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12:32 AM
Prepare an ensemble of identically prepared photon pair states, measure thrm to get interference information. Group all of them together to get the hologram formed by the wavefunction of a single photon
Will read the paper in more detail to see how they carry this out

[Something else about inteepretarions] We define the wavefunction as something that contains all the information we can learn about the quantum state. After reading the PSE answers about probability in quantum
mechanics, I am tempted to think that a hermitian operator $\hat{M }$ kind of filter out the probability amplitude that corresponds to the information of the observable $m$ only, which then, when the overlap of that with the wavefunction is computed via the inner product $\langle \psi \lvert \hat{M} \rvert \psi \rangle$
with give the sum of all values of $m$ weighted by the amplitude squared (i.e probability), thus giving the expectation value as expected. However there is an obvious question to ask. We knew that
while there are a lot of observables are noncommutative, there are some that are. This means there exists some state $\lvert K \rangle such that it is both an eigenvector of two observables$\hat{A}, \hat{B}$. Then the interpretation of hermitian operators "filtering" out relevant information from the wavefunction will be challenging in this special case: Suppose I have the ket$\lvert K \rangle$, does it mean that it contains information of both$a$and$b$...? In that case we cannot really say a hermitian operator$\hat{A}$filter out information of$a$only from the wavefunction, as the above ket also contains information of$b$Perhaps one way to look at it is similar to the mathematical phenomenon of some equation of motion having multiple solutions at some positions in phase space, it might just happened that under a certain basis, the state vector that corresponds to some observable$a$happened to have the same form as$b$And since it is true for one basis, it is true for all bases as eigenvectors of linear operators are independent of basis. So maybe, if we insist on the filtering interpretation, there are actually two$\lvert K\rangle$that look the same but contains diffetent information of the wavefunction that corresponds to the quantum state it started with. An alternate way to think about this issue is that most filters are not perfect (because they are linear hermitian operators, there are only so many possible states they can act on and not give the same state vector), therefore when a hermitian operator$\hat{A}$is trying to filter information of$a$it is more common that the filtered result will contain information of, say,$b$. This will then explain noncommutativity But even then, there is still an interpretational issue on that: 1:08 AM @HDE226868 I give up. @0celo7 Search for knowledge. I've spoken with graduate students fluent in QM, yet they were giving the same definition for i that I originally gave. They were just as surprised when I brought up the problems you showed me with the definition. Because any state vector is linear, they can be linear combinations of other state vectors, which may contains eigenvectors corresponds to some hermitian operator. Then assuming we have the filtering information interpeetation above, we will need to deal with the issue of how can information of$a$(say position) be added to information of$b$(say spin projected at the z axis) to give information of a completely unrelated$c$(say the mass of a particle) If the above interpretation is the correct one (which I doubt it) it will imply information has a weird property that they are linear, they give different physical qualities when arranged in certain ways and then, one may come to the conclusion that all the physical things like charge, momentum etc might be nothing but a result of different ways these information overlap with each other. And in the macroscopic scale, decoherence (may be one of thr things that) ensure we only ever observed one of the many, with some probability Unless there's an experimental way to probe this interpretarion about information, I don't think the filtering interpretation I had make sense Therefore, either information has such weird properties, or that the filtering interpretation is wrong, thus meaning that quantum probabilities and logic cannot simply be treated in the same way (with some modification such as linearity and superposition) of claasical probability in the notion of that propositions filter out relevant elements from the set (except that by definition it just happened that a vector space is a type of set) Or put it more directly, quantum probabilities cannot be thought of as some form of venn diagram 1:48 AM Who do you think will read that? 2:02 AM Shog9 2:12 