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7:47 AM
@EmilioPisanty Good morning sir !!

Commentary on the recent gigantic rep loss.

yesterday, 7 minutes total – 25 messages, 4 users, 0 stars

Bookmarked yesterday by Marshmallow

@Slereah Good Morning sir!
What are some sci-fi research areas for those who know Classical Mechanics?
Classical Mechanic can't work on cool stuff like Teleportation, Space Time Continnum, then what are the options left for him?
@EmilioPisanty so, presumably because of Mark Dennis leaving, that group no longer exists :/
8:02 AM
@AbhasKumarSinha Spaceshiiiips
@Slereah if someone made Teleporter, then no one will use spaceships
@Slereah A idea, first UFO Saucer for Humans, although UFO seems fictional, but they are more aerodynamic and much far better than normal ones... But again same problem, if someone made Teleporter, then no one will use that...
@AbhasKumarSinha Making teleporters is hard
by which I mean you can't make teleporters
@Slereah No one can say, we have more postdocs around the world than students in Asia, I believe someone can do it, might just be hidden for now
I mean sure
in the same way that maybe we can make a magic wand
but I wouldn't invest my money in it
@Slereah Sir, people love Teleporters more than Spaceships :(
@Slereah Using Spaceships, we can't even move out of solar system... :(
8:09 AM
We can't even get out of a room with teleporters we don't have
but we do have spaceships rn
@Slereah Okay that's makes sense, people love space tours then teleported over other planets directly...
I wonder if aliens exist or history.com/.amp/news/ufos-washington-dc-news-reports this newspaper report is real, then how aliens are coming here? Without teleportation? We can't even escape solar systems with that....
We need to work more hard in order to build something worth being showed to a different civilization (if they exist ofcourse)
@Slereah if you don't mind, can you tell me about this portion of the book
what portion of what book
8:24 AM
@Slereah take it as a fiction, it's Blue Planet Project, are the facts written about dimensions correct?
If they are wrong, then , this book is fake, otherwise there is something real about aliens
8:47 AM
@Slereah ..?
1 hour later…
9:55 AM
Whenever I do computer working I am reminded of Spaceballs
10:21 AM
@AbhasKumarSinha Seriously? I don't know what the rest of that book is like, but if that sample is representative, the book is rubbish. If that were text on a website, it'd be bad enough, with the random capslock, and the numerous grammatical errors. As a book, it's laughable. If an author cannot write grammatical English, I'm inclined to suspect that they don't have good comprehension of technical material that they've read.
@AbhasKumarSinha Yes, it's possible that there are extra space dimensions, but that doesn't mean they could ever be used for interstellar transport.
10:45 AM
@PM2Ring Is it true that, extra dimensions make a alien craft look bigger than it is now?
@PM2Ring Interesting thing is that I've tried to challenge the facts in the book (of course which is known) by asking others over internet, but nothing there is found to be fake (all the scientific stuff written there is factually correct)
The author claima, that the 5th dimension wrt to us has 5th dimension, in which we are living
Is that correct?
Ugh! That book contains pictures of dead alien bodies... That looks total fake book...
Can anyone believes? THIS BOOK EVEN HAS A WIKIPEDIA!!! en.m.wikipedia.org/wiki/Project_Blue_Book
This is another level insanity!! I think there must be something correct here, otherwise how it can even have a Wikipedia?
Good afternoon sir @JohnRennie
Sir have you written any book about Classical Mechanics?
11:03 AM
@AbhasKumarSinha hi. I haven't written any books.
@JohnRennie so do you have research papers?
If I were to write a book it would be about special relativity, though I suspect the world has enough books about special relativity already.
All the colloid science work I did was for Unilever so it's not published.
@JohnRennie Woooow!! So, what are your research now? Do you do researches?
@AbhasKumarSinha I quit science in 1998 and switched to IT.
Wooow! So you even write papers in IT... That's incredible!
11:07 AM
@AbhasKumarSinha no, I've never written any papers in IT. I just fix broken servers :-)
@JohnRennie hehehe XD
@AbhasKumarSinha If those extra dimensions could be manipulated, then yes, you could make a ship that has more space inside than what it looks like from the outside. However, as far as I know, the theories that postulate those extra dimensions say that the compact dimensions have always been compact, they didn't collapse, and nothing can expand them.
