Schrödinger local and deterministic?

Gerenuk
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There have been many QM Interpretation thread, but I haven't found this question answered:

Taking aside the fact that a complex probability amplitude is not something we can picture, is the Schrödinger equation local and deterministic at once?
 
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Yes, the Schrodinger equation and the evolution of the wavefunction that follows from it is local and unitary. Unitary means that the time evolution of the wavefunction is unique and completely determined by the initial conditions. It is therefore deterministic.

There is, however, a major practical obstruction that prevents us from actually calculating this time evolution for any macroscopic system. This is particlely because it is practically impossible to determine the initial state of a macroscopic system. But even if we did know this state or if we are somehow able to finetune it, the time-evolution itself is a many-body problem which is, again, computationably intractable.

We therefore always need to resort to some form of approximation, e.g. a statistical description of the system or ignoring a large number of degrees of freedom. Such a statistical description automatically introduces a degree of 'uncertainty' which manifests itself as a non-determinstic description of the system.

So even if you put the whole measurement problem aside, you still end up with a non-deterministic description of macroscopic systems due to practical limitations.
 
So how does that compare to the saying "QM can't be local and deterministic" by Bell's theorem and similar ones?
 
Bell's theorem plus the results of experiments testing it (insofar as one accepts those results and/or the validity of Bell's theorem with respect to those experiments, which is the source of the vigorous arguments here about the subject :smile:) support the statement that "QM can't be local and realistic", which is not the same thing as deterministic.
 
Yes sure. And I do not wish to start yet another Bell discussion.

But what's wrong about saying the Schrödinger equation is local and deterministic?
Mathematically it does look so.
 
Gerenuk said:
Yes sure. And I do not wish to start yet another Bell discussion.

But what's wrong about saying the Schrödinger equation is local and deterministic?
Mathematically it does look so.

I believe you said it, "...the fact that a complex probability amplitude is not something we can picture..." is the reason. Well, if it can only exist as math, it's not physics, just math. So, from the perspective of a mathematician... it is as you say. From the perspective of a Physicist... it is too, but it's not useful if it can't be made to do work. Hence all of the rest... so I'd say to answer your question: To avoid confusion.
 
I guessed so. Now I'm trying to get some ideas to understand how determinism gets lost... :)
 
Gerenuk said:
I guessed so. Now I'm trying to get some ideas to understand how determinism gets lost... :)

See, that's not too hard, because Determinism is lost when we have to calculate positions, velocities, etc... as probabilites. It all comes from the Heisenberg Uncertainty Principle (HUP), now backed up by the CMB surveys.
 
Frame Dragger said:
See, that's not too hard, because Determinism is lost when we have to calculate positions, velocities, etc... as probabilites. It all comes from the Heisenberg Uncertainty Principle (HUP), now backed up by the CMB surveys.

I don't agree that the UP supports an external indeterminism in events; The UP does highlight however our lack of knowledge on a system. Just because there is a lack of knowledge from our behalf should not suggest that the universe is not deterministic.
 
  • #10
ManyNames said:
I don't agree that the UP supports an external indeterminism in events; The UP does highlight however our lack of knowledge on a system. Just because there is a lack of knowledge from our behalf should not suggest that the universe is not deterministic.

The CMB would beg to differ, barring a superdeterministic uneven distribution of "stuff" at 360K years post-BB...
 
  • #11
Calculating probabilities is fine. The problem comes in when someone tries to make a theory that works with probabilities alone.

So if ppl wouldn't try to squeeze QM into basic probability theories, then QM would be local, deterministic and even linear?

Maybe some sophisticated ingredient can make even the probabilities logical again.
 
  • #12
Gerenuk said:
Calculating probabilities is fine. The problem comes in when someone tries to make a theory that works with probabilities alone.

So if ppl wouldn't try to squeeze QM into basic probability theories, then QM would be local, deterministic and even linear?

Maybe some sophisticated ingredient can make even the probabilities logical again.

Time to start building the AI's that can find that... maybe they'll even be nice enough to try and explain it to us! :wink:
 
  • #13
Frame Dragger said:
The CMB would beg to differ, barring a superdeterministic uneven distribution of "stuff" at 360K years post-BB...

No i beg to differ, because the UP is in light of what we can know - its a limitation of knowledge which does not impede determinism.
 
  • #14
ManyNames said:
No i beg to differ, because the UP is in light of what we can know - its a limitation of knowledge which does not impede determinism.

Ok... then how is it that something which is a limitation on KNOWLEDGE managed to effect the (should-have-been-EVEN) distribution of "stuff" in the early universe? The HUP explains that nicely, as does SUPERdeterminism. The HUP + Determinism = Horse****.
 
  • #15
Wasn't the fundamental problem that the equation didn't properly model the interaction between particle and wave as de Broglie envisioned? (It rather just models "some wave" of unknown origin and constitution)
 
  • #16
PhilDSP said:
Wasn't the fundamental problem that the equation didn't properly model the interaction between particle and wave as de Broglie envisioned? (It rather just models "some wave" of unknown origin and constitution)

Yeah... sadly yes... and the Bohmian interpretation replaces that issue with a Pilot wave of "unknown origin and constitution" as you put it so well. Welcome to QM... I need some aspirin. :wink:

EDIT: Hence us left with 50-50 chances, or worse, 50-50-1! Never good when you get 101% in a physical theory...
 
