werner's definition of his states. For d=2, it's $W=(2^3-2)^{-1}((2-\Phi)\mathbf{1}+(2\Phi-1)V)$ where $V$ is the exchange operator
where the projectors onto the $V=\pm 1$ eigenstates are $P_A =(\mathbf{1}- V)/2$, $P_S=(\mathbf{1}+V)/2$
the singlet state is antisymmetric under exchange, so the singlet state is just $P_A$ itself
the maximally mixed state is $1/4$ the identity, which is just a sum of the projectors. so $\mathbf{1}/4=P_A/4+P_S/4$
 
Nihar Karve
Nihar Karve
in any case I don't think even the original Ashtekar paper (or the 1990-1995 Rovelli/Smolin papers) claim that you can unambiguously construct the physical LQG Hilbert space
 
ACuriousMind
ACuriousMind
@bolbteppa what do you mean "a free particle isn't separable"?
 
Semiclassical
Semiclassical
oh, shoot. have i been using the wrong parametrization of Werner states
bleh. too many conventions
 
bolbteppa
bolbteppa
The $e^{i \mathbf{p} \cdot \mathbf{r}}$ solutions are not countable
 
ACuriousMind
ACuriousMind
3:25 PM
yeah, but they're not part of a Hilbert space
 
Nihar Karve
Nihar Karve
@NiharKarve when you try to impose the scalar Hamiltonian constraint as an operator equation there are some regularisation issues or something
 
ACuriousMind
ACuriousMind
whether they're "physical" or not is a matter of debate, but they're not elements of the Hilbert space since they're non-normalizable
 
bolbteppa
bolbteppa
Right, and in qft people seriously choose to ignore these solutions and impose separability instead
 
Nihar Karve
Nihar Karve
y'all have this conversation about once a month
 
ACuriousMind
ACuriousMind
@NiharKarve it's either this or "is GR diffeomorphism-invariant" :P
 
Semiclassical
Semiclassical
3:31 PM
@glS irritatingly, i can't seem to make mathematica produce a decomposition with only two terms in $\sum_k p_k \rho_k^A \otimes \rho_k^B$
so either i'm doing something wrong or one needs more terms
so even more parameters...yaaaay
going back to the werner parametrization. in terms of projectors we have $\mathbf{1}=P_A+P_S$, $V=P_S-P_A$, so $$W=\frac16[(2-\Phi)(P_A+P_S)+(2\Phi-1)(P_S-P_A)]=\frac16[3P_A+(1+\Phi)P_S]$$
blah, should've been $3(1-\Phi)P_A$
 
glS
glS
Watrous does it in a different way. They observe that (1) the channel $\Xi:X\mapsto \int dU (U\otimes U)X(U^\dagger\otimes U^\dagger)$ (the same one discussed before) always produces Werner states, and fixes Werner states
(2) that $\Xi$ is a separable channel
(3) that $\Xi(uu^\dagger\otimes (\alpha uu^\dagger+(1-\alpha)vv^\dagger))=\frac{1+\alpha}{2}\frac{P_S}{\binom{n+1}{2}}+\frac{1-\alpha}{2}\frac{P_A}{\binom{n}{2}}$ is a Werner state for all $\alpha\in[0,1]$.
(4) separable channels preserve separability, thus the separability result for Werner states
 
Semiclassical
Semiclassical
comparing with the Wikipedia page version, that gives $W=(1-\Phi)\cdot (P_A/2)+(1+\Phi)\cdot (P_S/6)=p\cdot (P_A/6)+(1-p)(P_S/2)$
hrm, factors of 2 how i hate thee
 
glS
glS
@Semiclassical honestly, just leave that notation. You can just write Werner states as mixtures of $P_S$ and $P_A$ with coefficients. I find writing $W$ in terms of its expectation value on the swap, that $\Phi$, unwieldy
 
Semiclassical
Semiclassical
sure, but i want to be able to compare with what Werner is doing
where his $\Phi=0,-1/4$ lines up
oh, i misread wiki. should be $p\cdot P_S/3+(1-p)\cdot P_A$
so $p=(1+\Phi)/2\implies \Phi = 2p-1$
which then makes sesne
 
glS
glS
@Semiclassical then you have from $\rho_W=p\frac{P_S}{\binom{n+1}{2}}+(1-p)\frac{P_A}{\binom{n}{2}}$, and ${\rm Tr}(P_S {\rm Swap})=\binom{n+1}{2}$, ${\rm Tr}(P_A {\rm Swap})=-\binom{n}{2}$, and thus finally ${\rm Tr}(\rho_W {\rm Swap}) \equiv\Phi= 2p-1$
 
