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amanuel2
amanuel2
1:19 AM
Anyone know how the prof got L/2 - D here?
 
Nobody recognizeable
Nobody recognizeable
1:42 AM
Hey guys is transmission and refraction of light same phonemena ?
 
 
3 hours later…
Sir Cumference
Sir Cumference
4:33 AM
Naïve physics
Naïve physics or folk physics is the untrained human perception of basic physical phenomena. In the field of artificial intelligence the study of naïve physics is a part of the effort to formalize the common knowledge of human beings.Many ideas of folk physics are simplifications, misunderstandings, or misperceptions of well-understood phenomena, incapable of giving useful predictions of detailed experiments, or simply are contradicted by more thorough observations. They may sometimes be true, be true in certain limited cases, be true as a good first approximation to a more complex effect, or predict...
Huh, TIL this was a phrase
 
Akash. B
Akash. B
4:44 AM
hi
hmm..
naive physics?
 
John Rennie
John Rennie
5:00 AM
@dmckee what did you think of Cloud Atlas? I quite enjoyed it but it seems to have a cult status that I'm not sure is totally deserved. If you're interested David Mitchell has written a novella Slade House that I really enjoyed, and a related novel The Bone Clocks that I enjoyed but is rather long.
I suspect 700 page novels and having small children don't work well together.
 
 
1 hour later…
Captain Bohemian
Captain Bohemian
6:07 AM
though email system is far easier than physical mail system, not many people really like to use it. Most mails in my mailbox are automatic mass mails rather than mails sent by a person who writes specific for me.
 
Sir Cumference
Sir Cumference
@CaptainBohemian I guess spam was more costly to send in the physical days
 
Captain Bohemian
Captain Bohemian
I think only professors really use email because only writing to them would get reply. Writing to any previous collaborator, peer, classmate, colleague seldom get a reply.
I actually didn''t write physical mails often even before email system is widespread. I usually only used physical mails for business purpose, rather for personal communication, like application, raffle.
 
 
2 hours later…
Novalium Company
Novalium Company
7:54 AM
Why an electron's spin has probabilities whether to be 0 or 1 (or a photon for example) an only when observed (say using a filter) we get t he final value. How are we not sure that the photon or the electron has been 0 or 1 the whole time?
 
 
1 hour later…
Loong
Loong
9:05 AM
 
Captain Bohemian
Captain Bohemian
9:31 AM
what is a dilation group? is it a renormalization group? What is a dilation operator?
 
 
2 hours later…
MartianCactus
MartianCactus
11:43 AM
so when we feel the warmth of the sun its because of photons hitting out face.
And giving their energy to us. But light gets its energy from electric and magnetic fields.
So that energy we feel actually is coming from the electric field and the magnetic field? Its the same energy that builds up between 2 magnets/in a circuit?
Thats soo cool
 
PM 2Ring
PM 2Ring
12:10 PM
@MartianCactus Light is electromagnetic energy. So an atom in the Sun's photosphere interacts electromagnetically with a molecule in your face by sending it a photon.
@dmckee Wow! I didn't realise that Varley had written a new 8 worlds story. I think he's a great writer, even though a lot of his stuff is a bit dark for my tastes. I've read a lot of his stories, mostly from the library, although I do have a few of his books on my shelves, including Steel Beach and the Blue Champagne collection.
 
 
3 hours later…
Semiclassical
Semiclassical
3:38 PM
@PM2Ring i'd quibble with that first statement a little, insofar as light also carries momentum
 
PM 2Ring
PM 2Ring
How can someone who knows what a derivative is even ask a question like this? :baffled: physics.stackexchange.com/questions/470749/…
 
Semiclassical
Semiclassical
but since $E=pc$ for a photon I can't object over-much
@PM2Ring never underestimate the human capacity for confusion
 
PM 2Ring
PM 2Ring
@Semiclassical Oh good. ;) MartianCactus's statement gave me an impression of a photon as a little ball with electromagnetic energy inside it, and I wanted to counteract that image.
 
Semiclassical
Semiclassical
(there's also the subtleties of what any particle "is" in QFT but ugh no thanks too much headache)
In classical electromagnetism, btw, I'd say that what light "is" is electromagnetic radiation i.e. electric and magnetic field propagating in spacetime
so it does depend on what theory you're talking about
 
PM 2Ring
PM 2Ring
I agree it's hard to not impose classical intuitions on this stuff. A particle in QFT is a quantized excitation of the field. You push on a field & up pops a little bubble. ;)
 
bolbteppa
bolbteppa
3:49 PM
@CaptainBohemian dilation group maybe means the subgroup of the conformal group consisting of just dilations?
Conformal symmetry
In mathematical physics, the conformal symmetry of spacetime is expressed by an extension of the Poincaré group. The extension includes special conformal transformations and dilations. In three spatial plus one time dimensions, conformal symmetry has 15 degrees of freedom: ten for the Poincaré group, four for special conformal transformations, and one for a dilation. Harry Bateman and Ebenezer Cunningham were the first to study the conformal symmetry of Maxwell's equations. They called a generic expression of conformal symmetry a spherical wave transformation. == Generators == The conformal group...
A particle in QFT is also just a particle depending on your pov
 
Secret
Secret
Now I get some very basic idea of the AdS/CFT correspondance
wtf is a penrose diagram doing in that question
and wtf is a "tensor network"
clearly google is not helping me to understand what that question is about lol
 
Semiclassical
Semiclassical
to be fair, tensor networks are a thing
I don't know them well myself but I'm at least aware of them
(by contrast, I am quite certain that the OP has absolutely no idea what they are)
the penrose diagram seems especially silly, given that if you stretch it out I think you just get a square with parallel red and blue lines :)
 
Secret
Secret
I have to say I am really patient with this guy when this penrose like diagram keep popping up in math chat in the past 2 years and yet he always failed to elaborate what it means and then keep raming questions after questions about it
but I am reaching my limit of weirdness-ness
 
Semiclassical
Semiclassical
you've got more patience than I
 
Secret
Secret
Weirdness is a strange thing, it allows you to capitalist even on pseudoscientists and climate deiners to get something productive from them
as long there is a promise of something truly unorthodox. But persistent inability to flesh out the unorothdox idea leads to repetition, and I really hate repetition
A lot of people get misclassified by me as RHVs because I see the same thing get repeated too long
 
Semiclassical
Semiclassical
4:03 PM
RHV's?
 