  • #17
Frame Dragger said:
Ok... then how is it that something which is a limitation on KNOWLEDGE managed to effect the (should-have-been-EVEN) distribution of "stuff" in the early universe? The HUP explains that nicely, as does SUPERdeterminism. The HUP + Determinism = Horse****.

You do realize that particles are simply statistical averages right? Physics in general is a statistical theory at best yes? It's statistical because we don't have all the knowledge on a quanum system, but this is because of our lack of knowledge, not because there needs to be an indeterministic world externally of our limited knowledges.
 
  • #18
ManyNames said:
You do realize that particles are simply statistical averages right? Physics in general is a statistical theory at best yes? It's statistical because we don't have all the knowledge on a quanum system, but this is because of our lack of knowledge, not because there needs to be an indeterministic world externally of our limited knowledges.

You keep saying this, but this is generally rejected as a viewpoint. The HUP is not about lack of knowledge, although at one time that was a common belief. It is generally held that particles have attributes only within the context of a measurement.
 
  • #19
ManyNames said:
You do realize that particles are simply statistical averages right? Physics in general is a statistical theory at best yes? It's statistical because we don't have all the knowledge on a quanum system, but this is because of our lack of knowledge, not because there needs to be an indeterministic world externally of our limited knowledges.

To paraphrase DrChinese in my own words, representing my own opinion, "No, I don't realize that, because observational data has shown the HUP is a physical law, not merely a statistal event horizon for observers."
 
  • #20
Gerenuk said:
There have been many QM Interpretation thread, but I haven't found this question answered:

Taking aside the fact that a complex probability amplitude is not something we can picture, is the Schrödinger equation local and deterministic at once?

Classical determinism: Repeating the same experiment many times always has the same result. Classical mechanics allows us to determine that result.
Quantum determinism: Repeating the same experiment many times yields a unique probability distribution of all possible results. Quantum mechanics allows us to determine that probability distribution.
Quantum mechanics does not predict the experimental result; it is not deterministic in the classical sense.

Locality is a property of the space-time of classical physics. It is classical in nature. The wavefunction (probability amplitude) is defined in a Hilbert space. It seems to me that locality is an issue only if the wavefunction propagates in space-time, as many believe.

In the classical sense, quantum mechanics is neither deterministic nor local.
 
  • #21
Gerenuk said:
I guessed so. Now I'm trying to get some ideas to understand how determinism gets lost... :)

Hello Gerenuk,
The answer to your question is simple and i am surprised that nobody has given it yet.

The quantum theory relies on two processes. One deterministic process, called U, like Unitary, governed by Schrödinger's equation, and a probabilistic process, called R, like Reduction, governed by Born's rule.

Both are needed for the theory to actually work. The loss of determinism occurs inside the R process, which has nothing to do with Schrodinger's equation.
 
  • #22
Pio2001 said:
Hello Gerenuk,
The answer to your question is simple and i am surprised that nobody has given it yet.
I'd be very, very careful with such a statement ;-)
Usually the guys crying out "it's so easy!", don't have the slightest clue what the problem is about.
This observation doesn't apply here, but it is one thing to remember :)

Pio2001 said:
Both are needed for the theory to actually work. The loss of determinism occurs inside the R process, which has nothing to do with Schrodinger's equation.
To my knowledge the R process is ill-defined, so it's hard to use it for arguments. I mean when is an observation an observation? Why don't we consider the human being as quantum objects and thus have U processes only?
And how does this R process lose locality or determinism?
For me it's very important not to just know a keyword, but to really understand where mathematically either locality or determinism is lost. Or why at all some people say it is lost, whereas all the theory seems to be based on local and deterministic concepts?
 
  • #23
Pio2001 said:
Hello Gerenuk,
The answer to your question is simple and i am surprised that nobody has given it yet.

The quantum theory relies on two processes. One deterministic process, called U, like Unitary, governed by Schrödinger's equation, and a probabilistic process, called R, like Reduction, governed by Born's rule.

Well, that is no longer agreed upon I think, since decoherence is now a well-established experimental and theoretical phenomenon that shows it is possible to have very rapid processes that proceed in a unitary fashion according to the TDSE, yet produce observations that are consistent with the original "collapse" (or reduction) theories. In fact, you will see the phrase "there is no collapse" thrown around a lot on this forum.

Both are needed for the theory to actually work. The loss of determinism occurs inside the R process, which has nothing to do with Schrodinger's equation.

I would say that it is very much an open question whether or not there is in fact a loss or determinism as you claim.
 
  • #24
Gerenuk said:
To my knowledge the R process is ill-defined, so it's hard to use it for arguments. I mean when is an observation an observation? Why don't we consider the human being as quantum objects and thus have U processes only?

