Two defining features of quantum mechanics never appear together

March 21, 2016 by Lisa Zyga, Phys.org feature

Representation of measurements that demonstrate the contextuality-nonlocality tradeoff. Credit: Zhan, et al. ©2016 American Physical Society
(Phys.org)—Two of the most important ideas that distinguish the quantum world from the classical one are nonlocality and contextuality. Previously, physicists have theoretically shown that both of these phenomena cannot simultaneously exist in a quantum system, as they are both just different manifestations of a more fundamental concept, the assumption of realism. Now in a new paper, physicists have for the first time experimentally confirmed that these two defining features of quantum mechanics never appear together.

The physicists, Xiang Zhan, et al., have published a paper on the nonlocality-contextuality tradeoff in a recent issue of Physical Review Letters.

In the everyday world that we observe, an object can only be affected by nearby objects (locality), and when we make a measurement, the outcome does not depend on other independent measurements being made at the same time (noncontextuality).

In contrast, the world is nonlocal, as demonstrated by where two objects can influence each other even when separated by large distances. And in the , measurements are contextual, so quantum systems do not have predetermined values but instead their values depend on how measurements are made.

To show that a is nonlocal or contextual, physicists have defined inequalities that assume a system is the opposite (local or noncontextual). Then they perform experiments that attempt to violate these inequalities to show that the system is not local or noncontextual. So far, these two types of inequalities have never been tested simultaneously.

Experimental setup demonstrating the contextuality-nonlocality tradeoff in a qubit-qutrit system. Credit: Zhan, et al. ©2016 American Physical Society

In the new study, the researchers have attempted to violate both inequalities at the same time, but have found that only one inequality can be violated at once. Their experiment uses entangled photons to generate photonic qutrit-qubit systems (a qubit is a superposition of two states, whereas a qutrit is a superposition of three states). By performing various measurements on these photons, the researchers could violate the inequalities separately, but not at the same time.

"The greatest significance of our work is that we provide experimental evidence of the assumption that quantum entanglement and contextuality are intertwined quantum resources," Peng Xue, a physicist at Southeast University in Nanjing, China, and one of the lead authors of the paper, told Phys.org.

As the explain, the reason for the nonlocality-contextuality tradeoff arises from the fact that both properties have the same root: the assumption of realism, which is the assumption that the physical world exists independent of our observations, and that the act of observation does not change it.

Since nonlocality and contextuality can be thought of as two different manifestations of the basic assumption of realism, then one of them can be transformed into the other, but both cannot exist at the same time because they are essentially the same thing.

"We think the contextuality-nonlocality monogamy suggests the existence of a quantum resource of which entanglement is just a particular form," Xue said. "The resource required to violate the noncontextuality inequality and that required to violate the locality inequality are fungible through entanglement. That is, to violate the locality inequality costs entanglement as a resource, while to violate the noncontextuality costs contextuality as a resource. In a quantum system, only one of the two inequalities can be violated because nothing is left to violate the other one."

The researchers hope that the new experiment will open the doors to further exploring the mutual resource in the future, as well as lead to potential applications.

"We plan to study contextuality as a resource for experimental , such as for quantum computation," Xue said.

Explore further: Physicists find extreme violation of local realism in quantum hypergraph states

More information: Xiang Zhan, et al. "Realization of the Contextuality-Nonlocality Tradeoff with a Qubit-Qutrit Photon Pair." Physical Review Letters. DOI: 10.1103/PhysRevLett.116.090401

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Hyperfuzzy
not rated yet Mar 21, 2016
Help me. This makes no sense to me. If a measurement of a field on the moon and the field affects the field upon the earth, each local measurement will be affected. Correct? So what are we trying to say. That this can be violated? Why do we even ask this question? The point is, QM only expresses possibility, realism is not assumed or defined. The idea then is why use a tool not defined for the task at hand to prove or provide a solution to the task? Anyway, the dependance and time, contextually, are not defined. So this is an exercise in futility, i.e. no causal effects or initial conditions.
Hyperfuzzy
not rated yet Mar 21, 2016
Note: The reference plane for both local and contextual may change. Only the change or the dynamic may be detectable if any measure is to be defined. It may or may not be possible to detect the change, depending upon the speed and value. Therefore the entire argument has no basis using QM since quanta essentially is minimalistic. The later redefines quanta as "less". Therefore seeking to holistically define realism is not possible with QM.
shavera
1 / 5 (3) Mar 21, 2016
[deleted]
shavera
3.4 / 5 (7) Mar 21, 2016
"The assumption of realism" as the article calls it, is not one physicists generally assume. Bell's theorem tells us that reality cannot be both real and local for the experimental results we observe. If reality is "not real" then quantum particles truly exist in superpositions of states (spin up and down simultaneously) until some interaction requires one specific state or another (this is often called "measurement" even though it does not require a conscious observer setting up some experiment).

The article misrepresents the idea of "non-real" quantum physics, and the extent to which it is quite common throughout the community.

Non-local quantum physics, where some kind of information about the "real" underlying quantum mechanics somehow goes around faster than light is also possible, but not as commonly supported in the community.
compose
Mar 21, 2016
This comment has been removed by a moderator.
freestylexplore
1 / 5 (6) Mar 21, 2016
The pattern is binary.. Circle and straight make sphere. Straight and sphere make triangle.. Triangle and sphere make square... so on.. easy to solve. They say one thing doesn't effect another at along distance. Well because all force uses the same code and the code never changes. It's the same blob or mass. There is no real distance. There is lack of code. Not distance. In math code (0,1)are code distance is based off of : none (equal to), zero (less than), negative (opposite of).. there is more but it might get too complicated for general users. Linguistics is key to knowing and understand higher math. Higher you go the more languages you have to deal with.
freestylexplore
1 / 5 (4) Mar 21, 2016
When deal with quantities it always the amount that can be stored and the amount that can unstore it. Look at gravity. The mass has to store (load) it's potential. The easiest path is (in) so we get a centralized mass. No bending of space required. Plus once matter (code) is connected it works both ways. So the suns output keeps earth orbit stable. Not a bend. No output. No orbit. Black holes are connected to the Galaxy. They are the same thing. One mass.
Noumenon
3 / 5 (6) Mar 21, 2016
If reality is "not real" then quantum particles truly exist in superpositions of states (spin up and down simultaneously) until some interaction requires one specific state or another (this is often called "measurement" even though it does not require a conscious observer setting up some experiment).


Actually it does require mind setting up the apparatus,… as the apparatus design determines the possible observable states to begin with, and the observable values are conceptual values, which are not resolvable via decoherence (quantum interaction).

Decoherence does not solve the 'measurement problem'. 'Measurement' is necessarily mind-dependent….

Realism refers to assumptions that our conceptual descriptions implicit in apparatus design and interpretation (locality, counterfactuality, particle, wave, determinism, etc] correspond to mind-independent objective reality. This is what is refuted by experiment via hidden variables that assume these concepts.
obama_socks
1 / 5 (2) Mar 21, 2016
When deal with quantities it always the amount that can be stored and the amount that can unstore it. Look at gravity. The mass has to store (load) it's potential. The easiest path is (in) so we get a centralized mass. No bending of space required. Plus once matter (code) is connected it works both ways. So the suns output keeps earth orbit stable. Not a bend. No output. No orbit. Black holes are connected to the Galaxy. They are the same thing. One mass.

- freestylexplore (Otto's sockpuppet)
Da Schneib
not rated yet Mar 21, 2016
Very interesting, kind of like the particle/wave duality.

Da Schneib
5 / 5 (1) Mar 21, 2016
Realism refers to assumptions that our conceptual descriptions implicit in apparatus design and interpretation (locality, counterfactuality, particle, wave, determinism, etc] correspond to mind-independent objective reality.
No, that's philosophical realism. Physics realism is the statement that particles have real values of their parameters even if we can't measure them, for example because one value is in superposition due to Heisenberg uncertainty.
Da Schneib
not rated yet Mar 21, 2016
I should also note that this type of realism is also called "counterfactual definiteness."
dan42day
1 / 5 (3) Mar 21, 2016
Neither time nor realism would exist without conscious observers, the entire history of the universe is but an instant to a photon. That is why I consider myself an agnostic rather than an atheist. There is definitely something spiritual about physics!
Noumenon
3.7 / 5 (6) Mar 21, 2016
Realism refers to assumptions that our conceptual descriptions implicit in apparatus design and interpretation [..] correspond to mind-independent objective reality.
No, that's philosophical realism.

The failure of Realism to hold, implies that.
Physics realism is the statement that particles have real values of their parameters even if we can't measure them,…

That is only one such condition.
…for example because one value is in superposition due to Heisenberg uncertainty.

Superpositions [SP's] are not due to Heisenberg uncertainty relations [HUR]. HUR's concern conjugate variables, not SP's. The quantum wavefunction (in SP) is not a SP of values. To extract the observable values requires a quantum operator acting on the wavefunction to render eigenvalues [the statevector is scaled without changing it's direction in Hilbert Space]. The quantum operator is in effect, a representation of the experimental apparatus.
Da Schneib
3.7 / 5 (3) Mar 21, 2016
Realism refers to assumptions that our conceptual descriptions implicit in apparatus design and interpretation [..] correspond to mind-independent objective reality.
No, that's philosophical realism.
The failure of Realism to hold, implies that.
I don't know what that even means.

Physics realism is the statement that particles have real values of their parameters even if we can't measure them,…
That is only one such condition.
Name others.

…for example because one value is in superposition due to Heisenberg uncertainty.

Superpositions [SP's] are not due to Heisenberg uncertainty relations [HUR].
Incorrect. For example a particle whose polarization has been measured in X has its polarization in any other axis in superposition. Known fact of quantum mechanics. Please try to stay away from philosophy and stick to physics. Thanks.

[contd]
Da Schneib
3.7 / 5 (3) Mar 21, 2016
HUR's concern conjugate variables, not SP's.
Measurement of a variable that is conjugate under Heisenberg uncertainty places the conjugate variable in superposition. Known fact of quantum mechanics. Please try to stay away from philosophy and stick to physics. Thanks.

The quantum wavefunction (in SP) is not a SP of values.
I don't even know what that means. Superposition means that one of a particle's parameters (the one that is superposed, specifically) has a probability to be two or more various values when measured. Superposition is precisely about values, period. Please try to stay away from philosophy and stick to physics. Thanks.

[contd]
Da Schneib
3.7 / 5 (3) Mar 21, 2016
To extract the observable values requires a quantum operator acting on the wavefunction to render eigenvalues [the statevector is scaled without changing it's direction in Hilbert Space]. The quantum operator is in effect, a representation of the experimental apparatus.
In other words the superposed value must be measured in order to stop being a superposed value and become a real value.

Which is exactly what I said and has nothing to do with your previous claims. Please try to stay away from philosophy and stick to physics. Thanks.
Noumenon
3.4 / 5 (5) Mar 21, 2016
Physics realism is the statement that particles have real values of their parameters even if we can't measure them,…


That is only one such condition.


Name others.


1) locality, ....2) counterfactual definiteness -[ that entities have objective values independent of measurement so that statements can be made of experiments not actually performed], ....and the 3) no conspiracy condition,..... if the Bell inequalities rule out one, then they rule out "local realism".

For example, if one interpretation maintains locality, then it must reject counterfactual definiteness and with it the notion of properties existing independently of measurements. If another interpretation maintains counterfactual definiteness, then it must reject locality. In either case Bell's theorem disproves local-realism.
Da Schneib
3.7 / 5 (3) Mar 21, 2016
Physics realism is the statement that particles have real values of their parameters even if we can't measure them,…


That is only one such condition.


Name others.


1) locality,
Wrong. Locality is the statement that local causes have local effects. Realism is the statement that parameters of particles always have a definite value even if we can't measure it.

Again, please try to stay away from philosophy and stick to physics. Thanks.
Noumenon
3.4 / 5 (5) Mar 21, 2016

Superpositions [SP's] are not due to Heisenberg uncertainty relations [HUR].

Incorrect. For example a particle whose polarization has been measured in X has its polarization in any other axis in superposition. Known fact of quantum mechanics.


You described a superposition of states. That is NOT the Heisenberg uncertainty relation.

I explained Hilbert Space formulation to you before, in which statevectors [superposition of states] evolve, …..with a Fourier transform required for its conjugate space representation. That is where the HUR comes in.

Please try to stay away from philosophy and stick to physics. Thanks.


I will school you in both topics simultaneously :). If you wish to be wrong and rude, I can put you on ignore, otherwise take advantage of my generosity.

Da Schneib
3.7 / 5 (3) Mar 21, 2016
As for the equivalence of contrafactual definiteness and realism, I already said that. And it's an equivalence; that is, the statements of contrafactual definiteness and realism are logically the same. So that's not an additional example.

Moving right along, the no-conspiracy condition is that there isn't a deity or supernatural being or superpowerful alien manipulating the results of our experiments. Realism is the statement that the parameters of a particle have real values even when they are unmeasured or unmeasurable, which is not at all the same thing.

Again, please try to stay away from philosophy and stick to physics. Thanks.
Da Schneib
3.7 / 5 (3) Mar 21, 2016
Superpositions [SP's] are not due to Heisenberg uncertainty relations [HUR].

Incorrect. For example a particle whose polarization has been measured in X has its polarization in any other axis in superposition. Known fact of quantum mechanics.
You described a superposition of states. That is NOT the Heisenberg uncertainty relation.
No, I didn't. I said that the Heisenberg uncertainty principle can generate a superposed state when a conjugate parameter is measured to have a definite value. Please don't lie about what I said. Thanks.

I explained Hilbert Space formulation to you before, in which statevectors [superposition of states] evolve, …..with a Fourier transform required for its conjugate space representation. That is where the HUR comes in.
We're not talking about the representation of a state in Hilbert space, we're talking about the actual values of real parameters. Please stop trying to bamboozle. It's embarrassing to watch. Thanks.
Da Schneib
3 / 5 (2) Mar 21, 2016
I will school you in both topics simultaneously :). If you wish to be wrong and rude, I can put you on ignore, otherwise take advantage of my generosity.
You are schooling no one. You are lying and playing logic games and it is totally transparent. Feel free to run away and hide at any time. That also will be transparent.
Noumenon
3.4 / 5 (5) Mar 21, 2016
The quantum wavefunction (in superpostion) is not a superpostion of values. To extract the observable values requires a quantum operator acting on the wavefunction to render eigenvalues

I don't even know what that means.


Then why are you insulting me? That is the basic mathematical formulation of QM. Do you not know what a quantum operator is? What a wavefunction is? What an eigenstate, eigenfunction, and eigenvalues are? This is not philosophy, it is the Schrodinger equation,… "quantum theory as a eigenvalue problem" was the name of his famous paper.

---->
Noumenon
4 / 5 (4) Mar 21, 2016
<----

A wavefunction, is NOT a wavefunction of VALUES. The measured values are the "eigenvalues" obtained from solving the Schrodinger equation for solutions that satisfy the eigenvalue problem [that scale the state-vector without changing its direction in Hilbert Space].

The normalized square of the wavefunction gives the probabilities, not the values.

The wavefunction Fourier transformed, can even be given a different representation of observable values,… another basis space,.. i.e. its conjugate.

We're not talking about the representation of a state in Hilbert space, we're talking about the actual values of real parameters.


The Hilbert Space formulation is the fundamental QM mathematical formulation !! The basis of that space is the actual values measured !!

Your rudeness and simultaneous ignorance is what is embarrassing here.

Da Schneib
1 / 5 (1) Mar 21, 2016
Then why are you insulting me? That is the basic mathematical formulation of QM.
Stop the BS, dude. Not impressed. You haven't answered anything I've said correctly so far.
Da Schneib
3.7 / 5 (3) Mar 22, 2016
Meanwhile, let's address wavefunctions. Wavefunctions are representations of the values of parameters on particles when they are in superposition.

Like I said, stop the BS. You're going to get nowhere with it.

You said,
The quantum wavefunction (in superpostion) is not a superpostion of values.
This is egregious.

1. A wavefunction is the representation of the value of a parameter on a particle that is in superposition. So your statement is explicitly untrue.
2. There is no necessity to state that the wavefunction is in superposition; there is no need to represent it as a wavefunction if it is not because it has a real value.

