How do entanglement experiments benefit from QFT (over QM)?

In summary, the conversation discusses two important points: the first being the difference between QFT and quantum mechanics (QM) and the second being the role of QFT in analyzing entanglement experiments. QFT is a more comprehensive theory than QM, and while it is commonly used in quantum optics papers, it is not often referenced in experimental papers on entanglement. The main reason for this is that QFT is primarily used when dealing with particle-number changing processes, which are not relevant in most entanglement experiments. In addition, while QFT helps to understand how entanglement should not be explained, it does not provide a significant advantage in explaining entanglement itself, and discussions of entanglement often focus on photons due to
  • #246
A. Neumaier said:
The main difference is that mathematicians are trained to use precise concepts and smell easily when something informal cannot be made precise. This is usually the case where informal mathematical arguments go wrong, hence our sensitivity to lack of conceptual precision. Physicists are much more liberal in this respect and aim for precision only when their informal thinking led them completely astray. Thus they tend not to notice the many subtleties inherent in the quest for good conceptual foundations.

But I asked for the precise physical definition of what you call an ensemble (as contrasted to state). In your description the term didn't occur: you only explained the meaning of states and observables.
This is only mock-mathematical, as preparation procedures are no mathematical entities, and the equivalence relation in question is not even specified.

According to which notion of ensemble? That's the question of interest in the present context.

If ''ensemble of colliding proton beams'' is just another phrase for ''several colliding proton beams'' then it is plain wrong to later claim subensembles by filtering according to spin, say. The magnet creates two beams from one, hence two ensembles from one.

So what else did you mean?
I don't see where the problem is. I use a magnet and choose one of the two ensembles by beam-dumping the other. Then I've a new ensemble. It's a preparation procedure in two steps.

Of course, a complete random experiment is only defined if also the measurement on the prepared state is given, and only then the Kolmogorov axioms make sense.

I also have nothing against mathematical regidity and refinements of physical definitions, but it's impossible to make sense of a mathematical abstract prescription like the POVM, if there's not a single example, where it is applied to a real-world experiment. It would be great, if there'd be a simple example, like the one I tried somewhere in this Forum about a position measurement with a detector. So far I've only seen very abstract descriptions with no reference to a real-world measurement.
 
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  • #247
DarMM said:
The state can be specified by a preparation procedure yes. However the state does not define an ensemble, only a state and a measurement choice.It doesn't. It is in a sense a pre-ensemble. One simply needs to look at the statistical properties of various measurements upon the preparation to see this.I gave it in #207Yes for a defined measurement choice and a preparation procedure the probabilities obey Kolomogorov's axioms. However the probabilities of various observables considered together do not. In essence QM is a bunch of entwined Kolmogorov theories.
Ok, so you only call an ensemble if a complete random experiment is defined. I can live with calling a preparation a "pre-ensemble".

A full determination of the prepared state is of course usually not possible with a single experiment (see Ballentine's textbook for a discussion about complete state determinations through measurements).
 
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  • #248
vanhees71 said:
I don't see where the problem is. I use a magnet and choose one of the two ensembles by beam-dumping the other. Then I've a new ensemble. It's a preparation procedure in two steps.
I know that you call it that. But I still do not know what properties of a physical setup makes it qualify as an ensemble in your sense. You are using the term without explaining what it should mean.
vanhees71 said:
it's impossible to make sense of a mathematical abstract prescription like the POVM, if there's not a single example, where it is applied to a real-world experiment. It would be great, if there'd be a simple example, like the one I tried somewhere in this Forum about a position measurement with a detector.
I provided a detailed example for momentum measurement in the POVM thread.
 
  • #249
An ensemble is a collection of independent equally prepared systems. What else is there to define? What else do you understand under an "ensemble"?
 
  • #250
vanhees71 said:
Ok, so you only call an ensemble if a complete random experiment is defined. I can live with calling a preparation a "pre-ensemble"
The interesting thing is that in classical mechanics the preparation alone would define an ensemble.
 