Because the R process makes experimental predictions that the U process doesn't. Example, that YOU will get this or that result when you measure a given system in a given way. If you keep the U process only and use it to built a many world interpretation, you loose the definition of "you", and the above experimental prediction is no more defined.

Gerenuk said:
And how does this R process lose locality or determinism?

The R process lacks determinism in its axiomatic definition, and says nothing about locality.

Later, Bell, CHSH, GHZ, and Mermin (excuse me if I forget some), have shown that locality and determinism could not coexist. In a larger context, we can say that locality, determinism and realism can't coexist in quantum mechanics.

Some people however have suggested workarounds. Mark Rubin, for example, in his article about local realism in the Heisenberg picture of operators in the MWI, or JesseM in this forum, with his idea about pasting parallel universe when their future light-cones meet (which is more or less the same idea, as far as I understand). These ideas deserve to be developed. I'm working on JesseM's idea in my spare time.

Gerenuk said:
For me it's very important not to just know a keyword, but to really understand where mathematically either locality or determinism is lost. Or why at all some people say it is lost, whereas all the theory seems to be based on local and deterministic concepts?

They are lost when you violate Bell's inequality in an EPR-like experiment. No modelization of the experiment have been given yet that
1) Describe what happens in terms of realistic objects
2) Predicts the violation of the inequality by means of the above description

SpectraCat said:
Well, that is no longer agreed upon I think, since decoherence is now a well-established experimental and theoretical phenomenon that shows it is possible to have very rapid processes that proceed in a unitary fashion according to the TDSE, yet produce observations that are consistent with the original "collapse" (or reduction) theories.

Consistent yes, but with not as much predictive power. They do not predict the violation of the inequality without completing decoherence with the last part of the R process, which consists in picking one of the possible results out of many, in an undeterministic way.

SpectraCat said:
I would say that it is very much an open question whether or not there is in fact a loss or determinism as you claim.

I don't disagree, but Gerenuk's question was simple, and I gave the simple answer, from which we can go on and start further discussions :smile:
 
  • #25
Pio2001 said:
Because the R process makes experimental predictions that the U process doesn't. Example, that YOU will get this or that result when you measure a given system in a given way.
I don't think this R process idea is a satisfactory explanation. And the many attempts for interpretations probably share the same view. It's not well defined when someone is measuring and when he isn't and what reality means.

Pio2001 said:
Later, Bell, CHSH, GHZ, and Mermin (excuse me if I forget some), have shown that locality and determinism could not coexist. In a larger context, we can say that locality, determinism and realism can't coexist in quantum mechanics.
I do not want to discuss their work. I'll go through it later, but I know they all make their own hidden assumptions. Anyway:

If I let the universe run for a long time governed by the Schrödinger equation, and if I make one measurement in the end, then everything was a local and deterministic U process to the very end and I can extract probabilities from this l&d process? Right?

And then someone else comes along and says, I'm only a stupid quantum process and he is the real observer and waiting for an even longer time than me before he does the measurement. So in his theory everything was l&d an even longer time?!

It seem everything is l&d at all times. (unless you insist on removing the wavefunction and introduce real probabilities)
 
  • #26
Gerenuk said:
I don't think this R process idea is a satisfactory explanation.

It depends on the explanation you're looking for.

If you want an explanation about the presence of probabilities in the theory in spite of Shrödinger's equation, then the R process is a good answer.

If you want an explanation of reality, then the R process shows severe limitations.

Gerenuk said:
If I let the universe run for a long time governed by the Schrödinger equation, and if I make one measurement in the end, then everything was a local and deterministic U process to the very end and I can extract probabilities from this l&d process? Right?

I'm not sure... If we try to modelize Alice and Bob's EPR experiment this way, we'll run into their decoherence into two different preferred basis. If we simplify their states enough, we don't have the information necessary to get the final probabilities giving the inequality violation. We must not simplify, but keep the initial entanglement scattered into the billions of particles of their environment.
I don't know exactly how it works, nor even if the description of the system in terms of density matrixes holds enough information in order to predict Alice and Bob's correlations.

And if you describe them as wave vector, it doesn't work, because you'd have to consider them as, for example, half-spin particle in order to derive the right probabilities, which they are not : Alice with her detector turned rightside, summed with Alice with her detector turned leftside does not equal Alice with her detector turned upside, because these three state vectors are orthogonal (they're three eigenstates of the position).
However, making this summation is mandatory in order to get the inequality violated !

Don't think in terms of Schrödinger's cats. You can describe them in local and deterministic ways.
Think about Alice and Bob measuring half-spin particles across alpha and beta directions and getting a correlation of -cos(beta-alpha). That's much trickier.
 
  • #27
...or maybe Alice's summation holds true thanks to constructive interferences scattered through the environment's microscopic state ?
 
  • #28
Pio2001 said:
nor even if the description of the system in terms of density matrixes holds enough information in order to predict Alice and Bob's correlations.

I'm not used to the density matrix formalism, but after a second thought, I think that it holds the necessary information. However, it is non-local.