Like I said, I don't even know what that means. It's a bunch of bamboozling BS.
jgmirl69
1 / 5 (1) Mar 22, 2016
How did all of you get so far away from the original topic. you think your geniuses, you all hide behind your presumptions of the supposition of quantum states, when the fact is NONE of can guess at the eventual outcome.
Noumenon
3.4 / 5 (5) Mar 22, 2016
In other words the superposed value must be measured in order to stop being a superposed value and become a real value.


NO, the wavefunction is not a superposition of "values". To get the possible measurable values you must apply a quantum operator in the Schrodinger equation to find solutions that satisfy an eigenvalue problem. This gives discrete results. On actual measurement gives only one such result. The square of the wavefunction gives the probability.

None of this is philosophy,.... you insulting ding-bat,... it is the mathematical foundation of QM.

Wavefunctions are representations of the values of parameters on particles when they are in superposition.


Particles? No. The 'quantum system' is represented by the wavefunction. Not the values. The values come in by solving the Schrodinger equation as a eigenvalue problem. You use a Quantum Operator,.... like the Hamiltonian (energy), or position or momentum operator,... to find the [eigen]values.
Da Schneib
3 / 5 (2) Mar 22, 2016
In other words the superposed value must be measured in order to stop being a superposed value and become a real value.


NO, the wavefunction is not a superposition of "values".
Sorry, you're lying again. Please stop lying.
Noumenon
4 / 5 (4) Mar 22, 2016
In other words the superposed value must be measured in order to stop being a superposed value and become a real value.


NO, the wavefunction is not a superposition of "values".
Sorry, you're lying again. Please stop lying.


You are ignorant of the mathematical foundations of quantum mechanics and how that formulation works.

You attempted to correct me, but clearly have some issue with being corrected yourself, so you degenerate into insults.

Da Schneib
not rated yet Mar 22, 2016
You are attempting to bamboozle us with BS. I'm calling you out, and you're whining about it.

It's transparent.

Claiming that parameters on particles don't have values, whether they are in superposition or not, is nonsense. It's carefully crafted nonsense, by all appearances. You are a liar and a troll.

You can prove me wrong at any time: go answer my original point that you are confusing philosophical realism with physics realism.
Da Schneib
not rated yet Mar 22, 2016
Oh and BTW nice try at pretending that the behavior of the real world is something in Hilbert space. Having a little map/territory differentiation trouble there, philosopher? Just askin'.
Da Schneib
not rated yet Mar 22, 2016
How did all of you get so far away from the original topic.
Ummwut? Did you comprehend that the original topic was realism?

you think your geniuses,
We think our geniuses what?

you all hide behind your presumptions of the supposition of quantum states, when the fact is NONE of can guess at the eventual outcome.
Err, perhaps you don't quite get what "realism" means. Noum isn't helping much. We're not talking about philosophical realism; we're talking about physics realism, and we knew that was broken when Heisenberg wrote his uncertainty relation.

As far as being able to guess at the eventual outcome, hate to tell you but the physics we're talking about here is all about quantifying superpositions in terms of probabilities, and knowing the exact probabilities isn't guessing. It's quantifying. It's as much as we can know about the eventual outcome; there simply isn't any more information available.
Noumenon
4 / 5 (4) Mar 22, 2016
You used the phrase "superposed values". This displays a lack of understanding of the mathematics of quantum mechanics.

The wavefunction is not a superposition of values. In fact the wavefunction is a complex valued function.

Oh and BTW nice try at pretending that the behavior of the real world is something in Hilbert space.


Again, profound ignorance. The Hilbert Space formulation is the "mathematically rigorous formulation of quantum mechanics, developed by John von Neumann", and is entirely compatible with Dirac's bra-Ket notation. This is not philosophy.

I'm bamboozling you with orders if magnitude more knowledge of the subject, while you are bamboozling me with Jerry-Springer insults and lack of humility.

Da Schneib
1 / 5 (2) Mar 22, 2016
You used the phrase "superposed values". This displays a lack of understanding of the mathematics of quantum mechanics.
No, it just displays that I use a different interpretation than you do.

And that displays that you know little of quantum mechanics. Just sayin'.

The wavefunction is not a superposition of values.
You're lying again.

In fact the wavefunction is a complex valued function.
Ummm, that's true but it doesn't prove that it doesn't have a value, nor does it prove that the value is not a superposition of real values. Do try to keep your eye on the ball, there, sport. Functions get solved and the solutions yield values.

On Earth.

This isn't even a map/territory confusion; it's a complete lack of understanding of the fundamental entities that quantum mechanics is all about.

[contd]
Da Schneib
1 / 5 (2) Mar 22, 2016
Oh and BTW nice try at pretending that the behavior of the real world is something in Hilbert space.
Again, profound ignorance.
No, you appear not to understand that your Hilbert space map is not the territory of particles moving in space.

The Hilbert Space formulation is the "mathematically rigorous formulation of quantum mechanics, developed by John von Neumann", and is entirely compatible with Dirac's bra-Ket notation. This is not philosophy.
Bra-ket notation isn't the territory either, philosopher.

I'm bamboozling you with orders if magnitude more knowledge of the subject
You're lying again.
Noumenon
4 / 5 (4) Mar 22, 2016
Locality is the statement that local causes have local effects. Realism is the statement that parameters of particles always have a definite value even if we can't measure it.


They are both conditions for "local realism". Locality [local relativistic causality], and counterfactual definiteness, etc.

"1) locality, ....2) counterfactual definiteness -[ that entities have objective values independent of measurement so that statements can be made of experiments not actually performed], ....and the 3) no conspiracy condition,..... if the Bell inequalities rule out one, then they rule out "local realism"." - Noumenon

@CaptainStumpy, are you qualified in down-rating my posts above? Care to tell me what you object to?
Da Schneib
1 / 5 (2) Mar 22, 2016
Here's the skinny: if you measure the polarization of a particle in an arbitrary axis, call it "X," and then measure it in another arbitrary axis, call it "Y," then the polarization in X becomes a superposition. That superposition has a complex value, which represents the probability that when you measure it again in X you'll get the same answer, and the probability that when you measure it again in X you'll get a different answer. It's just that simple. That's how QM works. Your BS is exposed. Stop flailing, dude. It's not gonna work this time.

Locality is the statement that local causes have local effects. Realism is the statement that parameters of particles always have a definite value even if we can't measure it.
They are both conditions for "local realism". Locality [local relativistic causality], and counterfactual definiteness, etc.
So? How does that contradict what I say?
Da Schneib
1 / 5 (2) Mar 22, 2016
"You're not smart enough to understand how it contradicts what you say" in

3...

2...

1...
Noumenon
4 / 5 (4) Mar 22, 2016
@Schnieb, the use of the mathematics of Hilbert Space as formulated by von Neumann for QM,.... the Schrodinger equation, ....and the Dirac Bra-Ket notation,... represents the mathematical foundation of quantum mechanics,... and if you don't understand this, then you don't understand quantum mechanics.

you appear not to understand that your Hilbert space map is not the territory of particles moving in space.


That's because quantum "particles" don't move in simple classical trajectories. There may be interference terms. This is why the wavefunction is a complex quantity and why the Hilbert Space formulation is a consistent mathematical description.

You are objecting to a more precise mathematical description of QM, and looking like a crank in doing so.

I see you didn't bother taking notes when you where schooled in This Thread, ...instead hide under your desk where you belong.

Da Schneib
1 / 5 (2) Mar 22, 2016
Noum, the mathematics of Hilbert space are you obfuscating. We aren't talking about representations here, we're talking about particles and their parameters, and the values those parameters can take on.

Real photons, polarized in a filter oriented with its axis pointing to X, are passed through another polarizer oriented with its axis pointing to Y. If we then add a third filter pointing to X again, then some of the photons make it through and some don't, indicating that they were repolarized in X by having their polarization measured in Y. No matter what math you use, no matter which interpretation you use, those are absolute facts, and the percentage that make it through the second X oriented filter is calculable and fixed, given the directions X and Y.

We're talking about stuff that basic, and you want to change the subject to Hilbert space or bra-ket notation. You're lying, Noum, and it's transparent.
Da Schneib
1 / 5 (2) Mar 22, 2016
Sharp folks will notice that I varied the usual setup for the three-polarizers experiment. You will find, however, after you think it over, that the variation will also be true, since if X and Y are orthogonal, no photons will make it to the third filter, and thus no photons will be passed by it, and you will also realize that this is entirely in agreement with all formulations and all interpretations of QM.

Also, polarization is a parameter, and it has two values. This is a real particle, and when it interacts with the environment, that interaction has consequences. This isn't some representation in Hilbert space, it isn't some equation in bra-ket notation, it's a real particle in the real world- but when its polarization in one axis has been measured, its polarization in other axes is uncertain, and subject to a quantum mechanical calculation to determine the probability of its polarization having one or the other value in the next axis you choose to measure.
Noumenon
4 / 5 (4) Mar 22, 2016
I brought up the mathematics underlying QM for two reasons,... 1) to retort against your repeated childish accusation of speaking "philosophy",..... 2) to help you understand the statement you made below (etc) is not correct, as the wavefunction is not a representation of parameter values, and there is no superposition of values.

If you reference "wavefunction" or "superpostion" you're de facto referencing the mathematical representation. The wavefunction hence its name, is any arbitrary complex valued mathematical function that is Fourier decomposable. To obtain the set of possible observable parameter values, requires solving the Schrodinger equation for eigenvalue solutions.

Wavefunctions are representations of the values of parameters on particles when they are in superposition. - DaSchnieb

Da Schneib
1 / 5 (2) Mar 22, 2016
I brought up the mathematics underlying QM for two reasons,
You're lying again. You brought them up for only one reason: to obfuscate.
Noumenon
4 / 5 (4) Mar 22, 2016
Two final points to make clearer why that statement was incorrect or at best sloppy,....

- a single wavefunction can represent more than one "particle",

- not all quantum states (represented in a superpostion of states) are observable.

I brought up the mathematics underlying QM for two reasons,
You're lying again. You brought them up for only one reason: to obfuscate.


If bringing up the mathematics of QM obscures my post for you, that is a statement about your state of knowledge, not one of my honesty.

Noumenon
4 / 5 (4) Mar 22, 2016
You repeated the following insult over and over again to antagonize, and my response was to bury you in a deeper mathematical understanding of the subject.

"Please try to stay away from philosophy and stick to physics. Thanks. - DaSchnieb"

This COULD have been a cordial discussion, but for your behaviour.
Zenmaster
5 / 5 (1) Mar 22, 2016
Are the authors implying that there is no theoretical framework that predicts that nonlocality and contextuality would be essentially the same thing?
Da Schneib
1 / 5 (2) Mar 22, 2016
Two final points to make clearer why that statement was incorrect or at best sloppy,....
No, Noum, they are not final points and there is nothing sloppy. You're lying again.

- a single wavefunction can represent more than one "particle",
So? Each particle in the ensemble has its own individual wavefunction, and they add up to a composite one. This has nothing to do with the central question, which is whether philosophical and physics definitions of realism are the same.

- not all quantum states (represented in a superpostion of states) are observable.
This is duh; for a given particle, when the wavefunction collapses only one of the superposed states will be realized. And this also has nothing to do with the central question, which is whether philosophical and physics definitions of realism are the same.

You're bobbing and weaving, Noum. Why are you doing that if your point is straightforward?
Da Schneib
1 / 5 (2) Mar 22, 2016
I brought up the mathematics underlying QM for two reasons,
You're lying again. You brought them up for only one reason: to obfuscate.
If bringing up the mathematics of QM obscures my post for you You're lying again, Noum. It obfuscates it for everyone. And that's your intent.

Just admit you're wrong and that philosophical realism and physics realism have nothing to do with one another.

You repeated the following insult over and over again to antagonize,
No, Noum, you lied and I pointed it out. Get over it, and while you're at it, stop pretending that philosophy is physics.
Da Schneib
1 / 5 (2) Mar 22, 2016
Are the authors implying that there is no theoretical framework that predicts that nonlocality and contextuality would be essentially the same thing?
Hmmm, I think so. And I have to say that I've never heard of one. Do you know of one?
Noumenon
4 / 5 (4) Mar 22, 2016
- a single wavefunction can represent more than one "particle",


So? Each particle in the ensemble has its own individual wavefunction, and they add up to a composite one.


Wrong again. Its a SINGLE superposed wavefunction. The mathematical representation space in which such a wavefunction evolves is a tensor product space, which is how it is "separated".

-----------

Your childish and very defensive behaviour was also in display in http://phys.org/n...cal.html where you were rude also to DarkLordKelvin who had shown more knowledge of the subject than you also.

You said something in that thread that was very telling about your personality which sums up this exchange as well....

"But being told I'm "wrong" is pretty much not OK. - DaSchnieb"

Da Schneib
1 / 5 (2) Mar 22, 2016
You're still talking about the representation, and avoiding the question. Obfuscating, and that's lying.

Admit you're wrong, Noum. I've proven it; now it's up to you to man up.

And stop trying to change the subject. It's transparent.
Noumenon
4 / 5 (4) Mar 22, 2016
- not all quantum states (represented in a superpostion of states) are observable.


This is duh; for a given particle, when the wavefunction collapses only one of the superposed states will be realized


No, you didn't understand. Not all quantum states are observable, even in principle.

Just admit you're wrong and that philosophical realism and physics realism


I can agree to discuss this point further once you firstly admit that....1) the bulk of my posts above were not about philosophy at all, but about the mathematics of QM, and.... 2) you admit that your objection to philosophical points made by me here is entirely made mute by my superior and deeper understanding of quantum mechanics.

Da Schneib
1 / 5 (2) Mar 22, 2016
- not all quantum states (represented in a superpostion of states) are observable.
This is duh; for a given particle, when the wavefunction collapses only one of the superposed states will be realized
No, you didn't understand. Not all quantum states are observable, even in principle.
As I said, in any given interaction, that's true, and even in any given ensemble, but in the totality of all interactions you're lying again.

Just admit you're wrong and that philosophical realism and physics realism
I can agree to discuss this point further once you firstly admit that 1) the bulk of my posts above were not about philosophy at all, but about the mathematics of QM,
Sure, but the math was not relevant and you were obfuscating to avoid admitting you were wrong. When you admit that, then we will be in agreement.

[contd]
Da Schneib
1 / 5 (2) Mar 22, 2016
2) you admit that your objection to any philosophical points made by me is washed out by my superior and deeper understanding quantum mechanics itself.
Not a chance. Get over it. Your philosophy and your failure to admit it's not physics is the entire reason I dislike you. Philosophy is not physics, and the cat has been out of the bag for decades on that. See the Sokal Affair.
Zenmaster
5 / 5 (1) Mar 22, 2016
Don't know of one. When they say "the contextuality-nonlocality monogamy suggests the existence of a quantum resource of which entanglement is just a particular form", to what "resource" are they referring? What does "resource" mean? If contextuality and nonlocality are two conjugate or reciprocal aspects of the same thing then does this "resource" serve as the basis for that thing?
Da Schneib
1 / 5 (2) Mar 22, 2016
@Zen, I think they're talking about quantum irrealism as if it were a principle, and trying to quantify it.
Da Schneib
1 / 5 (3) Mar 22, 2016
See, Noum, I think you're a troll who's using obfuscation and other logical flaws to try and get a response out of people, and I think it's pretty disgusting, and I aim to expose you and stake you out for everyone to look at.
Noumenon
4.2 / 5 (5) Mar 22, 2016
See, Noum, I think you're a troll


You calling me a "liar" over and over exposes your mentality and lack of sophistication. You started this discussion by antagonizing as point out.

2) you admit that your objection to any philosophical points made by me is washed out by my superior and deeper understanding quantum mechanics itself.


Your philosophy and your failure to admit it's not physics is the entire reason I dislike you.


The subject of "https://en.wikipe...physics" is contributed to by very prominent physicists, and is ubiquitous in the history of quantum mechanics in interpretations of theory, as well as other branches of physics.

but the math was not relevant


Math is relevant in QM, that is the point. The tables were turned on you, don't you understand that yet?

In this thread,.... you objected to me making philosophical statements,,,, AND objected to me making mathematical statements.

