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Einstein and "spooky actions"

LegionOnomaMoi

Veteran Member
Premium Member
Why not just take the equation at face value? Say that the object you're dealing with actually is this complex-valued self-interacting field that you get entangled with, and voila, no ontology issue. :p
Mainly because we'd be left with a very nice equation describing something that doesn't exist. Because if it did, and "you get entangled with" it, then you'd have to describe the way it existed in Euclidean space (or Minkowski), rather than some nth-dimensional complex space. In other words, you'd require something that takes the equation and relates it to whatever is interacted with such that it isn't meaningless. We can't very well have decoherence if all we have is a quantum system and no observables.
 

zaybu

Active Member
You didn't answer the question. In an ontological sense, is a particle "really" definitely spinning either up or down even when the state vector says its both?

That's because you still see the wavefunction as real. I don't. For me, it's just a mathematical function that allows the calculation of probabilities. Wave collapse, to me, means taking a measurement. So in a Stern-Gerlach experiment for instance, the wavefunction is a linear combination of ↓ ↑ such as:

|ψ> = 2 ^(-1/2 ) |↑> + |↓> )

When you take a measurement (wave collapse) you get either ↓ or ↑. And if you calculate what's the probabilty of getting a spin up, then:

P(&#8593;) = (<&#968; |&#8593;> )^2 = 1/2

I see nothing real in |&#968;> , |&#8593;> or|&#8595;>. These are state vectors, and QM is a probability theory.
 

PolyHedral

Superabacus Mystic
That's because you still see the wavefunction as real. I don't. For me, it's just a mathematical function that allows the calculation of probabilities. Wave collapse, to me, means taking a measurement. So in a Stern-Gerlach experiment for instance, the wavefunction is a linear combination of &#8595; &#8593; such as:

|&#968;> = 2 ^(-1/2 ) |&#8593;> + |&#8595;> )

When you take a measurement (wave collapse) you get either &#8595; or &#8593;. And if you calculate what's the probabilty of getting a spin up, then:

P(&#8593;) = (<&#968; |&#8593;> )^2 = 1/2

I see nothing real in |&#968;> , |&#8593;> or|&#8595;>. These are state vectors, and QM is a probability theory.
So what's reality? :D
 

PolyHedral

Superabacus Mystic
Reality is when you take thousands of measurement, and when you tabulate your results, you get 50% up, 50% down. And you are elated to know that QM works. ;)
If you are not measuring the wavefunction, (because that's not real) what are you measuring? What is "really" there? You're saying the wavefunction is not reality, but you're not replacing it with anything.

Mainly because we'd be left with a very nice equation describing something that doesn't exist. Because if it did, and "you get entangled with" it, then you'd have to describe the way it existed in Euclidean space (or Minkowski), rather than some nth-dimensional complex space. In other words, you'd require something that takes the equation and relates it to whatever is interacted with such that it isn't meaningless. We can't very well have decoherence if all we have is a quantum system and no observables.
Not quite - you have to show why it appears to take place in a almost-Euclidean space according to entities inside it. Entities appearing in 3/4D space and the actual universe taking place in this infinite-dimensional complex space aren't mutually incompatible.
 
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zaybu

Active Member
If you are not measuring the wavefunction, (because that's not real) what are you measuring? What is "really" there? You're saying the wavefunction is not reality, but you're not replacing it with anything.

Take measuring the spin of electrons for example, you would pass an electron into a magnetic field, and then to a detector, which will register an up spin or a down spin. Another example is measuring the energy levels of hydrogen: you would pass light through a sample of the gas, then through a prism. The resulting spectral lines would correspond to the energy levels, and QM will give you the same results with its mathematical apparatus. Observation confirming theory. What else do you Want? You're not measuring a wave - you're measuring spin in one case; energy levels in the second case.
 

PolyHedral

Superabacus Mystic
Take measuring the spin of electrons for example, you would pass an electron into a magnetic field, and then to a detector, which will register an up spin or a down spin. Another example is measuring the energy levels of hydrogen: you would pass light through a sample of the gas, then through a prism. The resulting spectral lines would correspond to the energy levels, and QM will give you the same results with its mathematical apparatus. Observation confirming theory. What else do you Want? You're not measuring a wave - you're measuring spin in one case; energy levels in the second case.
I want an explanation of what I'm measuring. What's the electron's state described by, if not a wave function? Does it have a defined spin before I measure it?
 

zaybu

Active Member
What's the electron's state described by, if not a wave function? Does it have a defined spin before I measure it?

My underlining.

Think about it: how would anyone know? You can speculate about it. But is that knowledge? The only reality we are concerned with is when it is measured. It's a combination of what the state is in before measurement + interaction with the measuring apparatus giving you the observed result.