I have already tried something like you suggest : keep the U evolution until the end and get the probabilities when Alice and Bob meet.
The problem is that in order to have the right probabilities, you need Alice and Bob to be themselves elementary particles with half spin properties, with the problem mentionned above.
Another solution is to have them decohered before they meet, but you then need to associate amplitudes to their decohered copies in order to predict the inequality violation, and in this case, the amplitude of Bob's copies must depend on the angle chosen by Alice and vice versa ! Which is non-local. Otherwise, the inequality is not violated.

I eventually found a solution, relying on the many-world formalism. It can be summarized in three additional postulates.
1)When a system splits into many worlds, its copies keep track of the wave function that caused it to split.
2)When a system splits into many worlds, it also keeps track of a unique label that makes it distinguishable from any other similar system.

-> One split = one wave function = one label = many worlds (each one described by a wave function).

3)When some local parts of a split system meet, its copies recognize each other by means of the label given in postulate 2. Then the probabilities for a local copy to meet with one or the other copy from another place is given by the usual born rule applied to the initial wave function that the copies carry with them according to postulate 1.

This way I think it works. I have still to write the whole calculus.
 
  • #29
First, I'd like to see mathematically how locality is lost. It's the R process? Is it possible to write it down, so that it looks non-local from the mathematical view?

I'm not sure I've understood your explanation fully. Why can't I say B knows about A's complete state and her measurement apparatus completely, and thus is able to model all of A's measurement from an external view with U processes only?
 
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  • #30
Hello Gerenuk,
I can do that. I'll try tonight (CET) with entangled photons. They are more interesting than spin 1/2 particles, because they are destroyed during the measurment.
 
  • #31
That'd be great. I just want to mention that I've seen Bell's-inequality-like proofs, but they use a different model/language, so that I still don't know at which point locality is lost. Ideally I'd like to see something which states why the mathematical form of the R process is non-local. Maybe f(x)=\sum_{x'}g(x,x')? Whatever the correct definition of non-local is :)
 
  • #32
Gerenuk said:
There have been many QM Interpretation thread, but I haven't found this question answered:

Taking aside the fact that a complex probability amplitude is not something we can picture, is the Schrödinger equation local and deterministic at once?
Schrodinger equation is certainly deterministic, there is no doubt on it. However, the question whether it is also local is more subtle. Since wave function lives in the configuration space and not in the ordinary 3+1 space, what does it even MEAN to be "local"?

However, the locality of the Hamiltonian (which is an important ingredient of the Schrodinger equation) is a well defined concept. So we can say that Schrodinger equation is "local" in the sense that realistic quantum systems are based on a local Hamiltonian. On the other hand, we can also say that the Schrodinger equation is "nonlocal" in the sense that the quantity it describes - the wave function - is not a local object. An even more precise terminology is not that the wave function is "nonlocal", but that it is NONSEPARABLE.
 
  • #33
Demystifier said:
On the other hand, we can also say that the Schrodinger equation is "nonlocal" in the sense that the quantity it describes - the wave function - is not a local object. An even more precise terminology is not that the wave function is "nonlocal", but that it is NONSEPARABLE.
Can you write this down mathematically? I believe in physics you should never trust words alone unless both parties know exactly which mathematics stands behind it. Because once I learned the maths for some advanced topics I started noticing how much rubbish me and other people were talking before, when we knew only words, i.e. popular science.

Edit: preferably using non-locality and not non-separability. Unless you can find a very good proof to draw the connection.
 
  • #34
Post # 7 by the OP
"Now I'm trying to get some ideas to understand how determinism gets lost."

Here is how Roger Penrose describes the issue, and it adds a bit of detail to Pio's post #21 above:
(from his talk commemorating Stephen Hawking 60th birthday)

Either we do physics on a large scale, in which case we use classical level physics; the equations of Newton, Maxwell or Einstein and these equations are deterministic, time symmetric and local. Or we may do quantum theory, if we are looking at small things; then we tend to use a different framework where time evolution is described... by what is called unitary evolution...which in one of the most familiar descriptions is the evolution according to the Schrodinger equation: deterministic, time symmetric and local. These are exactly the same words I used to describe classical physics.

However this is not the entire story... In addition we require what is called the "reduction of the state vector" or "collapse" of the wave function to describe the procedure that is adopted when an effect is magnified from the quantum to the classical level...quantum state reduction is non deterministic, time-asymmetric and non local...The way we do quantum mechanics is to adopt a strange procedure which always seems to work...the superposition of alternative probabilities involving w, z, complex numbers...an essential ingredient of the Schrodinger equation. When you magnify to the classical level you take the squared modulii (of w, z) and these do give you the alternative probabilities of the two alternatives to happen...it is a completely different process from the quantum (realm) where the complex numbers w and z remain as constants "just sitting there"...in fact the key to keeping them sitting there is quantum linearity...
 
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  • #35
Pio2001 said:
Consistent yes, but with not as much predictive power. They do not predict the violation of the inequality without completing decoherence with the last part of the R process, which consists in picking one of the possible results out of many, in an undeterministic way.


right, an adscititious "collapse" unlike of intrisic.
because superpositions persist, MWI is deterministic.


