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

[atyy]

Not in general, i.e. out of equilibrium. Thus the general probability theory of Bohmian Mechanics is quite different from QM. One example being the existence of a single sample space.

How about under the assumption of equilibrium?

 

[A. Neumaier]

How about under the assumption of equilibrium?

It is still not equivalent since it posits additional structure not present in QM. Thus it is a nontrivial extension of QM.

 

[A. Neumaier]

According to your definition it only defines a state. DarMM says that states and ensembles are not synonymous. Why else would one use two different names for it? It seems that you and DarMM use the same term with a completely different meaning!

How do you define ensembles on a precise/formal level?

The problem of mutual understanding between us is that I think in terms of physics and you in terms of mathematics.

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.

As I said the physics logic for me is the following. First there are states and observables, and these have to be clearly distinguished.

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.

On the physical operational level a state is a preparation procedure or, to put it mathematically, an equivalence class of preparation procedures.

This is only mock-mathematical, as preparation procedures are no mathematical entities, and the equivalence relation in question is not even specified.

In my example the LHC is a "preparation machine" for (unpolarized) proton beams with a quite well-defined momentum and energy. These beams are prepared such that they collide in specific points along the beam line. For me the preparation procedure delivers a well-defined ensemble of colliding proton beams.

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?

 

[Morbert]

Is the protean nature of ensembles in QM a weakness in the minimalist ensemble interpretation?

My understanding so far: The theory of a given system is the double $(H,\rho)$, the dynamics and the preparation. I.e. All physical content is contained in these terms. The triple $(H,\rho,\sigma)$ describes an ensemble in terms of possible outcomes of a measurement (or possible outcomes of a sequence of measurements), where $\sigma$ is the set of possibilities. The triple $(H,\rho,\sigma')$ describes an ensemble in terms of a different, incompatible set of measurement possibilities $\sigma'$.

Could we say the physical content of the triples $(H,\rho,\sigma)$ and $(H,\rho,\sigma')$ is the same, and the choice of one over the other is merely a choice of appropriate descriptive terms for a measurement context. I.e. A choice of measurement context does not change any physical content of the preparation. It merely constrains the physicist to use a description appropriate for that context.

[edit] - Added some clarification.

 

[A. Neumaier]

My understanding so far: The theory of a given system is the double $(H,\rho)$, the dynamics and the preparation. I.e. All physical content is contained in these terms. The triple $(H,\rho,\sigma)$ describes an ensemble in terms of possible outcomes of a measurement (or possible outcomes of a sequence of measurements), where $\sigma$ is the set of possibilities. The triple $(H,\rho,\sigma')$ describes an ensemble in terms of a different, incompatible set of measurement possibilities $\sigma'$.

In your terms,

• $H$, the Hamiltonian, characterizes (together with the Hilbert space and its distinguished operators) a ''system'', i.e., an ''arbitrary'' system of the kind considered (e.g., a beam),
• $(H,\rho)$, where $\rho$ is a state, characterizes a ''preparation'', i.e., a ''particular'' system of this kind (e.g., the beam prepared in a particular way specified by values of controls, means of generation, etc.), and
• $(H,\rho,\sigma)$ where $\sigma$ is a measurement setting (choice of POVM or, in the idealized case, of an orthonormal basis), characterizes an ''ensemble'', i.e., a statistical population measured or to be measured.

 

[vanhees71]

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.

 

[vanhees71]

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).

 

[A. Neumaier]

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.

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.

 

[vanhees71]

An ensemble is a collection of independent equally prepared systems. What else is there to define? What else do you understand under an "ensemble"?

 

[DarMM]

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.

 

[A. Neumaier]

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.

 

[vanhees71]

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.

 

[vanhees71]

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?

 

[DarMM]

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.

 

[vanhees71]

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".

 

[DarMM]

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.

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.

 

[RUTA]

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?

 

[vanhees71]

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.

 

[DarMM]

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.

 

[DarMM]

Instruction sets would be the classical case where the state preparation alone determines the sample space, right?

Correct.

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)

 

[Lord Jestocost]

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)

 

[vanhees71]

I see. The "potentiality interpretation" of the "wave function" (or more generally a quantum state) is due to Schrödinger.

 

[A. Neumaier]

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:

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.

 

[vanhees71]

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 ;-)).

 

[A. Neumaier]

Now I can define a subensemble with sz=+ℏ/2 by just "dumping" the other partial beam.

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!

 

[vanhees71]

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".

 

[A. Neumaier]

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''.

 

[atyy]

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.

 

[A. Neumaier]

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:

In the analysis of

• T. Norsen, The pilot-wave perspective on spin. American Journal of Physics, 82 (2014), 337-348.

Figure 2 suggests that rather than measuring spin it measures starting in the upper part of the SG arrangement, independent of spin!

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.

 

[DarMM]

then it should also break down for BM.

It doesn't. In the probability theory for Bohmian Mechanics the total law holds.