  • #251
vanhees71 said:
An ensemble is a collection of independent equally prepared systems. What else is there to define? What else do you understand under an "ensemble"?
The same. But your definition conflicts with your earlier usage of the word subensemble, which makes no sense with this meaning. Hence I wondered what you mean.

Or does your notion of ensemble have some sort of temporal permanence so that it remains the same when you change its momentum through a mirror and that it splits in a beam splitter? But then the state would not be associated with the ensemble (i.e., the independent equally prepared systems) but with their momentary mode of existence.
 
  • #252
What I called "subensemble" was simply to sort each measurement into the different outcomes of the measurement. I guess it's a misleading wording, and I'll avoid it henceforth.
 
  • #253
DarMM said:
The interesting thing is that in classical mechanics the preparation alone would define an ensemble.
I still do not understand what you mean by the word "ensemble". Obviously I could find some kind of agreement with @A. Neumaier . Why is for you the preparation of many independent systems not defining an ensemble in the QT case but in the classical case?
 
  • #254
vanhees71 said:
I still do not understand what you mean by the word "ensemble". Obviously I could find some kind of agreement with @A. Neumaier . Why is for you the preparation of many independent systems not defining an ensemble in the QT case but in the classical case?
Because different measurements cannot be considered as partitioning a common ensemble into alternate subensembles due to the failure of the total law of probability.
 
  • #255
This I never claimed, but the preparation procedure is independent of the meaurements you can do afterwards. So how can the "ensembles" defined by state preparation depend on what's measured afterwards? I guess, what was really misleading was my use of the word "subensembles".
 
  • #256
vanhees71 said:
This I never claimed, but the preparation procedure is independent of the meaurements you can do afterwards
Of course the preparation is. However the ensemble is not. The preparation procedure alone does not define an ensemble.

vanhees71 said:
So how can the "ensembles" defined by state preparation depend on what's measured afterwards?
Because unlike the classical case the preparation alone does not give a well-defined sample space of outcomes or lattice of events.
 
  • #257
DarMM said:
Of course the preparation is. However the ensemble is not. The preparation procedure alone does not define an ensemble.

Because unlike the classical case the preparation alone does not give a well-defined sample space of outcomes or lattice of events.
I've been reading these posts and trying to figure out where the mystery lies and how it's resolved according to this "statistical" interpretation. DarMM you seem to have a grasp of that, so let me ask you to explain it in terms of the Mermin device ... in the spirit of Dr. Chinese, who started this thread. For anyone who doesn't know the Mermin device, I've attached his original paper.

Fact 1 about the Mermin device states that the outcomes (Red or Green) are always the same when Alice and Bob choose the same measurement setting (both choose 1, both choose 2, or both choose 3). Mermin posits the existence of "instruction sets" to account for Fact 1. He says it's the only way he knows to guarantee Fact 1, since the outcomes can be spacelike separated from each other and the other person's measurement choice, and we don't want superluminal communication or causation. Instruction sets would be the classical case where the state preparation alone determines the sample space, right? That is, each trial of the experiment instantiates one of the possible instruction sets, 1R2R3G, 1R2G3R, 1G2G3R, 1R2R3R, etc., at particle creation independently of Alice and Bob's measurement choices. Mermin then shows that instruction sets entail an overall agreement of outcomes (for all trials, regardless of settings) of more than 5/9 (Bell's inequality for the Mermin device). But, Fact 2 of the Mermin device is that we have an overall agreement of outcomes (for all trials, regardless of settings) of only 1/2, in violation of Bell's inequality. So, the quantum preparation (as modeled by the Mermin device) does not define an ensemble ...

Would you please finish the translation from there?
 

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  • #258
DarMM said:
Of course the preparation is. However the ensemble is not. The preparation procedure alone does not define an ensemble.Because unlike the classical case the preparation alone does not give a well-defined sample space of outcomes or lattice of events.
I see. So an ensemble is mathematically only defined by the specification of the complete random experiment, i.e., the preparation procedure together with what's measured on the prepared systems, in the most general form defined by a POVM.
 