Da Schneib
1 / 5 (3) Mar 22, 2016
Noum, I really don't care if you like me, and I really don't care about your whining about how insulting it is to be exposed for what you are.

You're a philosopher masquerading as a physicist.

You'll use anything, including math, to try and bolster your point, and you don't care whether you're wrong or right.

I care, and I find you to be a pretty small souled person.

This thread is about physics, and neither math nor philosophy will save you from everyone figuring out you don't really understand physics.
Da Schneib
1 / 5 (3) Mar 22, 2016
Let me try to explain this in a manner most people will understand:

Physics says that very small and very fast things are qualitatively different from ordinary consensus reality you see around you, but also that these very small and very fast things, and their odd behavior, *lead to* that reality you see.

Among those differences, the most important one is that things can have properties that are neither one thing nor another, but a mixture of both. But they can only have properties like that when you can't, or don't, look at them. The properties that are in this state are said to be "in superposition." If the time comes that you look at these properties, then they turn out to be a particular thing, and if you look at a lot of the same kinds of events, then there will turn out to be some that go this way, and some that go that way, and the percentage of each will be well-known, but how an individual one actually goes is random.

[contd]
Da Schneib
1 / 5 (3) Mar 22, 2016
[contd]

When a particular property turns out, in an individual case, to be some particular thing, then the superposition is said to have "collapsed." And in fact it isn't even necessary that you have looked; if the thing with the property touches some other thing in any way, then this "collapse" will occur, and the property will be realized.

This is an essential difference from the way things around you behave; you are used to things having properties that remain static no matter what happens. Quantum things are essentially different from the things you deal with in your perception of reality in this manner. And the experiment that this thread is about is one that looks very carefully at this behavior of quantum things.

If you can follow the logic, even though it's different from what you're used to, you will understand this behavior much better, and this will help you understand quantum reality.

That's the skinny.
Noumenon
4.2 / 5 (5) Mar 22, 2016
The notion from you that I don't understand the physics is laughable after the above exchanges.

Quantum mechanics requires an understanding of the mathematical formulation. If you don't understand it then you don't understand QM.

I referenced the mathematical structure to be more precise in discussion which QM requires, while you display a discovery-channel understanding, and yet have the nerve to say I don't understand the physics. You're a insulting and rude fraud and anyone who has seriously studied the subject can see this , but not people like CS.

And in fact it isn't even necessary that you have looked; if the thing with the property touches some other thing in any way, then this "collapse" will occur, and the property will be realized.


Another example of your lack of understanding. Decoherence does NOT cause collapse of the wavefunction.
Mimath224
5 / 5 (2) Mar 22, 2016
@Da Schneib I am interested in your conversation with Noumenon. I don't know enough about QM (though I am familiar with operators, eigenstates, superpostion etc. at an introductory level) or qualified enough to enter your discussion so I'd like to ask a genuine question. On the philosophical side I've always problems with not separating 'realism' and 'reality' because I don't view them as being the same and as you point out, it is best to stay away from such things when doing QM or even physics in general.
My question is; when having more than one solution the Sch. equ. then adding the solutions, apparently, is also a solution and results in a 'wave packet'. Apparently it is the w.p. that leads to a representation of a physical particle and so I might understand that as being the way to a reality in terms of QM. But how far back (before these) should I be looking for the 'assumption of realism'? I am not sure I have put that correctly but any info would be appreciated. Thanks
shavera
4.8 / 5 (6) Mar 22, 2016
Mimath, There's a lot to unfold in your question, too much for the character-limits of this forum. 1) the simplest schr. equ. we solve is the one describing position and momentum. The momentum of a particle is related to the frequency of the wave, the phase (where the peaks and valleys of the wave are) determine likely position. Moreover, for all but the simplest cases, your solution is a set of waves that are a "Fourier transform" of the information. The more precise knowledge of momentum you have, the fewer frequencies allowed, the more spread-out the wave must be, and the less information you may have about position. The more precise knowledge of position, the more frequencies required, the less information of momentum allowed.

shavera
4.8 / 5 (6) Mar 22, 2016
Quantum realism says that all that wave stuff is just some mechanism that is blurring out a "real" particle with real position and momentum. Eg, They are "Pilot" waves that guide the particle's trajectory through the universe. Even if there are some rules that say we must *never* know about the 'true' particle, realism states that such a particle exists.

The problem is that if quantum realism is true, then the blurring mechanism, pilot waves or whatever, must somehow work faster than light, if not instantaneous and backwards in time. If there are *real* properties to a particle, then these waves have to somehow communicate how they are being interacted with in an experiment.

On the other hand, if you accept that the wave *is* the particle, that the very nature of everything that exists is not precisely definite in all senses, then you no longer require the waves to be faster than light. The superposition itself accounts for the state between two entangled particles
antigoracle
5 / 5 (3) Mar 22, 2016

You calling me a "liar" over and over exposes your mentality and lack of sophistication.
....................

Math is relevant in QM, that is the point.

The fact that Da Schneib does not realize that Maths cannot lie, just confirms his idiocy.
Hyperfuzzy
not rated yet Mar 22, 2016
In other words, realism is only relevant to the observer and the choice of reference; however, QM defines none. Also the choice of potential and kinetic may not be realizable for each state returned or observer. My response, "futile", implies, set the measurement correctly and measure accordingly. It will always be relative to the reference and the interpretation. I see a single "Truth" there exist an undefinable source of the "+" and "-", i.e. we may measure and use but ...

Hence, moot! Look deeper. Anyway, + and - is not what I gain or lose but is something else. Never reconciled. Potential to kinetic and kinetic to potential with fundamentals, relative to whom? juz say'n
Mimath224
5 / 5 (2) Mar 22, 2016
@shavera many thanks for your time and post. Sorry to ask again but if 'The problem is that if quantum realism is true, then the blurring mechanism, pilot waves or whatever, must somehow work faster than light, if not instantaneous and backwards in time....' is there a suggestion here that a similar process to photon entanglement (instantaneous) might be involved?
While 'On the other hand, if you accept that the wave *is* the particle, that the very nature of everything that exists is not precisely definite in all senses, then you no longer require the waves to be faster than light....' might deny photon entanglement (instantaneous) by FTL travel?
Although I am aware of superposition implications etc how would this function as FTL travel for entangled particles? (my books don't seem to mention these alternatives) Perhaps you know of a website that would be useful to me and be less bother to posters here (Ha!). Thanks.
Noumenon
4 / 5 (4) Mar 23, 2016
@Minmath, …. You implied, follow DaSchieb, that it's best to stay away from influences of philosophy when doing QM. This is true when "doing QM", that is 'using the theory'.

However, when discussing implications of experimental results of QM in an effort to understand the physical world, as is the ultimate point of science, one naturally is going to offer explanatory hypothesis for that purpose.

In QM, there are many such 'interpretations of quantum mechanics', but what they all have in commen is that they abide to current experimental results.

IOW, if they do not offer new predictions over and above core QM, then they are not new theories over and above QM,…. they are stickily speaking a branch of Philosophy of Physics.

Noumenon
3.7 / 5 (3) Mar 23, 2016
There are epistemological assumptions made that are conditions for scientific observation to be possible,... locality, separability, counterfactual definiteness, particle, wave, space, time, determinism, etc .... because experiments must be conducted at the macroscopic level, and that is where our minds evolved to synthesize experience in terms of those 'forms of thought'.

"There is no way to remove the observer us from our perception of the world, which is created through our sensory processing and through the way we think and reason. Our perception and the observations upon which our theories are based are shaped by a kind of lens, the interpretive structure of our human brains." - Stephen Hawking

In interpretations of QM experiment, one has the choice of making use of one conjugate concept at the expense of another.
Noumenon
3.7 / 5 (3) Mar 23, 2016
The essential point,… which I believe is more sensible than proposing objects that are in principal unobservable [i.e. metaphysical, as in pilot-wave, many-worlds]….. is to accept the wavefunction as not being a physical object, but rather a mathematical construct which represents information of the system,…. yet in need of 'conceptual form'.

The application of the quantum-operator to this mathematical wavefunction, represents 'adding conceptual form' to obtain conceptual values, as explained above. The reason is is that the quantum-operator represents the experimental apparatus in essence. The experimental apparatus is macroscopic and so must accord with our 'forms of thought',…i.e. the assumptions made as conditions for macroscopic observation…..
Noumenon
3 / 5 (4) Mar 23, 2016
….

The point here is that the 'underlying reality' is conceptually formless (as it exists independently of our knowledge of it). Experiment in effect projects a 'underlying something' into our a-priori conceptual framework [see Hawking's quote]…. so that… the non-intuitive nature of QM has established that ....

"The doctrine that the world is made up of objects whose existence is independent of human consciousness turns out to be in conflict with quantum mechanics and with facts established by experiment" - B. d'Espagnat

A physicist can have a Realist interpretation or a Positivist interpretation of QM. These are indeed philosophical positions,... however the former are refuted by experiment,.... once the elements of philosophical "Realism" are quantified as I enumerated above for "local realism".

shavera
5 / 5 (2) Mar 23, 2016
Mimath: I'll always recommend searching through r/askscience for more details on quantum stuff/asking your own.
Anyway: for details on the question of FTL, consider this. If there's a 'real' state, it may be that no physical process allows us to measure that 'real' state. Whatever my experiment is, it's made of other particles obeying the same quantum rules, so there may be no way to *actually* measure the 'real' state underneath. Therefore, if these pilot waves or whatever allow these 'real' states to communicate with each other FTL, they cannot communicate *useful information* faster than light, because they only talk to this "underlying" reality, and not to "our" reality. So the rules about "no information can pass faster than light" still hold more or less.
Noumenon
3.7 / 5 (3) Mar 23, 2016
......

A Realist in this sense believes that Reality is as we come to know it, independently of the way we must think and reason. In contrast, a Positivist, like Hawking, d'Espagnat, etc, and myself, believe that physics is not about ascertaining knowledge of Independent Reality, but rather of ascertaining knowledge of phenomena as experienced, with all the epistemological conditions implied by that.

This is the reason that experiment seems to create the attributes, and results change when we are looking!

Science studies phenomena. Phenomenal-reality has a necessary component that is mind-dependent due to the logical necessity of observation,.. the mind necessarily filters the underlying [though objective] reality through concepts, to produce Phenomenal-Reality.

Phenomenal-Reality = Mind (Noumenal Reality)
Noumenon
3.7 / 5 (3) Mar 23, 2016
Although I am aware of superposition implications etc how would this function as FTL travel for entangled particles?


Because a Single wavefunction description represents Two [or more] entangled particles, no mater how far they are subsequently observed. The wavefunction can spread arbitrarily in the space of possibilities**.

Once a measurement takes place on one of the two entangled particles, this wavefunction 'collapses' resulting in say two distinct positions.

** For multiple particles this is NOT regular 3D space, but rather a tensor space, however, a lower limit for the velocity that this collapse can be experimentally determined and if I recall something of the order of magnitude of 100,000 times c. It makes more sense to just say it is not a physical wave.

Noumenon
4 / 5 (4) Mar 23, 2016
The above rant is a 'philosophical interpretation' and may be entirely wrong and may not even be falsifiable, In principal, the mind and its operational conditions ARE investigable though,... in a way that 'pilot-wave's' and 'many-worlds' are not.

Protoplasmix
5 / 5 (4) Mar 23, 2016
Our perception and the observations upon which our theories are based are shaped by a kind of lens, the interpretive structure of our human brains." - Stephen Hawking
What are you saying? What are you saying Hawking's saying? Is it these obvious pitfalls? → "Previous experience (and even the act of experiencing itself) colors our perception. We see only what we want or expect to see. Prophecy easily becomes self-fulfilling from just a little bias." What?

I use a variety of lenses, both external and internal, and compare the results. How could I possibly know exactly, precisely how and what you (or anyone) see and feel? But if I can repeat your experiment and I get the same results, then there is consensus. And consensus is a thing greater than the sum of its parts. Maths and science – these are not rose colored glasses; they're perfect lenses if you know how to use them. At least that's been my experience. So what was your point, Noumenon?

cont'd >
Protoplasmix
5 / 5 (4) Mar 23, 2016
> cont'd

It wasn't too long ago people didn't know for sure if we would ever directly "see" gravitational waves. You can make a strong argument we didn't "see" them, we "heard" them. But then you'd make my point about no ill affect on our ever increasing knowledge and mastery over the environment from the manner in which we experience things. Does that make me a realist? Because I feel a whole lot more positive about experiencing things that a "positivist" says we can't, if I understand you correctly.
farstriderr
5 / 5 (3) Mar 23, 2016
"The assumption of realism" as the article calls it, is not one physicists generally assume.
The article misrepresents the idea of "non-real" quantum physics, and the extent to which it is quite common throughout the community.


They assume realism in order to develop experiments that violate realism. You can't assume "non-realism", then develop experiments that violate realism. That's illogical.
farstriderr
5 / 5 (3) Mar 23, 2016
Not sure anyone commenting here understands this experiment at all. All this is is an experiment that tests locality and contextuality simultaneously via entanglement. Previously, we had experiments violating locality via entanglement, and experiments violating contextuality via erasure. This combines both into what is essentially a delayed-choice quantum eraser with added entanglement (quite ingenious). Alice's path tests contextuality sort of like a DCQE by bringing her particle in and out of superposition before it hits the final detector. Bob's path tests locality by means of entanglement with Alice's particle. The idea was that the violations of contextuality on Alice's path should register on Bob's detector via a simultaneous nonlocal action. It did not. Therefore it is concluded that there must be some more fundamental "thing" that controls both the violation of contextuality and locality.
Noumenon
3.7 / 5 (3) Mar 23, 2016
There is no way to remove the observer us from our perception of the world, which is created through our sensory processing and through the way we think and reason. Our perception and the observations upon which our theories are based are shaped by a kind of lens, the interpretive structure of our human brains." - Stephen Hawking
..… I use a variety of lenses, both external and internal, and compare the results. How could I possibly know exactly, precisely how and what you (or anyone) see and feel?


My use of Hawking's quote is not meant to refer to personal subjectivity. My above several posts were meant to explain this further….
Hyperfuzzy
not rated yet Mar 23, 2016
OK, can't get through. QM will express the potential energy and the kinetic energy of a group of particles. Models typically based upon Bhor or Dirac. The electron and the proton produce the EM waves as defined. Other particles as interpreted are a misinterpretation. E = MC^2 is nonsense, for there does not exist a quantifiable Mass. h nu is a quantum measurement, Mathematical construct. The energy within the field about a particle is spherical and a point on the sphere has an e(r) field divided by the surface area, Gauss. The rest is simply mental masturbation or some nonsense to keep our enemies from developing nuclear weapons. It ain't real! It's only what we used to think, 19th century nonsense. Time to move on! Does anybody really understand possibility and Not Possible? Simply put, you can't create an experiment from stupidity and define anything other than stupidity or luck. A very stupid way to study without considering the facts. Wow, can we be in two places?
Noumenon
4 / 5 (4) Mar 23, 2016
[…]my point about no ill affect on our ever increasing knowledge and mastery over the environment from the manner in which we experience things.

Does that make me a realist? Because I feel a whole lot more positive about experiencing things that a "positivist" says we can't, if I understand you correctly.


But by "experiencing things", you are adding conditions for doing so, as pointed out by Hawking and my posts above.

A 'positivist' pointing this fact out, is in no way anti-science in any respect. In fact a Positivist is less likely to inject hypothetical metaphysical objects, than is a Realist. A Positivist would not expect QM to 'make intuitive sense', because he recognizes that our mind-dependent conceptual framework is in essence an artificial one.

I made the distinction between Realist and Positivist above.

A Positivist would confirm that physics is about acquiring predictive knowledge of experience (observation),.... just is not independent of it.

Hyperfuzzy
not rated yet Mar 23, 2016
Look at the stupid title. It does not define anything! Let's start with "defining features" and "QM" and "never". Why not "Defining features do not exist for QM." People, see! What are we talking about?
Da Schneib
1 / 5 (2) Mar 23, 2016
The notion from you that I don't understand the physics is laughable after the above exchanges.

Quantum mechanics requires an understanding of the mathematical formulation. If you don't understand it then you don't understand QM.
BS. Sorry I just plain don't believe this. I think this is you blowing smoke so you look smart, and failing to understand that the underlying reality is not any of the mathematical formulations.