AM Okay I'm reading about degenerate gases and energy levels, so I need to know Is energy discrete or continuous? The whole "energy level" thing makes it sound discrete depends your system oh my god i'm tired All right, say we're in a neutron star I know nothing about those AFAIK they have no practical purpose 2:19 AM Okay, give me an example of a system where energy is discrete @0celo7 They're the epitome of awesomeness @SirCumference harmonic oscillator Why is energy discrete there? I'm not a physicist What are you then? doing homework 2:22 AM Oh crud, I didn't realize I was talking to you I didn't check the username... @0celo7 Anyway, on the chat? you're dumb Perhaps right now i'm tired tho You know, it's funny how many people say "Schrödinger's cat is an example of quantum mechanical effects" When really it was made to ridicule QM Schrödinger would be rolling in his grave It was made to ridicule an interpretation of QM @0celo7 And yet people are talking about how great of an example it is for QM effects Most of them think it was made to support that interpretation of QM Oh, but you're smarter than them, I see 2:35 AM I just looked up the history I feel like it got bastardized very clever >_> You're being sarcastic...stop being sarcastic... I don't know why you would thing I'm being sarcastic Did you look up some dirty pictures yet Yeah in other news Here's a picture of a dog what about it 2:40 AM It's the beauty of photoshop what Do you not see the human smile? Sure, so? My dog smiles I hope not like that Exactly like that 2:42 AM Jesus christ get it exorcized no you're Jewish, what do you know about that I know that face isn't holy You're christian I assume, you should be worried Don't assume 2:44 AM Okay religion do you believe in (if any) it makes an ass out of you and me, but mostly you. You know what happens when you assume, you make a you out of ass and me @SirCumference I worship the Nine Divines. @0celo7 Ok I can't really respond to that Depends, do you support Talos? 2:45 AM "Support"? Infidel! Wat At least you don't worship pokemon like half the people I know You realize how many people are obsessed with Pokemon go? No. I need to show that$L(f)\cdot L(g)=L(f\times g)$... Well a pokemon appeared in Central park, there was a riot never heard of "Pokemon" 2:47 AM Wallets stolen, people stabbed @0celo7 That's a rock worth living under It's called having a job I don't see why that means you've never heard of pokemon Me neither Right... You see the wikipedia vandalism that google cached? This here is why I don't trust Wikipedia There's a Vaporeon near me By the university gym 2:51 AM Nice knowing you Prepare for a stampede People in the south are civilized Unlike the ghetto north HAH @0celo7 History says otherwise >_> Excuse me? <_< Solved the problem With lotsa blood... Speaking of which, I'd get out of there if there's a vaporeon 2:54 AM @SirCumference The North started the war You wanna go? breaks wine bottle bleeds to death oh please :runs: I have German warrior blood also American weapon fondness Says you. 2:57 AM which part do you doubt as he rus... what I hope (wish) you are not a German. @lucas based on your behavior, the feeling is mutual assuming that means I might be related to you user218912 3:18 AM lolwut roger penrose authored a book with deepak chopra? amazon.com/Consciousness-Universe-Quantum-Physics-Evolution/dp/… user218912 that's ridiculous user218912 if you don't understand why that's retarded watch this youtube.com/watch?v=tZXewVWUKAM user218912 "consciousness is defined as a superposition of possibilities" Penrose is insane @3750 3:24 AM I wonder, if spinors are a linear way to track the topologies of special orthogonal groups...? user218912 @0celo7 how do you know? given their path dependence that depends on only two nonequivalent rotational classes @3750 he's gone into old physicist mode which is related to the topology of the rotational group being not simply connected (i need to check whether SO(n) is also mot simply connected later...) SO(n) is not simply connected for$n>2$. 3:46 AM Ok, so it seems we are fine (for now) > Lorentz transformations include 3-dimensional rotations as well as boosts. A boost transfers to a frame of reference with a different velocity, and is mathematically like a rotation into time. (the above quoted block of text is unrelated to the spin discussion) I wonder, since lorentz boost mathematically act like a rotation in spacetime, surely there is a way to measure the rapidity, maybe it is tan (v)...? We can directly see a rotational angle (by measuring how much something is spatially displaced by a rotation). Do similar notions for direct measurement of rapidity exist...? > Additionally, the assumption (known as microcausality) that spacelike separated fields either commute or