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  • #259
vanhees71 said:
I see. So an ensemble is mathematically only defined
An ensemble is defined only when one has a well defined sample space. In a sense an ensemble is an approximate physical realization of a sample space.

In Classical Mechanics the preparation alone (of multiple copies) gives one a well-defined ensemble, since after the preparation one has a well defined lattice of events. An Observable is just a family of events and an observable outcome is one event/subspace of the sample space.

In Quantum Theory the preparation alone does not give a well-defined ensemble as you cannot consider the outcomes for different observables to be events on one common sample space. This is what prevents counterfactual reasoning. If you measure ##S_{z}## say, since there isn't a common sample space you cannot consider an ##S_{x}## event which may have occurred but you didn't measure. There is no common sample space containing both ##S_{x}## and ##S_{z}## events.

Another way of phrasing it is that the difference between the classical (stochastic) case and the quantum case is that in the classical case the observables you don't measure still had an outcome you just didn't observe it. In the quantum case they don't, only the observable you look at obtains a value/outcome. And since a sample space is a collection of outcomes you have to specify the observable to even define outcomes. And then further since an ensemble is an approximate realization of a sample space, we have to choose an observable to even speak about what the preparation has in fact prepared.
 
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  • #260
RUTA said:
Instruction sets would be the classical case where the state preparation alone determines the sample space, right?
Correct.

RUTA said:
Would you please finish the translation from there?
No problem. I'll just think on it a bit as I'd like to make it as concise as possible without rambling. I'll add a bit about your Relational Blockworld at the end as it has a simple enough explanation there (of course I imagine you already know this, but more for others)
 
  • #261
vanhees71 said:
Why is for you the preparation of many independent systems not defining an ensemble in the QT case but in the classical case?

“Ensembles” or “subensembles” are artificial contrivances based upon concepts of statistical thermodynamics. An “ensemble” is a collective of identically prepared systems which superficially seem to be identical but distinguish from each other on a deeper level; in that sense, an “ensemble” is a “statistical collective”. However, in the case of quantum mechanics, thinking of the post-measurement situation in a statistical way doesn’t allow to infuse statistical considerations into the thinking of the pre-measurement situation: “The deeper reason for the circumstance that the wave function cannot correspond to any statistical collective lies in the fact that the concept of the wave function belongs to the potentially possible (to experiments not yet performed), while the concept of the statistical collective belongs to the accomplished (to the results of experiments already carried out).”(V. A. Fock)
 
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  • #262
I see. The "potentiality interpretation" of the "wave function" (or more generally a quantum state) is due to Schrödinger.
 
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  • #263
vanhees71 said:
I still do not understand what you mean by the word "ensemble". Obviously I could find some kind of agreement with @A. Neumaier . Why is for you the preparation of many independent systems not defining an ensemble in the QT case but in the classical case?
The preparation of many independent systems may define an ensemble in the quantum case, but
then it is not consistent with what you describe here:
vanhees71 said:
What I called "subensemble" was simply to sort each measurement into the different outcomes of the measurement. I guess it's a misleading wording, and I'll avoid it henceforth.
But a Stern-Gerlach experiment does not constitute a measurement: it is a unitary operation.
Thus in such an experiment you are not sorting measurement outcomes into different groups.

This were the case if you'd perform the same experiment on each of you prepared independent systems, producing certain results for each system, including a spin up or down, and afterwards group the systems into those systems where spin was up and those systems where spin was down, and look at the other observables of the resulting subensembles.

But instead you:
1. take the ensemble of prepared systems, each in the state given by a symmetric superposition of (spin-up, momentum-up) and (spin-down, momentum-down);
2. change the system description by selecting the upper path, say, for further consideration only - not by measuring anything but by arrangement of your measuring equipment (no detectors at the down beam);
3. measure (at half the rate of the rate you'd have gotten with the original beam) a position on the upper beam;
4. declare the result as a spin-up measurement, invoking Born's rule for spin measurement.

