Is the dimension squared though? i forgot to account for the monogamy of entanglement and while there are sure many Hermitian matrices, shouldn't the monogamy butcher their dimension in order to produce valid DOs?
Okay, sorry i forgot to mention i was thinking of a scenario where two system...
I don't know. Actually now that you mention it, it reminds me quite a bit on Hamilton-Jacobi equation in classical mechanics where any complex problems is reduced to the triviality of a simple constant abstract rotation in some crazy complicated representation. I think HJE isn't employed often...
overcomplete in the Hilbert space of quantum states, yes. The DOs formed from those states by ##|\psi\rangle\langle\psi|## however are in a different space which has a has a higher dimension, thus a true basis of the Hilberts space basis just isn't enough to cover it all. The fact that those...
Thanks! Took me a little reading because Wigners pseudo-probability is quite a bit more special. However the articles on quasiprobability distribution does explain the general framework which i am looking for.
That article claims that coherent states form a overcomplete basis (of the Hilbert...
Hmm, indeed you are right a basis of Hermitian matrices should do the trick. With them we can make use that an operation like ##S \rho S^\dagger## is linear in ##\rho##. For the qubit case ##d=2## we can write ##\rho## as a 4 dimensional vector of that basis and use ##\tilde S \rho## to express...
Work & life is keeping me busy, so sorry i am unable to respond in a timely manner.
I have tried to find any indication that quantum probabilities are different from classical ones, but i couldn't find anything that would show that. All the no-go theorems are very interesting but all of them...
Fun fact: they actually are just those conditions. But it turns out that if you formulate them as naively as i did you run into weird Banach-Tarsky issues in infinitely enough spaces. which is why mathematicians came up with sigma algebras to make that simple concept work in every case.
And the...
That's a crude misinterpretation of those no-go theorems. All those are based on Bell's idea and derived from Bell's factorization assumption representing the classical concept of locality. Now the issue is classic probability doesn't have a concept like that, so you can just take the truth...
The decomposition i wrote in my post above is what i had in mind. For the qubit case the basis has 3 elements.
Indeed the Stokes parameter or Bloch spheres are a proper full map of the DO space which i was looking for, but both aren't linear. The decomposition i had in mind is linear but the...
Hmm, i figured that ##\rho_1 = \frac 1 2\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}##, ##\rho_2 = \frac 1 2\begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}##, ##\rho_3 = \frac 1 2\begin{pmatrix}
1 & -i \\
i & 1
\end{pmatrix}## are all valid DOs and all other DOs lie within their linear span...
I understand that. But you are aware that the mapping of DOs onto a Bloch ball is not an isomorphism? it is bijective but not a linear mapping. While it covers the whole space, It does not preserve the structure of its origin (that's what isomorph literally means: preserve the structure).
Same...
Oh, sorry, i missed that you claimed that the DO manifold is a cone before. I am not sure if it's the decomposition into a Bloch sphere you meant by it since for the vector we have ##|a|\leq 1##, hence it's a cone.
But note that this decomposition always require the adding of a ##id## operator...
Since convex condition is a "##\geq##" a simplex ends up being both convex and concave. It's also sometimes described as the convex hull of its k + 1 vertices.
##\rho = \sum_i p_i |\phi_i\rangle\langle \phi_i|## assuming the ##\phi_i## are orthogonal is a valid density operator if and only if...
Of course the Hermitian operators form a linear space and the subspace the DOs span within it is called a simplex. This structure is generated by the probability condition that the sum over all possibilities must be one. There are a lot of well known methods for this particular subspace -...
This is weird. I don't know what to look for to get some literature on usual decomposition methods for density matrices. You would think that when you have a limited dimensional linear space of objects and everything done with those objects is linear in them, people would jump to apply standard...
Lol, i played around with the same measure. There are even some older posts here with a few of my fails to build it properly. I would say this is the canonical approach, which is why all of us stumbled onto it naturally.
But the issue that the density matrix does not uniquely specify the...
But i am not looking for an analogy but rather want to identify if there are probability measures which produce the same probabilities as calculated in QT. So if QT just uses a different representation of classic measures on some different space.
The thing is that DOs are formulated over the...
I am looking for a way to compare the handling of probability in QT with how it's done in classic PT (probability theory) - and their interpretations. QT does have it's own formalism that works, so there isn't much motivation to bring it into a usual representation which makes it hard to find...
Well, if i have a question i always make thoughts of my own about it and often combine both things in my posts. Isn't that even a rule of these forums? Anyhow, the post from @A. Neumaier just shows a good answer and a way to do it, though taking a somewhat different angle.
That seems like a...
Since i was talking about the correspondence, i used "observable" strictly in the technical sense of QM and "variable" in whatever it is held against in the correspondence limit. I think that's one process that enables us to determine the interpretation of an observable in QM.
whether...
I attempted to differentiate between the formal "observable" definition as in QM (as a linear operator) and "variable" as a general concept shared between all physical theories.
Which i would find is a very odd situation.
If there are observables that can be measured simultaneously (as in...
Hmm, yeah now i see. The expectations would only agree for every state if the basis is the basis for which the operator is diagonal, in which case this yield just the diagonalized representation of it... yeah, ##\langle \psi_i |O_{\psi}| \psi_j \rangle## doesn't null out.
Hmm. Which observable...
Oups, you are right, thanks. It should be ##O_{\psi} :=\sum_i \langle \psi_i |O| \psi_i \rangle | \psi_i \rangle \langle \psi_i|##
Don't know why but latex preview didn't work when i originally posted this (now it seems to work) and so i didn't see it. Fixed my original post.
So generally in most literature observables are represented via self-adjont (or equivalently real-valued) linear operators in QT. But that definition leaves it open for a wide variety of operators that can be view as observables. I was always a bit uncertain how to understand this freedom, so i...