Pio2001 said:
Because the R process makes experimental predictions that the U process doesn't. Example, that YOU will get this or that result when you measure a given system in a given way1 If you keep the U process only and use it to built a many world interpretation2, you loose the definition of "you", and the above experimental prediction is no more defined.

1.-Forced.
2.-or you can reduce R.
 
  • #36
And here's how Carlo Rovelli describes a related issue: QM vs GR:

(http://www.cpt.univ-mrs.fr/~rovelli/book.pdf) thanks to Marcus...

QM was formulated using an external time variable (the “t” in the Schrodinger equation) or a FIXED non dynamical spacetime (the spacetime on which Quantum field theory is defined). But this external time variable and fixed background spacetime are incompatible with general relativity. In turn GR was formulated in terms of Riemannian geometry assuming that the metric is a smooth and deterministic dynamical field. But QM requires that any dynamical field be quantized: at small scales it manifests itself in discrete quanta and is governed by probabilistic laws. We have learned from GR that spacetime is dynamical and we have learned from QM that that any dynamical entity is made by quanta and can be in probabilistic superposition states.
 
  • #37
I don't see a sandbox in the forums, and LaTeX behaves strangely in the preview form. Let me try and see if it works.

Gerenuk, it is easier for me to describe the EPR experiment with spin 1/2 particles because I have alreay all the equations written in LaTeX.

I start from the following initial state vector :

|\Psi\rangle = \frac{1}{\sqrt{2}} ( {|+\rangle_1} \otimes {|-\rangle_2} - {|-\rangle_1} \otimes {|+\rangle_2} )

Is it displayed properly ?
 
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  • #38
You have to use small letters for the (tex] tag.

Let me try again:
|\Psi\rangle = \frac{1}{\sqrt{2}} ( {|+\rangle_1} \otimes {|-\rangle_2} - {|-\rangle_1} \otimes {|+\rangle_2} )

But really don't forget to introduce a definition of non-local and show why the R process explicitely is non-local. Because I know the Bell-type derivations. I just don't know where the non-locality creeped in.
 
  • #39
Here is a description of an EPR experiment with two particles of spin 1/2.

I'll point out the non-locality mathematically. However, I won't use Schrödinger's equation at all and show that it is not even possible to give a local objective description of what happens using the usual conventions. The non-locality will appear in the wave function itself, even before the R process takes place.

1 Notations, framework.

The two observers, Alice and Bob, are labeled 1 and 2. The measurement results are noted + and - (using \hbar/2 units). No angle is mentionned when the measurement occurs along the Oz axis. The measurement devices can rotate inside the yOz plane. Their angle with Oz is noted \phi when the observer is not mentionned, \alpha when it is Alice, who is observer number 1, and \beta when it is Bob, who is observer number 2.

|M+\phi\rangle_i and |M-\phi\rangle_i are the state vectors of the measurement devices of observer i (i=1 for Alice and i=2 for Bob), when they form an angle \phi with Oz and show the results + and - respectively.

We may also use the letter O for observers who read the result on the measurement device and give them the state vectors
|O+\phi\rangle_i and |O-\phi\rangle_i

Here is what you proposed :
Why can't I say B knows about A's complete state and her measurement apparatus completely, and thus is able to model all of A's measurement from an external view with U processes only?

I won't follow this proposition, because I can't check for Bell's inequality with one measurment only. I prefer to adopt the point of view on an omniscient observer. Getting the probabilities for all possible outcomes, I can check for Bell's inequality at once.

When you say "with U processes only", I assume that you modelize a measurement by intricating the measurement device with the particle measured.

2 The EPR Experiment

At time t0, a source emits two particles in the state

|\Psi\rangle = \frac{1}{\sqrt{2}} ( {|+\rangle_1} \otimes {|-\rangle_2} - {|-\rangle_1} \otimes {|+\rangle_2} )

This answers the question you were asking to Demystifier above. When you take two particles, labeled 1 and 2, you can write their wave function by taking the tensorial product of their respective wave functions.
The above wave function is non-separable because it can't be written as a tensorial product with only states of particle 1 on one side and only states of particle 2 on the other side. This case only occurs with quantic superpositions, but it does not involve non-locality in itself.

The particles reach the detectors at time t1.

After t0 and before t1, Alice turns her measurment device until it forms an angle \alpha with Oz, and Bob does the same until his device forms an angle \beta with Oz.