Step 2 looks like taking a subensemble (since you lose in step 3 half the rate) but is not associated
with measurement but with the choice of a subset of the basis in which to measure. Thus it does not fit your explanation of what an ensemble is. Effectively you simply changed the preparation and
prepared a new state.

Step 4 makes sense only if you interpret Step 2 as having collapsed the system to the state spin-up, momentum-up) by projecting it on the upper eigenspace of the momentum. For only then you are guaranteed to find spin up (as you claim having obtained).

But you always said that collapse is not needed. This is why I still find your terminology confusing if not misleading.
 
  • #264
A. Neumaier said:
The preparation of many independent systems may define an ensemble in the quantum case, but
then it is not consistent with what you describe here:

But a Stern-Gerlach experiment does not constitute a measurement: it is a unitary operation.
Thus in such an experiment you are not sorting measurement outcomes into different groups.

This were the case if you'd perform the same experiment on each of you prepared independent systems, producing certain results for each system, including a spin up or down, and afterwards group the systems into those systems where spin was up and those systems where spin was down, and look at the other observables of the resulting subensembles.
But that's precisely what I meant. I don't necessarily need to perform other measurements on the subensemble, though I could of course do so, and at least in gedanken experiments one does this by measuring the spin component in another direction demonstrating that it is not determined though the spin state is completely specified as a pure state.

In an SGE(z) through the magnetic field the spin-z-component gets entangled with the z-position-component of the atom, i.e., it becomes split into two partial beams, each with a (almost perfectly) determined spin-z component being either ##+\hbar/2## or ##-\hbar/2##. So far it's a unitary operation, which can in principle be reversed (though not in practice). Now I can define a subensemble with ##s_z=+\hbar/2## by just "dumping" the other partial beam. Now I could perform of course other measurements like an SGE(x) with the well-known result that I get with 50% probability either of the two possible results ##s_x=\pm \hbar/2##.

If I understood it right, according to @DarMM only this complete operation defines a subensemble, i.e., after the preparation through the filtering ("partial-beam dumping") I also have to specify the observable I want to measure on it afterwards to completely specify the subensemble. I still don't understand, why this should be necessary, but I can live with it. I'll just avoid the "word subensemble", though I find it very helpful when describing things like "quantum-erasure delayed-choice experiments".

I don't understand the difference between what you summarized in bullets 1-4 to my description. Also nowhere I need a collapse, except you call using a beam dump a collapse ;-)).
 
  • #265
vanhees71 said:
Now I can define a subensemble with sz=+ℏ/2 by just "dumping" the other partial beam.
vanhees71 said:
nowhere I need a collapse, except you call using a beam dump a collapse ;-)).
Why does the beam dump define a subensemble with sz=+ℏ/2?
Only because you collapse the superposition to something pure!
 
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  • #266
Sure, and if you call a beam-dump a collapse, fine with me. Indeed a "filter measurement" is FAPP a collapse, but it's still a local interaction of the beam with the material it hits, no "spooky action at a distance".
 
  • #267
vanhees71 said:
Sure, and if you call a beam-dump a collapse, fine with me. Indeed a "filter measurement" is FAPP a collapse, but it's still a local interaction of the beam with the material it hits, no "spooky action at a distance".
Suppose a beam is split into a superposition of two beams. At positions where the two beams are very far apart, a beam dump collapse is obtained if one destroys one of the resulting beams (by position measurements there) and makes measurements on the other one. This is a bilocal activity created by coordinated local interactions at two far apart places.

Such activities, together with a comparison of the joint measurement statistics at a later time,
are at the heart of all nonlocality experiments. It is not ''spooky action at a distance'' but ''spooky passion at a distance''.
 