@PM2Ring wooooooow incredible, that book contains alien facts!!!! :O
Can anyone verify the fact that our dimension is 5 th dimension to others in oher parallel universe?
Or dimensions are relative... Like our 5th dimension is 3rd dimension to others in 5th dimension...
@PM2Ring @JohnRennie Can this be verified?
@AbhasKumarSinha You can probably discuss the geometry of higher dimensional spaces in the Math chat room. We have the mathematics to discuss higher dimensional geometry, but that doesn't imply that spaces with that geometry are physically real in any way. For example in 9 dimensional Euclidean space, a hypercube with sides of length 1 unit has a long diagonal of 3 units.
@PM2Ring hehehe, sorry but that seems funny..
11:16 AM
@AbhasKumarSinha Sorry, that doesn't make much sense.
@PM2Ring are dimensions relative?
@PM2Ring If mathematicians discuss Space-time, then what physicist do?
Okay, I'll be reading something useful now, seems that these alien things are over my mind, anyways thanks for the time :)
@AbhasKumarSinha I don't know what that's supposed to mean. In 3 dimensional space, we commonly use 3 mutually orthogonal directions, X, Y, and Z, to describe the locations of points. But we are free to choose any 3 directions for X, Y & Z, as long as they are all perpendicular to each other.
@AbhasKumarSinha you need to understand what manifolds are to understand what physicists mean by dimensions.
@PM2Ring Okay thanks :) I think that book is not worth reading
@JohnRennie oh okay
@AbhasKumarSinha Definitely not! If you're interested in learning a little bit about higher dimensional geometry, take a look at en.wikipedia.org/wiki/Polytope and the links there.
11:26 AM
What's the smallest size of nuclear Bomb that can be safely made and tested in Home? (Assuming that it's small enough to have my home safe and small enough to be detonated easily)
I mean to ask from that what is the need to make bigger Nuclear Weapons and test, if they can easily test the smaller version
Bigger weapons = more sanctions, smaller = no sanctions and safe and secure
That'd be definitely the best idea overall to do testing and keep everyone safe and also less cost
@AbhasKumarSinha There's no such thing as a safe nuclear explosion! See en.wikipedia.org/wiki/Critical_mass
I guess you could have a couple of atoms of uranium or plutonium, and force them to undergo fission by bombarding them with neutrons, but that's not the same as testing an actual bomb.
Oh okay
Bye, I'll go to sleep now
Sweet dreams.
2 hours later…
1:41 PM
@JohnRennie are there really that many books about special relativity? Most of what I've seen is it getting crammed into a chapter of a classical mech or EM book, which is a bit disappointing
2:41 PM
@danielunderwood gourghoulon
Is what you need
It is 100% special relativity
Special Relativity in General Frames?
3:13 PM
6 hours ago, by Abhas Kumar Sinha
@Slereah ..?
@ Abhas If he knew something is too ridiculous, he will not respond
Took me 2 years and too many bans to figure this out
But anyway, I will not use a scifi book to answer physics questions
@Secret hey bro, sup?
@Secret need help
@Secret second theorem
Are + infinity, - infinity and only infinity different?
I am not sure how your book teaches you, but I am guessing the unsigned infinity might mean the sequence diverges
@Secret so series converge at infinity?? XD
Probably some kind of russian notations
Or what?
I think that can be a question in math SE
3:30 PM
well go ahead and ask that, don't forget to quote the book
In mathematics, the affinely extended real number system is obtained from the real number system ℝ by adding two elements: + ∞ and − ∞ (read as positive infinity and negative infinity respectively). These new elements are not real numbers. It is useful in describing various limiting behaviors in calculus and mathematical analysis, especially in the theory of measure and integration. The affinely extended real number system is denoted R ¯ {\displaystyle {\overline {\mathbb...
The reason they don't use $\infty$ is because they added two new well-defined numbers, $+\infty$ and $-\infty$, to $\mathbb{R}$, whereas saying $\infty$ when working in $\mathbb{R}$ is just a shorthand for multiple possible situations
@bolbteppa Sir, I'm afraid, extended Number Line system (surreal, hyperreal number) are out of syllabus, this book was given to us to be solved as a 11th grader
@bolbteppa So, standard Infinity means both positive and negative infinity, not sure if that is what you mean?