These actions restrict the possible states of Alice, Bob and their devices after t1 to the linear combinations of the following state vectors :

Alice :
\{ |O+\alpha\rangle_1, \, |O-\alpha\rangle_1 \}

Bob :
\{ |O+\beta\rangle_2, \, |O-\beta\rangle_2 \}

Alice's device :
\{ |M+\alpha\rangle_1, \, |M-\alpha\rangle_1 \}

Bob's device :
\{ |M+\beta\rangle_2, \, |M-\beta\rangle_2 \}

In order to intricate these states with the measured system, we must rewrite the system in the new basis.
The first step is easy. Since the initial wave function is symmetric under any rotation around the Ox axis, we can rewrite it

|\Psi\rangle \, = \, \frac{1}{\sqrt{2}} ( {|+\alpha\rangle_1} \otimes {|-\alpha\rangle_2} - {|-\alpha\rangle_1} \otimes {|+\alpha\rangle_2} )

We can start with Alice's intrication (measurement without reduction). The order has no importance. It gives

<br /> \frac{1}{\sqrt{2}} <br /> ( <br /> {|+\alpha\rangle_1} \otimes {|M+\alpha\rangle_1} \otimes {|O+\alpha\rangle_1} <br /> \otimes <br /> {|-\alpha\rangle_2} <br />

<br /> - <br /> \,<br /> {|-\alpha\rangle_1} \otimes {|M-\alpha\rangle_1} \otimes {|O-\alpha\rangle_1} <br /> \otimes <br /> {|+\alpha\rangle_2} <br /> )<br />

But now, in order to account for Bob's intrication, we must rewrite the vectors of his particle using the follwing substitutions :

<br /> {|+\rangle}<br /> =<br /> cos{\frac{\phi}{2}} <br /> {|+\phi\rangle}<br /> -<br /> sin{\frac{\phi}{2}} <br /> {|-\phi\rangle} <br />

<br /> {|-\rangle}<br /> =<br /> sin{\frac{\phi}{2}} <br /> {|+\phi\rangle}<br /> +<br /> cos{\frac{\phi}{2}} <br /> {|-\phi\rangle} <br />

which gives

<br /> \frac{1}{\sqrt{2}} <br /> (<br /> {|+\alpha\rangle_1} \otimes {|M+\alpha\rangle_1} \otimes {|O+\alpha\rangle_1} <br /> \otimes <br /> (<br /> sin{\frac{\beta-\alpha}{2}} <br /> {|+\beta\rangle_2} <br /> +<br /> cos{\frac{\beta-\alpha}{2}} <br /> {|-\beta\rangle_2}<br /> )<br />

<br /> - <br /> \,<br /> {|-\alpha\rangle_1} \otimes {|M-\alpha\rangle_1} \otimes {|O-\alpha\rangle_1} <br /> \otimes <br /> (<br /> cos{\frac{\beta-\alpha}{2}} <br /> {|+\beta\rangle_2} <br /> -<br /> sin{\frac{\beta-\alpha}{2}} <br /> {|-\beta\rangle_2}<br /> )<br /> )<br />

Then, Bob's intrication leads to

<br /> \frac{1}{\sqrt{2}} <br /> (<br /> {|+\alpha\rangle_1} \otimes {|M+\alpha\rangle_1} \otimes {|O+\alpha\rangle_1} <br /> \otimes

<br /> (<br /> sin{\frac{\beta-\alpha}{2}} <br /> {|+\beta\rangle_2} \otimes {|M+\beta\rangle_2} \otimes {|O+\beta\rangle_2}<br /> +<br /> cos{\frac{\beta-\alpha}{2}} <br /> {|-\beta\rangle_2} \otimes {|M-\beta\rangle_2} \otimes {|O-\beta\rangle_2}<br /> )<br />

<br /> - <br /> \,<br /> {|-\alpha\rangle_1} \otimes {|M-\alpha\rangle_1} \otimes {|O-\alpha\rangle_1} <br /> \otimes

<br /> (<br /> cos{\frac{\beta-\alpha}{2}} <br /> {|+\beta\rangle_2} \otimes {|M+\beta\rangle_2} \otimes {|O+\beta\rangle_2}<br /> -<br /> sin{\frac{\beta-\alpha}{2}} <br /> {|-\beta\rangle_2} \otimes {|M-\beta\rangle_2} \otimes {|O-\beta\rangle_2}<br /> )<br /> )<br />

Rearranging the different terms, we get

<br /> \frac{1}{\sqrt{2}} <br /> (\,<br /> <br /> sin{\frac{\beta-\alpha}{2}} <br /> ({|+\alpha\rangle_1} \otimes {|M+\alpha\rangle_1} \otimes {|O+\alpha\rangle_1}<br /> \otimes <br /> {|+\beta\rangle_2} \otimes {|M+\beta\rangle_2} \otimes {|O+\beta\rangle_2}) <br />

<br /> +<br /> \,<br /> cos{\frac{\beta-\alpha}{2}}<br /> ({|+\alpha\rangle_1} \otimes {|M+\alpha\rangle_1} \otimes {|O+\alpha\rangle_1} <br /> \otimes <br /> {|-\beta\rangle_2} \otimes {|M-\beta\rangle_2} \otimes {|O-\beta\rangle_2} )<br />

<br /> - <br /> \,<br /> cos{\frac{\beta-\alpha}{2}}<br /> ({|-\alpha\rangle_1} \otimes {|M-\alpha\rangle_1} \otimes {|O-\alpha\rangle_1} <br /> \otimes <br /> {|+\beta\rangle_2} \otimes {|M+\beta\rangle_2} \otimes {|O+\beta\rangle_2} )<br />

<br /> +<br /> \,<br /> sin{\frac{\beta-\alpha}{2}}<br /> ({|-\alpha\rangle_1} \otimes {|M-\alpha\rangle_1} \otimes {|O-\alpha\rangle_1}<br /> \otimes <br /> {|-\beta\rangle_2} \otimes {|M-\beta\rangle_2} \otimes {|O-\beta\rangle_2} )<br /> <br /> )<br />
 
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  • #40
3 Non locality and Bell's inequality violation

This final expression is the key. You can see both non-locality and the inequality violation in it.