Secret
Secret
Told you that before: Residential help vampires
 
bolbteppa
bolbteppa
If only AdS/CFT was as simple as this :p
 
Semiclassical
Semiclassical
gotcha
 
Secret
Secret
Perhaps the most repetitive thing I am aware of is human history (has a cycle of roughly 50 according to sociologists and anthropologists). This sometimes caused me to do crazy things in political meetings, hoping that small events pile up enough to undermine all of society so that history will stop repeating itself lol
Other more interesting quirks include I will prefer to draw one big diagram, then to draw 4 sections of them in 4 pages, because that is less repetative because I only need to put everything into one page. Technically the former takes a lot more effort, and thus people tend to mistaken me as hard working because of that
but mathematically speaking, it really because we use a different cost function to evaluate effort so to speak
Anyway...
I am going to reread that "quantum mechanics cannot consistently apply to itself" paper again, because it appears that even many worlds failed to contain the contradiction according to an updated report by the team when they get interviewed by NewScientist
which is pretty surprising since observer dependent outcomes of observables that contradict each other (Alice see X, but Bob deduce Alice see anti X) should be pretty compatible to having different branches of the global wavefunction
In fact, I am also suspecting that interpreting quantum mechanics as it is (since Acuriousmind said my following way of looking at quantum mechanics is not an interpretation, but quantum mechanics as it is)
Mar 9 at 16:47, by Secret
I am suspecting, whether the correct way to interpret any quantum experiment, is to recognise that the whole setup itself is in some very complicated superposition (with different events all correlated with each other in some complicated way in both space and time), such that not just the observable itself, but all the which path information is only determined when a measurement occurred
may not be able to contain the contradiction. If that's true, it will support more the objective collapse interpretation, or that quantum mechanics is self inconsistent and something else need to be figured out
Either way, its exciting times for the philosophy of quantum mechanics
 
bolbteppa
bolbteppa
4:22 PM
What does this wave function collapse stuff even mean
Wave function collapse
In quantum mechanics, wave function collapse is said to occur when a wave function—initially in a superposition of several eigenstates—appears to reduce to a single eigenstate due to interaction with the external world; this is called an "observation". It is the essence of measurement in quantum mechanics and connects the wave function with classical observables like position and momentum. Collapse is one of two processes by which quantum systems evolve in time; the other is continuous evolution via the Schrödinger equation. However, in this role, collapse is merely a black box for thermodynamically...
 
Secret
Secret
Good question, its not being called the measurement problem for nothing. We absolutely have no idea what measurement is actually doing in quantum experiments
 
bolbteppa
bolbteppa
It seems as silly as having issues with 'flip a fair coin, heads or tails, can't know until you flip it, once you flip the outcome is known instantaneously'
 
Secret
Secret
well, its actually slightly different. After the coin is flipped, it is clear it lands either heads or tails. Even if the outcome is not disclosed, it does not changed the fact that it has a well defined outcome that can be extracted without being disturbed
Quantum states on the other hand, the outcome literally does not exist until you measure it (and what constitute a measurement and how it collapse the wavefunction is basically the measurement problem)
Things get even stranger for entangled states where you can have a degree of correlation between subsystems higher than their classical analogues (Bell Theorem in a nutshell), such as flipping two loaded coins at the same time so that if one is heads, the other must be tails.
ok the ordering of sentence is a bit weird. Let me said that again:
Things get even stranger for entangled states where you can have a degree of correlation between subsystems higher than any classically correlated systems (Bell Theorem in a nutshell)
an example of a classically correlated system is flipping two loaded coins at the same time so that if one is heads, the other must be tails.

Bell Inequality and PR box correlations simplified

Jun 15 '18 at 4:21, 1 hour 59 minutes total – 200 messages, 7 users, 0 stars

Bookmarked Jun 15 '18 at 6:24 by Secret

Semiclassical actually has a very good analogy in terms of football matches
 
Novalium Company
Novalium Company
4:50 PM
I'm still trying to understand how exactly is Earth's gravity intensity distributed? Maybe someone can draw something up :\
 
Secret
Secret
it's bumpy
 
Semiclassical
Semiclassical
@bolbteppa something something projection postulate
 
knzhou
knzhou
@bolbteppa Well, yeah. That would be a nice picture, but it also is exactly the picture Bell inequality violation tells us doesn't work.
 
Semiclassical
Semiclassical
it's also worth noting that, at least as far as the Bell inequality goes, you're not going to violate anything if you allow only two settings for each observer
(I have in mind the version of the Bell inequality where you send pairs of entangled electrons through a pair of Stern-Gerlach devices with particular orientations)
In order to test Bell's inequality, you need to have three settings available
e.g. Bell's original derivation allows Alice to use settings a,b and Bob to use settings b,c. (I may be mis-remembering the labels he used)
 
Novalium Company
Novalium Company
@Secret I'm talking about the cross section. (Like if you cut it in half)
 
Secret
Secret
5:05 PM
ah then I don't know
(do we even have the gravity map of our planet interior?)
 
Semiclassical
Semiclassical
probably depends on how far down you go
however, you do care about the internal gravitational field structure in places like the petroleum industry
as well as the internal magnetic field
since both get used to help accurate track the location of boring sensors
(one of my uncles is a petroleum engineer who works on this kind of thing)
 
Secret
Secret
True
 
Semiclassical
Semiclassical
there's interesting physics there, though not really my cup of tea
 
Physics Meta
Physics Meta
0
Q: Why "Leave Open" votes don't cancel the "Close" votes?