I referenced the mathematical structure to be more precise in discussion which QM requires, while you display a discovery-channel understanding, and yet have the nerve to say I don't understand the physics. You're a insulting and rude fraud and anyone who has seriously studied the subject can see this , but not people like CS.
More smoke.

[contd]
Da Schneib
1 / 5 (1) Mar 23, 2016
And in fact it isn't even necessary that you have looked; if the thing with the property touches some other thing in any way, then this "collapse" will occur, and the property will be realized.


Another example of your lack of understanding. Decoherence does NOT cause collapse of the wavefunction.
I wasn't discussing decoherence. Decoherence is not interaction; I was discussing interactions.

So much for how much physics you understand.
Da Schneib
1 / 5 (1) Mar 23, 2016
Noumenon is spot on with his QM knowledge. Da Schneib, on the other hand, demonstrates time and again a poor understanding of the subject and, with his continual insults and ranting "liar" to anybody disagreeing with him, is a foul-mouthed street-corner layabout. End of.
Apparently you don't know the difference between decoherence and interaction either.

Just sayin'.
Da Schneib
1 / 5 (1) Mar 23, 2016
@shavera, nice explanation of "wave packets."

Question on
If there's a 'real' state, it may be that no physical process allows us to measure that 'real' state. Whatever my experiment is, it's made of other particles obeying the same quantum rules, so there may be no way to *actually* measure the 'real' state underneath.
: how do you see this interacting with
a) the subject article of this thread,
b) the Transactional Interpretation's Wheeler-Feynman absorber theory,
c) the Bohm interpretation,
d) the Bell's Theorem proof that all hidden variable theories are ruled out, by the Aspect experiment, etc.?
Da Schneib
1 / 5 (1) Mar 23, 2016
IOW, if they do not offer new predictions over and above core QM, then they are not new theories over and above QM,…. they are stickily speaking a branch of https://en.wikipe...chanics.
No, because they relate the mathematical formalism to the experimental results, which is not philosophy. They do give rise to various parts of the philosophy of physics, but they aren't part of it; they are one of its subjects of study. You're taking a bigger bite for the philosophy of physics than it actually deserves. And this is what I've always claimed you're doing.
Da Schneib
3 / 5 (2) Mar 23, 2016
The above rant is a 'philosophical interpretation' and may be entirely wrong and may not even be falsifiable, In principal, the mind and its operational conditions ARE investigable though,... in a way that 'pilot-wave's' and 'many-worlds' are not.
We don't know that. Just because no one's come up with a way to differentiate between interpretations doesn't mean no one will. You're essentially claiming that we know for sure we never will, and that's possible, but unproven.

More smoke.
Da Schneib
3 / 5 (2) Mar 23, 2016
@Protoplasmix: Well done.

@farstriderr:
Not sure anyone commenting here understands this experiment at all. All this is is an experiment that tests locality and contextuality simultaneously via entanglement. Previously, we had experiments violating locality via entanglement, and experiments violating contextuality via erasure. This combines both into what is essentially a delayed-choice quantum eraser with added entanglement (quite ingenious).
Nice analysis! You're correct, this experiment has many similarities with the DCQE. And the added entanglement lets the experimenter choose between the standard DCQE approach, which violates contextuality, and the later derivative experiments of the DCQE that violate locality, and shows you can't violate both.
Da Schneib
3 / 5 (2) Mar 23, 2016
My use of Hawking's quote is not meant to refer to personal subjectivity. My above several posts were meant to explain this further….
Bob duck and weave. More smoke.

But by "experiencing things", you are adding conditions for doing so, as pointed out by Hawking and my posts above.
You mean like "having a brain." Or "being made of matter." Because that's what Hawking means, not the smoke you were spouting. And what he said has nothing to do with the interpretations of QM.

A 'positivist' pointing this fact out, is in no way anti-science in any respect. In fact a Positivist is less likely to inject hypothetical metaphysical objects, than is a Realist. A Positivist would not expect QM to 'make intuitive sense', because he recognizes that our mind-dependent conceptual framework is in essence an artificial one.
We're discussing quantum realism, not Realism vs. Positivism. I defined quantum realism above. Maybe you forgot.

[contd]
Da Schneib
3 / 5 (2) Mar 23, 2016
[contd]
I made the distinction between Realist and Positivist above.
And it's completely immaterial because we're not discussing Realism, we're discussing quantum realism, which you won't admit is not the same thing. And you still haven't admitted it.

A Positivist would confirm that physics is about acquiring predictive knowledge of experience (observation),.... just is not independent of it.
Still immaterial. More smoke.
Da Schneib
1 / 5 (1) Mar 23, 2016
So, @Noumenon, got an explanation for your confusion of decoherence and interaction for us?

And how about admitting that Realism isn't the same thing as quantum realism? You as much as admitted it above, but then you fell off the wagon.

How about an explanation for the fact that you claim that interpretations of QM are philosophy of physics when they're not, or for the fact that they relate the formalisms to experimental results, neither of which is philosophy?

I also found it amusing that you downvoted my compliments to others. Now THAT is childish.
Noumenon
4 / 5 (4) Mar 23, 2016
And in fact it isn't even necessary that you have looked; if the thing with the property touches some other thing in any way, then this "collapse" will occur, and the property will be realized. - DaSchnieb


Decoherence does NOT cause collapse of the wavefunction. - Noumenon


I wasn't discussing decoherence. Decoherence is not interaction; I was discussing interactions.


Decoherence is not interaction? Are you sure about this?

Decoherence occurs when a QM system INTERACTS with its environment, which results in the loss of (off-diagonal) interference terms by the loss of phase coherence. Interaction.

You have already proven your character to me above, and so I have no interest in having a 'jerry-springer' quality debate with you. The posts I made today were not in response to anything you posted.

Da Schneib
3 / 5 (2) Mar 23, 2016
Decoherence is not interaction? Are you sure about this?
Yep. Interaction collapses the wavefunction. As you yourself said, decoherence does not.
Noumenon
4.2 / 5 (5) Mar 23, 2016
Decoherence is not interaction? Are you sure about this?
Yep. Interaction collapses the wavefunction. As you yourself said, decoherence does not.


Are you sure that interaction collapses the wavefunction? Are you sure that decoherence is not interaction,..... because it is caused by it.

The problem of how or even if wavefunction collapse occurs is at the heart of the presently unsolved 'measurement problem'. I offered an epistemological solution above, I don't expect you to follow. See "von Neumann's cut" from his preeminent text on QM.

In principal quantum reality operates according to the deterministic Schrodinger equation, and if that system was allowed to evolve without an observation, then that QM system would INTERACT with the environment,.... i.e. decoherence would occur AND yet there would be no collapses of the wavefunction.

farstriderr
5 / 5 (3) Mar 23, 2016
I made a booboo on my previous post by the way. It isn't like a DCQE with entanglement added. Of course the DCQE already incorporates entanglement. More accurately it's like a DCQE with added mechanisms along Alice's path to violate noncontextuality whilst Bob tries to measure those violations. The article doesn't really try to explain how the experiment works, unfortunately.
Da Schneib
1 / 5 (1) Mar 23, 2016
Decoherence is not interaction? Are you sure about this?
Yep. Interaction collapses the wavefunction. As you yourself said, decoherence does not.


Are you sure that interaction collapses the wavefunction? Are you sure that decoherence is not interaction,..... because it is caused by it.
It doesn't matter how many times you repeat this lie. Go read the article: https://en.wikipe...oherence
Da Schneib
not rated yet Mar 23, 2016
I made a booboo on my previous post by the way. It isn't like a DCQE with entanglement added. Of course the DCQE already incorporates entanglement. More accurately it's like a DCQE with added mechanisms along Alice's path to violate noncontextuality whilst Bob tries to measure those violations. The article doesn't really try to explain how the experiment works, unfortunately.
Well, technically, depending on exactly how you interpret the DCQE you can work your way around to an interpretation that violates locality, or (much more difficult) violates quantum realism... so, technically I'm not sure you're wrong depending on how you interpret the DCQE.
Noumenon
4.2 / 5 (5) Mar 23, 2016
… if the thing with the property touches some other thing in any way, then this "collapse" will occur, and the property will be realized.


Decoherence does NOT cause collapse of the wavefunction.


I wasn't discussing decoherence. Decoherence is not interaction; I was discussing interactions.


Decoherence occurs when a QM system INTERACTS with its environment, which results in the loss of (off-diagonal) interference terms by the loss of phase coherence. Interaction. Decoherence is not interaction? Are you sure about this?


Yep. Interaction collapses the wavefunction. As you yourself said, decoherence does not.

Are you sure that interaction collapses the wavefunction? Are you sure that decoherence is not interaction?

It doesn't matter how many times you repeat this lie. Go read the article: [..]

It confirms what I already posted. Interaction/Decoherence does not collapse the wavefunction.
Da Schneib
1 / 5 (1) Mar 23, 2016
It confirms what I already posted. Interaction/Decoherence does not collapse the wavefunction.
But my objection wasn't that you were wrong in saying that, my objection was that you were wrong in claiming I said it does.

Note that I never mentioned decoherence in what you quoted. Just sayin'.

More smoke.
Noumenon
4 / 5 (4) Mar 23, 2016
IOW, if [interpretations of quantum mechanics] do not offer new predictions over and above core QM, then they are not new theories over and above QM,…. they are [strictly] speaking a branch of Philosophy of Physics.

No, because they relate the mathematical formalism to the experimental results, which is not philosophy.


That is what philosophy of physics IS,… interpretations [whether they have mathematical form or not] of the same experimental results. Did you read the link provided? It said this …. "Quantum mechanics is a large focus of contemporary philosophy of physics, specifically concerning the correct interpretation of quantum mechanics".

Are you then speaking philosophical above even while berating it, because you just uttered this gem…..
Da Schneib
1 / 5 (1) Mar 23, 2016
No, because they relate the mathematical formalism to the experimental results, which is not philosophy.
That is what philosophy of physics IS,…
No, it's not. Relating mathematical formalisms of QM to classical experimental results is absolutely not philosophy.

More smoke.
Noumenon
4.2 / 5 (5) Mar 23, 2016
Does this mean you're speaking philosophical then....

Quantum mechanics requires an understanding of the mathematical formulation. If you don't understand it then you don't understand QM. - Noumenon

BS. Sorry I just plain don't believe this. I think this is you blowing smoke so you look smart, and failing to understand that the underlying reality is not any of the mathematical formulations. - DaSchnieb


Please try to stay away from philosophy and stick to physics. Thanks. \\sarcasm.

It confirms what I already posted. Interaction/Decoherence does not collapse the wavefunction.


But my objection wasn't that you were wrong in saying that, my objection was that you were wrong in claiming I said it does.


I see, but you said "collapse" and "interaction",.... I don't know how else to have interpreted that but as, decoherence or that interaction causes collapse of the WP.

If I read that wrong, then I apologize.
Noumenon
3.7 / 5 (3) Mar 23, 2016
I also found it amusing that you downvoted my compliments to others. Now THAT is childish.


Because your behavior above certainly wasn't childish, right?

You 1 rated everyone of my posts in this thread, even while responding. Had you not done that I would not have down rated a single post of yours.

Da Schneib
1 / 5 (1) Mar 23, 2016
Does this mean you're speaking philosophical then...
Quantum mechanics requires an understanding of the mathematical formulation. If you don't understand it then you don't understand QM.
BS. Sorry I just plain don't believe this. I think this is you blowing smoke so you look smart, and failing to understand that the underlying reality is not any of the mathematical formulations.
Please try to stay away from philosophy and stick to physics. Thanks. \\sarcasm.
1. I stay as far from philosophy as I can get.
2. That's a statement of fact, not a statement of philosophy. "The map is not the territory" is about as basic a statement of fact as you can find. There's no water, or dirt, or rock, or hills, or forests, or anything else but ink and paper in a map. And there aren't any photons, or quarks, or leptons in a mathematical theory of physics. Just numbers.

You're not even all that good a philosopher.

[contd]
Da Schneib
1 / 5 (1) Mar 23, 2016
[contd]
It confirms what I already posted. Interaction/Decoherence does not collapse the wavefunction.
But my objection wasn't that you were wrong in saying that, my objection was that you were wrong in claiming I said it does.
I see, but you said "collapse" and "interaction",.... I don't know how else to have interpreted that but as, decoherence or that interaction causes collapse of the WP.

If I read that wrong, then I apologize.
Since I never said "decoherence" you read it wrong. Now you need to address the other points, starting with forgetting that interaction causes the collapse of the wavefunction (if you buy Copenhagen-with-collapse, as the largest plurality of physicists do).
Da Schneib
1 / 5 (1) Mar 23, 2016
I also found it amusing that you downvoted my compliments to others. Now THAT is childish.
Because your behavior above certainly wasn't childish, right?

You 1 rated everyone of my posts in this thread, even while responding. Had you not done that I would not have down rated a single post of yours.
I 1 rated them because you're confusing philosophy with physics and you won't admit it and you won't stop and it's stupid.

Get over it. Say something smart.

Oh and BTW you went around to other threads and downvoted posts there, without ever providing any argument to say there was anything wrong, to "get me." Since we're on the subject. Retaliation is stupid, too, and quite transparent-- and quite childish.

Stop trying to "win," Noum. It's pretty disgusting to watch. Just stick to the facts-- one of which is that physics is not philosophy.
Whydening Gyre
5 / 5 (2) Mar 23, 2016
Decoherence does NOT cause collapse of the wavefunction. - Noumenon


I wasn't discussing decoherence. Decoherence is not interaction; I was discussing interactions.


Decoherence is not interaction? Are you sure about this?

Decoherence occurs when a QM system INTERACTS with its environment, which results in the loss of (off-diagonal) interference terms by the loss of phase coherence. Interaction.

No. It's the RESULT of interaction. Decoherence IS the wave "collapsing"...
Don't over think it...
BTW - WB, The DA (you crotchety ol' curmudgeon...:-))
Da Schneib
2.3 / 5 (3) Mar 23, 2016
Heh, thanks Whyde.

More to the point, decoherence is the loss of entanglement, caused by interaction with the environment. In other words, photon A interacts with proton B, causing photon A's polarization in some axis, call it X (and entangling it with proton B, which has also had its polarization in X measured); later, photon A interacts with electron C, causing photon A's polarization in some other axis, call it Y, which (under Heisenberg uncertainty) causes the polarization in X to decohere, by causing it to lose its entanglement with proton B. That's what decoherence is.
Mimath224
5 / 5 (2) Mar 23, 2016
To those who answered my questions, thank you and to fastriderr I freely admit my ignorance (Ha!) but I am fascinated by QM. There are some things in QM that I readily accept and understand (duality) that some others seem to have a problem with (I see it all around us and have posted my ideas on other threads before). However, I am a layman and need all the help I can get (another Ha!).
@Noumenon, you need to be careful about implying what I am implying. I did post that I was not going to enter your discussion with Da Schneib on this article and that remains. But when you start talking about 'Positivism' you are really diving into heavy waters. First you need to define which branch of Logical Positivism you are coming from...they aren't all the same. If you take the 'empiricism' route then likewise you need to define your stance, 'restrictive' or 'lesser restrictive'. The latter could be such that it requires; (cont)
Mimath224
5 / 5 (2) Mar 23, 2016
(cont) Every individual statement of a theory must refer to an empirical state of affairs. If you follow that then every sentence becomes a target of analysis rather than the descriptive terms (of the theory) in the sentence. I am not suggesting that you, me or anyone else are for or against this idea but just an example of how bogged down one can get and thus is best left to other forums that discuss such topics.
However, basically most science started way back as a kind of philosophical question and just as Homo Sapiens carry the genes of their prehistoric beginnings I suppose science must do so also. But just as Homo Sapiens are different so is the evolved science and we need to be aware of that and act accordingly.
@Whydening Gyre & Da Schneib 'More to the point, decoherence is the loss of entanglement,....I thought that was in all basic texts on the subject, slit experiments, sunlight being only partially coherent (and incoherent) etc. Noumenon not understand the difference?
Da Schneib
3 / 5 (2) Mar 23, 2016
It's another of those map/territory confusions, @Mimath.
Noumenon
3.8 / 5 (5) Mar 23, 2016
@Gyre, Minmath224,...