The non locality is mathematically represented by the fact that the amplitude of Alice's state vectors depends on \beta, and the amplitude of Bob's state vectors depends on \alpha, in a non-separable way, while \alpha was chosen in a space-time region spatially separated from Bob, as he is represented here, and \beta in a space-time region spatially separated from Alice, as she is represented here.

To put it short, Alice's complete description is a function of \beta, and Bob's complete description is a function of \alpha.

You may argue that this is only one possible representation among others, and that we may find another one that is local. I challenge you to find one, and then to calculate Bell's inequality violation from it.

Here how Bell's inequality violation is calculated from the above. Since you're already familiar with it, I skip the normalization part, the local mean values, and directly get to the calculus of &lt;{S_{1 \alpha}}\otimes{S_{2\beta}}&gt;

With or without involving an R process, we must assume that the frequency at which we observe a given pair of result is proportional to the squared modulus of the amplitude of the matching state vectors.

For (+,+), the product of the two measurements is 1, and the probability to get it is

(\frac{1}{\sqrt{2}} <br /> \,<br /> sin{\frac{\beta-\alpha}{2}})^2 <br /> \,<br /> =\,\frac{1}{2}sin^2{\frac{ \beta-\alpha}{2}}

For (+,-), the product is -1, and the probability is

(\frac{1}{\sqrt{2}} <br /> \,<br /> cos{\frac{\beta-\alpha}{2}})^2 <br /> \,<br /> =\,\frac{1}{2}cos^2{\frac{ \beta-\alpha}{2}}

For (-,+), the product is -1, and the probability is

(-\frac{1}{\sqrt{2}} <br /> \,<br /> cos{\frac{\beta-\alpha}{2}})^2 <br /> \,<br /> =\,\frac{1}{2}cos^2{\frac{ \beta-\alpha}{2}}

For (-,-), the product is 1, and the probability

(\frac{1}{\sqrt{2}} <br /> \,<br /> sin{\frac{\beta-\alpha}{2}})^2 <br /> \,<br /> =\,\frac{1}{2}sin^2{\frac{ \beta-\alpha}{2}}Thus &lt;{S_{1 \alpha}}\otimes{S_{2\beta}}&gt; equals :

sin^2{\frac{ \beta-\alpha}{2}} - cos^2{\frac{ \beta-\alpha}{2}}

Which, after trigonometric simplification, leads to

&lt;{S_{1 \alpha}}\otimes{S_{2\beta}}&gt; \,=\, -cos(\beta-\alpha) \,=\,-cos(\alpha-\beta)

Which is enough to get

S = 2\sqrt{2}

For \alpha = 0°, \beta = 45°, \alpha&#039; = 90°, et \beta&#039; = 135°
 
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  • #41
Note : in the above conclusion, S is Bell's parameter and the inequality violated is S <= 2.
The S in &lt;{S_{1 \alpha}}\otimes{S_{2\beta}}&gt;, on the other hand, are the projections of the spin operator.
 
  • #42
I'll go through it in detail tomorrow.

Could you just point out how the full wavefunction would look like including position coordinates and time-dependence? Maybe you can also confirm that all the derivation stays valid if you do it correctly with a full wave-function. A spin function alone might not mean much. Some people were even surprised that it's impossible to find a fully antisymmetric three-spin state.

Is there a derivation which does not use spins only? Because otherwise that suggests that the funny spin states are the reason for trouble.

At first guess I even think it's hard to do such a derivation with position wavefunctions only and still use some notion of particles clearly flying apart?!

Anyway, I try to work through it.

PS: I once had the feeling that non-locality in QM could be connected with permutations of particles only somehow. Has anyone made a theory about this?
 
  • #43
Pio2001 said:
I don't see a sandbox in the forums, and LaTeX behaves strangely in the preview form. Let me try and see if it works.

Gerenuk, it is easier for me to describe the EPR experiment with spin 1/2 particles because I have alreay all the equations written in LaTeX.


i know a model with spin 1/2 particles (but not MWI)
http://arxiv.org/PS_cache/quant-ph/pdf/9505/9505025v1.pdf

....A parametrized model, "Q", for the state vector evolution of spin-1/2 particles during measurement is developed...is local, deterministic, nonlinear and time asymmetric...that Q is not constrained by Bell’s inequality, locality and determinism notwithstanding.
 
  • #44
Gerenuk said:
Can you write this down mathematically? I believe in physics you should never trust words alone unless both parties know exactly which mathematics stands behind it. Because once I learned the maths for some advanced topics I started noticing how much rubbish me and other people were talking before, when we knew only words, i.e. popular science.