Dvij MankadOnce the question has been closed, one can cast re-open votes and a question requires exactly as many re-open votes as close votes in order for it to be re-opened. Given the homogeneity of time, I suppose this certainly means that a re-open vote, in effect, nullifies a close vote. If so, why a le...

discussion feature-request reopen vote-to-close moderation
 
bolbteppa
bolbteppa
Hmm... Measurement (assuming discrete spectra here) consists of a classical apparatus, represented by a quasi-classical wave function, which on interacting with a quantum system (represented by a wave function) is such that the state of the quasi-classical wave function changes, or rather the wave function of the whole system changes, and basically using linearity we then infer the state of the quantum system was in a certain state during that measurement,
but after measuring one gets a wave function for the system that depends on how the measurement was performed, but this wave function you get from measuring has to be related to the actual possible wave functions for the quantum system, and so should be expanded in terms of them, so that on measuring a second time there's no reason why you should get the same result
We really have no idea what's going on with the quantum system, we have no idea if it's doing anything, all we know is it does something to a classical apparatus and the possible things it can do to a classical apparatus are less or more likely using symmetry principles
 
Novalium Company
Novalium Company
5:16 PM
I want to make sure my understanding of light is right. So when a light source (such as the sun) sends light waves to Earth, things begin to reflect that white light while also acting as a filter (that's why we see a leaf green and blood red...). So the electrons in a green leaf receive the energy from the photons, the electron goes up and falls down while releasing energy in the form of light again but only at the specific green frequency? Is my understanding correct?
 
bolbteppa
bolbteppa
I don't know why this notion of wave function collapse even matters
Measurement in quantum mechanics
The framework of quantum mechanics requires a careful definition of measurement. The issue of measurement lies at the heart of the problem of the interpretation of quantum mechanics, for which there is currently no consensus. The question of how the operational process measurement affects the ontological state of the observed system is unresolved, and called the measurement problem. == Measurement from a practical point of view == Measurement plays an important role in quantum mechanics, and it is viewed in different ways among various interpretations of quantum mechanics. In spite of c...
 
Secret
Secret
actually, it is red and violet light that absorbed, not green light, as it is reflected (that's why the leaf looks green)
 
bolbteppa
bolbteppa
From this article:

"The above is completely described by the Schrödinger equation, and there are no interpretational problems with this. Now the problematic wave function collapse does not need to be understood as a process | ψ ⟩ → | ψ n ⟩ on the level of the measured system, but can also be understood as a process | ϕ ⟩ → | ϕ n ⟩ on the level of the measuring apparatus, or as a process | e ⟩ → | e n ⟩ on the level of the environment.
Studying these processes provides considerable insight into the measurement problem by avoiding the arbitrary boundary between the quantum and classical worlds [10], though it does not explain the presence of randomness in the choice of final eigenstate. "
So it seems like the issue is the randomness of the values the system takes, which goes back to the fundamental randomness of QM itself, which is not going to be rationalized away by formalism, seems like this measurement problem stuff is pretty much nonsense
 
Novalium Company
Novalium Company
@Secret So you if you strip away violet and red from white, you get green? The electron levels determine the color the atom will project?
 
bolbteppa
bolbteppa
I like that the article randomly throws in a dig at BM and many worlds for good measure :p :

"the appearance of collapse can be generated by either the Bohm interpretation or the Everett interpretation, which both deny the reality of wave function collapse. Both of these are stated to predict the same probabilities for collapses to various states as the conventional interpretation by their supporters. The Bohm interpretation is held to be correct only by a small minority of physicists, but there are no scientific reasons for that."
 
Secret
Secret
5:23 PM
well if you want to be predantic:
 
Semiclassical
Semiclassical
@bolbteppa actually, that last line seems more like an advertisement for BM to me
i.e. that there's no scientific reason why it's only held to be correct by a small minority of physicists
(not saying you should agree with it, but that's how I read their assertion)
 
bolbteppa
bolbteppa
Right, but it could also be read as saying it's a fringe interpretation which is completely unjustified haha
 
Secret
Secret
 
Semiclassical
Semiclassical
Not really. You can believe that's what they should have said, but I think the statement itself is pretty clear as to their opinion.
"The Bohm interpretation is held to be correct only by a small minority of physicists, but there are no scientific reasons for that."
 
Secret
Secret
@NovaliumCompany yes, though it is much more complicated than levels because you have basically a protein that does the absorption, thus the electronic wavefunction is very complicated
 
bolbteppa
bolbteppa
5:26 PM
I would read it as implying it's a fringe interpretation held by an ideological minority just by the way it's written
 
Semiclassical
Semiclassical
What else can "that" be in the sentence besides "the bohm interpretation is held to be correct only by a small minority of physicists"?
 
bolbteppa
bolbteppa
@Semiclassical have you thought about a PD in BM e.g. that Goldstein guy
 
Novalium Company
Novalium Company
@Secret Ok but proteins are bigger than atoms, so how can there be a protein inside the atom (which will do the absorption...)
 
Semiclassical
Semiclassical
plus, why would you use "but" in that sentence otherwise?
 
bolbteppa
bolbteppa
Right, the 'but' is implying it's like a debated issue
 
Semiclassical
Semiclassical
5:28 PM
Right, and in doing so it's claiming that the lack of support among physicists is not for scientific reasons
 
Secret
Secret
@NovaliumCompany No I mean the electron in the protein absorb the photons, it get promoted to an excited state of the protein to be used in photosynthesis, whereas all the green light, because the energy gap does not match, passes straight through and does not get absorbed
 
Semiclassical
Semiclassical
@bolbteppa PD = PhD?
 
bolbteppa
bolbteppa
Yeah
Post-Doc
 
Semiclassical
Semiclassical
oh
 
Secret
Secret
and meanwhile some green light get reflected back after some complicated absorb and reemission process I do not quite understood in detail
 
Semiclassical
Semiclassical
5:29 PM
hadn't really thought about that
 
Novalium Company
Novalium Company
@Secret Alright, thanks. :))
 
Semiclassical
Semiclassical
main problem is that Goldstein and co. are more philosophers than physicists, and I'm not trained as the former
 
bolbteppa
bolbteppa
Ah really
 
Semiclassical
Semiclassical
(though I could come up with a hundred different scenarios, and dismiss each of them...soooo)
 
Secret
Secret
(In the background and why I had not gone asleep yet) Waiting for a 2 hour long facebook video to upload which is 3/4 done
Australia internet really sucks
 
Semiclassical
Semiclassical
5:32 PM
the place that'd be great to work at as far as that goes would be the Perimeter @bolbteppa
 
bolbteppa
bolbteppa
Yeah, they even have a few QM interpretation video courses online on youtube
 