And QM gives the theoretical result that agrees with experiment, as in the case of the spin, with P(&#8593;) = (<&#968; |&#8593;> )^2 = 1/2. The mystery is why QM works that way but not in some other way. There are those who believe that QM is incomplete and are looking at ways to formulate it differently. But after 80 years, that other formalism hasn't come through. And with my studies in QFT, what was developped after QM, and with resounding successes, I believe that this search for a new QM is a waste of time.
 
Again you're putting words in my mouth I didn't say. The correlation in our scenario is one thing, and Bell's theorem is another. His inequality, which can be derived by Venn's diagram, was trying to prove whether there are hidden parameters or not. You are trying to prove that the wave function is real, and when it collapses, a spooky action at a distance makes sure the law angular momentum is conserved.
I'm not putting words in your mouth. Here is what you said:

zaybu post #11
zaybu said:
Bell made two assumptions: 1) realism or logic, and 2) locality. If his inequality is violated, which was proved experimentally subsequently by Aspect, it means one of those two is wrong. You have focused solely on the second assumption, but most physicists have realized that it is the first assumption, which I have mentioned at least twice: classical logic fails to describe quantum system.
zaybu post #411
In this case, you need QM only for the observation that the spins are either up or down. In classical physics, the spin of an object, be it planet earth or a top, can have any value. But at subatomic scale, it is quantized along a given axis. For that, you need QM to explain this result. But for the correlation of Alice's data and Bob's data, plain old fashion classical physics can explain that.
This isn't even wrong, it's just self-contradictory at worst, confusion at best.

zaybu said:
Susskind has a different purpose in mind. His was to show that a quantum system will violate Bell's theorem on theoretical grounds, never mind on experimental grounds. In no way this prove your twisted theory of spooky action.
Susskind's purpose is irrelevant, you were wrong about how to correctly write down the singlet state. Why can't you admit it?

zaybu said:
There's no contradiction, but more of your misunderstanding that is at play. In the first scenario, it hadn't been spelled out clearly if Alice knew that Bob was going to measure an entangled state. So for a while I was asking you from which POV your questions were assuming. That is why I came up with the second scenario so that it would be very clear that she didn't know.
What on Earth are you talking about? I said very clearly from the beginning we are talking about 2 particles in an initial singlet state, IOW an entangled state. If they aren't entangled then obviously there's no need for "spooky action", we all agree there.

zaybu said:
Again, this shows you are still hung up on the "wave collapse" misconception. All we can say is before making the measurement, we don't know in what state it is, that's why we write as a linear combination of &#8593;&#8595;. After measurement, it will be either this &#8593; or that &#8595;. What it was in the real world beforehand, we don't know. The state vector does not represent a real thing in itself. Just the probabilities.
Here you describe your interpretation, and that's certainly one possible interpretation. Griffiths calls it agnosticism. I don't think it is the most parsimonious interpretation assuming QM is correct and complete. I think the violations of Bell's theorem force you to pick a side, either QM is incomplete or we really, really have to believe what it says at face value (essentially). I would ask you a series of questions to demonstrate this, but past experience suggests you will dodge them.

Therefore, allow me to simply explain my own preferred (orthodox) interpretation, since i.m.o. you haven't done it justice.

The particles exist before we measure them. They therefore exist in some kind of quantum state before we measure them. This state might be difficult for primate brains like ours to imagine. But it does exist, and it can be represented mathematically by the formalism of QM, i.e. the state vector or wavefunction |Psi>. Furthermore, we can know exactly what that state is. In the case of 2 entangled fermions with total spin zero, we can know exactly what state the particles are in: they are in the singlet state. Now, just as a wave may not have a well-defined position, in the singlet state state each particle does not have well-defined spin. Only particular states, called spin eigenstates have well-defined spin.

I prefer to emphasize the particles in a singlet state do not have well-defined spins, although it is also of course valid to say we do not know their spins. Sometimes the latter can be a useful shortcut. But fundamentally, I believe the former is simpler, leads to less confusion, and is less compatible with classical physics. The former is also more compatible with the assumption of an objective reality. The latter could be, in principle, compatible with classical physics. Or it could be interpreted as leading to some kind of indeterminate or subjective reality. This leads to confusion, i.m.o.