Well, the symbol $\infty$ is really just a shorthand, it's not well-defined, it means whatever you want it to mean, usually it means 'does not exist' or $1/0$ or something, it can mean both $+\infty$ and $-\infty$, it's a context-dependent shorthand, which is why people prefer to just add $+\infty$ and $-\infty$ to $\mathbb{R}$ formally so we can distinguish what's going on with a given limit better
3:46 PM
@bolbteppa Okay, thanks sir, I've got the point here. Thanks :)
4:31 PM
@PM2Ring forums.xkcd.com/… Looks like they shut it down when he tried the same attitude as he used on SE. Thankfully for the sake of scientific advancement, it looks like he posted a similar thread today
> If you have read through this thread, mark yourself as one of the first to bear witness to the largest scientific discovery since the double slit experiment.
The same guy said he would "make sure we are held accountable" for closing a question that was off-topic, and not even worded as a question physics.stackexchange.com/posts/481871/revisions They are quite confident that they found something paradigm-shifting
Yeah I got to see that one in action. Now you have me wondering if "largest scientific discovery" or "make sure we are held accountable" is worse
hmmm Baez gives 10 points to paradigm shift and potentially 70 for the held accountable one
5:03 PM
@JMac I just noticed the new thread. It won't last long if he doesn't behave himself.
I was thinking of calculating his Baez crank index, too. I'm sure he's got quite a few points already. He hasn't quite compared himself to Newton or Einstein, but he's come close. And claiming to be the biggest thing since the double slit experiment is rather grandiose.
Another mod has just locked it. Oh well. forums.xkcd.com/viewtopic.php?f=18&t=126667#p4458097 I love Second Talon's colourful analogies. :)
My favourite line is: "Weirdness is when an atom can only be described as math."
5:26 PM
One of the reasons I didn't run for mod is that "objective", civility-based rules are simply not enough to shape the kind of community you want.
That guy's posts are perfectly civil, but he's still lowering the quality of that forum.
5:38 PM
Calling people snobs & twats isn't exactly civil. But I guess he felt provoked, and therefore his insults were justified, in his mind.
5:54 PM
I just voted to close this optics question as a dupe: physics.stackexchange.com/questions/482724/… The existing answers are not good. :(
6:06 PM
hello people. So I just asked this question. I didn't expect it to be too controversial of a question, but from the comments I got I guess I was wrong. I would love to hear opinions from some people here about the objections of G.Smith in the comments.
Q: Do the eigenvalues of observables have direct physical meaning, or do they only refer to post-processing quantities?

glSIt is well known that in quantum mechanics any Hermitian operator $A$ can be thought of as an observable. Given any (pure) state $\lvert\psi\rangle$, measuring such observable gives an average measured value of $\langle A\rangle_\psi\equiv \langle\psi\rvert A\lvert \psi\rangle$. Being $A$ Hermit...

I don't really get what people mean when they say that "measurements produce eigenvalues". Any measurement will produce one of a number of possible outcomes, not an "eigenvalue". If anything, I would argue that one measures one of the eigenvectors of the observable that corresponds to the measurement, or more precisely, a state that we conventionally associated with one of the eigenvectors of the observable.
2 hours later…
8:07 PM
@knzhou hi. Sorry, it's probably better if we discuss about this here rather then in the already too long comment section in the question.
@knzhou I wonder, do you not agree that it is wrong to say that one "measures the eigenvalues", or similarly that the eigenvalues of an observable have direct physical meaning?
8:32 PM
@glS the question seems to evoke the copenhagen interpretation vs alternatives. suggest you look into weak measurements. also Neumaiers thermal interpretation. in short there are standard and new/ unorthodox schools of thought on the question and also new experimental aspects/ findings.
@vzn wait what? How is this about interpretations of qm? I am talking about how things work at an extremely practical level here. Like, about what one actually observes in an experiment. I'm saying that when you in your favourite quantum optics lab, you observe one detector or the other "clicking", not the photon being "$+4$" or "$+123$", whatever that means
@glS this sentence sounds very similar to (new) ideas/ conclusions by Neumaier.