Secret
Secret
When the maths in QM is overinterpreted:
One started to wonder just what kind of "stuff" the quantum state is made of that allows all values of physical quantities to be represented as eigenvalues of some linear operator and treating the quantum state as another linear map
It's is actually amazing when you think about this, because by giving the relevant linear map, you can get momentum distributions from it, or position distribution or charge distribution etc.
and it is very easy to get lured into the thinking that if the wavefunction is a real stuff, then what exactly does momentum, charge, position etc. had in common that can be encoded by the eigenvalue of some linear operator acting on that wavefunction
 
bolbteppa
bolbteppa
Yeah
My sense is it stems from the HUP saying 'there is no path' so therefore we embrace randomness and emracing the idea that all we can do is work with randomness and so try to use probability to deal with this randomness. Also we know CM must exist in some to-be defined 'classical limit', so we need a probability that somehow reduces to CM in some to-be defined limit.
With this alone you could try and start from CM and end up with randomness somehow (as Schrodinger did to end up with wave functions and the SE), or keep exploring probability to directly set up QM
If you don't instantly accept the HUP I don't think anything quantum makes sense, hence my issues with BM
 
Semiclassical
Semiclassical
5:47 PM
I'd weaken that a little bit to "paths have no operational meaning" in the sense that you definitely don't get a meaningful path if you try to do what you would classically: measure the position of a specific particle at a sequence of times and infer a progression
that indeed has no hope of surviving the HUP
 
bolbteppa
bolbteppa
Right, and what is the reason why this happens?
 
Semiclassical
Semiclassical
simplest answer is probably: by measuring the particle's location, you've interacted with it, so the experimental context is now different than what it'd have been if you had not interacted with it
hence "what the particle would be doing if you didn't interact with it" is a different question than "what will the particle do if I periodically measure its position"
(if you're asking why HUP should occur in BM, it's because they still would accept the existence of a wavefunction governed by the SE. they'd just say that there's also a particle with a given configuration, which is being guided by that wave)
 
bolbteppa
bolbteppa
We would be able to correct for the interaction interfering and still predict where it would end up if it was just the measurement process causing deviations from an otherwise well-defined path
 
Semiclassical
Semiclassical
Why should we?
I will note that, in BM, there does remain an uncontrollable randomness. But it's an uncontrollable randomness in the initial position of the particle. We're coming up on stuff I don't really know as well as I'd like tho
 
bolbteppa
bolbteppa
Because all you'd be doing is adding an extra potential to $F = ma$ i.e. changing $F$ and getting a result which would reduce to a derivable result by taking away that potential and so you'd be able to predict the value of the measurement each time you do it, which is not what happens
 
Semiclassical
Semiclassical
5:57 PM
sigh
 
bolbteppa
bolbteppa
haha
 
Semiclassical
Semiclassical
How many times do I have to say that pilot wave theory is first-order dynamics rather than second-order
Closest you get to F=ma in BM is the quantum potential in Bohm's formulation of it
and the quantum potential is determined by the wavefunction
once you've measured the particle, you've changed how the wavefunction will evolve
and that'll change the quantum potential
I feel like we'd have to talk about how position measurements act on the wavefunction to get much farther tho
 
bolbteppa
bolbteppa
It really makes no sense to say the reason CM doesn't hold is because the measurement process inherently interacts with what you're measuring and changes the outcome, and just ignoring that interactions are just extra terms to $F = ma$ that we could reverse engineer and rationalize away to predict exactly the deviation a certain measurement process produces, and to then make a giant leap and just throw away CM for no reason to get BM when the justification to do so would actually save CM
 
Semiclassical
Semiclassical
I'll also object that, as I noted earlier, it's a first-order dynamical theory not a second-order theory
 
Secret
Secret
hmm... now that I think about that oil drop analogous system to BM, I actually don't think it can violate bell inequality. There's clearly a local hidden variable in the form of the waves propagating between two oil drops, mediated by the oil bulk itself
 
bolbteppa
bolbteppa
6:03 PM
This first order stuff is ridiculous, where does it even come from haha
 
Semiclassical
Semiclassical
that means you can't specify the initial velocity separately from the initial position
the latter determines the former. (at least as long as the wavefunction is non-singular---there's some annoying issues there)
@Secret i'd buy that, which is why I find it hard to really like the oil-drop stuff
 
bolbteppa
bolbteppa
In other words, the reasoning BM'ers give to need BM are contradictory
 
Semiclassical
Semiclassical
at best, it's analogy
@bolbteppa not really. you may not agree with their conclusions, but the reasoning is entirely consistent
the first-order business does get you into trouble, to be sure. i think you can trace the failures of relavistic BM back to that?
being sympathetic to BM would be a lot easier if relativity weren't a thing, frankly
 
Secret
Secret
In entangled states meanwhile, assuming one is using a nonlocal hidden variable theory, the hidden variable is encoded within the quantum state itself, and for theories that does not have hidden variables, it is the property of the quantum state itself that give rise to these correlations, or in a more epistemic and/or operationist approach, it is how the outcomes turns out to be correlated and no deeper reason on that other than one is actually dealing with one system
rather than n subsystems on their own
 
Semiclassical
Semiclassical
for instance, I think it's hard to sustain even something as familiar as the relativity of simultaneity in BM
the story only works if you can say that there really is some well-defined time-ordering to the sequence of measurements
 
bolbteppa
bolbteppa
6:09 PM
I just don't see how you can even leave the realm of classical mechanics if the claims of BM are to make any sense and this is made worse given that they simply steal special cases of QM equations and call them axioms, it's just a big contradiction when you actually look at it tbh
 
Secret
Secret
the reality however is indefinite time ordering is a thing proved by quantum experiments
You can have A->B and B->A in a superposition for example
 
bolbteppa
bolbteppa
I need to learn about this Bell inequality stuff and entanglement and EPR properly
It's amazing that Bell translated Landau's QM and ended up as the biggest champion of a theory which contradicts it's most fundamental claims (BM)
 