Hmm, I would have though that is pretty basic QM stuff directly based on how the theory treats observables. On the one hand correlations are values which can be measured, hence they are observables in term of the theory and in a experimental sense, no? On the other hand QM definition of an...
And there you have it: the measurement has a new "bringing" part with isn't part of any local measurement. You forget that if no one tells Alice and Bob to signal their results to a specified location... then the the correlation becomes unmeasurable everywhere. Hence, the signaling part is...
However, the measurement of the correlation between Bob and Alice is another matter. This event differs drastically from other events as it does not form a spacetime point but rather a hypersurface where it can obtained initially. This is because it cannot be measured at a single spacetime point...
i used the "" for a reason and its a word @stevendaryl used which i originally replied to.
Then you clearly misunderstand. The difference is that before the measurement Alice would have described the system to be in ##|HV\rangle+|VH\rangle## whereas afterwards she would says its...
But no such thing is required. If you think of events, the "current state" partly contains the event of calculating the correlation between Alice and Bob - but partly. The current state changes in what we can say about the possible correlations but since Alice "current state" does not yet...
So Alice measurements affect the "current state". This isn't contrary to relativity, since this change of current state isn't locally measurable. Relativity makes no statements for such things. It restricts anything that can be used as a signal which are inherently locally accessible.
where the "current state" is unusually a collection of values relating to various locations, i.e. non local data.
So in the thinking EBR such a "current state" would only be allowed to contain data from within the specific light cone? I suppose that EBR had something like a particle picture in...
The way you formulate it sounds deeply flawed. Like take the statement and apply it to any deterministic theory which in theory makes predictions about the future with 100% certainty. This view would strictly imply the future to be already preexisting for such a theory.
But surely we can agree...
If you present it like that, then i am sure i follow the argument there.
The argument against an emergent behavior can only be made on the grounds of the spatial separation and nothing else - and as such it already based on an assumption only.
On the other hand, any attempt to understand the...
No, i have reduced the structure i am querying about to the following two states in a similar 4 beam setup but a different source emitting a different initial state instead of two independent Bell pairs.
But you'll need to read a little more about the following discussion to follow the context...
You could say the same for Bell inequalities which violation was derived solely on the basis on the existing math formalism of QT and only after it was tested it became a physical reality that surprised of quite a few people. Isn't it how theory works in general? you look in on what the know...
Because the math formalism dictates it. If we write a Hilbert spaces as ##\mathcal {H}_{AB} \otimes \mathcal {H}_{A'B'}## for the four beams 2 or 4 photon states, then this mathematical structure incorporates states like the ones i have written down. And the formalism allows to make consequent...
That is in part for a reason. This kind of signaling entrenches on the very nature of relativity as "instantaneous" isn't exactly relative. The paper is to be understood as a response to the no-signaling theorems available which chose a similar framework. It's noteworthy that it is, while far...
Yes, and my problem is that i see absolutely nothing in QT that would prevent the reduced density matrix of part A being affected by operations done outside that part, albeit it obviously won't happen for simple states and i couldn't put my finger on it yet, what i was exactly looking for...
Let me first start that i am sorry for my unclear and ambiguous writing. the discussion is frustrating for us both when i don't make it clear enough what i am talking about. I try my best to get better.
Oh... actually now that you say it out loud, i just noticed i was being an idiot not to...
last few words of your quote i was answering to?
you quote a part of my post which start with me explicitly saying that in that paragraph i talk about the single photon interference, why do you have to ask? But i admit i had trouble with some terminology since the other ensemble case is rarely...
Communication isn't what is interesting about single photon interference. It's a good example to study some basic properties of quantum states and their measurement in QT - particularly what information they store and what can be measured and what cannot.
Look at just the situation i have...
It's quite likely i get quite a few things wrong about how it works, but there is a reason i ask. What i want for now is to see how to calculate the HOM interference for two independent Bell states. For as long as i don't understand that, i won't be able understand what is possible or not - and...
Yeah, of course that is everyone's intuition, but this isn't exactly how the calculus in QT works. Whether two photons are in one way or another indistinguishable cannot be determined entirely locally since not all states are separable. However in the calculus this makes a difference in...
Oh, sorry I was referring to the example you provided in your previous post you linked (the "Experimental delayed-choice entanglement swapping" paper). I was also using their Alice, Bob, Victor terminology and their setup and then tried to explain the difference to the setup i have in mind and...
Thanks, and now we are getting somewhat closer. But in the example in the quoted paper is still a simpler case and not yet there. I have to say though, reading how the delayed choice of Victor is interpreted there sounds kind of spooky. I mean looking at the raw result and how the calculation...
This is exactly what i was looking for, thanks. I haven't read it all yet but the sections you mention are just what i was looking for... except that they stop when it gets interesting.
IV.A a) is indeed exactly what i had in mind and good to know the proper terminology for it: entanglement...
Ah, thanks. I thought as much that HOM effect would still appear but i got uncertain because the formalism didn't naturally distinguish between the cases. Okay, so the trick is to rewrite everything in terms of the modes emitted by the original sources.
The problem i initially started with...
Assume we have a photon pair prepared just so they are suited for a two photon interference. Let's call the corresponding beams A and B. Then we run each beam individually through a 50:50 beam splitter such that we get beams A1, A2, B1, B2 such that we can symbolically write the state of the...
Looking at the formulation of the assumptions and the theorem, i don't think it makes more sense to view Bell as being about Einstein-locality of all the information bearing variables relevant for predictions of the model (Bell factorization condition on the random variables). While the...
Force being directly observable is a hard oversimplification. If we go into detail, it's way more complicated and nothing is really directly observable all on its own.
forces represent the interaction between different entities, in this case the field and a particle. But we can measure neither...