anticommute can be made only for relativistic theories with a time direction. Might need to ask Acuriousmind later on why microcausality means spacelike separated fields can only be of the form$[/phi_1, /phi_2]=/pm 14:06 AM @0celo7 do you study all day or something? @Oblivno 1 hour later… 5:18 AM Ah, erm, yes, my latest badge. Erm, perhaps I shouldn't be too proud of this one :-) Congrats, I guess. :-) Are you the first? @JohnRennie Yes, though I'm not boasting about it. The (gold) badge of shame :-) 5:57 AM en.m.wikipedia.org/wiki/Spin_(physics) I might need to revise again why spin of a particle is always aligned with magnetic moment of a particle Also if spin basically describe the topology of SO(n) and in 3+1 spacetime, lorentz invariance result in spacelike separated field operators to tranform in a symmetric or antisymmetric way, then how is one of the two classes of action of SO(3) being assigned to what field operator. @0celo7 what's wrong with it? For example, why do field operators that creates something with lepton number 1, charge -1,mass m_e$(i.e. Electron) will have a SO(3) topology that is of the class that corresponds to "rotations by 4$\pi$"...? I might get my answer when I revise how the value of spin of particles are calculated, such as the Higgs, the photon etc. 6:35 AM Hi @JohnDuffield @Skill patrol: hi. I thought I'd look in while I wait for the wife to bring me my morning cup of tea. Ah, here it is. I am English you know. Yes, I know :-) What kind of tea? @JohnDuffield Not as obviously English as some. You won't be getting too many toodle pips from me. @skillpatrol : Tetley! Golly good :D Enjoy your Tetley. Hi @lucas Hi Nice day! @skillpatrol 6:44 AM @skillpatrol : would you believe it, I have to go already. :-) user116211 23 Many string theorists would like to know more algebraic geometry. There are a few of us who know algebraic geometry at a pretty high level (not me) but many more who would like to learn more and feel it would help with their research but find the literature very difficult. I think the optimal sol... user116211 I think the optimal solution would be to find such a string theorist and agree that you will teach them algebraic geometry if they will teach you string theory. user116211 7:22 AM Holy Lord! user116211 Bourbaki did write a treatise on Algebraic Geometry!! user116211 90 Wrong! Here is Bourbaki document on algebraic geometry, taken from the now available Master's Archives: click on Autres rédactions, then on Chap.I Théorie globale élémentaire (91 p.) This preliminary draft was apparently written (according to a penciled annotation on the first page) for Bourb... 8:19 AM Morning all (In england anyway) 9:17 AM What does mean the "PTSD"? Post traumatic stress disorder? Thanks! This means he/she has had an accident? Anything can cause PTSD; Common among soldiers, in accidents, victims of abuse also internet bloggers Go on..? 9:28 AM Oh a lot of internet people pretend to have PTSD Well! When something makes them upset fair enoug @NoahP Is English your native language? Yeah 9:31 AM May you please teach me some short phrases for thanking. Typically I'd say: Thanks! Cheers Or put thanks a lot, not many variations I want to use them for those people that post me answer or comment. Thanks or thanks a lot But I need more, because someone said me duplicate phrases aren't nice. I always say: Thank you. I cant really think of anymore, with saying thanks it doesn't really matter about repeating yourself I do it all the time 9:35 AM But sometimes more than one comment posted to me. There's no need to thank them all Shouldn't I answer all comments? I think this make the poster uncomfortable. I woudlnt expect a thanks for a comment OK. Thank you very much! :-) No worries! :) user116211 9:41 AM Something is weirdo here: user116211 1 MAFIA36790 thought it was inappropriate for a person (myself) with PTSD to ask people on the internet not to be abusive to them (as I did in the original post and he/she removed) so I abandoned this question immediately since honestly I think I ought to be able to ask people to not be abusive to ... user116211 Weirder fact is that it got a upvote too ;P user116211 I've left a comment though: user116211 I really don't know why you are doing all these. I removed an unnecessary part of your question; we have Be nice policy on SE; so you don't have to explicitly mention that. I'm not bullying you in any way. The comment was generated from the review as it didn't answer the original question. Also, if you have any grievance, you can post it Meta Physics SE. — MAFIA36790 21 mins ago It's got a downvote and a flag now too user116211 9:44 AM she seems to be adamant, though. He looks like be a boy. In any case, the question is not the place to discuss the issue user116211 @NoahP