> But then, if this is true, I wonder: why do standard treatments of quantum mechanics put so much emphasis on observables, rather then simply explaining that what is of direct physical significance is the probability associated with a state collapsing in a given basis, with the concept of observables just adding some post-processing on top of this?
@vzn link?
May 2 at 1:15, by vzn
@PM2Ring btw real explosions involve cooling also. which reminds me, a sharp cohort cited Neumaier a long time ago but havent seen him cited in ages. think very worthwhile to look at for anyone who puzzles over interpretations/ foundations/ bohmian/ fluid mechanics etc Foundations of quantum physics II. The thermal interpretation / Neumaier https://arxiv.org/abs/1902.10779
@vzn but that is about alternative interpretations of QM. I'm not in any way dealing with interpretations of quantum mechanics here, I'm simply questioning what is actually directly observed in any given experiment, or equivalently, what is the physical meaning attached to an "observable". I honestly sort of thought the answer to this question would have been that it is totally obvious that eigenvalues are only about how the data is post-processed.
8:53 PM
@knzhou I consider those issues one of the biggest frustrations of the job.
@glS historically the word "observable" probably comes from german, it was likely bohrs terminology, and yes its maybe not the best word in retrospect, and neumaier has a strong point about that at the end of his paper, others have made similar points.
There are, however, few alternatives. Math Overflow has moderators who, according to the Shog make the local gang look like "kittens", but I don't think there is a consensus for that here.
And I like to see the community reject bad ideas rather than counting on the argument from moderator authority. Even when we are talking about topics where I am arguable a narrow gauge authority.
@gls quantum mechanics produces eigenvalues as expected values of a given measurement, e.g. the measuring the x coordinate of a particle gives a real number, you're not measuring eigenvectors...
@bolbteppa I'm saying that you measure one of the possible outcomes, each one of which is associated with one of the eigenvectors. "produces eigenvalues as expected values of a given measurement" how so? If anything, the eigenvalues are the weights which are attached to each outcome's probability: $\langle A\rangle=\sum_k \lambda_k \underbrace{\lvert\langle \lambda_k\rvert\psi\rangle\rvert^2}_{p_k}$.
@glS I'm afraid I don't really understand the question - couldn't you similarily claim that the results of classical measurement are "post-processing"? Do we really measure the "value" of e.g. angular momentum or do we see a particular state of the system and "post-process" this phenomenon to obtain a value?
9:05 PM
@ACuriousMind yes! that is my point. In classical physics this is (should?) be obvious: one simply conventionally attaches numbers to different possible experimental outcomes, and then computes quantities out of these. What I'm trying to argue is that the same is essentially true in QM: the eigenvalues are values we conventionally decide to attach to the possible experimental outcomes. (...)
(...) In this case, however, the distinction can be of greater relevance, because it means that there is no "quantumness" associated with the eigenvalues of an observable, because those are simply classical quantities that we attach to the experimental outcomes after all the quantum shenanigans happened.
I don't get how that's relevant.
@ACuriousMind define "relevant" =)
@glS You said: "In this case, however, the distinction can be of greater relevance"!
And I don't get for what this is supposed to be relevant
@vzn The founders of qm were strongly influenced by the philosophy of Logical Positivism, and "observable" is a term that logical positivists would heartily approve of. FWIW, Bohr was a Dane, and although he could speak German it wasn't unusual for Germans to poke fun at his accent & weird grammar.
@ACuriousMind well, among things it means that there is really no need for the concept of "observable" in QM, at a foundational level. One can, in principle, fully get away with dealing only with states and the probability of finding them in other states. If what characterises a measurement is only the measurement basis, and not the eigenvalues of the observable, then why do we bother with the concept of observables at all?
9:10 PM
@gls what you've written, $\sum_k \lambda_k p_k$, is the expected value of a random variable $\lambda$ with values $\lambda_k$, i.e. the value we expect to find on measuring, i.e. the expected measurement, how is this in any way essentially the same as in classical mechanics?
@glS While the specific numeric value is not significant, there is significance to the relations between the eigenvalues. Take e.g. energy: It is very relevant what the differences between the eigenvalues are, as this determines e.g. the spectral lines of an atom.