  • #268
DarMM said:
By "determined value" I assume you mean that there will be an observable with a completely predictable outcome, not "already has that value prior to measurement" in line with your agnosticism on the issue.In a sense yes and no.

A quantum state is a sort of a pre-ensemble (not a standard term, I'm just not sure how to phrase it), Robert Griffiths often uses the phrase "pre-probability". When provided with a context, the state together with the observables of that context will define an ensemble.

A basic property of an ensemble is something like the total law of probability which says that if I have two observables to measure on the ensemble ##A## and ##B## with outcomes ##A_{i}## and ##B_{i}##, the for a given ##A## outcome:
$$
P\left(A_{i}\right) = \sum_{j}P\left(A_{i} | B_{j}\right)P\left(B_{j}\right)
$$
which just reflects that ##A## and ##B## and their outcomes just partition the ensemble differently. This fails in Quantum Theory and is one of the ways in which it departs from classical probability. Thus quantum observables cannot be seen as being drawn from the same ensemble.

Thus to define an ensemble in QM you have to give the state and the context of observables, not the state alone.

Streater explains it well in Chapter 6 of his text, as does Griffith in Chapter 5 of his Consistent Quantum Theory. There are explanations in Quantum Probability texts, but I think you'd prefer those books.

If the total law of probability breaks down for QM, then it should also break down for BM. However, BM is described by classical probability, so I'm not sure it's quite correct to say that a breakdown of the total law of probability is not consistent with classical probability.
 
  • #269
atyy said:
If the total law of probability breaks down for QM, then it should also break down for BM. However, BM is described by classical probability, so I'm not sure it's quite correct to say that a breakdown of the total law of probability is not consistent with classical probability.
BM has additional variables that makes the position basis preferred, and if one only looks at probabilities for position alone, these integrate to 1. In contrast to standard quantum mechanics, these are the only probabilities that matter in BM.

Indeed, BM has no probabilities for spin or momentum measurements. The latter are only illusions, in reality being position measurements in disguise:
A. Neumaier said:
In the analysis of
Figure 2 suggests that rather than measuring spin it measures starting in the upper part of the SG arrangement, independent of spin!
Demystifier said:
From the Bohmian perspective it's indeed silly to call it measurement of spin. [...] Bohmians speak to "ordinary" physicists by saying something like this: The procedure that you call measurement of spin is really a measurement of position and I will tell you what is really going on when you think you measure spin.
Whether Born's rule for arbitrary observables follows from BM (with quantum equilibrium assumption) is unclear to me.
 
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  • #270
atyy said:
then it should also break down for BM.
It doesn't. In the probability theory for Bohmian Mechanics the total law holds.
 
  • #271
DarMM said:
It doesn't. In the probability theory for Bohmian Mechanics the total law holds.

How can that be if the predictions of QM and BM are the same? For example, if A and B are the outcomes of position and momentum measurements in QM, then the probabilities of the outcomes should be the same in QM and BM. What is different in the formula between QM and BM?
 
  • #272
atyy said:
How can that be if the predictions of QM and BM are the same? For example, if A and B are the outcomes of position and momentum measurements in QM, then the probabilities of the outcomes should be the same in QM and BM. What is different in the formula between QM and BM?
Bohmian Mechanics in equilibrium is equivalent to QM because of a posited strict restriction on epistemic reasoning within its probability theory. In Bohmian Mechanics in general the total law holds.

When we demand equilibrium we impose a very specific restriction on access to/ability to reason about the hidden variables. Provided this epistemic block holds always the probability theory effectively reduces to that of QM. When blocked in this absolute way the effective Bayesian reasoning about observations is a probability like QM's. The resultant probability theory then does break Kolomogorv's axioms. Breaking the total law is not consistent with classical probability theory. Blocking statistical inference in a very specific way on a theory that normally has classical probability can cause it to not have classical probability.

Note that equilibrium exactly holding cannot be true but must be a thermalisation effect, so in essence if Bohmian Mechanics were true one should be able to see the total law restored.