Decoherence does not result in wavefunction collapse. Decoherence is the result of interaction with the environment.

"A dynamical collapse of the wave function would require nonlinear and non-unitary terms in the Schrödinger equation [not standard QM] Since nonlinear terms in the Schrödinger equation lead to observable deviations from conventional quantum theory, they should at present be disregarded for similar reasons as hidden variables." - Heinz-Dieter Zeh, the discoverer of decoherence. [see my post also wrt the Schrodinger equation and decoherence]

"Decoherence does not generate actual wave function collapse. It only provides an explanation for the observation of wave function collapse, as the quantum nature of the system "leaks" into the environment. [...] Specifically, decoherence does not attempt to explain the measurement problem." - [Link

Noumenon
3.6 / 5 (5) Mar 23, 2016
It gives an explanation for the 'appearance of wavefunction collapse', but does not itself generate actual mathematical wave-function collapse in the theory of QM, without a modification of the Schrodinger equation to accommodate this specifically. This would be an alternative objective collapse theory.

@Whydening Gyre & Da Schneib 'More to the point, decoherence is the loss of entanglement,....I thought that was in all basic texts


No, its the gain of entanglement [coupling with the environment] such that there is a loss of phase coherence and thus loss of interference terms that are responsible for quantum behavior.

Da Schneib
5 / 5 (1) Mar 23, 2016
But the whole point, Noum, is that I never mentioned decoherence.

You forgot again.
Noumenon
4.6 / 5 (5) Mar 24, 2016
Since I never said "decoherence" you read it wrong.


Now you need to address the other points, starting with forgetting that interaction causes the collapse of the wavefunction (if you buy Copenhagen-with-collapse, as the largest plurality of physicists do)


The Copenhagen interpretation which I accept, does not propose that wave function collapse occurs on account of quantum interaction. It is an additional layer of interpretation over and above the mathematical theory itself.

There is an incompatibility between the deterministic evolution of the Schrodinger equation and the state-reduction (WP collapse) upon observation. C.I. just states that incompatibility as a consequence of observation....

I posted my extended take about why this is so above. See also 'von Neumanns cut'.

Noumenon
4.4 / 5 (5) Mar 24, 2016
Since I never said "decoherence" you read it wrong.


Perhaps, but since you said "interaction" and "collapse", I'm not sure how else I was expected to take that. In any case, quantum interaction according to the deterministic Schrodinger equation does result in wave function collapse either.

Da Schneib
4 / 5 (1) Mar 24, 2016
The Copenhagen interpretation which I accept, does not propose that wave function collapse occurs on account of quantum interaction.
OK. I like CH myself, on odd dates; on even ones I like Jack Cramer's TI. Wikipedia claims that TI has wavefunction collapse, I'm not sure I agree because it's not much like Copenhagen. Wikipedia says: "According to the Copenhagen interpretation, physical systems generally do not have definite properties prior to being measured, and quantum mechanics can only predict the probabilities that measurements will produce certain results. The act of measurement affects the system, causing the set of probabilities to reduce to only one of the possible values immediately after the measurement. This feature is known as wavefunction collapse." https://en.wikipe...retation

[contd]
Da Schneib
not rated yet Mar 24, 2016
[contd]
Wikipedia also says Copenhagen doesn't have decoherence: "The discontinuous 'wave function collapse' postulated in the Copenhagen interpretation to enable the theory to be related to the results of laboratory measurements now can be understood as an aspect of the normal dynamics of quantum mechanics via the decoherence process. Consequently, decoherence is an important part of the modern *alternative to the Copenhagen interpretation*, based on consistent histories." https://en.wikipe...surement (Emphasis mine.)

So if you're a strict Copenhagen advocate you shouldn't even be *talking* about decoherence. You don't need decoherence in Copenhagen; you got direct literal wavefunction collapse, but it's limited to the particle's contact with the measurement equipment.

[contd]
Da Schneib
4 / 5 (1) Mar 24, 2016
[contd]
Now, I'm willing to be a little fuzzy and say that collapse and decoherence have something to do with one another; but get right down to it, I really don't think collapse is, and I don't mean this in any physics sense, I guess I'm doing a little philosophy here, realistic. By which I mean, I think it's so divorced from how things really happen that it's not a good enough description unless you just wanna calculate. As far as actual particles undergoing real wavefunction collapse, sorry, man, I don't believe in ghosts either. Decoherence makes a great deal more sense.

And wavefunction collapse and decoherence don't occur in the same interpretations. You don't need one if you have the other. See how that works?

Sorry if I was being a little imprecise so everyone else could kinda understand quantum mechanics.

Meanwhile,

[contd]
Da Schneib
not rated yet Mar 24, 2016
[contd]
There is an incompatibility between the deterministic evolution of the Schrodinger equation and the state-reduction (WP collapse) upon observation. C.I. just states that incompatibility as a consequence of observation....

I posted my extended take about why this is so above. See also 'von Neumanns cut'.
And CH and Bohm use decoherence to do substantially the same thing- to explain the apparent "wavefunction collapse" as the result of interactions with environmental particles, step by step. And we know by experiment that a single interaction, if it happens to measure a complementary property of an entangled property, decoheres the entanglement.

Now everybody's confused.
Da Schneib
not rated yet Mar 24, 2016
Since I never said "decoherence" you read it wrong.


Perhaps, but since you said "interaction" and "collapse", I'm not sure how else I was expected to take that. In any case, quantum interaction according to the deterministic Schrodinger equation does result in wave function collapse either.
You can't have decoherence and collapse in the same interpretation. And you're not even supposed to be talking about decoherence if you're a strict Copenhagen advocate.

You at least thought about these ones a little more, but your claims to know a lot about QM have just been exposed; you know only one interpretation and you don't appear to have properly studied any of the others. And you don't know enough about Copenhagen to know it doesn't have decoherence in it, either.
Da Schneib
not rated yet Mar 24, 2016
Oh and BTW the current Wikipedia article on CH is crud. It doesn't even mention the Born Rule.
Protoplasmix
5 / 5 (3) Mar 24, 2016
But by "experiencing things", you are adding conditions for doing so...
Easy enough to evaluate the conditions, whatever they may be, and include them in the calculations. I think I understand the distinction between positivist and realist better now, though, thanks Noumenon.
Noumenon
4.5 / 5 (4) Mar 24, 2016
You can't have decoherence and collapse in the same interpretation. And you're not even supposed to be talking about decoherence if you're a strict Copenhagen advocate.


Since the element of the measurement-problem that explains why a particular outcome is realized as opposed to some other possibility, is NOT resolved by decoherence, it is still necessary to postulate an additional layer of interpretation,… in the case of C.I. this would be a non-objective collapse or state reduction. It is compatible retrospectively.

This is why Everett postulated many-worlds, which developed into consistent histories.

Here is the rest of the quote you posted….

"However, decoherence by itself may not give a complete solution of the measurement problem, […] To present a solution to the measurement problem in most interpretations of quantum mechanics, decoherence must be supplied with some nontrivial interpretational considerations."
Noumenon
4.5 / 5 (4) Mar 24, 2016
And wavefunction collapse and decoherence don't occur in the same interpretations. You don't need one if you have the other. See how that works?


You need to make an important distinction here. Are you speaking of an objective-collapse, as if the wavefunction is a physical thing? There ARE alternative objective-collapse theories. As pointed out above they require non-standard modification to the Schrodinger equation. In this case, decoherence (just a consequence of the standard Schrodinger equation), would be made compatible with wavefunction collapse (not a consequence of the standard Schrodinger equation).

If you are referring to standard QM theory instead, then they are not incompatible either due to the fact that decoherence leaves the important element of the measurement problem unresolved, thus still requiring the extra postulate of collapse (in the C.I. and extensions).

In C.I. it is an extra postulate just as the Born rule is.
Noumenon
4.5 / 5 (4) Mar 24, 2016
The C.I. emphasizes the unavoidability of classical-concepts in measurement,..... while decoherence emphasizes the notion of classicality as dynamically emergent phenomena from an otherwise underlying quantum reality.

The effective interface between these two is macroscopically evolved mind-dependent observation [insert Hawking quote]. Because decoherence does not explain why particular conceptual values are observed, this requires one to step outside the subject and into epistemology imo.

Some interpretations choose to step outside the subject, into metaphysics,... pilot-wave, Everett,... epistemology is at least investigable in principal.

This is what I attempted to convey above. Again it may not be right, but the essence of the C.I. was this, and not to make presumptions about reality that cannot be observed.

Ryan1981
not rated yet Mar 24, 2016
Don't ask me why but I feel this is related:

https://en.wikipe..._paradox
Hyperfuzzy
not rated yet Mar 24, 2016
Again we babble. Exactly WTF is the dynamics of "collapse." I think it is the discontinuity between reality and QM. That is, reality rules. So finish your little conversation while I define my model with "reality."
compose
Mar 24, 2016
This comment has been removed by a moderator.
Whydening Gyre
not rated yet Mar 24, 2016
And CH and Bohm use decoherence to do substantially the same thing- to explain the apparent "wavefunction collapse" as the result of interactions with environmental particles, step by step. And we know by experiment that a single interaction, if it happens to measure a complementary property of an entangled property, decoheres the entanglement.

Now everybody's confused.

Disentangled tri-angles...
(Sung to the tune of "Decomposing Composers" by Ian Durry)
Da Schneib
not rated yet Mar 24, 2016
You can't have decoherence and collapse in the same interpretation. And you're not even supposed to be talking about decoherence if you're a strict Copenhagen advocate.
Since the element of the measurement-problem that explains why a particular outcome is realized as opposed to some other possibility, is NOT resolved by decoherence, it is still necessary to postulate an additional layer of interpretation,…
Or you can use an interpretation like CH or Bohm or TI that doesn't require collapse of the wavefunction to account for measurements. (Note, Wikipedia thinks TI needs wavefunction collapse, I disagree, we can talk about that later if you like; it's basically for the same reason Bohm doesn't, based in TI's case on Wheeler-Feynman absorber theory.)

in the case of C.I. this would be a non-objective collapse or state reduction. It is compatible retrospectively.
So you favor non-traditional CI without objective wavefunction collapse?

[contd]
Da Schneib
not rated yet Mar 24, 2016
[contd]
This is why Everett postulated many-worlds, which developed into consistent histories.
Hmmm, actually I would argue that Everett's reasoning for developing the MWI was
acceptance of the objective truth of the universal wavefunction. Decoherence was an *outcome* of MWI, not its founding principle. I would also argue that the reasoning behind CH was the acceptance of the truth of both the universal wavefunction and Born's Rule, and that decoherence merely provided a path that allowed this synthesis. I would also argue that Bohm's interpretation gave as much impetus to decoherence as the MWI. As for TI, I would say that it's an attempt to move Bohm's interpretation into the mainstream, driven by Wheeler-Feynman absorber theory.

[contd]
Da Schneib
not rated yet Mar 24, 2016
[contd]
The approach of traditional CI to wavefunction collapse is to treat it as a real phenomenon; this has some pretty serious problems in terms of how we see real systems behave. In particular, the ability to create what appear to be objective superposed states in quantum optics experiments seems to indicate that interpretations that contain objective wavefunction collapse are at odds with experimental evidence. I will refer you to the DCQE and its derivatives, and I will also point out that Wikipedia statements in the article on decoherence mention ions in superposed states.

Non-traditional CI that treats wavefunction collapse as a mathematical procedure rather than an objective reality avoids this problem, but introduces the EPR-based contention that it must necessarily be incomplete; that is, it cannot model a free particle in open space, specifically the radioactive decay of an unstable nucleus unaffected by its surroundings.

[contd]
Da Schneib
not rated yet Mar 24, 2016
[contd]
Here is the rest of the quote you posted….

"However, decoherence by itself may not give a complete solution of the measurement problem, […] To present a solution to the measurement problem in most interpretations of quantum mechanics, decoherence must be supplied with some nontrivial interpretational considerations."
I would argue against "most." In fact, most interpretations yield decoherence rather than accepting it as a postulate. Most modern interpretations avoid the measurement problem, which is the result of wavefunction collapse, by substituting non-classical logic; it is the attempt to deal with QM using classical logic that gives rise to the need for wavefunction collapse in the first place, and most modern interpretations use quantum logic rather than classical logic.

[contd]
Da Schneib
not rated yet Mar 24, 2016
And wavefunction collapse and decoherence don't occur in the same interpretations. You don't need one if you have the other. See how that works?
You need to make an important distinction here. Are you speaking of an objective-collapse, as if the wavefunction is a physical thing?
Don't attribute this to me; I am not an advocate of objective wavefunction collapse. I am saying that traditional CI has objective wavefunction collapse, and non-traditional CI has it as a mathematical procedure that the interpretation claims is impenetrable. My position is that it is not impenetrable, that decoherence is the objective reality, and that wavefunction collapse is a mathematical procedure that happens to work but that is not the final word, merely a procedure.

[contd]
Da Schneib
not rated yet Mar 24, 2016
[contd]
There ARE alternative objective-collapse theories.
Traditional CI is not an "alternative" interpretation; objective wavefunction collapse is the original interpretation.

As pointed out above they require non-standard modification to the Schrodinger equation.
Actually, von Neumann was the one who added objective wavefunction collapse to CI; Bohr's approach was the same as Feynman's: shut up and calculate. We don't know what happens there.

In this case, decoherence (just a consequence of the standard Schrodinger equation), would be made compatible with wavefunction collapse (not a consequence of the standard Schrodinger equation).
Errr, no. Decoherence isn't a consequence of the Shroedinger equation. It's a consequence of the development of the Everett and Bohm interpretations. And it's not a feature of CI; it's an add-on, bolted on in order to "look under the covers" of wavefunction collapse.

[contd]
Da Schneib
not rated yet Mar 24, 2016
[contd]
If you are referring to standard QM theory instead,
No. We're not talking about QM here, we're talking about interpretations of QM. The map is not the territory. We're also talking about how we, our classical selves, can come to an understanding of the meaning and implications of QM. But we are not talking directly about QM; that's why I chided you for bringing Hilbert spaces and bra-ket math into the conversation.

then they are not incompatible either due to the fact that decoherence leaves the important element of the measurement problem unresolved, thus still requiring the extra postulate of collapse (in the C.I. and extensions).
But all the interpretations that include decoherence don't have the measurement problem, and don't include objective collapse. Wavefunction collapse is just a mathematical procedure, just like the use of Hilbert spaces or matrices. It works; that doesn't mean it actually happens.

[contd]
Da Schneib
not rated yet Mar 24, 2016
[contd]
In C.I. [objective wavefunction collapse] is an extra postulate just as the Born rule is.
I disagree, on two grounds:
1. Objective wavefunction collapse is silly; it leads to all sorts of paradoxes, that are avoided if it's viewed simply as a mathematical procedure that happens to work out correctly rather than an objectively verifiable fact.
2. The Born Rule is an essential fact of quantum interactions, and accepting it as reality requires nothing more than accepting that superposed states are real, but unmeasurable. Accepting the Born Rule as objective fact doesn't lead to paradoxes, as objective wavefunction collapse does, and furthermore we are actually able to generate these superposed states in real quantum optics experiments. That we cannot measure them directly is immaterial.

[contd]
Da Schneib
not rated yet Mar 24, 2016
The C.I. emphasizes the unavoidability of classical-concepts in measurement,.....
But measuring equipment is unavoidably constructed from quanta. And I disagree that measurement requires classical physics; look at the traditional three-polarizers experiment for an example. These results cannot be explained by classical physics; the fan has slipped and the green whiskers are evident.

while decoherence emphasizes the notion of classicality as dynamically emergent phenomena from an otherwise underlying quantum reality.
Now that I agree with, but that is not all it does.

Also, decoherence just plain makes sense. It says that quantum states, i.e. entanglement, an objectively non-classical state, i.e. superposition, another objectively non-classical state, are *real*, but cannot be described or treated using classical mechanics. This is an obvious fact if you consider real quantum optics experiments.
Da Schneib
not rated yet Mar 24, 2016
Now, see there? I didn't give you 1s for those. I gave you 3s; you're wrong, but at least you're not playing logic chopping games any more.
Da Schneib
5 / 5 (1) Mar 24, 2016
Disentangled tri-angles...
(Sung to the tune of "Decomposing Composers" by Ian Durry)
Actually most things seem to be about dualities.
Noumenon
3.7 / 5 (3) Mar 24, 2016
That is not so. The traditional C.I. did NOT regard the wavefunction as representing a physical wave [nor do I, as I thought I had made clear]. Heisenberg in particular regarded the WP as representing information of the system.

