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J. Delaney

 Discussion on question by Rasmus: If …

Imported from a comment discussion on physics.stackexchange.co...
19
J. Delaney
16:12
@TobiasFünke I guess you can see by the votes that the question as stated doesn't make sense to a lot people. A more accurate way to state it, in my opinion, is to say that $\psi$ that satisfies the given condition is either the vacuum ($N=0$) or the zero vector.
J. Delaney
16:12
@TobiasFünke I see what you mean, but this is not the definition of an "$n$-particle state", at least not as used by physicists: of course $\psi=0 \in H_n$ for every $n$, but to be called an "$n$-particle state" $\psi$ has to have a unique $n$ associated with it, i.e. it needs to be a non-zero vector in the eigen-subspace of the number operator corresponding to the eigenvalue $n$.
J. Delaney
16:12
@TobiasFünke what is the meaning of saying that $\psi=0$ is an $N>0$ particle state ? which $N$ does this state correspond to ?
 

 Discussion between MathsGuy and J. De…

Imported from a comment discussion on physics.stackexchange.co...
J. Delaney
Feb 6 10:52
If you'll write it only with r1,r2,r3 it will become immediately clear to you
J. Delaney
Feb 6 10:41
Yes. If you assume r1=x1 , you don't need additional symbols. Reformulate your analysis using only r1,r2,r3
J. Delaney
Feb 6 10:36
If they are the same then why did you use two different letters for them ?
J. Delaney
Feb 6 10:33
@MathsGuy what's the difference between $r_1$ and $x_1$ ?
J. Delaney
Feb 6 10:33
@MathsGuy the functions $r_{1,2,3}(t)$ . You are substituting the 2-body solutions into the 3-body equations
J. Delaney
Feb 6 10:33
@MathsGuy No, your solution doesn't satisfy the differential equation. You forget that $r(t)$ are different functions in the 3 vs 2 body problems when you make the substitutions. Try using the same "trick" on a similar system of 3 linear equations and you will immediately see where you went wrong.
 

 Discussion on question by Andrzej: Un…

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J. Delaney
Feb 1 21:26
@Andrzej Yes. Hopefully this discussion contributed a bit as well !
J. Delaney
Feb 1 21:26
@Andrzej Assuming that systems can interact that way is equivalent to assuming that your Hamiltonian contains non-local interaction terms. In a sense you are right that QM by itself does not prevent this, but we know that QM is just the classical limit of a relativistic QFT, in which all interactions are local ... if you allow for non-local interactions, then it follows pretty much by definition that you will get FTLC
J. Delaney
Feb 1 21:26
@Andrzej That's right, so you just showed that $A$ and $B$ are interacting with each other. which means they can't be arbitrarily far apart ...
J. Delaney
Feb 1 21:26
@Andrzej That's what I meant by too many details obscuring the truth. You don't need to analyze every small component of an engine in order to know that total energy is conserved. Equivalently, you know from first principles that an isolated system must evolve unitarily. Conversely if it doesn't, that it must be in interaction (i.e. communicating) with another system
J. Delaney
Feb 1 21:26
@Andrzej Those are different concepts. Isolated systems means that the full Hamiltonian can be written as $H = H_A + H_B$ , which implies that the sub-systems don't interact and evolve separately unitarily. The density matrix describes entanglement. Two systems can be isolated and entangled at the same time.
J. Delaney
Feb 1 21:26
@Andrzej "local operations" means that the systems are isolated from each other. The time evolution of an isolated system is unitary, so yes "local operations" means "local unitary". You can have non-unitary local time evolution only if the systems are not isolated, i.e. interacting with each other. But obviously interacting also means transferring information in the usual way.
J. Delaney
Feb 1 21:26
@Andrzej There is a basic fact in QM that local operations on entangled systems do not communicate information (no matter what speed). Complicated experimental setups are usually just of way of hiding this simple truth behind confusing details. If you can't point to what exactly in your setup makes FTLC possible, then you probably just managed to confuse yourself
J. Delaney
Feb 1 21:26
@Andrzej The fact that a unitary operation on a full space induces a non-unitary operation on a subspace is quite trivial. I don't see how it is related to FTLC. You didn't write anything about it in your question so it is hard to understand what you mean
J. Delaney
Feb 1 21:26
@Andrzej what you describe is exactly the "quantum eraser" setup that I mentioned before ("erasing" the which-path information). It is well known that it does not violate causality or allow FTLC. See in particular this section
J. Delaney
Feb 1 21:26
Hi @Andrzej, I think everything you wrote is correct, but why would that allow FTLC ? Also the idea seem very similar to "delayed choice quantum eraser" experiment
 