Edit: preferably using non-locality and not non-separability. Unless you can find a very good proof to draw the connection.
I know that you already know a lot about this, and I don't want to write you the stuff that you already know. Thus, you would help me if you could specify what EXACTLY you want me to write down mathematically.
 
  • #45
yoda jedi said:
i know a model with spin 1/2 particles (but not MWI)
http://arxiv.org/PS_cache/quant-ph/pdf/9505/9505025v1.pdf

....A parametrized model, "Q", for the state vector evolution of spin-1/2 particles during measurement is developed...is local, deterministic, nonlinear and time asymmetric...that Q is not constrained by Bell’s inequality, locality and determinism notwithstanding.

Thanks for the link.
I read the abstract and the introduction. This model is anti-realistic. Therefore the statement I wrote earlier stands true, until proven otherwise :

No modelization of the experiment have been given yet that
1) Describe what happens in terms of realistic objects
2) Predicts the violation of the inequality by means of the above description
 
  • #46
Oops ! My mistake.

Interpretations that include backwards causality fit in the above statement :)
 
  • #47
Gerenuk said:
Could you just point out how the full wavefunction would look like including position coordinates and time-dependence?

For this purpose, I suggest to imagine ion traps positionned at the output paths of the spin measurment devices. We would use 1/2 spin ions, and they would be trapped into a given box if they get out of the measurment device from one way, and into another box if they get out the other way. The boxes would isolate completely the ions from their environment, preventing them to decohere.
This way, position would be trivially entangled with spin and all the math stuff would be essentially the same.

By the way, when you measure spin with a Stern-Gerlach device, you don't measure the spin of the particule, you measure its position at the output of the device !

Gerenuk said:
Maybe you can also confirm that all the derivation stays valid if you do it correctly with a full wave-function.

I don't know how to write a full wave function. So I can't confirm that. However, the calculus that I made seem correct to me.

Gerenuk said:
Is there a derivation which does not use spins only? Because otherwise that suggests that the funny spin states are the reason for trouble.

There is at least the one with photons polarisation, used to modelize 1982 Aspect's experiment.
And it seems to me that the superposition principle, combined with Heisenberg's inequalities, allows to do this with any pair of measurment.

For example in the delayed choice experiment, a Bell inequality must have been violated measuring the position of the impacts on the screen on one side (Alice), and the quantum eraser output on the other side (Bob). But the two events did not occur in spatially separated regions, so I guess it was not worth checking for Bell's inequality.
 
  • #48
Gerenuk said:
I'd be very, very careful with such a statement ;-)
Usually the guys crying out "it's so easy!", don't have the slightest clue what the problem is about.
This observation doesn't apply here, but it is one thing to remember :)

To my knowledge the R process is ill-defined, so it's hard to use it for arguments. I mean when is an observation an observation? Why don't we consider the human being as quantum objects and thus have U processes only?
And how does this R process lose locality or determinism?
For me it's very important not to just know a keyword, but to really understand where mathematically either locality or determinism is lost. Or why at all some people say it is lost, whereas all the theory seems to be based on local and deterministic concepts?


Right ! irrelevant refutations or endorsing or spurious justifications...
or things like "is an old study, outdated" (Special Relativity dates back to 1905 ! so ?)

or the guys say solemnly "SHEER VOLUME OF EVIDENCE !", "is NOT generally accepted", "In probably 500+ papers in the past year alone" (ah ! then truth is matter of the number of papers, more papers, more truth), "ALL (and I mean 100%) of the experiments !", "I don't see the issue as being relevant"

hell ! they are cheer leaders or parrots ?



Pio2001 said:
Thanks for the link.
I read the abstract and the introduction. This model is anti-realistic. Therefore the statement I wrote earlier stands true, until proven otherwise :

No modelization of the experiment have been given yet that
1) Describe what happens in terms of realistic objects
2) Predicts the violation of the inequality by means of the above description

what is a realistic object ?
 
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  • #49
Hmm, that question seems really interesting to me:
What is the full (position and time dependent) wave-function of that singlet state, where the particles fly apart?
 
  • #50
yoda jedi said:
...or the guys say solemnly "SHEER VOLUME OF EVIDENCE !", "is NOT generally accepted", "In probably 500+ papers in the past year alone" (ah ! truth is matter of the number of papers, more papers, more truth), "ALL (and I mean 100%) of the experiments !", "I don't see the issue as being relevant"
hell, thay are cheer leaders or parrots ?

As Einstein said his critics, and to paraphrase: it only takes 1.

But your criticism misses the mark soundly (since I "parrot" all of the above frequently). This board is not for alternative science. End of the story. I have my own web site, and you can too. On that site, I say whatever I want. So could you. But this site is intended to discuss mainstream science. It is not a place to trumpet your homespun agenda.

Further: it DOES matter what 500+ papers say. It is patently absurd to claim otherwise. They could still be wrong, but the likelihood is low. Hey, they thought the human body had a temperature of 98.6 degrees for years. That has been adjusted. Were all those others "wrong" ?

Make a good argument and the rest will follow.
 
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