John Von Neumann, in formulating a consistent mathematical foundation for QM, followed Heisenberg in introducing the discontinuous change in the deterministic Schrodinger evolution of the WP upon a measurement,.... without attempting to explain state-reduction dynamically.

In fact, you may recall that Schrodinger was quit disappointed that the Copenhagen guys, with the Max Born rule [Btw, Grandfather of Olivia Newton John], ….rejected the notion that the wavefunction represented a physical wave. Schrodinger desired it to be Real as he wished to maintain a classical description.

The Born rule is an extra step of interpretation, just as state-reduction is,... precisely because neither is mathematically derivable via Schrodinger/Matrix Equations.
Noumenon
3.7 / 5 (3) Mar 25, 2016
My above post was meant to respond to the following post...

The approach of traditional CI to wavefunction collapse is to treat it as a real phenomenon


---------------------------------------------------------
Objective collapse (dynamical collapse) theories require a modification to the Schrodinger equation. That is non-standard and was definitely not part of the traditional C.I. as C.I. did not treat the WP as a physical wave.

"A dynamical collapse of the wave function would require nonlinear and non-unitary terms in the Schrödinger equation [...] lead to observable deviations from conventional quantum theory." - Heinz-Dieter Zeh discoverer of decoherence.

From this it must mean that decoherence does not result in collapse of the wavefunction, but must result from the evolution of the Schrodinger equation itself.

Noumenon
4 / 5 (4) Mar 25, 2016
but at least you're not playing logic chopping games any more.

These baseless accusations are unnecessary.
Da Schneib
1 / 5 (1) Mar 25, 2016
That is not so. The traditional C.I. did NOT regard the wavefunction as representing a physical wave
Now you're logic chopping again. More smoke.

I never claimed that CI regarded the wavefunction as physical. I claimed it regarded the collapse of the wavefunction as physical.

Why do you keep trying to get away with this crap? Do you still have the illusion that I won't notice because you're a genius and I'm some sort of lower life form?

When will you stop insulting me subtly and pretending you didn't?

but at least you're not playing logic chopping games any more.

These baseless accusations are unnecessary.
Errr, you just did it again.

Just stop. You're not going to "win." You're just making yourself look stupid.
Da Schneib
1 / 5 (1) Mar 25, 2016
The Born rule is an extra step of interpretation, just as state-reduction is,... precisely because neither is mathematically derivable via Schrodinger/Matrix Equations.
Oh, BS.

The Born Rule is, quite simply and basically, the statement that quantum phenomena are *intrinsically different* from classical phenomena in that they are not necessarily of definite value, but may be represented by probabilities that they may, when they interact, turn out to be one value or another. The motivation is Heisenberg uncertainty and superposition.

To put it another way, accepting the Born Rule as real means you accept that quanta can exist in real experiments in superposed states, and the experiments that prove it have been done. See Bell experiments, see Aspect, see DCQE, see superposed ion experiments.
Mimath224
5 / 5 (3) Mar 25, 2016
@Da Schneib having done some searching it seems to me that Noumenon is arguing from a Gleason's Theorem standpoint in as much that say, QM probability or measurements, is the result of a set of logical steps that can be set out as propositions. Obviously more to it than this but to be honest, I don't completely understand its range or why Hilbert Space is the domain. Perhaps you or Noumenon do. Since this theorem can be a path to quantum logic I suspect it might rapidly turn philosophical and not a place I wish to go. Having said that, the failure of the p & (q or r) of propositional calculus in QL does interest me. Welcome your response.
Noumenon
4 / 5 (3) Mar 25, 2016
The approach of traditional CI to wavefunction collapse is to treat it as a real phenomenon
That is not so. The traditional C.I. did NOT regard the wavefunction as representing a physical wave [nor do I, as I thought I had made clear]. Heisenberg in particular regarded the WP as representing information of the system.


Now you're logic chopping again. More smoke.

I never claimed that CI regarded the wavefunction as physical. I claimed it regarded the collapse of the wavefunction as physical.


There is no physical wavefunction collapse if the wavefunction if not considered a physical wave. Who's logic chopping here?

I went out of my way to explain the difference between objective dynamical [i.e. physical] wavefunction collapse and the discontinuous state-reduction [mathematical, informational] implicit in the C.I. above because you had asked this….

So you favor non-traditional CI without objective wavefunction collapse?

Noumenon
4.3 / 5 (3) Mar 25, 2016
The traditional C.I. does not attempt to explain the state-reduction at all dynamically/physically/objectively. The C.I. is represented in von Nuemann's 'mathematical foundations of quantum mechanics' text, where "state-reduction" [The term "collapse" came much later not by C.I.] is a mathematical component with no dynamical/physical/objective attempt at explanation. In fact it is a postulate,.... the "projection postulate".

A "collapse of the wavefunction as physical" would require a dynamical explanation somehow. Many Q.M.I. avoid this altogether by avoiding the collapse. Decoherence does not explain why one state-reduction occurs as opposed to another of the many possibilities,... so it fails as a dynamical explanation linking the mathematical foundation [Schrodinger / Matrix Mechanics] to the experimental conceptual valued results.

Noumenon
4 / 5 (3) Mar 25, 2016
The Born rule is an extra step of interpretation, just as state-reduction is,... precisely because neither is mathematically derivable via Schrodinger/Matrix Equations.
Oh, BS.

[...] To put it another way, accepting the Born Rule as real means you accept that quanta can exist in real experiments in superposed states, and the experiments that prove it have been done. See Bell experiments, see Aspect, see DCQE, see superposed ion experiments.


The Schrodinger Equation was formulated as an eigenvalue problem, and so already implied quanta,... and Schrodinger did not even like the Born Rule.

More to the point, the Born Rule derives probabilities via the square of the [normalized] wavefunction (i.e. for the eigenvalue solutions). This is not easy to derive from the core mathematics of any Q.M.I.

As presented by C.I. (Max Born) it is an extra layer of interpretation. Another postulate.

Noumenon
4.8 / 5 (4) Mar 25, 2016
But by "experiencing things", you are adding conditions for doing so...
Easy enough to evaluate the conditions, whatever they may be, and include them in the calculations.


Not so much, as there always has to be an observer/apparatus who is outside the quantum system.

This was pointed out in the Schrodinger Cat paradox. See also Heisenberg / von Neumann's cut (Schnitt).

If you are interested I outline this argument in This Thread, .... search for the text 'Mathematical Foundations of Quantum Mechanics'.

Hyperfuzzy
not rated yet Mar 25, 2016
Gosh guys slow down, when we have multiple solutions we choose one, i.e. the observed, collapse
Da Schneib
5 / 5 (2) Mar 25, 2016
@Da Schneib having done some searching it seems to me that Noumenon is arguing from a Gleason's Theorem standpoint in as much that say, QM probability or measurements, is the result of a set of logical steps that can be set out as propositions. Obviously more to it than this but to be honest, I don't completely understand its range or why Hilbert Space is the domain. Perhaps you or Noumenon do.
Pretty sure Noum does, but here goes: Hilbert space is used because it's a way of representing parameters that have one through infinity degrees of freedom. For example, the spin of a particle on a particular axis has only one degree of freedom; it can only be positive or negative, since for any given particle, the spin is a fixed absolute value, like 1/2, 1, 3/2, 2, etc. And for example, the phase of a particle has infinitely many degrees of freedom; it can be anything from zero to plus the amplitude to minus the amplitude, and phase is not quantized.

[contd]
Da Schneib
5 / 5 (2) Mar 25, 2016
[contd]
Both these types of parameters can be represented in a single vector in Hilbert space, which makes Hilbert space a very convenient way to organize all the parameters (in practice, all of the parameters of interest) in the same mathematical entity. Changes can then be applied that modify the vector, or modify the particular chosen basis of the Hilbert space, and the new vector worked out, yielding the parameters that the particle will have after the change. If the vector is changed, then the particle has undergone an interaction; if the basis of the Hilbert space is changed, then the particle is being observed differently. This turns out to work very well with the ways that physicists generally deal with interactions, and changes to the means, position, etc. of observation.

[contd]
Da Schneib
5 / 5 (2) Mar 25, 2016
[contd]
Since this theorem can be a path to quantum logic I suspect it might rapidly turn philosophical and not a place I wish to go. Having said that, the failure of the p & (q or r) of propositional calculus in QL does interest me. Welcome your response.
I'm not familiar with the theorem, so I'll have to look it up and figure it out. Gimme a while.
Da Schneib
3 / 5 (2) Mar 25, 2016
The approach of traditional CI to wavefunction collapse is to treat it as a real phenomenon
That is not so. The traditional C.I. did NOT regard the wavefunction as representing a physical wave [nor do I, as I thought I had made clear]. Heisenberg in particular regarded the WP as representing information of the system.
Now you're logic chopping again. More smoke.

I never claimed that CI regarded the wavefunction as physical. I claimed it regarded the collapse of the wavefunction as physical.
There is no physical wavefunction collapse if the wavefunction if not considered a physical wave. Who's logic chopping here?
Ahhh, but you see, there *is* a wavefunction collapse: the measurement!

As for who's logic chopping, it's Bohr. The CI basically says, wavefunctions are a mathematical tool, we don't really know what's going on there, but wavefunction collapse is real, it's the measurement.

Yes, really. Read the Wikipedia article.

[contd]
Da Schneib
3.5 / 5 (2) Mar 25, 2016
[contd]
I went out of my way to explain the difference between objective dynamical [i.e. physical] wavefunction collapse and the discontinuous state-reduction [mathematical, informational] implicit in the C.I. above because you had asked this….

So you favor non-traditional CI without objective wavefunction collapse?
Yes, but it appears you have a pretty odd conception of the CI; you appear to think that the wavefunction collapse is not real. There's nothing wrong with that, but it's not traditional CI and you're claiming it is. *That* is the problem.
Da Schneib
3.5 / 5 (2) Mar 25, 2016
The traditional C.I. does not attempt to explain the state-reduction at all dynamically/physically/objectively. The C.I. is represented in von Nuemann's 'mathematical foundations of quantum mechanics' text, where "state-reduction" [The term "collapse" came much later not by C.I.] is a mathematical component with no dynamical/physical/objective attempt at explanation. In fact it is a postulate,.... the "projection postulate".
But it *is* objective: it's the measurement!

Now do you see what the "measurement problem" is?

[contd]
Da Schneib
3.5 / 5 (2) Mar 25, 2016
[contd]
A "collapse of the wavefunction as physical" would require a dynamical explanation somehow.
It has one: the measurement. Or, in interpretations that avoid dealing with measurement as a separate thing, the interaction. That's where decoherence comes in.

Many Q.M.I. avoid this altogether by avoiding the collapse.
The ones that do use decoherence- that is, removal of entanglement and resolution of the superposed state to a single state.

Decoherence does not explain why one state-reduction occurs as opposed to another of the many possibilities,...
Not for a single interaction, but it does for an ensemble of interactions; the probability predicted by the Schroedinger equation is always observed in the ensemble. And according to our best understanding of state reduction from a superposed state to a single state all we can know before the reduction is the probability.

[contd]
Hyperfuzzy
not rated yet Mar 25, 2016
[contd]
I went out of my way to explain the difference between objective dynamical [i.e. physical] wavefunction collapse and the discontinuous state-reduction [mathematical, informational] implicit in the C.I. above because you had asked this….

So you favor non-traditional CI without objective wavefunction collapse?
Yes, but it appears you have a pretty odd conception of the CI; you appear to think that the wavefunction collapse is not real. There's nothing wrong with that, but it's not traditional CI and you're claiming it is. *That* is the problem.

You guys realize that the observer of the space you define is outside the space, thus ad nauseum, an infinite Hilbert Space. I know, you don't get it! Let it collapse, or rethink it!
Da Schneib
3.5 / 5 (2) Mar 25, 2016
[contd]
so it fails as a dynamical explanation linking the mathematical foundation [Schrodinger / Matrix Mechanics] to the experimental conceptual valued results.
I disagree. Measurement yields only one value, and an ensemble of measurements yields the probabilities of the wavefunction. Both are dynamical. The fact that we cannot predict, for an individual interaction, which state will result from the resolution of the superposed states is an inherent feature of QM. All we can know is the probability of the various possible outcomes.

This is why I say that the Born Rule is an essential feature of QM, and this is why QM is essentially different from classical mechanics.
Da Schneib
4 / 5 (2) Mar 25, 2016
The Born rule is an extra step of interpretation, just as state-reduction is,... precisely because neither is mathematically derivable via Schrodinger/Matrix Equations.
Oh, BS.

[...]To put it another way, accepting the Born Rule as real means you accept that quanta can exist in real experiments in superposed states, and the experiments that prove it have been done. See Bell experiments, see Aspect, see DCQE, see superposed ion experiments.
The Schrodinger Equation was formulated as an eigenvalue problem, and so already implied quanta,... and Schrodinger did not even like the Born Rule.
How the wavefunction is represented is immaterial, and that's not the only thing Schroedinger turned out to be wrong about.

Just sayin'.

[contd]
Da Schneib
4.5 / 5 (2) Mar 25, 2016
[contd]
More to the point, the Born Rule derives probabilities via the square of the [normalized] wavefunction (i.e. for the eigenvalue solutions). This is not easy to derive from the core mathematics of any Q.M.I. Err, interpretations don't have core mathematics. QM does. The interpretation relates the core mathematics of QM to the observed results in experiments.

As presented by C.I. (Max Born) it is an extra layer of interpretation. Another postulate.
Sure, but this postulate appears to be required. It's a feature of all interpretations; in some (CI) it's a bolt-on, in others (CH) it's a postulate of the interpretation itself.

My argument is, since it's an intrinsic feature of QM, any interpretation that bolts it on after the fact is inherently incomplete. That's why they say CH is "CI done right."
Da Schneib
5 / 5 (1) Mar 25, 2016
But by "experiencing things", you are adding conditions for doing so...
Easy enough to evaluate the conditions, whatever they may be, and include them in the calculations.
Not so much, as there always has to be an observer/apparatus who is outside the quantum system.
But the observer/apparatus *is* a quantum system! And any interpretation (see CI) that claims the observer/apparatus is not a quantum system is therefore suspect/incorrect/incomplete. That's why I don't like the CI.

This was pointed out in the Schrodinger Cat paradox. See also Heisenberg / von Neumann's cut (Schnitt).
Schroedinger's Cat relies upon confusion of two incompatible views: the Born Rule is applied to a classical system. There is no confusion if one merely states that one can either look at the trajectory of the cat/box system in time, or look at the probabilities of the outcome without considering the state of the cat while its state is in superposition.

[contd]
Da Schneib
5 / 5 (1) Mar 25, 2016
[contd]
This is logically consistent; if at any time you open the box, the cat is either dead or alive. It is only in superposition while you cannot look at it, and this is consistent with the states of photons in superposition in a DCQE experiment. You can't measure a superposition directly; you can only measure an ensemble of superpositions to determine the probabilities of various outcomes. The probability of the cat being dead or alive is equal to the probability that the radioactive isotope has decayed or not decayed.

I gave you a 4 for that one because you asked a good question.
Mimath224
5 / 5 (2) Mar 25, 2016
@Da Schneib, many thanks for your time and help. Am beginning to get the gist but that gives me another problem and to my mind, much more fundamental. One way out, for me that is, would be to deny any form of 'QM realism' but then I'd be doing it for personal reasons and that wouldn't be scientific. Can't post it here because it would be off topic...I think...and in any case I'll need time to put it into words. So once I've researched the net a little more and can ask a question that's not ambiguous, maybe I'll write you a private note, if that's ok with you.
Da Schneib
5 / 5 (1) Mar 25, 2016
Oh, I think it's totally relevant. You're doing fine. Other people will learn from watching you ask good questions, @Mimath. Quite frankly you ask some of the best questions on this forum.