There is a much broader point here that most of the interpretations of QM are not really interpretations but actually different theories. All hidden variable theories replicate QM under some kind of epistemic restriction that cannot hold in general and in some scenarios even with that restriction will have divergent predictions. Other views such as Many Worlds make conjectures about the formal structure of the theory that have yet to be verified.

The only actual interpretations proper are things like Quantum Bayesianism vs Copenhagen where really it's a purely philosophical thing, e.g. how do you view probabilities. You'll see a similar remark from Rudolf Peierls in "The Ghost in the Atom" from Cambridge University Press.

EDIT:
Note also that even in the underlying Bohmian theory preparations do not prepare ensembles of most quantities, we still have contextuality afterall. Only a position ensemble is prepared.
 
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  • #273
DarMM said:
Bohmian Mechanics in equilibrium is equivalent to QM because of a posited strict restriction on epistemic reasoning within its probability theory. In Bohmian Mechanics in general the total law holds.

When we demand equilibrium we impose a very specific restriction on access to/ability to reason about the hidden variables. Provided this epistemic block holds always the probability theory effectively reduces to that of QM. When blocked in this absolute way the effective Bayesian reasoning about observations is a probability like QM's. The resultant probability theory then does break Kolomogorv's axioms. Breaking the total law is not consistent with classical probability theory. Blocking statistical inference in a very specific way on a theory that normally has classical probability can cause it to not have classical probability.

Note that equilibrium exactly holding cannot be true but must be a thermalisation effect, so in essence if Bohmian Mechanics were true one should be able to see the total law restored.

There is a much broader point here that most of the interpretations of QM are not really interpretations but actually different theories. All hidden variable theories replicate QM under some kind of epistemic restriction that cannot hold in general and in some scenarios even with that restriction will have divergent predictions. Other views such as Many Worlds make conjectures about the formal structure of the theory that have yet to be verified.

The only actual interpretations proper are things like Quantum Bayesianism vs Copenhagen where really it's a purely philosophical thing, e.g. how do you view probabilities. You'll see a similar remark from Rudolf Peierls in "The Ghost in the Atom" from Cambridge University Press.

EDIT:
Note also that even in the underlying Bohmian theory preparations do not prepare ensembles of most quantities, we still have contextuality afterall. Only a position ensemble is prepared.

Yes, I agree. Maybe the only difference is that I would say that all the routes here are also open to the minimal interpretation, so we don't have to say that the minimal interpretation goes beyond classical probability, any more than BM does. We could also say the minimal interpretation is contained within classical probability.
 
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  • #274
atyy said:
Yes, I agree. Maybe the only difference is that I would say that all the routes here are also open to the minimal interpretation, so we don't have to say that the minimal interpretation goes beyond classical probability, any more than BM does. We could also say the minimal interpretation is contained within classical probability.
QM mathematically violates classical probability, thus is not contained in it. It is contained in a classical theory with a certain kind of epistemic restriction such as Bohmian Mechanics at equilibrium. However note that in equilibrium we are violating Kolmogorov's axioms anyway due to how the epsitemic restriction functions. I'll say more about this in a while as it links into Spekkens model. Bohmian Mechanics and other hidden variable theories replicate much of QM by this restriction alone, you only need the nonlocality/retrocausality to violate CHSH or Bell inequalities.

So in no sense is QM contained in classical probability. Mathematically classical probability is a subset of quantum probability not the other way around. It's like saying curved spacetime is contained in flat spacetime because the latter might turn out to be the correct description of nature.

I think more so one should say that a truly minimal view is neutral to there being a deeper theory where classical probability theory holds. However it would have to acknowledge that as far as we can tell now and operationally in labs preparations do not constitute ensembles. Regarding your previous statement:
atyy said:
so I'm not sure it's quite correct to say that a breakdown of the total law of probability is not consistent with classical probability
Breaking the total law is not consistent with classical probability mathematically. It may be the case that there is a deeper theory which uses classical probability but that is a separate statement. Similarly a Newton-Cartan bundle is not consistent with a Lorentzian metric theory, but the deeper gravitational turned out to involve such.

