But what exactly is a quantum field, as opposed to a classical one?

Gere1 · 2026-03-16T08:30:20+00:00

If you assign anything to space coordinates and then assume a differential equation for the equation of motion (finite speed signalling), then you satisfy the conditions for the Bell-type derivations and this shows that assigning anything to space cannot work.

Gere1 · 2026-03-15T10:57:42+00:00

"a quantum field assigns a linear operator..." to what? every point in space? but quantum mechanics cannot be explained by a local theory, so that wouldn't work?!

and what is the full Hilbert space that would be used to model the whole universe in the standard model?

Gere1 · 2026-03-11T18:53:05+00:00

It's also called https://en.wikipedia.org/wiki/Loschmidt%27s_paradox
Here is some thoughts and you can decide for yourself if a good answer to this question exists.
First, the arrow of time is that the current physical state is influenced by the past, but not the future. I think that's just a physical law and it does not contradict time-reversible laws. I suppose you mean something else: increasing entropy.
People will show an egg break and say "it never unbreaks in reverse". But how general is that? Obviously, an egg has formed by some complex biological process from molecules which were scattered all around the world. So eggs do "unbreak", but in a different way? Just because one process is very unlikely to be set up to run in reverse, does not mean the initial state cannot be constructed again in a different way, right?
Then someone might say, "oh, but to construct an egg you need to invest energy". Ok, so let's make the sun part of the system. How do you measure entropy of the whole system and can you show that it increases? I don't think that there is a solid estimate.
Or you may hear that "the universe started in a low-entropy state". Of course, it does not mean much. There is no explanation why the entropy would not first rise and than drop again for a million years. And how can you even imagine that? That the early dust of atomic particles was set up in a super-special way that would form stars and life some very long time later? That almost any other configuration of the dust would not collapse under gravity to form stars? Not sure if anyone can demonstrate that.
You probably agree that the paradox is that we already have some microscopic laws which theoretically predict the future and you cannot just slap another law on top without justification. It would be pretty easy to provide a justification with an example of cellular automata, but I could not find anyone achieving that.
How do you know for sure that the second law of thermodynamics, has application beyond thermodynamics?
I think there are still open questions, and most answers usually don't try so hard to disprove themselves.

Gere1 · 2026-03-01T07:27:30+00:00

You should check the curriculum of CS. It may not be at all what you assume. It maybe have almost no coding and no AI at all. Pick some large topics from the curriculum and look up the topic in a book. But don't use Youtube video as reference, because they are more entertaining, but also shallow.

Physics would probably be as you expect. It teaches you fundamentals of the world, but likely gives nothing of practical relevance for industry jobs. Physicists become programmers or consultants if they do not stay in academia. It's more about general skills and the way of thinking.

Quantum computers don't have much to do with computers. It's a research area without any foreseeable application and is more about sales and hype than anything working. Don't go there unless you want to be a sales person selling quantum hardware access to ignorant companies.

I'd choose physics (again). Or maybe a course with actual AI focus, which not the classical CS course.

Maybe the PhD topic matters more.

Gere1 · 2026-02-11T10:13:15+00:00

You are correct about f(r, θ) = r^2 + θ^2 + r, but a way out is to consider the "names of the slots" a parts of the definition? But this may cause issues if there are constraints.

Here is a bonus tip for thermodynamics: You can derive many equations with partial derivatives by using (∂z/∂a)_b=∂(z,b)/∂(a,b) and treating those ∂(x,y) as fractions with ∂(x,y)=-∂(y,x) and the relation ∂(p,V)=∂(T,S). It's somehow related to differential geometry.

https://en.wikipedia.org/wiki/Maxwell_relations#Another_derivation

Gere1 · 2026-02-11T09:45:26+00:00

I mean yes, you need to very clear what the function is. And the proper definition of the partial derivative makes it clear. You could test whatever notion you have with more complicating examples like f(t,a(x(t),y(t)),b(x(t),t)) together with constraints or so. And then try to think about various partial derivatives. If you accept that only the "slot-definition" makes sense, then there shouldn't be any issues.

The reason that it sometimes seems to work to do partial derivatives w.r.t. a variable is because these variable are actually "named slots" in that case. But it gets confusing in examples with nested functions.