If we're going to talk about what reality is and does, we have to acknowledge that what we perceive as the most salient features of, call it "reality," don't appear to hold below a certain size, somewhere around the size of an atom. The salient features of things smaller than atoms appear to behave fundamentally differently from the everyday things we encounter. For example, Born's Rule; for example, Heisenberg uncertainty; for example, entanglement. These things simply are not so when we use classical mechanics, and simply are so when we use quantum mechanics.

The really mysterious part is how the classical behaviors arise from the quantum behaviors, and not only arise from them but do so inevitably, and by mathematical theorems. The epitome of this is the Fluctuation Theorem which I
TehDog
5 / 5 (2) Mar 25, 2016
"The epitome of this is the Fluctuation Theorem which I ...."

I have a vision of black helicopters descending...
:)
TehDog
5 / 5 (2) Mar 25, 2016
Crap, now they know I'm reading this thread...
(Ducks and runs, grabbing a roll of tin foil...)
Da Schneib
3.7 / 5 (3) Mar 25, 2016
Bah, which I recommend to your attention. Stephen Curry made a great shot and my wife yelled and I hit post and went and rewound it and watched it. DVRs rule.
Da Schneib
1 / 5 (1) Mar 25, 2016
"The epitome of this is the Fluctuation Theorem which I ...."

I have a vision of black helicopters descending...
:)
But wait I don't even
Mimath224
3 / 5 (3) Mar 26, 2016
@Da Schneib, thanks once again for FT and your kind words, which I don't deserve. My problem goes more micro rather than QM to macro. To be honest, I don't possess the necessary expertise to really confront the problem or even formulate the appropriate question so I ask that you grit your teeth in trying to educate my thick skull.
Without going into one camp or another (opinions etc), let's say we have an unobserved state which might have some mathematical representation (say a wave function ) if we could know without disturbing (observing) the system. Now I know this is more De Broglie than QM but I'm trying an analogy here, but suppose the QM entity in question is an electron. We can theorize what a typical e- wave function might be because basically all e- in some given state appear identical. Then the environment reacts with this state, say through the moving of needle on a volt meter. What is the difference between the wf now 'collapsed' state and its continuing state (cont)
Mimath224
4 / 5 (2) Mar 26, 2016
(cont) That is to ask will there be a similar wf as the e- moves on, or does the e- become something else because of being observed? That's my first part. (I am aware of Pauli excl. etc)
Going down further into the QM world; during experiments gamma photons are produce which then disappear in pair production of say, e- and e+. If more than one pair production occurs then I assume that the e-'s are identical and e+'s are too. If my assumption is close to being correct then it seems to me that the gamma photon either combined with some other, perhaps unknown, entity to produce e- or that the photon was governed by some fundamental law to combine in a certain with itself to produce e-. Conclusion, for me such a law might fill my 'QM realism' gap. I do appreciate this then begs other questions like why isn't space filled with e- instead of gamma rays? does this law operate on other particles as well? Over simplistic but I hope not too nonsensical and maybe not a good question this time.
Captain Stumpy
3.7 / 5 (3) Mar 26, 2016
To be honest, I don't possess the necessary expertise to really confront the problem or even formulate the appropriate question so I ask that you grit your teeth in trying to educate my thick skull
@Mimath224

may i also suggest some links?
this one is free - http://ocw.mit.ed...echanics

this one will cost you, but it is a good book considering it may well have the information you seek, or at least will point you in the right direction
http://www.barnes...71455466

Not saying reading about QM above won't help, mind you
just saying that you may find those links helpful
Mimath224
5 / 5 (3) Mar 26, 2016
@Captain Stumpy thank you I have several books on QM including, similar to the one you cite, from beginners to QT, Path integrals etc. The link you provide certainly has a wide range but as an OAP I'm not if I have enough time available, Ha! However, I will view the various courses & content to see if they anything different to what I already have (my PC desktop included).
Mimath224
5 / 5 (3) Mar 26, 2016
@Captain Stumpy forgot to mention that the first lecture begins almost confirming what I mention about electrons in my previous post. Ha, I suppose that shows I have learned something in the past...not a lot...but something.
Noumenon
4 / 5 (4) Mar 26, 2016
The approach of traditional CI to wavefunction collapse is to treat it as a real phenomenon
That is not so. The traditional C.I. did NOT regard the wavefunction as representing a physical wave. Heisenberg in particular regarded the WP as representing information of the system.

I never claimed that CI regarded the wavefunction as physical. I claimed it regarded the collapse of the wavefunction as physical.
There is no physical wavefunction collapse if the wavefunction if not considered a physical wave.
Ahhh, but you see, there *is* a wavefunction collapse: the measurement!

You don't appear to understand the difference between a Postulate and a Dynamical explanation, the latter of which is not included in the basic equations of QM thus the measurement problem, and the discontinuity between unitary evolution of WP and discontinuous measurement result, …. the former of which does not attempt to explain the discontinuity.
Noumenon
4 / 5 (4) Mar 26, 2016
As for who's logic chopping, it's Bohr. The CI basically says, wavefunctions are a mathematical tool, we don't really know what's going on there, but wavefunction collapse is real, it's the measurement.


By "real phenomenon" is properly meant Objective with a Dynamical explanation possible, as expressed in equations, in detail.
Yes, "collapse" occurs 'during the experiment', but it is not understood how. C.I. does not offer a dynamical explanation,… they put forth a POSTULATE instead.
Yes, but it appears you have a pretty odd conception of the CI; you appear to think that the wavefunction collapse is not real. There's nothing wrong with that, but it's not traditional CI and you're claiming it is.


Well it is, because C.I. does not even consider the WP as a Real entity! The problem is you don't seem to understand what a postulate is.
Da Schneib
2.3 / 5 (3) Mar 26, 2016
@Da Schneib, thanks once again for FT and your kind words
Your humility does you credit.

let's say we have an unobserved state which might have some mathematical representation (say a wave function ) if we could know without disturbing (observing) the system.
Can't do it. There's only one way to observe a quantum system: bounce other quanta off it. And once you do that you've disturbed it. This is a basic property of quantum systems.

Now, that said, there have been some recent advances in detecting the states of entangled particles without disturbing those states, but those are kind of outside the scope of this discussion, so far anyway.

suppose the QM entity in question is an electron. We can theorize what a typical e- wave function might be because basically all e- in some given state appear identical.
Well, that depends on whether you include position in the state; if so, then they are not actually in the same state.

[contd]
Da Schneib
2.3 / 5 (3) Mar 26, 2016
[contd]
Then the environment reacts with this state, say through the moving of needle on a volt meter. What is the difference between the wf now 'collapsed' state and its continuing state
It's a new wavefunction, related to the old one by the interaction that caused the measurement/collapse. In other words, the wavefunction has been modified by the interaction. At least in interpretations that use decoherence.

That is to ask will there be a similar wf as the e- moves on, or does the e- become something else because of being observed?
It depends on the interaction, but mostly an e- remains an e- unless it interacts with a W or Z. And there aren't many measuring devices that use Ws or Zs.

[contd]
Noumenon
4 / 5 (4) Mar 26, 2016
[Decoherence] fails as a dynamical explanation linking the mathematical foundation [Schrodinger / Matrix] to the experimental conceptual valued results.

I disagree. Measurement yields only one value, and an ensemble of measurements yields the probabilities of the wavefunction. Both are dynamical. The fact that we cannot predict, for an individual interaction, which state will result from the resolution of the superposed states is an inherent feature of QM.


You are mixing QM probability and classical probability there, and indeed in CH there is that dualistic issue.

That's why they say CH is "CI done right."


I'm ok with CH as well, however it inherits a similar issue as decoherence combined with collapse, as pointed out in the above quote,… about which set of consistent histories will actually occur.
Noumenon
4 / 5 (4) Mar 26, 2016
But by "experiencing things", you are adding conditions for doing so...
Easy enough to evaluate the conditions, whatever they may be, and include them in the calculations.
Not so much, as there always has to be an observer/apparatus who is outside the quantum system.
But the observer/apparatus *is* a quantum system! And any interpretation (see CI) that claims the observer/apparatus is not a quantum system is therefore suspect/incorrect/incomplete. That's why I don't like the CI.


You didn't understand my response. I am not claiming there that the observer is not a quantum system. I'm simply pointing out the obvious fact [see Schrodinger's cat] that the experimenter does not have such a QM mathematical description for himself and the apparatus,.. hence "outside the system".
Da Schneib
2.3 / 5 (3) Mar 26, 2016
[contd]
Going down further into the QM world; during experiments gamma photons are produce which then disappear in pair production of say, e- and e+. If more than one pair production occurs then I assume that the e-'s are identical and e+'s are too.
You need to be very careful using the word "identical" here. Position is part of the state of a particle, and no two fermions of the same kind can have the same position, due to Pauli exclusion (as you previously noted).

[contd]
Noumenon
4.2 / 5 (5) Mar 26, 2016
suppose the QM entity in question is an electron. We can theorize what a typical e- wave function might be because basically all e- in some given state appear identical. Then the environment reacts with this state, say through the moving of needle on a volt meter. What is the difference between the wf now 'collapsed' state and its continuing state


The WF starts to evolve again from the collapsed state,…. say position [represented as a Dirac delta function]. If a subsequent experiment is conducted fast enough, the same collapsed state would be observed. The rest of the wavefunction superposition is simply gone in C.I. or exists in "many-worlds". CH does not propose that "histories" correspond to alternative realities.
Da Schneib
2.3 / 5 (3) Mar 26, 2016
[contd]
If my assumption is close to being correct then it seems to me that the gamma photon either combined with some other, perhaps unknown, entity to produce e- or that the photon was governed by some fundamental law to combine in a certain with itself to produce e-.
No, this is incorrect. A gamma photon of the correct energy simply decays into a particle pair (or quad, there is a process for a gamma of the correct energy to make three electrons and a neutrino, with appropriate charge conjugations). That *is* the interaction; it's a self-interaction, like the decay of a radionuclide. No other particle is involved. This existence of self-interactions is part of why physicists have sought better interpretations than CI, which doesn't handle these types of self-interactions reasonably.

[contd]
Noumenon
4.2 / 5 (5) Mar 26, 2016
during experiments gamma photons are produce which then disappear in pair production of say, e- and e+.

A gamma photon of the correct energy simply decays

What experiment are you referring to? I didn't know photon's disappear into production of electrons. If you're referring to QED diagrams, that refers to 'virtual photons'. Photon's have no mass so do not decay.

If my assumption is close to being correct then it seems to me that the gamma photon either combined with some other, perhaps unknown, entity to produce e- or that the photon was governed by some fundamental law to combine in a certain with itself to produce e-.


QED does not attempt to explain the dynamics here. For example, in an accelerator, it is enough that E = mc² to justify saying that the colliding particles no longer exist at all, and that the resulting energy of 1) the colliding particles and 2) the relativistic kinetic energy, produced the subsequent shower of particles.
Da Schneib
2.3 / 5 (3) Mar 26, 2016
[contd]
Conclusion, for me such a law might fill my 'QM realism' gap.
I don't quite understand, but I just realized on re-reading that my explanation fits your statement after the "or," that is, "...the photon was governed by some fundamental law to combine in a certain with itself to produce e-..." That should be "e- and e+," but other than that, yes, that is what happens.

I do appreciate this then begs other questions like why isn't space filled with e- instead of gamma rays?
Because the probability that the gamma ray will decay like this is low, and only a gamma ray of exactly the right energy will do so. The answer then is, "because space is not filled with gamma rays of the exact right energy."

[contd]
Da Schneib
3 / 5 (4) Mar 26, 2016
[contd]
does this law operate on other particles as well?
Yes. It operates on any particle that has the right energy and other parameters to decay into another particle or particles; there is always a possibility that this decay will occur for such a particle. Physicists refer to these possibilities as "decay channels."

Over simplistic but I hope not too nonsensical and maybe not a good question this time.
No, you're still asking good questions. Keep 'em coming!
Da Schneib
1 / 5 (2) Mar 26, 2016
You don't appear to understand the difference between a Postulate and a Dynamical explanation
I'm sorry but your explanation of why I don't appears defective, and I'm getting bored with your logic chopping and constant zero-sum gaming.

Start over. Try to do better before I get bored enough to put you on ignore.
Da Schneib
1 / 5 (2) Mar 26, 2016
Measurement yields only one value, and an ensemble of measurements yields the probabilities of the wavefunction. Both are dynamical. The fact that we cannot predict, for an individual interaction, which state will result from the resolution of the superposed states is an inherent feature of QM.


You are mixing QM probability and classical probability there,
Yes, precisely: the wavefunction yields the ratios of outcomes in an ensemble. That's what it's for: relating QM results to actual measured results over an ensemble.

You're still doing boring philosophy and ignoring interesting quantum physics.

Sorry, gotta go, got pizza which is much more interesting than you are.
Noumenon
4 / 5 (4) Mar 26, 2016
Measurement yields only one value, and an ensemble of measurements yields the probabilities of the wavefunction. Both are dynamical. The fact that we cannot predict, for an individual interaction, which state will result from the resolution of the superposed states is an inherent feature of QM.


You are mixing QM probability and classical probability there, and indeed in CH there is that dualistic issue.


Yes, precisely: the wavefunction yields the ratios of outcomes in an ensemble. That's what it's for: relating QM results to actual measured results over an ensemble.


I meant you're conflating QM probability and classical probability in your quote.

Since CH is an interpretation of the experimental results and does not make differing predictions from that of standard QM,... it is "Philosophy of Physics".

Da Schneib
3.7 / 5 (3) Mar 26, 2016
A gamma photon of the correct energy simply decays
What experiment are you referring to?
How about the γ->tt- channel at the LHC? That do ya?

I'm amazed I'm talking to someone who claims to know physics who doesn't know that gamma ray photons of the appropriate energy have a decay channel to quark/antiquark and lepton/antilepton pairs, not to mention W/W+ and Z/Z pairs. This is kind of like talking to someone who claims to know relativity who's never heard of gravity.

I meant you're conflating QM probability and classical probability in your quote.
No, I'm pointing out that quantum probability in the wavefunction must yield classical probability in the ensemble to be a correct description of QM.

Are you seriously arguing that QM doesn't yield classical physics?

Get a grip, dude.
Da Schneib
1 / 5 (3) Mar 26, 2016
You're no physicist. You don't even know about the decay channels of gamma rays. Stop posing, Noum.
Da Schneib
1 / 5 (3) Mar 26, 2016
Here's the Wikipedia article on this. It's usually called "pair production" but detector physicists call it "gamma decay," and a "channel" is a decay to a particular pair of particles, for example an electron and a positron, or a muon and an antimuon, as noted in the article.

https://en.wikipe...oduction

Note that "gamma decay" is the more general term, since as I said above these decay channels can produce more than pairs. Quartets can also be formed, for example two electrons and two positrons.
Noumenon
4.3 / 5 (6) Mar 26, 2016
You're no physicist. You don't even know about the decay channels of gamma rays. Stop posing, Noum.


You're an idiot and not even an honest one. You stated this .....

A gamma photon of the correct energy simply decays - DaSchieb


....Photon's do NOT decay as they are massless. I lost count of how many times you stated things carelessly and outright wrongly. FineStructureConstant's post was spot on.

Show me a reference where gamma photons decay, that is not virtual photons and that is not the result of interaction with matter.

I already pointed out E=mc² above wrt particle production in colliders,.... in which it is the ENERGY that produces particles. Photon's don't decay,... ever!

Noumenon
4.3 / 5 (6) Mar 26, 2016
Here's the Wikipedia article on this. It's usually called "pair production" but detector physicists call it "gamma decay," and a "channel" is a decay to a particular pair of particles, for example an electron and a positron, or a muon and an antimuon, as noted in the article.

https://en.wikipe...oduction


Where in that link does it say "a gamma photon of the correct energy simply decays". This is patently false. If you think photons decay you are not physicist.

What it DOES say is what I had already confirmed myself wrt E = mc²... "the ENERGY OF a photon can be converted into an electron-positron pair" - your wiki source.

Now is this logic chopping? NO, it is a deeper understanding of physics. Photon's do not decay as they're massless.

My "crime" before your ignorant and rude response was to ask what experiment is being referred to so that I can understand what you meant to say, but instead wrongly stated.

Noumenon
4.2 / 5 (5) Mar 26, 2016
A gamma photon of the correct energy simply decays into a particle pair [...] it's a self-interaction, like the decay of a radionuclide. No other particle is involved.