@bolbteppa I'm afraid I don't follow here. You can still deal with expected values of random variables in classical mechanics, can't you?
Forget about wave functions etc... lets just treat the $\lambda_k$'s as possible values of say the velocity/momentum of a given particle we could expect to find, along with probabilities $p_k$ for measuring those values - quantum mechanics says this is the best we can ever do, ask for 'expected values' for a given measurement because the value we get will fundamentally be random, classical mechanics says we fundamentally can know the value at each moment in time,
the concept of random variables is fundamentally irrelevant, at best a fiction arising from experimental error that someday (according to classical mechanics) we should be able to correct for, in QM it's fundamental however
And once you are at the point where you have a measurement basis and these relations, you can arbitrarily pick a zero and get the basis and (one choice of scaling for the) eigenvalues, and a basis and eigenvalues in turn define a self-adjoint operator, so all you're complaining about is that the data is usually given in one form (an operator) and not an equivalent one (basis + eigenvalue relations), which is purely aesthetic.
@ACuriousMind ah, yes the Hamiltonian is indeed a weak point in this argument, I know. It is also largely the reason I'm even asking this here, as it confuses me. It does seem that the eigenvalues of $H$ (or rather, the ratio between them) do matter for what concerns the dynamics. But putting this specific case aside for a second, do you agree that the choice of eigenvalues of an observable (that is not the Hamiltonian) is only a matter of convention (trivial as this statement might be)?
if not, then it would mean that measuring a 1-qubit observable with diagonal $(1, -1)$ involves doing something physically different than measuring an observable with diagonal, say, $(123, 987)$. If so, why so?
9:19 PM
@glS Again, the specific numerical values are arbitrary, but their relations are not. It is not only the Hamiltonian that has significance for the dynamics: Consider that one can famously measure (by interferometry) the relative phase incurred by rotating one part of a particle pair through application of a magnetic field, and that the magnitude of this relative phase (for a fixed apllication of magnetic field) depends on the spin eigenvalue of the rotated particle.
You can try to play word games where it is not "the eigenvalue" but a nebulous property of the specific states, but in the end it is a well-demonstrated fact that a straightforward way to encode this property is by representing the state as a vector in Hilbert space and the nebulous property as the eigenvalue of the standard spin operator.
@ACuriousMind I'm not complaining or trying to play word games. I'm merely trying to understand if my understanding is right here, and if not, what I'm missing.
@ACuriousMind I am not questioning the practical usefulness of using observables, either
@glS No, again, you're focusing on the specific numerical values there, and they are arbitrary (simply by our freedom to choose units we like). But this is no different in classical physics - the specific numeric values of position don't matter, and yet you don't complain that classical physics focuses on the importance of $x(t)$ and not the "post-processing" by which we obtain specfiic values of $x$.
The eigenvalues of the angular momentum operator in the hydrogen atom are not some matter of convention, neither are the energy levels...
@glS Sorry, "why do we bother with the concept of observables at all" sounded to me pretty much as if that was what you were doing :P
@bolbteppa Of course they are, at least their specific numerical values (in the absense of gravity the choice of zero potential and hence energy is arbitary). What is not arbitrary is dimensionless values, such as the ratio of two energy eigenvalues.
I don't usually get involved in discussions of quantum foundations because I am very much out of my depths, but ...
9:29 PM
But still, I don't think there is anything here specific to quantum mechanics. The ontological disconnect between the "actual" results of measurements and specific numeric values is there in classical mechanics as well. Which is why I say I don't really understand the question here.
@ACuriousMind I'm not complaining there simply because it is obvious in that case. But I'm a bit confused now because you say that specific numeric values don't matter, to which I agree, but then you also say that their relations are not. But then, what if instead of measuring the position of a particles whose position is in $(0, 1)$, I decide to label the positions is some super-awkward way (applying some discretisation for convenience) by permuting the numbers associated (...)