yes. Though im wary to point that out, so flagged it for a mod @MAFIA36790 If I was trying to find the temporal dependence of a variable, a, would that be da/dt? user116211 @NoahP Think so. 9:52 AM So I've got$a=a(t)$And I want the temporal dependence of that I know the initial condition is$a(t=0)=0$So I'm looking for$\dot{a}$? @MAFIA36790 user116211 @NoahP I'm not sure then; if they want the derivative. Because you said you have the function of a in terms of t. Yeah, but there's also$\dot{a}$'s all over the place I'm solving the Friedmann equation user116211 @NoahP Don't know about that; Is it a differential equation? user116211 If yes, then solve it for$a$; then use the initial condition to get the temporal dependence. user116211 10:01 AM @NoahP I can google ;P How would I do that exactly? user116211 @NoahP Show me your work then. I've got$k=0$and$\rho=Ca^{-6}$, where$C$is a constant from integration, not the speedy$c$So$(\frac{\dot{a}}{a})^2=\frac{8\pi G C}{3a^6}+\frac{\Lambda c^2}{3}$you suggest solving for$a$? user116211 @NoahP No time to go through the wiki article; if$a$is a function of$t$and you have got the initial conditions, solve it then using integration. I could marry you for pointing that out user116211 10:09 AM @NoahP ;D I've got some maths to crunch then user116211 @NoahP: One advice from me; learn how to solve differential equations and hence integration; they are easy; you can easily self-teach yourself. Ok, will do - Actually solving them is a new concept to me user116211 @NoahP solving means solving the differential equation which means integration ;) Do these symbols have different meanings? :-) ;-) :D ;D :P ;P :/ ;/ :| ;| , etc. 10:12 AM Ugh Can i integrate$\dot{a}^2$? user116211 @NoahP of course? I don't need to get rid of the$^2$somehow? user116211 I'm busy right now; would be back within an hour. Okay @MAFIA36790 @lucas when you use " ; " it means you are winking your eye. 10:22 AM And what does that mean? A wink usually means you are joking. Well! What about "D" and "P"? The first one is a smile, and the second one is sticking out your tongue. And have especial meanings? Nah They are what you would call body language 10:27 AM Thanks very much! You are so patient with me. on the net np = no problem etc 10:40 AM @JohnRennie I need you$(\frac{da/dt}{a})^2=\frac{8\pi G C}{3a^6}+\frac{\Lambda c^2}{3}$, where$a=a(t)$, and$a(t=0)=0$. Anyone got any ideas how to solve this for$a$? Well, solve it - I think for a, but this question states for the "temporal dependence of$a$" 10:57 AM I have passed differential equations some years ago and I have forgotten that completely. I am not sure, but maybe this help that differentiate both side of the equation with respect to the t for getting rid of power of 2. Is that not just going to get me further from the answer if I need to get rid of the derivatives? @NoahP Multiplying by$a^4$might help $$a^2 (\frac{da}{dt})^2=\frac{8\pi G C}{3a^2}+\frac{\Lambda c^2}{3}$$ @NoahP Sorry, as I said I have forgotten. It was a suggestion. @Slereah Don't you need$a^4$next to$\Lambda$then? @Slereah @lucas No worries" Then you can do$y(t) = a^3(t) \to \dot y(t) = 2a^2 \dot a ...$Nah I have no idea 11:01 AM Im lost Try "Polyanin A.D., Zaitsev V.E. Handbook of exact solutions for ordinary differential equations" It's a good place when you're lost on ODEs It looks horrific Math reference books all are but sometimes you gotta jump in I certainly can't do that on an empty stomach, so it's gonna have to wait a bit I'd say if you have to solve shit regularly You need that book The integral book And the function book 11:12 AM :( 0 Through substituting for values of$\rho$and$k$, I have:$H^2=(\frac{\dot{a}}{a})^2=\frac{8\pi G C}{3a^4} + \frac{\Lambda c^2}{3}a=a(t)$, and$a(t=0)=0$. Note that$C$is a constant of integration, introduced from the substitution to eliminate$\rho$- it is not the same as$c$. I'm not re... Might make this all a bit easier. I'm off for lunch now, back in a bit user116211 11:44 AM @NoahP You asked it at Physics? Why? It would be closed as in the present form, it is surely off-topic. user116211 @NoahP Better ask it as Math SE. user116211 Oh... it's already closed. @MAFIA36790 why would a question about finding the temporal dependence of the scale factor be off-topic for physics? user116211 @Jim Because it's homework based question. @MAFIA36790 homework based questions are not off-topic by default. They merely need to be worded properly. The question has potential so long as it isn't asked like "Show me what to do" user116211 11:50 AM I'm seeing there are two Friedmann equations. user116211 @Jim Sure, they can be. also, it's funny that people immediately turn to john rennie for cosmology. He is the top user in the tag, but I'd wager he's top in every tag because he answer probably 33% of all