Semiclassical
Semiclassical
3:46 PM
right
 
glS
glS
@Semiclassical i.e. we agree
 
Semiclassical
Semiclassical
Werner finds $\Phi=-1/4$ for his $d=2$ case
which gives $p=(1-1/4)/2=3/8$...wtf
and $p=3/8\leq 1/2$ would indeed be entangled.
so...huh.
that'd agree with what you said
 
glS
glS
@Semiclassical I mean... could be? though I don't really trust I understand what Werner is saying when mentioning $\Phi=-1/4$. Some of the discussion in (arxiv.org/pdf/1303.2849.pdf) is relevant btw. At page 21 they give another derivation of Werner result for two qubits
 
Semiclassical
Semiclassical
but not with other calculations i've done
well, what it -seems- to be saying is that when $d=2$ you get correlations that mimick the $p=3/8$ state
which would indeed be entangled and not separable
 
glS
glS
 
Semiclassical
Semiclassical
3:53 PM
@glS that is a much nicer statement than mine
for my own sanity, $2\cos^2(\alpha_B/2)-1=\cos \alpha_B$
so $p_B(0|\hat{n}_B,\lambda)=1$ if $\cos\alpha_B<0$
okay, i think i see what i was doing wrong
and it's the stooopid parametrization issue
Wikipedia writes it (for d=2) as $W=p(P_S/3)+(1-p)P_A$
whereas in that source they do $W=p |\phi_+\rangle\langle \phi_+|+(1-p)(\mathbf{1}/4)$
and those are very much not the same parametrization
i mean, they're parametrized in opposite senses for one: Wikipedia has $W(p=0)=P_A$ whereas the source has $W(p=1)=|\phi_+\rangle\langle \phi_+|$, and these are both the singlet state
but that's not so important. the more pressing point $W(p=1)=P_S$ for Wikipedia whereas $W(p=0)=\mathbf{1}/4=P_S/4+P_A/4$
so they're describing different ranges to begin with
in particular, $$W=\frac12 |\phi_+\rangle\langle \phi_+|+\frac12(\mathbf{1}/4) = \frac12 P_A+\frac18(P_A+P_S) =\frac{3}{8}(P_S/3)+\frac{5}{8}P_A$$
okay, that clinches it. $\Phi=-1/4$ in Werner's original notation indeed is $p=3/8$ in the Wikipedia notation whereas it's $p=1/2$ in the "Bell nonlocality" notation
whereas $\Phi=0$ should be $p=1/2$ in the Wikipedia notation and $p=1/3$ in the paper
(i sorta hate that they use "p" in the paper when you can't interpret $p$ as a probability. i think Preskill uses $\lambda$ for that reason)
anyways, it's the 1/2 that was confusing me. in the wikipedia case, p=1/2 demarcates separable vs. entangled states. in the Bell nonlocality notation--which is close to what Preskill uses in the problem this all started with--p=1/2 instead demarcates local vs nonlocal
and i must've mixed them up in my head
 
Bohemian relativist
Bohemian relativist
4:17 PM
under time-reversal, does the momentum change the sign? if we take $p_x=\frac{dp_x}{dt}$ in classical mechanics. it seems to be the case. But in quantum mechanics, $p_x=-i\hbar \frac{\partial}{\partial x}$, then how do we see $p_x$ changes sign under time reversal?
 
Semiclassical
Semiclassical
something-something time-reversal is antiunitary
 
Bohemian relativist
Bohemian relativist
sorry, $p_x=dx/dt$ above.
 
Semiclassical
Semiclassical
one (overpowered) way to think of this is that physics phenomena should be CPT symmetry (at least if you're not doing anything crazy)
so doing time-reversal should be equivalent to doing CP (conjugation and parity)
parity makes $\partial_x \to -\partial_x$, and conjugation makes $-i\to i$
which...hrm
oh
momentum operator is Hermitian, so $p^\dagger =p$
so $p$ is symmetric w/r/t conjugation and antisymmetric w/r/t parity, so it must be antisymmetric w/r/t time-reversal
which is what you said to begin with, so it matches
 
Semiclassical
Semiclassical
4:36 PM
Okay, so, correct statement now: The state that Werner constructs is entangled but still local, so we can't mimick this with product states but we can nevertheless give a LHV model. @glS
This does put me in a mind of something I want to understand eventually. The usual resource theory one uses is 'local operations and classical communication' (LOCC). But I've also seen discussion of 'local operations and shared randomness' (LOSR).
and they have different notions of what it means for a state to be entangled
Given how the LHV that Werner constructs precisely involves the two parties having some shared randomness, I wonder if Werner's "local entangled state" would still count as such under LOSR
hmm. looking briefly at what i remember, probably it would. LOSR-entanglement coincides with separability for states
on an entirely different note: arxiv.org/abs/quant-ph/0702225 is a lot of Horodecki's
 
Bohemian relativist
Bohemian relativist
@Semiclassical I think your argument should be T=-CP and thus $CP(p_x)=CP(-i\hbar\frac{\partial}{\partial x})=p_x$, so $T(p_x)=-p_x$.
 