Is it useful to make a distinctions between most real measurements (which are weak in the sense of "weak measurements") which result in a pointer reading that is only loosly coupled to the eignevalue and a idealized measurement in which the pointer is left in a state that exactlhy reflects the eignenvalue associated with the eigenstate that obtains at the end of the measurement.
with the different positions? Does this change anything in the way I interact with the particle? No, because what I actually observe are probabilities associated with different positions, and the eigenvalues of $\hat x$ only matter when I want to compute something like an "average positions" or other functions of the probabilities
@glS You merely complicate the relations - when you want to use the standard formulae for your permuted positions to compute something else, you have to undo the permutation, and then apply whatever weird transformation you chose for the quantity that is the result of the formulae.
@ACuriousMind let me then put the question in another way then. Can you in principle reformulate QM, and in principle derive all results, without dealing with observables at all, but only dealing with states and the associated probabilities in different bases?
9:33 PM
@glS What, exactly, do you mean by "states" there?
@ACuriousMind sure, that would be a totally stupid thing to do, but you could do it, right? But then this means that I can freely scramble even the relations between the eigenvalues of the observable, and in doing so not really lose anything (except possibly my mental sanity because I am then only making the other formulas needlessly involved)
@ACuriousMind $\lvert\psi\rangle$? Let's stick with pure states here, I don't think considering mixed states would change things much anyway (and also let's not go into quantum fields and such)
You can't really get things like the general uncertainty principle without considering operators.
It is the commutation relation of the two obserables that appears on the r.h.s., I don't really see what contortions one could do to avoid that
@ACuriousMind yes that is something I was thinking about. However, you don't strictly need the uncertainty principle formulated in terms of observables, you can work simply dealing with a state $\lvert\psi\rangle$ and the associated collapse probabilities in different bases $\{\lvert k\rangle\}_k$. You still get the same information that the uncertainty relations give you while dealing with the relations between different measurement bases
I don't know if I'm getting what glS means, but as far as I know most of the time when you formulate uncertainty principles the observables are used to represent measurements, those you can deal with without observables, just define appropriate basis and unitary Stinespring dilations
@glS No, you're just encoding the relations in more inconvenient relations! Where you had $f(\lambda_i) = g(\lambda_i)$ before, you now apply a scrambling $\phi$ and get the relation $f(\phi(\lambda_i)) = g(\phi(\lambda_i))$ Really, I don't get how what you want to do is anything other than choosing new units.
9:40 PM
@ACuriousMind well, the "non-commutativity" of observables would then simply translate into the fact that measuring $\lvert\psi\rangle$ into a basis $\lvert k\rangle$ affects the results of measuring into a different basis $\lvert k'\rangle$ in which $\lvert k\rangle$ and $\lvert k'\rangle$ point in different directions
you can treat a measurement as "what if I restrict to this subalgebra instead" without really mentioning observables
(by the way, sorry if I'm not answering to other people's comments in the discussion, I simply can't follow the different arguments all at the same time)
ah don't worry I just jumped in mid conversation
@glS Yes, and instead of having the general UP you'd have to write that down for an infinity of different possible measurement bases and there'd be no easy to see pattern.
This strikes me as the equivalent of saying that one doesn't need to use classical Hamiltonian mechanics if one simply writes down a table that contains the values for all quantities of interest at times of interest
Which...is true, but that's not what physics is for. If you just describe your system by brute-force tabulating the possible outcomes, you haven't gained anything that would resemble "understanding"
@ACuriousMind eh, not really. $[A,B]\neq0$ is the same as $\{\lvert\lambda^A_k\rangle\}_k$ being inequivalent to $\{\lvert\lambda^B_k\rangle\}_k$ (with some appropriate definition of "inequivalent"). But again, I don't care about practicality here, I get that observables are useful in practice. I'm only trying to understand if in principle one really needs the concept, and cannot instead get away without it (impractical as that might be)
9:45 PM
When you ask whether one "in principle needs the concept", what is the purpose you have in mind there?
Needing X implies there is a purpose I need X for
What is the goal that you want to achieve with this alternative description of a quantum system?
@ACuriousMind say, I need to define a group is to talk about group theory. I don't need to define what a subgroup is to derive results about group theory (that do not directly involve a subgroup), it is just incredibly inconvenient to do so.
@glS The content of the UP is not just that some uncertainties are not zero, but that there is a very specific lower bound for them! E.g. knowing the spread in the position values of a Gaußian allows you to very specifically predict the spread in the momentum values. Where is this quantitative prediction encoded in your approach?