 Discussion between PM 2Ring and J. De…

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J. Delaney
Jan 28 14:00
@PM2Ring I don't think that it is inconceivable that sometime in the not so far future, an LLM trained for example on the entire physics literature, would be able to answer physics questions at an expert level
J. Delaney
Jan 28 13:15
Conway's Game of life is cool, I particularly admire the guy who made this, I just don't see any meaningful contribution that Wolfram made to this (or any other) field
J. Delaney
Jan 28 13:12
Balancing parentheses with 7 nesting levels is hard even for humans! that's why we have sophisticated IDE's that find those errors for us. I agree that integrating LLM's with external tools is probably the way to go, but that's what everyone already seem to be doing now anyway with all the talk about agents. (if you ask chatGPT for example to multiply two numbers it will already now use an external calculator)
J. Delaney
Jan 28 12:15
@PM2Ring By the way, speaking of who/what generates complete nonsene ...
J. Delaney
Jan 28 12:15
@PM2Ring There are approaches to continual training of LLM's, so I don't think that's a major difficultly. As for logical reasoning, if the LLM sees in its training data for example patterns like "A is true, A->B, therefore B is true" repeated many times, then logical consistency can emerge from statistical plausibility (if A is true and A->B, then probably B is true). The layered structure of those transformers seems to give them a very good abstraction capability, that allows them to detect such patterns.
J. Delaney
Jan 28 12:15
@PM2Ring "Yes, it can say true things, but it can also say complete nonsense, and it can't tell the difference" - Unfortunately this is not unique to LLM's. People who say nonsense obviously don't think they are saying nonsense, and undoubtedly there is a lot of nonsense being said by a lot of people. Current AI models might not be at human level yet, but there is nothing fundamental that prevents them from getting there. After all, The human brain is also just a neural network.
 

 Discussion between Ryder Rude and J. …

Imported from a comment discussion on physics.stackexchange.co...
J. Delaney
Jan 24 17:50
I fail to see the conceptual problem. Yes, the state |psi(t)> evolves deterministically, but that doesn't mean the theory is deterministic. The QM state |psi> is not a physical state, it describes a probability distribution over physical states. QM is a probabilistic theory and therefore it only provides probabilities for all possible outcomes. I think the difficulty you are seeing is just because you insist on assigning physical meaning to |psi> which it is not supposed to have.
J. Delaney
Jan 24 13:55
Well, even in classical physics you can't create a simulation that includes yourself (because it will need to simulate the computer that runs the simulation and so on...) so those absurdities are not unique QM. When we are talking about modeling the entire universe we are obviously talking about a hypothetical universe that does not include ourselves, so that's beside the point. But we can definitely model a universe that operates by the same physical laws as ours, and use that to predict properties of our own universe.
J. Delaney
Jan 24 13:55
@RyderRude You could then compare this distribution to the number of stars we actually observe in the universe, and deduce whether our observation is consistent (statistically) with the QM prediction.
J. Delaney
Jan 24 13:55
@RyderRudeThat's not what I said. I said that the wavefunction of the universe could be simulated deterministically (by Schrodinger's eq.). If you'll run such a simulation, you'll find a superposition of many states corresponding, for example. to different star densities. So you get a probability distribution $|\langle \ n | \psi \rangle|^2$ over $n$, where $|n\rangle$ is a state with definite number of stars and $ | \psi \rangle$ is the wavefunction of the universe. This is just an application of Born's rule.
J. Delaney
Jan 24 13:55
@RyderRude In fact , people actually do treat the universe as a quantum system when modeling, for example, primordial quantum fluctuations that led to the the observed power spectrum of the CMB.
J. Delaney
Jan 24 13:55
@RyderRude Applying QM to the whole universe is not equivalent to MWI in the interpretational sense. For example you could just simulate the wavefunction of the universe on a computer and use it to study various observable characteristic (such as the average lifetime of stars as I mentioned before). There is absolutely nothing wrong or "absurd" about this.
J. Delaney
Jan 24 13:55
@RyderRude "any quantum system behaves deterministically when not being measured" - It is the wavefunction that evolves deterministically, but the wavefunction encodes probabilities. Applying QM to a system means using the wavefunction to calculate the probabilities of different outcomes. You can apply it to an atom and find the probability that it will decay, or apply it to a star and find the probability that it will explode, or apply it to the whole universe and find the probability that it will end up in whatever state you want.
J. Delaney
Jan 24 13:55
@RyderRude This is not true. Take for example protons colliding in the center of the sun. You can calculate - using Born's rule - the probability that they will fuse into Helium. From this you can deduce characteristics (such as brightness or lifetime) of typical stars in a universe governed by QM. There is no external measurement device looking at those protons. (Of course you can always consider the rest of universe as an "external" device, but that also leads to applying Born's rule)
J. Delaney
Jan 24 13:55
@RyderRude Whether or not the Born rule can be derived from MWI is a separate issue. Either way, it is an essential part of QM, so "applying QM to the whole universe", means also applying the Born rule. Your answer seem to suggest that applying QM to the whole universe somehow eliminates the Born rule. I don't see any basis for this claim.
J. Delaney
Jan 24 13:55
"The Schrodinger equation does not reproduce the Born rule." - This is a strange statement. Schrodinger's equation is not supposed to do this. The fact that $|\psi|^2$ represents probabilities is what relates QM to observations in the physical world. This remains true whether you apply QM to the whole universe or not.
 