I feel compelled to correct this patently false comment for MinMath224 (though I suspect he understands this).

Pair-Production is not the same thing as particle-decay. Radio-active atoms ("radionuclide") do spontaneously decay with "no other particle is involved",... but photons will not**,.. they require some interaction event with an atom.

A gamma photon of itself will never produce an electron/positron pair. It requires interaction with matter (an atom), and then it is only the ENERGY of the gamma photon that produces the electron/positron pair. Please see my comment above wrt the energy producing particles in colliders.

A deeper understanding of physics is necessary to avoid getting tripped up by phrases they use.

**I already qualified about virtual particles above.

Noumenon
4.3 / 5 (6) Mar 26, 2016
And btw, it does not have to be a gamma photon of the "correct energy", as the atom can recoil to absorb the excess. More language you use on account of the lack of understanding.

This existence of self-interactions is part of why physicists have sought better interpretations than CI, which doesn't handle these types of self-interactions reasonably.


C.I. is an interpretation of QM,.... quantum field theories such as QED are built upon the framework of QM.

.........................................

I probably won't continue with you because of your blatant rudeness and dishonesty, which is entirely unnecessary. I would not have been rude to you except on account of your behavior.

Da Schneib
1.5 / 5 (4) Mar 26, 2016
....Photon's do NOT decay as they are massless.
Then explain pair production.

On edit:

Never mind, Noum. You just lied one time too many. G'bye.
Noumenon
5 / 5 (5) Mar 26, 2016
....Photon's do NOT decay as they are massless.
Then explain pair production.

On edit:

Never mind, Noum. You just lied one time too many. G'bye.


I just told you, it requires photons to interact with matter.

And you insulted without provocation 30 times too many.
Noumenon
5 / 5 (5) Mar 26, 2016
I do appreciate this then begs other questions like why isn't space filled with e- instead of gamma rays?


Asking such questions leads me to think you understand more than what you lead on.

The answer is because gamma photons don't decay, period. In interaction with matter only, could the photon's energy produce the electron/positron pairs. If the gamma photon has more than enough energy to produce a electron/positron pair then the rest goes into the recoil of the atom it interacted with.

Now, you have a choice between my explanation and DaSchieb's, though you did not ask me.

Mimath224
4 / 5 (4) Mar 26, 2016
@Noumenon, unfortunately some physicists do call pair production, at least the type that I mentioned, as 'decay products' just as they say the same thing when e- & e+ annihilate each other to produce gamma rays. I was once heavily criticized on another forum for using that same terminology so I do try to avoid it but Da Schneib is correct when he uses it here.
You write 'Asking such questions leads me to think you understand more than what you lead on....' is a bit like saying that you suspect me as being, at very least, misleading and at most untruthful. I can assure this is NOT the case. As I have mentioned many a time on other threads, that I DO have a scientific background in my working life but that was always in industry of consumer products and never as a researcher at university installations or other similar types of labs. In fact although I went to tech college I did NOT study formally at a university and do not have BSc, ASc PHD and the like. (cont)
Noumenon
5 / 5 (5) Mar 26, 2016
Asking such questions leads me to think you understand more than what you lead on....' is a bit like saying that you suspect me as being, at very least, misleading and at most untruthful


It was meant as a compliment. You asked a very appropriate question at the right time.

The differences between my answer to your questions above and DaSchnieb's are very different, and is not simply a matter of terminology.

"Decay products" is fine. Even "gamma decay" is fine as long as one is not thus mislead to the description given by DaSchnieb, which is incorrect.

Mimath224
4.3 / 5 (3) Mar 26, 2016
What I did was to study up to a level that was required by my work (mostly in QC) at Day Release and evening classes. But I am fascinated by 'nature' in all its forms from the Cosmos down to QM so what I'm saying is that I am nothing more than a inadequately 'self taught' layman and I envy those who are involved in projects. I am only too glad and very pleased that some posters here are willing to impart their knowledge, and opinions, similarly experts in the various fields for writing and using the net to pass on their knowledge. I do have equivalent qualifications in another discipline but it is not a science subject although it uses a lot of practical classical physics in its application. I hope that satisfies you.
@Da Schneib once again, thank you. Yes, I do understand and appreciate your comments. I suppose what I'm really getting at is that 'particles' might consist of some kind of "photonic' [my own word] structure (cont)
Mimath224
4.3 / 5 (3) Mar 26, 2016
(cont) that obeys some fundamental law of combination. As for example, e- & e+ might say consist of 'photonic energy' [Ha, another made term] where the law dictates that the 'right energy level of this photonic energy must 'move', 'revolve', 'spin' etc to be e- or e+'. This is purely hypothetical and not intended to be an argument to claim that this is so. My next question would then be 'is that particular 'law' within in the 'photonic energy', or does it exist independently. (Oh boy, I do make a rod for my own back)
@Noumenon I appreciate your remarks and as I've said before I am interested in the discussion between you and Da Schneib though I'm not qualified to enter it.
Da Schneib
3 / 5 (4) Mar 26, 2016
Well, @Mimath, you're pretty close to string physics there. String physics says that all elementary particles-- force bosons like photons, leptons like electrons, and the quarks that make up protons and neutrons-- are actually tiny strings vibrating in 10 dimensions, only four of which are directly visible to us. So I think you should explore string physics; if you're not familiar with the literature in this area I can give you some good titles. You will find as much answer as anyone knows to your question there.
PhysicsMatter
not rated yet Mar 27, 2016
Quantum particle is nonlocal because is devoid of identity and is described mathematically as a wave function associated with probability density of certain quantum state. It is by definition nonlocal in "reality" unless we raise unlimited potential barriers to confine it, impossible feat.

What they call contextuality is an assumption, a hypothesis that so-called "undetermined" quantum states exist and hence idea of entanglement is realizable i.e. may exits in reality although will never be experimentally verified. And hence in contextual reality one measurement affects the other through entanglement, so called spooky action on distance [Einstein].

More on that, de-entanglement of QM terms may be found here:

https://questforn...-quanta/

Mimath224
4.3 / 5 (3) Mar 27, 2016
@Da Schneib, yes I thought you might say that,Ha! I do have books on SST, SUSY both popular and math. But no, I am not, as yet, a SST advocate and consider myself more mainstream. I would agree SST is attractive but imho it has some serious issues both theory and research status that it needs to confront. Even so, my question equally applies to Strings too. Obviously I haven't expressed myself as clearly and that shows up my 'layman-ship' here. I hope you'll forgive me but I'll try again still on the particle-wave as previous. I apologise, but I'll have be simplistic again. Let's just suppose that all e- are the same because mass and charge have never been found different (with accepted properties, spin, moments etc.).[ Wasn't it this that led to the J. Wheeler's idea of there being only one electron?]. Anyway, let's say my e- is formed in a pair production in the presence of excess energy what is 'it' that determines an e- will be produced (cont)
Mimath224
4.3 / 5 (3) Mar 27, 2016
(cont) and not some other 'heavier' stable electron (am aware of mesons) or for that matter, a 'lighter' electron? Another way of putting it would be if in the gamma ray pair production (decay) g has E = hf and e- so produced always have mass (m) then what is the law that says hf (-+)E must always produce e-. Is that 'law' intrinsic to the g=hf or does it exist in the environment (vacuum or other). Again, completely hypothetical, suppose the g ray decays into an e- by moving in ever decreasing circles, wraps itself up into a ball of energy that becomes my e-, is there some law that dictates the way the g ray should behave/move so it becomes e-? In the same way we might extend this to other 'stable' particles and particles that have a very brief life exist that way because this 'law' must not be violated for stable particles. Might we call such a governing law as 'QM realism'?
I won't bother you with this again, as sure you'll get bored and put me on 'ignore'. Thanks anyway.
Da Schneib
2.6 / 5 (5) Mar 27, 2016
yes I thought you might say that,Ha!
Well, I was gonna mention rishons too, but forgot. ;)

Let's just suppose that all e- are the same because mass and charge have never been found different (with accepted properties, spin, moments etc.). Anyway, let's say my e- is formed in a pair production in the presence of excess energy what is 'it' that determines an e- will be produced and not some other 'heavier' stable electron (am aware of mesons) or for that matter, a 'lighter' electron?
The simple answer is mass/energy. Electrons are very light, so producing an e+/e- pair is one of the easiest things a photon carrying enough energy can do; that energy must be greater than the mass of the two electrons (about 1.022 MeV total). Quarks can't be formed because the pair have to exit the interaction moving away from each other to conserve momentum, and quarks are confined by the strong force so we never see one alone.

[contd]
Da Schneib
3 / 5 (4) Mar 27, 2016
[contd]
Up until there's enough energy in the photon to produce a pair of muons, then, a pair of electrons is all that *can* be produced. The electrons produced by a gamma photon with more mass/energy than the required 1.022 MeV have more and more kinetic energy, but until it's enough mass/energy to form a muon, all that can be formed is muons. Once it's above the formation energy for muons, then it might form a pair of electrons, or a pair of muons; a little bit more, and there's enough to make a pair of pions, too, so then it might make a pair of muons, a pair of electrons, or a pair of pions. And so forth, on up the range of particle masses.

Now, there are two questions I can see you asking at that point:
1. Why do electrons and muons and pions have those masses?
or
2. Why does a *particular interaction* capable of producing, say, a pair of muons *or* a pair of electrons, choose one or the other?

[contd]
Da Schneib
3 / 5 (4) Mar 27, 2016
[contd]
The answer to 1. is, nobody knows! SST says it's because those are the possible resonances in our particular Calabi-Yau geometry; I'm not aware of any other physics that gives a reason. And even SST hasn't actually found the CY geometry that gives the masses, they just assert there is one and we'll find it eventually.

The answer to 2. is also, nobody knows! But if we accept the Born Rule as an axiom, then no further explanation is necessary; this is the attraction of interpretations that *do* accept the Born Rule as an axiom.

[ Wasn't it this that led to the J. Wheeler's idea of there being only one electron?].
Actually I thought it was Feynman's idea; you're speaking of the idea that all the electrons we think we see are actually just a single electron, moving back and forth in time (and when it's moving back, we see it as an e+), I think.

[contd]
Da Schneib
3 / 5 (4) Mar 27, 2016
[contd]
Another way of putting it would be if in the gamma ray pair production (decay) g has E = hf and e- so produced always have mass (m) then what is the law that says hf (-+)E must always produce e-. Is that 'law' intrinsic to the g=hf or does it exist in the environment (vacuum or other).
Well, one could say that both the CY geometry and the Born Rule are intrinsic features of the vacuum. That is, no different fundamentally than the existence of 3+1 dimensional spacetime.

Might we call such a governing law as 'QM realism'?
Maybe. But I think that's a type of realism that is more philosophical than physical.

We might have a chat about that, too.

Keep those questions coming, @Mimath; this is the interesting part of things!
Da Schneib
3 / 5 (4) Mar 27, 2016
I just told you, it requires photons to interact with matter.
No, it requires photons to interact with electromagnetic fields. That implies some matter somewhere to generate the field, but it does not imply that the photons interact with the matter- only with the EM field.

A gamma ray, traveling along in open space, through a sufficiently strong magnetic field, can interact to form an electron/positron pair. There isn't any matter in light years. Its probability to undergo this interaction-- this *decay*-- is governed by its wave equation and that of the virtual photons that the magnetic field expresses. The momentum transfer required against the magnetic field propagates back to the matter light years distant, after years of time.

We have detected these events. This isn't speculation, it's observation. It's astrophysics. Spectrograms don't lie.

Like I said, you don't know anywhere near as much physics as you pretend to.
Mimath224
4 / 5 (2) Mar 27, 2016
@Da Schneib, many thanks and I understand your points. You also anticipated correctly, Ha! You might be right about Wheeler or Feynman because, if memory serves, it apparently came out in a telephone conversation between the two...I think.
'Now, there are two questions I can see you asking at that point:
1. Why do electrons and muons and pions have those masses?
or
2. Why does a *particular interaction* capable of producing, say, a pair of muons *or* a pair of electrons, choose one or the other?
The answer to 1. is, nobody knows! ....The answer to 2. is also, nobody knows!'
Rather hoped it might have been the reverse, but that's life! Do you know if research is being done to find out or is it a question of available technology?
I don't really want to get philosophical if I can help it and usually stop at so called 'many valued Logic(s)' which is enough for me to handle. I 'kind of' saw QM realism as being a name for something we don't yet 'see' (similar to astro DM) (cont)
Mimath224
4 / 5 (2) Mar 27, 2016
(cont) Glad you didn't mention 'rishons' cos they remind me of Logic trees, TTT, TTV etc. Ha!
And finally...I think...I've come full circle. If there are intrinsic properties to either photons or the environment which basically says 'okay that's enough energy (g ray)to produce an electron' and so it happens, would that intrinsic value take us back to EPR and Bell's inequality? Or is that entanglement etc is not involved in pair production so EPR/Bell ine doesn't apply?
Da Schneib
2.3 / 5 (3) Mar 27, 2016
Do you know if research is being done to find out or is it a question of available technology?
I think the mass question is still open. I'm not so sure about Born's Rule. But in both cases, I think any advances depend more on theory than on technology.

I doubt either one has much to do with entanglement, EPR, or Bell's Theorem. I think that some sort of "answer" to why Born's Rule is about like an "answer" to why entanglement; they're both features of QM that are distinctly non-classical, and the tendency is to say, well, that's just how it is. As for the mass question, I think that's been a goal of physics for a long time and we're more likely to find some sort of answer to that.
Mimath224
4 / 5 (2) Mar 28, 2016
@Da Schneib, many thanks. 'I doubt either one has much to do with entanglement, EPR, or Bell's Theorem...' I was inclined to think that also but just am not confident of my feelings on this and the reason for that is I have a particular view of Time which might be at least part of the problem. However,this would be Time not in the normal sense and would be completely off topic to go further. It's been great 'talking to you' and perhaps we'll have other interesting conversations when we're both on another thread and I look forward to learning a bit more from you. Have good week.
Noumenon
5 / 5 (5) Mar 28, 2016
it requires photons to interact with electromagnetic fields.


Photons ARE the [quantum form of] electromagnetic fields. According to QED, photons cannot interact with other photons [EMF]. A photon does not decay into an e/p pair, as photons do not decay, as they are massless.

A gamma ray, traveling along in open space, through a sufficiently strong magnetic field, can interact to form an electron/positron pair.


Photons have no charge so therefore cannot interact with magnetic fields directly.

I asked if you were referring to QED diagrams and virtual process above.......i.e. through higher-order VIRTUAL processes (Feynman diagrams) within the constraints of the HUR; a photon can momentarily change into a virtual e/p pair [unobservable] which interacts with the other photon, then the virtual e/p annihilate.This effect on the photon is indirectly observed via further non-virtual interactions. The Real photon does not decay in this process.

Noumenon
5 / 5 (5) Mar 28, 2016
A gamma ray, traveling along in open space, through a sufficiently strong magnetic field, can interact to form an electron/positron pair. There isn't any matter in light years. Its probability to undergo this interaction-- this *decay*-- is governed by its wave equation and that of the virtual photons that the magnetic field expresses. The momentum transfer required against the magnetic field propagates back to the matter light years distant, after years of time.


Do you have a source for this?

Sounds more convoluted than your first attempt, that... "A gamma photon of the correct energy simply decays into a particle pair [.via..] self-interaction".

The honorable thing to do, especially in trying to help other posters here, would have just admitted to getting it. Instead you again rudely attack me. At least you fooled them into giving you 5's, so you're the tallest midget, congrats. :)

Noumenon
5 / 5 (5) Mar 28, 2016
EDIT: "The honorable thing to do, especially in trying to help other posters here, would have just admitted to getting it [wrong]. Instead you again rudely attack me."
Da Schneib
not rated yet Mar 29, 2016
@Noum, it doesn't matter how many times you post your BS.

Shouting the loudest doesn't make you right, and if you had any courage or personal integrity you'd admit you screwed up and move on instead of trying to "win."
Noumenon
5 / 5 (2) Mar 30, 2016
The facts make me right. I didn't "win" because you don't seem to have learned them. Lets not forget your rude behavior toward me right away unprovoked.

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