Semiclassical
Semiclassical
or maybe you're just ignoring what Bell actually did in favor of what you -think- he did
you see a sign that he may been smart enough to understand the situation, and you take it as a wonder that he still believe something silly
 
bolbteppa
bolbteppa
Yeah
haha
 
Secret
Secret
I wonder, if we can encode the time ordering this way: Instead of considering just one or a few measurements, consider all possible permutations of it (that effectively forms some kind of path in some parameter space) and the final pilot wavefunction then becomes a functional of such paths
 
Semiclassical
Semiclassical
6:13 PM
(I'm blatantly stealing from TS Eliot there: 'Signs are taken for wonders. “We would see a sign”')
the way I think of paths in BM, in an experimental sense, is like this
 
Secret
Secret
I.e. something like let $X$ be the jth measurement in a seequence of measurements $M$,and let $\psi_i$ be the initial wavefunction, then we have something like $$\psi_f = \psi_i[X \in M] = \int_{X \in M} \psi d[X]$$ thus $\psi$ is a linear functional of an ordered sequence of measurements and their superpositions of course
 
Semiclassical
Semiclassical
Suppose you prepare a particle by measuring its position and finding it at zero. You're then allowed to measure that particle once, at a later time.
You repeat that many times, looking for instances where the particle was initially found at zero, and then seeing where it's found at a later time.
 
bolbteppa
bolbteppa
@Secret looks path integral-ey
 
Secret
Secret
yeah, it does, though I have very little background of path integrals so I cannot formulate this properly yet
 
Semiclassical
Semiclassical
Setting aside any question of what theory governs the particle's motion: Can you use measurements like that to infer the particle's trajectory?
 
Secret
Secret
6:19 PM
but that is the best I can think to take account of the time ordering problem and hence relativity
 
Semiclassical
Semiclassical
You certainly can't use measurements like that to infer the trajectory of a single particle. One measurement per trajectory doesn't give you much info
But if you keep doing that over and over again, can you glean information about the trajectory from the statistics?
 
bolbteppa
bolbteppa
Looks like your $\psi$ being integrated against $d[X]$ where $X$ is a measurement i.e. an eigenvalue is similar to just saying $\psi = \sum_n c_n \psi_n$ where $\psi_n$ is the $n$'th eigenstate
@Semiclassical if CM existed in any sense of the word at all we would be able to derive rule by which the paths deviated from what we would expect by CM which would amount to determining the extra potential term we add to $F = ma$, this is what people did before QM was even developed e.g. Wilson cloud chambers
 
Secret
Secret
no, $X$ is an operator, think of it as some hamiltonian of the system corresponding to different interactions that the wavefunction can potentially experience for each instance. Thus integrating them all you are integrating all possible strings of $X$
 
Semiclassical
Semiclassical
I feel like you're setting up the following inconsistent conditions:
 
bolbteppa
bolbteppa
The fact there are absolutely no determinable rules leads to the HUP i.e. no paths exist, a single assumption from which basically everything else follows
 
Semiclassical
Semiclassical
6:24 PM
If BM is a version of CM, then it can't be consistent with reality. If it's not a version of CM, it's not worth considering in the first place.
 
bolbteppa
bolbteppa
That's what I'm saying yeah
Thats what I'm saying BM'ers are saying rather
 
Semiclassical
Semiclassical
Well, I certainly don't think BM is a version of CM
 
bolbteppa
bolbteppa
It has to be, this first order stuff says paths exist we just can't ever find them because we only have first order equations yet paths need 2nd order info to determine them
Therefore randomness because all we have is first order info
 
Semiclassical
Semiclassical
Only if you suppose that the initial velocity is determined independently of the initial position.
If you prepare a particle so that the initial wavefunction is a real Gaussian, then the initial Bohmian velocities are all zero. the initial positions have randomness, but the initial velocities do not in BM
 
bolbteppa
bolbteppa
But the point is you're saying the idea of initial velocity and initial position are definable concepts
This is an inherently classical idea
 
Secret
Secret
6:26 PM
So for example, Let $M$ be $\{H_{\uparrow}H_{\downarrow}\}$ which corresponds to hamiltonians that describes applying a downwards pointing electric field and applying an upwards pointing electric field respectively, then possible $X$ s we can have include:
$H_{\uparrow}H_{\downarrow}$,$H_{\downarrow}H_{\uparrow}$, $H_{\uparrow}+H_{\downarrow}$, $\frac{1}{2}H_{\downarrow}H_{\downarrow}$ etc.
 
Semiclassical
Semiclassical
What wavefunction you've prepared specifies the initial velocity in BM.
 
bolbteppa
bolbteppa
The whole thing is, QM says they don't even exist, if they did we'd have to be able to use some form of $F = ma$ because if paths exist we can describe them with ode's and we can then call those ode's 'Newton's new law'
 
Secret
Secret
Thus what one is integrating is really all possible outcomes when you do $X \psi$ where $X$ is the evolution induced to the wavefunction due to a measurement procedure
hence the "paths" are all hamitonians
 
Semiclassical
Semiclassical
Let me put it like this. In CM, you have randomness in initial conditions by having some smeared-out point in phase space
So a spread in initial position and initial velocity/momentum.
 
Secret
Secret
That.. should take care of all possible time orderings, and then we can define a lorentz transformation on the span of $M$ itself and it is then required to be invariant
thus the $X$ in $d[X]$ is operator valued
 
Semiclassical
Semiclassical
6:31 PM
In BM, you only have configuration space. There is spread in the initial position, but there’s no spread in the initial velocity
 
bolbteppa
bolbteppa
@Secret not sure that makes sense
 
Semiclassical
Semiclassical
To treat those scenarios as though they’re identical seems utterly wrong-headed
 
bolbteppa
bolbteppa
$\int \psi d[\frac{\hat{p}^2}{2m} + \hat{V}(x)] = ????$
 
Secret
Secret
right this is very weird, I think I need to study about path integrals a bit more before I can tidy this up
otherwise, I am basically trying to integrate all possible process that can take place on the wavefunction including all possible permutation of the time ordering of such events
 
ACuriousMind
ACuriousMind
@Secret There is no unique "evolution induced to the wavefunction due to a measurement procedure" in the formalism. A strong measurement happens instantaneously and it "collapses" the state non-deterministically into one of the eigenstates of the operator.
 