Finally, the last piece is measurement. If we take the mathematical formalism of QM seriously at face value, then a measurement (operator) acts on a quantum system (state vector). The state instantaneously "collapses" into one of the eigenstates, and the measurement obtains the corresponding eigenvalue, with probabilities given by the initial state. We are justified in being initially skeptical (as Einstein was) of all this. After all, we know from the beginning that this "measurement" must really be somehow an idealization of an interaction between a quantum system and a macroscopic system. We might expect that there are "partial measurements" where a semi-macroscopic system interacts with a semi-quantum system, and so forth, and maybe in such cases the predictions of QM break down. Or maybe a deeper theory (such as QFT) will show QM is fundamentally wrong. But as far as I know, neither of those things have happened. Aspect's measurements could have disproved the QM picture of "measurement" and "collapse" if he had obtained different results. He could have obtained results consistent with Bell's theorem. Or, he could have shown that the "collapse" is not instantaneous, it actually takes time to travel to Bob's particle. It is the incredible failure to be disproved, in face of many tests, that forces us to really, truly take the formalism of QM seriously, even if that is not the beautiful or intuitive picture we would have liked. (Actually, I find that intuition is fickle; to me, special relativity was counter-intuitive until I fully understood and accepted it as a fact; then it became "intuitive" and no longer "ridiculous").

Let me wrap up with a final comment about spooky action. To me, there is always spooky action going on with any measurement, e.g. the two-slit experiment, it's just that with two particles in an entangled state, we can experimentally rule out other possibilities. In the two-slit experiment the particle really was in a state without a definite position, before you measured it. When the particle interacted with your detector it collapsed into a position eigenstate. To me Alice and Bob's experiment is just an extension of the same quantum reasoning to two particles, which are also in a nonlocally "spread out" state, in a certain sense. It's just the spin of two particles which is "spread out" instead of the position of one particle.
 
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By the way zaybu I notice you (wisely) dodged my questions in post #425:
Mr Spinkles said:
Please show the calculation explicitly. After Alice measures, (2) = Total spin = spin of Alice's particle + spin of Bob's particle = ...? Spin of Alice's particle = up. Spin of Bob's particle = ... ? (Does Bob's particle have a definite spin? At one point you were saying Bob's particle remains in the singlet state!)
You say your argument is based on conservation of angular momentum. Please show the calculation and answer these questions, explicitly.
 

PolyHedral

Superabacus Mystic
My underlining.

Think about it: how would anyone know? You can speculate about it. But is that knowledge? The only reality we are concerned with is when it is measured. It's a combination of what the state is in before measurement + interaction with the measuring apparatus giving you the observed result
The problem is that "state." Classical physics says "This is a particle's state..." Quantum physics says, "This is a particle's state..." You're saying that both of these are wrong, but not providing any alternative explanation for what a particle's state "really" is. As far as the physics is concerned, the particle's state is what the wavefunction says - a superpositioned particle really is in a combination of both states at once. But according to you, that's not true - so what is? :D


Also, here's a puzzle for all the non-realists: If you swap one of the mirrors for a beam splitter, what happens to the output?
Bonus question: Compared to the previous interferometer setup, we have sent less of the light towards the A/B split - and yet more is detected at A! How is this possible if the photon is a single particle?

Mach-Zender_interferometer_paradox_zpsb3d4cf25.png


EDIT: I just noticed. I forgot to square the final modulus values. If you do that, you find that A and C detect with 9% probability each, and B with 81% probability, which is even less intuitive than the diagram.
 
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The problem is that "state." Classical physics says "This is a particle's state..." Quantum physics says, "This is a particle's state..." You're saying that both of these are wrong, but not providing any alternative explanation for what a particle's state "really" is. As far as the physics is concerned, the particle's state is what the wavefunction says - a superpositioned particle really is in a combination of both states at once. But according to you, that's not true - so what is? :D
Thank you. You said it much more concisely than I did. Unfortunately I'm out of frubals.
 

idav

Being
Premium Member
The problem is that "state." Classical physics says "This is a particle's state..." Quantum physics says, "This is a particle's state..." You're saying that both of these are wrong, but not providing any alternative explanation for what a particle's state "really" is. As far as the physics is concerned, the particle's state is what the wavefunction says - a superpositioned particle really is in a combination of both states at once. But according to you, that's not true - so what is? :D
The particle is not in two states it is just an illusion. If it is not an illusion then we should have no problem measuring two states but of course that won't happen cause reality isn't that wonky. What's being measured is something other than the particle, it's field.
 

zaybu

Active Member
What on Earth are you talking about? I said very clearly from the beginning we are talking about 2 particles in an initial singlet state, IOW an entangled state. If they aren't entangled then obviously there's no need for "spooky action", we all agree there.

They are entangled from your POV, as you are playing God who knows everything. But to Alice, or Bob, unless they are told of the set-up of this experiment before they take their measurement, they don't know. How hard is it for you to comprehend this? And so they can't write a singlet state. They are unlike you not pretending to be gods. It's only after they compared their the data they will figure that those two particles were indeed entangled.