Also note that from contextuality even still in such a deeper theory a preparation does not constitute an ensemble for most observables. Although classical probability is restored we cannot view our preparation as an ensemble for observables like angular momentum, but only the hidden ##\lambda##.
 
  • #275
A. Neumaier said:
Suppose a beam is split into a superposition of two beams. At positions where the two beams are very far apart, a beam dump collapse is obtained if one destroys one of the resulting beams (by position measurements there) and makes measurements on the other one. This is a bilocal activity created by coordinated local interactions at two far apart places.

Such activities, together with a comparison of the joint measurement statistics at a later time,
are at the heart of all nonlocality experiments. It is not ''spooky action at a distance'' but ''spooky passion at a distance''.
What do you mean by "superposition of two beams"? If you talk about superposition you have to tell the basis, according to which the state ket is a superposition.

The beam dump is due to a local interaction between the particles in the one partial beam dumped and the material you bump the particles in. This is with very good will something like a position measurement though nobody cares about the precise position where the beam is dumped ;-)). Then you do experiments with the other beam, which is also due to usual local interactions of the particles in this beam with the various elements of the experiment (in the SGE the magnet and the particle detector like in the original experiment the glas plates on which the silver atoms where catched and then developed to be measured under a microscope afterwards). Of course, in principle the beam dump and the experiment with the other beam can be as far apart as you wish. This has nothing to do with spooky actions at a distance.

One should clearly distinguish between nonlocal interactions, which according to standard relativistic QFT (aka the standard model) does not exist and correlations between far-distant parts of quantum systems described by entanglement. What you mean by "spooky passion at a distance" I can't say.
 
  • #276
vanhees71 said:
What do you mean by "superposition of two beams"? If you talk about superposition you have to tell the basis, according to which the state ket is a superposition.
I had done so in an earlier post to the same topic:
A. Neumaier said:
take the ensemble of prepared systems, each in the state given by a symmetric superposition of (spin-up, momentum-up) and (spin-down, momentum-down);
A. Neumaier said:
Such activities, together with a comparison of the joint measurement statistics at a later time, are at the heart of all nonlocality experiments. It is not ''spooky action at a distance'' but ''spooky passion at a distance''.
vanhees71 said:
What you mean by "spooky passion at a distance" I can't say.
It is a meaningful play with words. It means that something happens at a distance - namely that nature cooperates globally at long distance to ensure that the perfect nonclassical correlations predicted by quantum mechanics in certain experiments actually happen. But it cannot be controlled hence is a passive happening (a ''passion'') rather than an active one (an ''action''). In spite of (and consistent with) the locally induced interactions!
 
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  • #277
vanhees71 said:
I see. The "potentiality interpretation" of the "wave function" (or more generally a quantum state) is due to Schrödinger.
It's important to note that this is interpretation neutral. Due to contextuality some of the quantities we measure have to arise during interaction with the measurement device and can only be taken as properties of the device-system pair. Thus the state preparation has not prepared an ensemble for these quantities.
 
  • #278
vanhees71 said:
Sure, and if you call a beam-dump a collapse, fine with me.
Situations like these are precisely what induced Heisenberg 1927 to talk about state reduction (aka reduction of the state vector, aka collapse). That you don"t like the commonly used words for it doesn't mean that you don't make use of the same concept.
 
  • #279
A. Neumaier said:
I had done so in an earlier post to the same topic:
It is a meaningful play with words. It means that something happens at a distance - namely that nature cooperates globally at long distance to ensure that the perfect nonclassical correlations predicted by quantum mechanics in certain experiments actually happen. But it cannot be controlled hence is a passive happening (a ''passion'') rather than an active one (an ''action''). In spite of (and consistent with) the locally induced interactions!
This is the gibberish, I fight against. What do you mean by "nature cooperates globally".