bolbteppa
bolbteppa
6:34 PM
@Secret I said you were integrating against the eigenvalues of those operators acting on $\psi$ which makes sense (I wrote the discrete version) but it's not path-integraley at all
 
ACuriousMind
ACuriousMind
Strong measurement is not a time evolution.
 
bolbteppa
bolbteppa
@Semiclassical why do we just have configuration space
 
ACuriousMind
ACuriousMind
That's the whole crux of the measurement problem, you can't make that go away by writing down ill-defined equations :P
 
Secret
Secret
ah right... I completely forgot that measurement is nondeterministic
 
Semiclassical
Semiclassical
Because that’s what BM takes as primitive
 
bolbteppa
bolbteppa
6:36 PM
But why do we just ignore velocity if CM is based on knowing both position and velocity
 
Semiclassical
Semiclassical
Because BM isn’t CM
I mean, the velocity exists. But it isn’t an independent variable
 
bolbteppa
bolbteppa
So the position and velocity both exist, but we can't use CM because magic...
 
Secret
Secret
@bolbteppa right, and as Acuriousmind pointed out, I have a misconception about measurements thus actually those equations of mine makes no sense
 
bolbteppa
bolbteppa
It's like saying we can't use math to describe curves because magic...
QM very explicitly says these concepts don't exist as physical realities (otherwise we could clearly use math to describe the paths)
 
Semiclassical
Semiclassical
The reason why we treat position and momentum as independent variables in CM is because it’s a second-order theory. F=ma establishes the initial acceleration but not the initial velocity or position
 
ACuriousMind
ACuriousMind
6:39 PM
@bolbteppa You're being uncharitable - you can't use CM to explain quantum phenomena because it doesn't work for that. Physicists never have needed any other reason to not use a theory other than it not working for their use case.
 
Semiclassical
Semiclassical
For better or worse, BM is first-order dynamics. It gives up on velocity as being independent from position in order to save the latte
 
ACuriousMind
ACuriousMind
@Semiclassical That's...a bit Newtonian :P Hamiltonian mechanics is a first-order theory and still treats position and momentum as independent!
 
Semiclassical
Semiclassical
It’s a first-order system tho
 
ACuriousMind
ACuriousMind
Hm?
 
Semiclassical
Semiclassical
Yeah, that was cryptic. I’m on mobile so limited in that regard
 
bolbteppa
bolbteppa
6:42 PM
@ACuriousMind right - CM doesn't work; and the reason it doesn't work is because CM is wrong and no fix can be found to the force laws, people tried to fix the force laws over 100 years ago but the experiments found total randomness for which no fix was possible - if the concepts of position and velocity (i.e. curve and it's tangent) existed in any sense then such curves could be described as solutions of ODE's, and these ODE's become Newton's New 2nd Law.
The issue here is BM is saying the curves actually do exist but we can't use math to describe them because magic...
 
Semiclassical
Semiclassical
Ffs
 
bolbteppa
bolbteppa
Just because the tangent to a curve depends on the position of the curve (i.e. velocity is not independent of position), that doesn't mean we still can't describe the curve...
 
Semiclassical
Semiclassical
@ACuriousMind What I meant was: Newtonian mech is a second-order theory on configuration space. Hamiltonian mech is a first-order theory on phase space
BM is first-order in configuration space
 
Secret
Secret
no it goes more than that, BM is basically saying that the velocity is not the tangent of the "curve" hence position. The positions, if I read semiclassical correctly, are all smeared out, thus there isn't really a curve to begin with
 
Semiclassical
Semiclassical
Nah, not saying that (I think)
 
Secret
Secret
6:47 PM
15 mins ago, by Semiclassical
In BM, you only have configuration space. There is spread in the initial position, but there’s no spread in the initial velocity
 
Semiclassical
Semiclassical
The tangent vectors would still be the velocity by definition
 
ACuriousMind
ACuriousMind
@Semiclassical So...giving an initial position fixes the initial velocity? How do they explain that this relation gets lost in the classical limit?
 
Semiclassical
Semiclassical
More precisely: giving the initial wavefunction fixes the initial velocity.
Note that the Bohmian velocity is determined as v=j/rho (current density over probability density)
And since the latter are computed from the wavefunction, they determine the velocity
 
ACuriousMind
ACuriousMind
Ahh, I think "first order in configuration space" might be a bit misleading here - the dynamical object here is the wavefunction together with the particle guided by it, not just the particle, right?
 
Semiclassical
Semiclassical
Yeah. Mostly I was trying to stress why F=ma isn’t a useful way to think of it
(Subject to the constraints of mobile of course)
On that note, I’m going to search out a power outlet so I can get back on laptop
Bbl
 
Secret
Secret
6:55 PM
Hmm... is it correct to say that in BM, the wavefunction basically generates a velocity field which then the particle get guided by it?
 
Semiclassical
Semiclassical
Yep
Hence why it’s called pilot-wave theory
 
Secret
Secret
right, I am more comfortable with delocalised things like fields and blobs of stuff lol
 
bolbteppa
bolbteppa
Basically, lets say we know the coordinates of a particle at one instant of time - what information do we need in order to predict where it will be at the next instant? We need the velocity at that instant. That's all we need. Knowing this info for all instants of time lets us know the path of a particle.
Now, mechanics is all about saying that hey, if we had equations for acceleration, and we solved them, doing so would specify the position and momenta at each point, and the initial conditions (for each choice of IC's) gives a distinct but equally allowable path. But really, the equations are just a crutch for us to find out what those positions and velocities at each moment in time are, the real thing is just knowing those positions and velocities.
The gigantic claim QM makes is that the very notion of position and velocity as simultaneously knowable quantities does not exist. BM'ers are saying these things DO exist as concepts, but the ad-hoc math equations we use to discover the positions and coordinates for a given system are just first order so we can't ever constrain both the position and coordinates from our equations.
That absolutely does not upset the fact that particles still follow true paths, and we inherently could discover the true path by repeatedly doing experiments. The equations failing to describe things is completely irrelevant to the physical reality of those paths existing. It's just denying basic mathematics to say paths exist but we are banned from describing them (why are we banned? magic).
 