The state instantaneously "collapses" into one of the eigenstates, and the measurement obtains the corresponding eigenvalue, with probabilities given by the initial state.

It's an interpretation I don't subscribe to, which I have said before. And no matter how many times you repeat this nonsense, it won't change my position on this. THE WAVEFUNCTION IN QM IS NOT A REAL WAVE. Can you get that straight?
 

LegionOnomaMoi

Veteran Member
Premium Member
Not quite - you have to show why it appears to take place in a almost-Euclidean space according to entities inside it. Entities appearing in 3/4D space and the actual universe taking place in this infinite-dimensional complex space aren't mutually incompatible.
I don't care whether one's interpretation of QM is irreducibly statistical or a multiverse interpretation (or any relative state interpretation), everyone still has to deal with the fact that the "system" represented by the e.g., Schroedinger's wave function does not correspond to a physical system in any known way except through the use of another mathematical operation (not just an actual physical measurement):

"As stated above, the description of the state and its evolution does not constitute the entire quantum formalism. The wave function provides only a formal description and does not by itself make contact with the properties of the system. Using only the wave function and its evolution, we cannot make predictions about the typical systems in which we are interested, such as the electrons in an atom, the conduction electrons of a metal, and photons of light. The connection of the wave function to any physical properties is made through the rules of measurement. Because the physical properties in quantum theory are defined through measurement, or observation, they are referred to as ‘observables’."

from 1.3.1 of D. L. Hemmick & A. M. Shakur. (2012).Bell's Theorem and Quantum Realism: Reassessment in Light of the Schrödinger Paradox (SpringerBriefs in Physics).
 

LegionOnomaMoi

Veteran Member
Premium Member
If it is not an illusion then we should have no problem measuring two states but of course that won't happen cause reality isn't that wonky.

It has happened. We've had several experimental realizations of Wheeler's delayed choice thought-experiment covering all manner of possible "but what if..."'s. We have shown that "it is possible to freely and a posteriori decide which type of mutually exclusive correlations two already earlier measured particles have." (Ma, X. S., Zotter, S., Kofler, J., Ursin, R., Jennewein, T., Brukner, &#268;., & Zeilinger, A. (2012). Experimental delayed-choice entanglement swapping. Nature Physics, 8(6), 480-485.) We've combined the double-slit experiments (i.e., the version that reveals interference and the type that does not) into a single measurement capable of tuning back and forth at will. We've detected molecules of hundreds of atoms in two distinct and mutually exclusive states.

What's being measured is something other than the particle, it's field.

Calling it something else doesn't make any difference. Call them particles, phlogiston quanta, complexions, or whatever you wish, it doesn't change what empirical evidence demonstrates.
 

zaybu

Active Member
The wave function provides only a formal description and does not by itself make contact with the properties of the system. Using only the wave function and its evolution, we cannot make predictions about the typical systems in which we are interested, such as the electrons in an atom, the conduction electrons of a metal, and photons of light. The connection of the wave function to any physical properties is made through the rules of measurement. Because the physical properties in quantum theory are defined through measurement, or observation, they are referred to as &#8216;observables&#8217;."

Well, for all our disagreements and unpleasant exchanges in the past, I agree with that statement.

I apologize if I have overstated my case in the past and might have offended you. Can we bury the axe and move on?
 
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zaybu

Active Member
EDIT: I just noticed. I forgot to square the final modulus values. If you do that, you find that A and C detect with 9% probability each, and B with 81% probability, which is even less intuitive than the diagram.

It's more like 6.25% each (1/4)^2, and B with the remaining 87.5%. If you consider that in the previous case, it was 100% for B, it makes sense that most photons will still go to B, with some "leaks" to A and C.

Math rules.
 

idav

Being
Premium Member
Calling it something else doesn't make any difference. Call them particles, phlogiston quanta, complexions, or whatever you wish, it doesn't change what empirical evidence demonstrates.
The experiments demonstrate that they make physical changes to an experiment and it influences photon. So the experiment is choosing for the photon. The how is more than important.
 

zaybu

Active Member
The experiments demonstrate that they make physical changes to an experiment and it influences photon. So the experiment is choosing for the photon. The how is more than important.

zaybu-albums-physics-picture4412-hup.jpg


The Heisenberg Uncertainty Principle states that if you want to measure position and momentum simultaneously, there is an error ruled by planck constant,
&#916;x &#916;p &#8776; &#8463;In case (1), the error &#916; x is small, but the error &#916; p is larger. Vice-versa in (2) So just the act of measuring -- shining light on the particle --causes an error in both position and momentum. The HUP regulates that error, imposing a condition on it that you can't reduce one without increasing the other.

 
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