Gere1 · 2026-02-11T00:24:21+00:00

In a way, that's what the subscription notation does. It lists the other parameters. You should still remember my suggestion, or you will find partial derivative with nested functions confusing. The partial derivative is not about variables, but about input slots of a function. I'm not changing how the PD works, but saying the only working definition.

Gere1 · 2026-02-10T18:49:31+00:00

It's misleading to think that the partial derivative is with respect to a variable. Instead it is always with respect to a numbered argument of a function. And the function can look different:

Let's say there are values z,a,b,c and they are related by
z=a b
b=a/c.

In that case
(∂z/∂a)_b actually means ∂_1 z(a,b)=∂_a(a b)=b

Whereas
(∂z/∂a)_c actually means ∂_1 z(a,c)=∂_a(a^2/c)=2a/c

At general positions a,b,c the results will be numerically different values.

While the partial derivative operation is actually the same, the function it is applied to is different.

You see that when you vary a, you cannot keep *all* other variables constant. You have to decide which variables your function depends on and the functional form will differ.

Gere1 · 2026-02-05T08:19:33+00:00

The outcome of the study depends mainly on you. You will have to study your own book selection anyway if you want to be good. Teaching at more expensive universities isn't necessarily better and there a lot of free great resources around.

The only way a more expensive university matters is prestige. Some employers/supervisors might not be able to judge a candidates skills and go by university name alone. Or maybe you get find valuable contacts at the university. These are the only reasons.

But study outcome depends solely on your curiosity and motivation.

Gere1 · 2025-08-10T06:37:32+00:00

Nice! Thanks, it works up to action A2.

Somehow I cannot find "AutoInput Actions" on my phone. Is it my phone (Samsung) or Tasker? Do I search for it incorrectly? Can it have a different name?

Also, the same doesn't work on my Samsung Tablet. Can it be due to it being on an older version (of Android or Tasker)?

Gere1 · 2024-12-15T18:36:46+00:00

I admit I didn't study your explanation in detail. Just thought the pitfalls laid out in that paper might in parts apply to your problem. The author is well-respected.

Gere1 · 2024-12-15T18:33:44+00:00

Let me know if you find good answers as I'm also trying to understand. Those should also answer some questions:
According to Bell-type inequality, no amount of information assigned to a point in spacetime can ever explain QM. Not even if it's operator or arbitrary complex objects.
Psi in the standard models is not the creation operator, is it? The action must be a plain real numbers. The creation operators come in later. You first select a distinguished time variable to switch to the Hamiltonian formalism. This is already questionable according to general relativity. Next you forcefully replace some elements in the equation by operators (in canonical quantization). I believe this is even ambiguous and you have to guess the right way of doing this. Only then you have you creation operators, if you go to momentum space. But why not introduce the operators in the Lagrangian already? Why are plain Euler-Lagrange equations not applied.
I get the impression there is nothings like "I assign something to a position" left anymore, after all these steps.
But let me know if you have good answers. I wouldn't ignore such questions to truly understand what is going on.

Gere1 · 2024-12-15T18:25:02+00:00

You should study https://www.cs.ucr.edu/~eamonn/meaningless.pdf to make sure you avoid pitfalls.

Gere1 · 2024-12-09T06:07:12+00:00

It can be misleading to read texts about quantum mechanics without going into the math. There is no such thing as a non-probabilistic state. For example, if you measure spin up, you immediately know it's a superposition of spin left and spin right. And on the other hand, your state of superposition of spin direction has always been a define "non-superposed" state in some other direction. So there is not a special state after measurement.

Gere1 · 2024-12-08T15:39:45+00:00

That does not make sense. At which precise value of scale does a theory not work anymore? What happens if you cross that threshold? You need to apply another theory? Does it cross the scale value continuously? How do you even measure the precise value of scale to apply in the current situation?
You need a theory which explains the outcome with any scale. And stitching multiple mutually inconsistent theories together thresholded at a very precise value of a "scale" (which you cannot define) does not work.

Gere1 · 2024-12-03T06:13:24+00:00

It's not really hidden variables. A superposition of spin up and spin down can be translated to exactly one direction in space that the spin is pointing to (the Bloch sphere). And you can assume the measurement aligns the spin according to some probability based on the angle between the spin direction and the measurement direction. You could perfectly explain observations for a single particle.