Semiclassical
Semiclassical
You continue to create straw-men and burn them down
 
bolbteppa
bolbteppa
Unfortunately there's no straw men there
 
Semiclassical
Semiclassical
7:01 PM
Here's how one gets trajectories in BM. (on laptop now)
Here's the necessary ingredients: 1) There is a wavefunction $\Psi(x,t)$ evolving in some potential according to the Schrodinger equation. 2) The probability of the particle being at position $x$ at time $t=0$ is distributed as $\rho(x,0)\,dx=|\Psi(x,0)|^2\,dx$.
 
bolbteppa
bolbteppa
In CM we could measure the path of a given system each time and completely ignore the equations of motion and equally describe things. The fact BM equations fail to do this is meaningless, since we could equally just perform repeated measurements to measure the true path, and if measuring upsets the true path by interacting we can still just repeatedly measure to discover the potential we were creating and correct for it, from them on accurately predicting measurements
 
Semiclassical
Semiclassical
By making repeated measurements, you repeatedly change the wavefunction. Why should I accept that the particle should keep doing the same thing if I keep changing the wavefunction which guides it?
 
bolbteppa
bolbteppa
I would happily defend BM if it made sense, I see what I've described above as really fatal and basic issues, it's amazing how airtight Copenhagen QM is
 
Semiclassical
Semiclassical
Except that every time I actually try to use any equations you immediately shout down how they can't possibly make sense
 
bolbteppa
bolbteppa
and on Bohm, pretty sure he was questioning SR's legitimacy and we can guess why (BM) haha
 
Semiclassical
Semiclassical
7:10 PM
If you want to object to BM on grounds of SR, I've already noted that I think that's indeed quite bad for BM
But within the grounds of nonrelativistic theory? I think the 'contradictions' you assert reflect on what you -think- BM should be, rather than what it actually is.
 
bolbteppa
bolbteppa
Using equations in BM is fundamentally unjustifiable, we can use them out of practical necessity sure, but at least one experiment could be repeated over and over to eventually discover something predictable if paths existed in any sense of the word, and we could extrapolate. This is experimentally not the case, there's just absolutely no evidence to justify the fundamental claims of BM
 
Semiclassical
Semiclassical
Okay, this is pointless.
 
bolbteppa
bolbteppa
I think I am being super fair and open minded here tbh, I'm very open to the idea of it making sense, but it really just doesn't
 
Semiclassical
Semiclassical
BM is not experimentally different from QM. It can't be, because all that you're doing is interpreting what you get from the SE etc in a specific way.
and the fact that you respond to "you won't let me use equations" with "no, and doing so is being open-minded"
 
bolbteppa
bolbteppa
It's a shame I can see such basic basic problems with it, and I'm not even quoting other people's issues with BM, of which there are many claims
I said you can use equations out of practical necessity, but by BM's own logic the equations it uses are fundamentally irrelevant, I'm being so fair to it I'm following it's own logic
 
Semiclassical
Semiclassical
7:15 PM
basically tells me all I need to know about how fruitful any further conversation on this point is.
@bolbteppa WTF?
 
ACuriousMind
ACuriousMind
@bolbteppa From the outside, it looks like this: Semiclassical keeps telling you that what you think it is saying is not what it says, but you keep insisting that you got it right before he has elaborated what he was going to say. I'm no adherent of Bohmian mechanics, but it doesn't look like you're "being super fair".
 
bolbteppa
bolbteppa
@ACuriousMind well lets see - in these last few sentences, SC wants to use BM and examine what they imply, I said it's fine to use them out of practical necessity, but on a fundamental level they should be irrelevant because paths exist and so repeated measurements should be able to discover them, and I get no answer and am called closed minded rather than addressing my point that they should be irrelevant, clearly joking but still, I need an answer :p
the usual claim is we are banned from seeing the true path because a guiding wave is blocking us, or, as SC seems to be hinging on, that this first order stuff being the reason why we can't find the path, but I've argued why the first order stuff is irrelevant because the equations are irrelevant. The problem here really is my point is so airtight it can't be answered :p
 
Semiclassical
Semiclassical
In other words, you've already determined that you got it right and therefore what I have to say is entirely irrelevant.
Seems like ACM had it entirely right.
 
bolbteppa
bolbteppa
If we look at CM, I wrote above why the equations are just a crutch to discover something we could equally discover by repeated experiments, because those variables exist, in BM they exist too, there is absolutely no reason why we can't figure out what the real position and velocity in BM are, there is zero reason
 
krauser126
krauser126
Whats the chat room for physics related computation problems?
 
ACuriousMind
ACuriousMind
7:24 PM
@krauser126 You're probably looking for the Problem Solving room
 
krauser126
krauser126
Ah yes. Thanks
 
bolbteppa
bolbteppa
@Semiclassical can you say what you were going to say without using differential equations, you should be able to, just speak in terms of measurements and repeated experiments
 
Semiclassical
Semiclassical
Fine.
 
bolbteppa
bolbteppa
There's no fundamental reason why we can't figure out the true path in BM
Maybe I'm missing something
 
Semiclassical
Semiclassical
Prepare a particle with some initial wavefunction and in some potential. For simplicity, let me take the initial wavefunction to be a real Gaussian wavepacket and zero potential.
That wavepacket will diffuse in time as per the SE. As such, the probability distribution for finding the location of the particle later on will also spread out in time.
 
bolbteppa
bolbteppa
7:29 PM
I should have said can you say what you were going to say without using wave functions
 
Semiclassical
Semiclassical
No, and why should I? Like I've said, the whole reason it's called a pilot-wave theory is because the wave guides the particle
To speak of only the former is standard QM. But BM speaks about both.
Suppose we compute the probability to find the particle to the left of some $X$ at time $t$. (I am assuming a 1D example because that makes things less annoying.)
 
bolbteppa
bolbteppa
But if we forget about wave functions for the moment, and just look at an experiment directly, say we lives in 1897 and the electron was just discovered and found to exhibit weird behavior, what would you say as to why the experiments are not showing any notion of a path existing, even though you say the position and velocity as concepts inherently do exist
 
Semiclassical
Semiclassical
Again, why the hell -should- I forget about wavefunctions?
BM is not particle alone. It is wave and particle.
(Or, more generally: It is a wave in configuration space, and a point in configuration space.)
 