@ACuriousMind the goal is to understand what in an observable relates to properties inherent to quantum mechanics, and what instead involves classical reasoning on top of the quantum mechanics
@ACuriousMind correct me if I'm wrong, but the notion of observables is rarely used in quantum information theory as far as I know. The notion of conjugate observables is expressed in a way more similar to what glS describes with stuff like mutually unbiased bases
@ACuriousMind yes, and that you can get also classically by considering the relation between, say, time and Fourier domains, can't you? It is the same in QM, I don't need to talk about Heisenberg's uncertainty relation, as it is inherent in the definitions of $\hat x$ and $\hat p$, that is, in their eigenbases being one the Fourier transform of the other
9:50 PM
@glS If I pick any other observables beside position and momentum (such that they are not Fourier conjugate) your classical analogy is of no help anymore.
Yet the quantum uncertainty principle still holds, entirely unrelated to the Fourier transform.
@ACuriousMind you mean considering the generalised uncertainty relation $\sigma_A^2\sigma_B^2\ge\lvert\langle AB\rangle-\langle A\rangle\langle B\rangle\rvert^2$? There is no foundamental difference, this relation is essentially equivalent to specific relations between the eigenvectors of $A$ and $B$
@user2723984 Sure. That's because quantum information really is generally unconcerned with the values of observables! There aren't many interesting observables on a qubit.
@user2723984 that's where I'm coming from actually, and yes, I would confirm that in quantum information you rarely ever have to deal with observables. You can deal with POVMs, which are a generalisation of "simple" observables. Mutually unbiased bases are related but not quite the same as what I'm talking about, as they are bases with very specific relations among the elements
@ACuriousMind you do quantum information also with continuous variables, with position and momentum operators (which often are not actually to be interpreted as position/momentum of some particle, but that is another matter)
quantum information deals more generally with the information content of quantum systems, be them qubits or infinite dimensional systems
but I'd have to think about whether observables are relevant there and if not why not
Ah, all the "information information" sits in the density matrix. I will happily agree that observables are not very relevant for such considerations
9:56 PM
@user2723984 well, they are used of course, and one also deals with variations of uncertainty relations in that context as well. But this does not mean that they are strictly necessary, only that they are useful
@ACuriousMind sorry I don't get what you mean there. I am purposely avoiding density matrices here because that would only complicate things. We can stick with pure states for this discussion
I was just saying why I'm fully on board with saying that quantum information doesn't have a pressing need to involve many observables.
@ACuriousMind first, let me appreciate a German that writes Gaussian with ß =). Second, you surely can describe the spread without dealing with observables: you have a state $\lvert\psi\rangle$ which is some superposition of positions, say $\lvert\psi\rangle=\sum_x c_x\lvert x\rangle$ or the continuum version of this. You then apply Schrodinger's (sorry for the incorrect "o" here) equation, which gives you everything you would have otherwise seen using the operators (... cont)
(...) This is of course because I am still using "observables" in the definition of the Hamiltonian $H$, which determines how $H$ acts on $\lvert\psi\rangle$. Nevertheless, I can still say that except for the case of the Hamiltonian, I never need to talk about observables. In other words, I can define a single observable, which determines the dynamics, and dispose of observables anywhere else (again, that would be stupid, but still...).
Here's a question: How does your approach deal with degenerate eigenvalues?
It is a fact that if there is a degenerate eigenspace, then when I measure the corresponding eigenvalue, the resulting ("collapsed") state will be the projection to that space, not to a single vector.
So you'd need to do not only lists of measurement basis vectors, but allow for more general spaces. I.e. you're encoding the operator in terms of its eigenspaces.
@PM2Ring have read a lot of the history of QM... it does seem there is some valid point that so-called "observables" in QM are not directly "observed" and neumaier has tried ("valiantly") to carefully disentangle all this but it seems he has few acolytes so far...
10:11 PM
And then we're really at the point where the operator is fully encoded in the data except for a freely chosen bijection (the "scrambling" from earlier) between the labels of the eigenspaces and the actual eigenvalues.