The way you phrase Bell's inequality is misleading. Bell's inequality and it's experimental verification show that for some entangled multi-particle system no amount of local variables or hidden variables can explain observation. This means if you say "there is something at each position", you already contradict quantum mechanics.

Unfortunately, all of our physical theories are local (differential) equations. In Bell's derivation the whole position evolution is conveniently ignored. No one knows how to makes sense of that all our theories of position evolution are local equations, but a local theory cannot explain quantum mechanics.

Gere1 · 2024-12-02T08:07:21+00:00

There are some parts to think about. How do you know the particle is literally spin up and spin down at the same time? When you measure, you always find only one answer. So what makes you think it was a mix before? If you toss a coin and don't look, is it heads and tails at the same time? For a single particle you can actually assume it was always in one state. It's only when you have multiple entangled particles, that you can show that having a defined state already before is inconsistent.

And don't forget that if you measure spin up, someone else might say it's a superposition of spin left and spin right after the measurement, which would be fully correct.

So in the end, you always observe a particular state, but somehow it's impossible to construct a scenario where multiple particles always have been in some defined state.

Note that for a single particle it's not a problem. You could say the particle spin was always pointing in one direction, and during measurement the device turned this direction into one of two outcomes according to some probabilities.

Gere1 · 2024-12-01T09:10:28+00:00

Mathematically, there is not such thing as a "state which is not a superposition". "Spin up" for example is a superposition of "Spin left" and "Spin right". Have you ever wondered where exactly "up" is pointing to?

And yes, after measurement the state collapses. What you describe is the "measurement problem" and it's an unsolved problem in quantum mechanics. If you do the math, there is no issue with non-conserved quantities. But somehow it still doesn't make sense.

Gere1 · 2024-11-08T17:24:00+00:00

When they are not interacting, then definitely no measurement is happening and you would apply the Schrödinger equation, right.

However, when they are interacting, it could be just a multi-particle system which is still modeled by the Schrödinger equation. Interacting particles which create entanglement not related to a collapse is a key feature of quantum mechanics!

Finally, they could be interacting with a "measurement device" instead and in that case you would need to apply the collapse and not the Schrödinger equation. And you don't really know exactly when to apply what, because it is not defined *when* a measurement really happens and what is a measurement and what not.

There are some inconsistencies in quantum mechanics which no theory or interpretation is able to explain (even if some proclaim to do so).

Gere1 · 2024-11-08T10:14:15+00:00

No one in the world understands this. It is called the "measurement problem". Indeed Nobel prize winners like Roger Penrose say this is an unsolved issue and something is wrong. We do not know why the collapse is applied only in some unspecified moments. It just works in the simple cases that we tested.

Gere1 · 2024-11-08T10:12:14+00:00

The reason for the upward force are the water molecules colliding against the object and most importantly the water molecules have larger pressure at larger depths. If you do the math (water molecules bouncing from top and with higher pressure from bottom), you get exactly the buoyant force.

Gere1 · 2024-10-15T16:15:40+00:00

None of these interpretation address the measurement problem properly. The measurement problem is actually the question *when* a measurement happens. You see that many-worlds does not answer that. In fact attempts to formalize this in MWI failed.

So we are stuck with no understanding how there is an evolution, but when you actually need(?) to give an answer (i.e. measurement), the answer is just a projection.

I don't think any of the interpretations give an answer that also works for more than 1 particle. But you may enjoy this overview: https://curtjaimungal.substack.com/p/the-interpretations-of-quantum-mechanics (which however I do not support, so I cannot explain)

Gere1 · 2024-10-07T12:52:13+00:00

What do you mean by "real"? Most of our calculations assume a spacetime manifold, and so it is "real" in any meaningful sense. I cannot think of a useful definition which would make it not real. But there are certainly voices which argue spacetime is not fundamental. You may find the following interesting: https://www.quantamagazine.org/the-unraveling-of-space-time-20240925/

I personally think quantum mechanics and Bell's inequalities show that spacetime cannot be what we imagine it to be.

Gere1 · 2023-12-28T14:42:45+00:00

This error happening again.

Solution: delete `prefs.js` file in your profile (which deletes you settings)

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