questions asked. What people should do is turn to the top users with high score/answer ratios highest would appear to be ted bunn, then pulsar. Lubos doesn't count because his presence on chat and the site in general is too unreliable user116211 Still think the question is more competent for Maths SE. Is "Ingvild Lorentzen" a chick or a dude? chick user116211 11:55 AM @Jim nods ingvild sounds like a chick name Hi, Maths SE aren't much help as they say its a physics equation - hence why I put it here It's currently got 2 re-open votes, and I explained my reasons in a comment @MAFIA36790 @Jim @Jim John Rennie was the one who replied to me first on here, hence the mention John Rennie. It's good to know he's consistent. First responder to questions both on the main site and in chat Do you think there is a chance it'll be re-opened? yes, there's a chance 12:06 PM Ok #:) user116211 @NoahP: Is$\omega$a function of$t\;?$you could help it along if you edit it to ask more about the concepts of what you are doing and less about the specific mathematical method of doing it user116211 i.e.$p = \omega(t)\rho\;?$No, its not a function of t @MAFIA36790 sometimes but not always 12:07 PM It's not in this case$\omega$is a constant usually it isn't a function of time. But I've seen cases where the observational data has shown it to change slightly through history Ah ok user116211 @Jim Then it could have been converted to single linear second order differential equation. I haven't considered the second Friedmann equation for acceleration, which is precisely that user116211 You could have add the two. By suitable choice of the parameter; that could be made to Riccati equation. 12:09 PM That isn't going to look nice probably, but I'll come back to you with it in a sec yeah, you kinda need it. Otherwise, it'd be possible to claim that at late times$a$approaches a linear function of$t$user116211 The rest would be easier then. Ok, I get:$\frac{\ddot{a}}{a}=\frac{-16 \pi GC}{3a^6}+\frac{\Lambda c^2}{3}$now take a look at the late times, when$a$should grow very large Is it not important that$a(t=0)=0$? 12:14 PM$\frac{\dot a}{a}\to C$and$\frac{\ddot a}{a}\to C^2$Youve lost me there, how do they tend to that? @NoahP it is if you define it that way. Makes the early times messier. I have to define it that way @NoahP if$a$grows large,$\frac{1}{a^n}$drops off to zero Ok 12:16 PM$C$is a random constant Not random, but yes no, I'm using it to signify "approaches a constant value" There is already a$C$in the equation Are you talking about the same one no ahhh ok 12:19 PM from that relation, it's easy to see that at late times,$a(t)\to e^{Ht}$Where$H=\frac{\dot{a}}{a}$? yup And this all considers that$a(t=0)=0$? Thats pretty crucial at early times, it's a bit more complex Ok 12:21 PM but you could start with$a(t)=F(t)e^{Ht}F$? Since when do you know about cosmology, Jim? @NoahP random unknown function @0celo7 since I'm a cosmologist @0celo7 @Jim meow @Jim Okay I did my degree on early-universe cosmology 12:24 PM @Jim How do extended objects move in GR? Do you have a good reference? @0celo7 extended objects? Is this more about your metal sheet metric? Metal sheet metric? Whut? Do you have me confused for @Slereah ... there's a chance I'm thinking about someone else nevertheless, it might be useful to clarify your question You can treat extended objects as families of geodesics With constant distance between them okay, but how extended is extended? 12:35 PM @Slereah Can you prove that$\mathrm{GL}(n)$has two connected components @Jim extended enough for your mom :) I can prove at least two connected components @0celo7 okay, but the extension I'm sure my mom would be fine with is probably still treatable as a point in GR. Let's face it, less than a foot is hardly worth mentioning 40 I've noticed that$\mathrm{GL}_n(\mathbb R)$is not a connected space, because if it were$\det(\mathrm{GL}_n(\mathbb R))$(where$\det$is the function ascribing to each$n\times n$matrix its determinant) would be a connected space too, since$\det$is a continuous function. But$\det(\mathrm{G...

@KNP It's not even continuous

@0celo7 Ah, you want to know math. Ask someone else. I'm a cosmologist that was formerly an engineer. I know the cosmology concepts very well and how to perform the math I might be called on to use, but I never studied the pure mathematics. Math terminology and identities get me every time

12:42 PM
What
I'm asking him an unrelated question

Ok, then my go-to reference for anything GR is Carroll. Beyond that, I have no standard reference; I just look it up as I need it

12:56 PM
In figure a how to find pressure at pallet of mercury
And the air is present in column 20 cm
Sorry the picture which is uploaded is inverted
please help

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