Here you have a very clear preparation procedure consisting of entirely local physics: A beam of silver atoms comes through a hole from an oven, which gives a beam of unpolarized particles. Then it runs through a magnetic field such that you get an entanglement between the measured spin component and position. The entanglement refers to one and the same particle, and thus it's a "local property" of the single particle. Here you thus don't even have the long-distant correlations via entanglement as in the Bell experiments with two photons!

Of course you cannot control which spin state a single particle in the beam takes, that's the irreducible randomness of QT, but it allows you to prepare states with a definite spin component in the measured direction by selection of the wanted partial beam, thanks to the spin-component-position entanglement.

It's of course right that everything is consistent with local interactions. Otherwise we'd have to find a new theory instead of the Standard Model, which is difficult, because the Standard Model works better than wanted by the majority of particle physicists who are dissatisfied with it for various reasons.
 
  • #280
DarMM said:
It's important to note that this is interpretation neutral. Due to contextuality some of the quantities we measure have to arise during interaction with the measurement device and can only be taken as properties of the device-system pair. Thus the state preparation has not prepared an ensemble for these quantities.
Are you saying the protons in the LHC do not have the very well determined momentum? This claim contradicts the very functioning of the entire device!
 
<h2>1. What is the difference between QFT and QM?</h2><p>QFT (Quantum Field Theory) is an extension of QM (Quantum Mechanics) that combines the principles of quantum mechanics with the principles of special relativity. It allows for the description of particles as excited states of a quantum field, rather than individual particles.</p><h2>2. How do entanglement experiments benefit from QFT?</h2><p>QFT provides a more complete understanding of entanglement, as it allows for the description of particles as excitations of a quantum field. This allows for a better understanding of how entangled particles interact and how they are affected by their environment.</p><h2>3. Can you give an example of an entanglement experiment that benefits from QFT?</h2><p>One example is the Bell test, which measures the entanglement between two particles. QFT provides a more accurate description of the entangled particles and their interactions, allowing for a more precise measurement of their entanglement.</p><h2>4. How does QFT improve our understanding of entanglement?</h2><p>QFT allows for a more comprehensive description of entanglement, as it takes into account the effects of the surrounding environment and interactions between particles. This leads to a more complete understanding of how entangled particles behave and how they can be used in applications such as quantum communication and computing.</p><h2>5. Are there any practical applications of entanglement experiments using QFT?</h2><p>Yes, there are several practical applications of entanglement experiments that utilize QFT. These include quantum cryptography, quantum teleportation, and quantum computing. QFT allows for a better understanding and manipulation of entangled particles, making these applications more efficient and reliable.</p>

1. What is the difference between QFT and QM?

QFT (Quantum Field Theory) is an extension of QM (Quantum Mechanics) that combines the principles of quantum mechanics with the principles of special relativity. It allows for the description of particles as excited states of a quantum field, rather than individual particles.

2. How do entanglement experiments benefit from QFT?

QFT provides a more complete understanding of entanglement, as it allows for the description of particles as excitations of a quantum field. This allows for a better understanding of how entangled particles interact and how they are affected by their environment.

3. Can you give an example of an entanglement experiment that benefits from QFT?

One example is the Bell test, which measures the entanglement between two particles. QFT provides a more accurate description of the entangled particles and their interactions, allowing for a more precise measurement of their entanglement.

4. How does QFT improve our understanding of entanglement?

QFT allows for a more comprehensive description of entanglement, as it takes into account the effects of the surrounding environment and interactions between particles. This leads to a more complete understanding of how entangled particles behave and how they can be used in applications such as quantum communication and computing.

5. Are there any practical applications of entanglement experiments using QFT?

Yes, there are several practical applications of entanglement experiments that utilize QFT. These include quantum cryptography, quantum teleportation, and quantum computing. QFT allows for a better understanding and manipulation of entangled particles, making these applications more efficient and reliable.

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