 Discussion on question by Igarashi: T…

Imported from a comment discussion on physics.stackexchange.co...
J. Delaney
May 28, 2023 08:45
@safesphere I can assure you that anyone else reading your comments will come to the same conclusion.
J. Delaney
May 27, 2023 08:07
@safesphere Comments like these make me feel like I'm talking to a 12-years-old. The ability to admit one's mistakes is a sign if maturity, that you are clearly lacking. Without it, you will never be able to truly learn. Hopefully you will realize it when you grow up. I just hope it is not too late for you.
J. Delaney
May 26, 2023 16:58
It is more accurate to say that in GR all coordinates systems are equally valid. In any case, yes it is about definitions, so I'll take it as an admission on you part that your original comment on the Schwarzschild coordinates was wrong, even if you don't want to explicitly say it. That's settles the issue.
J. Delaney
May 26, 2023 16:29
If you think there is a better definition for a "source" of gravity other than the conventional one, you are free suggest it. I doubt if you can come up with a consistent definition, but you are free to try (nothing you said so far can be regraded as a general definition)
J. Delaney
May 26, 2023 16:26
@safesphere As I said, you are free to use whatever definitions you want as long as you explicitly define them, but you didn't. Saying that $r^2\equiv x^2+y^2+z^2$ certainly doesn't mean anything because it also holds for the conventional coordinates. Also, "spacelike" can also refer to the "t" coordinate inside the horizon.
J. Delaney
May 26, 2023 10:51
@safesphere
J. Delaney
May 26, 2023 10:48
lastly, as also mentioned in TimRias's comment, you can define boundary conditions on any Cauchy surface you like. If you want, you can define a coordinate system where the surface of the earth is at r=0, and similarly claim that this is the "source" of earth's gravity. But this will be meaningless precisely because this is an arbitrary choice.
J. Delaney
May 26, 2023 10:47
(of course, the earth's gravitatinal field also exists outside earth,
so your statement that the energy momentum tensor is "responsible only for how gravity changes inside a mass" is obviously false).
J. Delaney
May 26, 2023 10:47
The same applies for your usage of the term "source". Again, the fact that the energy momentum tensor is considered the source of gravity simply reflects the conventioal
meaning of the term "source". Just as the mass of the earth is the source of the gravitational field of the earth
J. Delaney
May 26, 2023 10:46
If you want to use unconventional definitions you can of course, but then you should
clarify precisely what you mean. You can't simply use the term "Schwarzschild coordinates" and then blame others for not realizing that you mean by it something different than the rest of the world.
J. Delaney
May 26, 2023 10:45
@Safesphere It is well known that the Schwarzschild coordiante "r" is timelike inside the event horizon. (In fact, I mentioned this already in my very first comment).
But this is completly irrelevant - your exact comment was: "In the Schwarzschild coordinates, $r^2\equiv x^2+y^2+z^2$ is zero already at the horizon",
which is a simple factual error given the conventional definition of Schwarzschild coordinates.
J. Delaney
May 24, 2023 10:01
(2) in GR, the "source" of gravity is the energy-momentum tensor, i.e. the RHS of Einstein's field equation. The Schwarzschild metric is a vacuum solution,
namely the energy-momentum tensor is zero everywhere outside the singularity (where it is not defined), so clearly it is also zero at the horizon. Furthermore
the Schwarzschild metric can be derived entirely (as exactly done in the original paper) from the requirements of being a static, sphericaly symmetric, and vacuum solution.
The event horizon then emerges as a property of the solution, it is not imposed as a boundary condition. It c…
J. Delaney
May 24, 2023 10:01
Physics is not a matter of opinions, it is about facts. While some issues are open for interpretation, there are also hard truths.
regarding the specific issue:
(1) In the modern conventional form of the Schwarzschild metric, which can be found in practically every textbook, the radial coordinate "r" is zero at the
singularity and equal to the Schwarzschild radius r_s at the horizon. This is an undisputable fact.
This coordinate is the same as "R" in the original paper you referred to. While it is true that Sch. also used a different "r" coordinate that is zero at the horizon,
 

 Discussion on question by nop nop: Mo…

Imported from a comment discussion on stats.stackexchange.com/...
J. Delaney
May 26, 2023 11:00
People shouldn't have to decipher your code in order to understand the question. You should explain more clearly what is the input data and what you are trying to estimate from it.
J. Delaney
May 26, 2023 11:00
You need to explain what you mean be generating data from this differential equation, what the actual data looks like, and if there is any statistical model involved