Semiclassical
Semiclassical
right. $C(p)=+p$ (symmetric), $P(p)=-p$ (antisymmetric), so $T(p)=-p$ (antisymmetric)
yeah, i had it wrong the first time
@Bohemianrelativist oh, no. $CPT=1\implies CP=T$
the point is that $C(p)=+p$, not $-p$
because the Hermitian conjugate of $p$ is itself
it's conjugation of an operator, not complex-conjugation
 
Bohemian relativist
Bohemian relativist
4:58 PM
@Semiclassical why isn't $CPT=1 \Rightarrow CP=T^{-1}$?
 
Semiclassical
Semiclassical
it is, but $T^2=1$
so $T^{-1}=T$
 
Bohemian relativist
Bohemian relativist
OK
@Semiclassical I think C in CPT is charge conjugate. Is charge conjugate just Hermitian conjugate?
 
Semiclassical
Semiclassical
hmm. am i being silly
entirely possible
 
Bohemian relativist
Bohemian relativist
I mean the antiparticle conjugate.
 
Physics Meta
Physics Meta
0
Q: How can I create circuit diagrams for use on this site?

Emilio PisantyI am writing a question regarding a specific electrical circuit, and which would be significantly clearer if I can include a diagram of the circuit. What software tools can I use to quickly, simply and efficiently create and add a circuit diagram? Ideally, such tools should (i) be completely fr...

discussion
 
Nihar Karve
Nihar Karve
5:06 PM
@Semiclassical warning: this is not always the case. For instance, on the spin-1/2 Hilbert space, $T^2=-1$
 
Semiclassical
Semiclassical
ugh, spinors
 
Bohemian relativist
Bohemian relativist
C changes a particle into its antiparticle.
 
Nihar Karve
Nihar Karve
this is due to a subtlety in considering projective representations, and it leads to Kramers degeneracy
 
Semiclassical
Semiclassical
yeaqh
 
glS
glS
5:39 PM
@Semiclassical haha first time? Yes they're a very renown (sur)name in the field. I think there's just one still active, but honestly I can't never remember which one did what
also, that review is really not the easiest to follow, although it's very complete to get a more "historical" account. I personally find arxiv.org/abs/0811.2803 much clearer to get an introduction on the topic
@Semiclassical yes.. as long as with "local" you mean "produces correlations explainable via a local model"
@Semiclassical also, it's not just that you can't mimick it with a product state. You cannot mimick it with a separable state, which is a much stricter condition
a great way to appreciate the differences between separability and nonlocality (and steerability, which is yet another one) is Wiseman et al's notation: arxiv.org/abs/quant-ph/0612147. They are also different assumptions on the way the correlations produced by the state can be written.
1) Nonlocality means being unable to write correlations as $P(a,b|A,B;W)=\sum_\xi p(a|A,\xi)p(b|B,\xi)p_\xi$ ($W$ underlying state, $\xi$ hidden variable).

2) Nonseparability means being unable to wrote $P(a,b|A,B;W)=\sum_\xi P(A|A;\sigma_\xi) P(b|B;\rho_\xi)p_\xi$ for some "hidden states" $\sigma_\xi$ and $\rho_\xi$.

3) Steerability means being unable to write $P(a,b|A,B;W)=\sum_\xi p(a|A,\xi)p(b|B,\rho_\xi)p_\xi$
these are all conditions one strictly stronger than the other (not necessarily in the order given above)
 
Physics Meta
Physics Meta
5:55 PM
0
Q: What happens to the forces acting on a bung in a horizontal circle when the radius of the circle is increased

M G(Note: the tension - centripetal force - stays constant) I understand that the angle to the horizontal of the string holding the bung will increase meaning the centripetal force or Tcos(theta) will decrease. This will therefore increase the vertical component of the force which is Tsin(theta). A...

bug
 
 
3 hours later…
Semiclassical
Semiclassical
8:45 PM
@glS yeah, that's all i meant
i'm not familiar with steerability
i have a bit of a hard time wrapping my head around mixed-state separability
probably because determining whether a given mixed state is separable is hard
so, for instance: I know that the state $\frac13 |\phi_+\rangle\langle \phi_+|+\frac23 (\mathbf{1}/4)$ is a separable state, because you get it by applying random unitaries to $|0\rangle|1\rangle$ and averaging over such unitaries
but i don't know if there's a simpler decomposition than "add up all possible $(U\otimes U)|01\rangle\langle 01|(U\otimes U)^\dagger$"
 
PM 2Ring
PM 2Ring
@53Demonslayer You can have 2 (or more) accounts on any Stack Exchange site, as long as you don't abuse them. See physics.meta.stackexchange.com/q/5657/123208 "Non-interacting" isn't sufficient. Your accounts must not be used to cast multiple votes on the same post.
 
Semiclassical
Semiclassical
9:17 PM
hmm. it looks like Watrous proves that Werner states are separable precisely by arguing that they're generated by averaging $(U\otimes U)\rho(U\otimes U)^\dagger$ over random unitaries
so i guess he doesn't simplify that further
 
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