bolbteppa
bolbteppa
Because if you forget about the wave function and just look at an experiment, it's necessity should arise, otherwise it's 'irrelevant'
 
Semiclassical
Semiclassical
I don't think that will entirely work, on the grounds that you still need to be able to talk about an initial wavefunction in order for the story to make sense.
That's part of the initial conditions of the problem, just as the initial position of the particle is.
I'm prepared to drop that, though, if you'll grant the following: Experimentally, I can reproducibly prepare a particle in "the same way" while still getting a well-defined distribution of initial positions.
So experimentally I do, at least, have some $\rho(x,0)$.
 
bolbteppa
bolbteppa
7:42 PM
Well, I am trying to say it that it either has to entirely work without a wave function, or it will unavoidably lead us to the notion of a wave function
We have an experiment, it's giving us results, the results completely contradict classical mechanics and absolutely no fine tuning of the experiment is giving us results that would tell us a path exists, everything is so random it's telling us the very idea of a path makes no sense when you zoom in far enough, I'm pretty much asking you why we can still assume the concepts of position and velocity are legitimate if the experiments seem to show they are not.
What some BM people (I think not you) say is there is this weird ghost thing called a wave blocking us from measuring the true particle and we call it a wave function, and ignore the fact we could reverse engineer what it does to block us and so still discover the true path.
 
Semiclassical
Semiclassical
and again, you immediately give up on actually hearing what I'm saying
 
bolbteppa
bolbteppa
What I think I see you doing, based on previous discussions, is skipping over all of this and just using new concepts like wave functions out of thin air, 'because they work', and they do work... I am simply trying to argue this is something one can't skip over, and the reason it works actually comes from analyzing this issue (leading to normal QM), though it does look like one could skip over this stuff because it works, you know what I mean?
 
Semiclassical
Semiclassical
I am not invoking the wavefunction. I am invoking only the idea that, if we prepare the particle repeatedly in the same way, then there remains some initial distribution of particle positions.
 
bolbteppa
bolbteppa
Okay
 
Semiclassical
Semiclassical
That's all I'm doing here.
 
bolbteppa
bolbteppa
7:45 PM
Okay lets say we can reproducibly prepare a particle in "the same way" while still getting a well-defined distribution of initial positions
The subsequent measurements give random results right?
 
Semiclassical
Semiclassical
Right. Note, though, that this isn't a case of 'repeated measurements' yet. For that you'd prepare the particle, then measure its position, then wait a moment and measure again, wait and measure again, etc.
Here we're preparing the particle, measuring its position, then re-preparing it
 
bolbteppa
bolbteppa
Right
As in, we prepare it, measure it subsequently, then repeat all this over again
 
Semiclassical
Semiclassical
Right
Obviously that's rather tedious experimentally
And it's subject to annoying points about the resolution of your measurement device
 
bolbteppa
bolbteppa
This is exactly what we should do to test BM
 
Semiclassical
Semiclassical
I will also take it as a fact that, if I were to repeatedly prepare the particle in the same way and wait for a time $t$ before measuring, that we'll get some distribution $\rho(x,t)$.
And this distribution will in general be different than the one I started with.
 
bolbteppa
bolbteppa
7:48 PM
Wait, what do you mean
 
Semiclassical
Semiclassical
As in: Instead of preparing the particle and immediately measuring its position (and recording that as x(t=0))
we could instead prepare the particle, wait a time t, and then measure (recording it as x(t))
and then re-initialize, doing that repeatedly
 
bolbteppa
bolbteppa
I'm not sure what you mean, lets say we prepare the particle at the same initial position over and over again, and then measure it subsequently, we will get a distribution of positions of where the particle will be measured at the subsequent instant right
 
Semiclassical
Semiclassical
Well, I'm taking for granted that you can measure the particle at the very instant after preparing it.
i.e. that it makes sense to talk about an empirical position distribution at t=0.
(where t=0 is the moment in which it's prepared)
And then I'm saying that, instead of measuring it at the earliest instant after preparing it
you could instead wait a time t
in which case your measurement sequence would be: Prepare the particle at t=0, measure at time t=t' (recording this as (x(t'),t') ) and then starting over.
In that manner you'd get a position distribution $\rho(x,t)$ for whichever time $t$ you choose.
 
bolbteppa
bolbteppa
When we say preparing the particle at $t = 0$, that means we first measured it at $t = 0$, then at $t = t'$ we measured it again, right?
 
Semiclassical
Semiclassical
Hmm. Yeah, I guess I'd have to do that.
So we're doing a bunch of two-time measurements
The trickiness, as I think through it, is that you want to have the same initial state
 
bolbteppa
bolbteppa
7:54 PM
So we first measure an electron at $t = 0$ at the same place over and over again, and then measure again at $t = t'$ to see where the particle ended up
 
Semiclassical
Semiclassical
What makes it a little subtler than that, I think, is that in order to get the same initial state each time
 
bolbteppa
bolbteppa
We could do a million repeats firing an electron through a gas chamber by emitting it from some atom by shining light on the atom, 100,000 of them we measured at the position $x$ at time $t = 0$ and then measured again at $t = t'$ to find where it went to in the gas chamber
 
Semiclassical
Semiclassical
you have to prepare the particle in the same way each time. but if you measure its position at t=0 and get different results, then that's not really preparing it in the same way.
Maybe something more like this?
 
bolbteppa
bolbteppa
I mean, the real experiments like the Wilson cloud chamber felt they had corrected for this right
(I've never been in a lab so what would I know, it's all theoretical :p)
 
Semiclassical
Semiclassical
You start with your particle in a narrow parabolic potential. At t=0, you turn off the potential so that the particle is free.
 
bolbteppa
bolbteppa
7:58 PM
I mean the Rutherford Gold Foil experiment was done thousands of times right, they know how to credibly claim they have repeated the same IC's within a margin of error
 
Semiclassical
Semiclassical
I guess we'd need need to ensure that the particle is in the ground state of the potential.
Hell if I know.
The other annoying thing is that a typical scattering experiment is not 1D :P
That's not a big impediment but it is annoying
Maybe it'll help if I say where I want to go with this
 
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