@ACuriousMind indeed, measuring an observable with degenerate eigenvalues would correspond to performing projections with trace greater than $1$. Or more simply put, you have a measurement with a smaller number of possible outcomes. For example, you measure a three-dimensional qudit $\lvert\psi\rangle$ by only probing the probabilities (for example) $p_1=\lvert\langle1\lvert\psi\rangle\rvert^2$ and $p_{23}=\lvert(1/\sqrt2)(\langle2\rvert+\langle3\rvert)\lvert\psi\rangle\rvert^2$.
So I'm willing to believe that yes, if you are masochistic enough you could probably get all of QM this way carefully avoiding the notion of "operator", since there's a bijection between the operator data and this "projection space data".
But, fun fact: There is a way to go the exact opposite direction and throw away the notion of states as vectors!
The C*-algebraic approach takes the abstract relations of the operator algebra as fundamental and gets "states" at first only as a consistent assignment of expectation values to these algebras. It is non-trivial but true that this turns out to be equivalent to the standard Hilbert space story (GNS construction) under reasonable assumptions.
So we can go "full operator" or "full state" and the standard formalism is somewhere in between. Seems legit.
@ACuriousMind I'm aware, and that is another interesting point indeed. My point with all of this is simply that, to try to understand inherently quantum properties (such as those that might lead to quantum advantages and such), one should not really focus on the eigenvalues of observables as those are not important at a fundamental level, and being only related to post-processing of data cannot be associated with quantum advantages or other quantum weirdnesses.
sorry I've got to run now. Let's continue later (or not, if people are too bored lol)
@glS I still think this talk about "post-processing" or "eigenvalues are not important at a fundamental level" is not really a valid interpretation of the situation. That there are different mathematically equivalent formulations of a physical theory does not mean that the objects that appear in one of these formulations are "fundamental" or not.
The abstract C*-algebras don't really know what an "eigenvalue" is, either! And if you try and look at everything in the projective Hilbert space, even all the linearity disappears!
None of the mathematical objects that appear in our models is guaranteed to have some sort of fundamental ontological relationship with nature. As much as you protested that, I think this part of your idea is still firmly in the realm of quantum interpretation.
@ACuriousMind if it was, then pilot-wave theory would be liked a lot more than it is: The velocity field generating the trajectories is just the probability current density / probability density. That's a perfectly sensible mathematical object, but that doesn't mean one is -compelled- to treat it as having ontological significance. (A Bohmian would, but that's not the default position.)
i guess the wavefunction itself (and whether one should be psi-ontic or psi-epistemic) is a good example of that
10:29 PM
I haven't yet found a good reason to care either way :P
I guess I'm $\psi$-apathetic
you're psi-sigh
@ACuriousMind would it be better if I replaced "fundamental" with "necessary" here? I then am trying to assess/ask/say that observables (with the possible exception of the Hamiltonian, I'm not sure about that yet) are not strictly necessary to do QM, though they might be very useful
@ACuriousMind they don't? I'm not particularly knowledgeable of C*-algebras, but as far as I remember you do define the spectrum of an operator $a$ as $\sigma(a)=\{\lambda\in\mathbb C:\,\,\lambda I-a\notin\mathrm{Inv}(\mathcal A)\}$, so you do have a notion of eigenvalues
10:45 PM
TIL Remember that you need protons for gravity as we understand it. From physics.stackexchange.com/a/482818/123208 I guess that question might need protection...
@PM2Ring interesting
@RyanUnger I'm slightly curious as to how he came to that conclusion, but I'm afraid to ask. :)
In other words, although I concede that talking about what is ontologically fundamental is ultimately a matter of interpretation. However, if one is interested in understanding what properties of nature are distinctively quantum (in the sense that they can be harnessed to elaborate information in ways not accessible via classical information processing), then it might be very useful to understand which parts of the formalism are directly related to the way we interact with physical systems,
and which parts are instead about us looking for a practical way to do calculations
@PM2Ring I asked him
I'll let you know
@ACuriousMind did you get the new total war
@bolbteppa sorry, just read this now. It is not really true though. You don't just have access to expectation values. Given a state you can sample from the corresponding probability distribution, that is, observe which one of the possible outcomes you get. This is the "most fundamental" question we can ask according to QM. Collecting enough samples we can eventually compute the associated statistics and with that compute expectation values

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