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

[DarMM]

The proof only applies to projections not measurements.
What is the point in assigning values to irrelevant and unknowable properties ?
No mathematical theorem can prove the existence or not of a real thing.

Measurements in labs have the structure of POVMs that the KS theorem applies to. Thus if quantum theory is correct, which seems to be the case, the theorem applies to the real world.
They're not "irrelevant and unknowable properties", they are how quantum theory represents actual measurements in labs.

 

[vanhees71]

That's not the point. The point is that the ensembles found in each of the four measurement choices have mathematical features preventing them from being understood as selections from one common ensemble. If you measure $A, C$ and you measure $B, D$ they cannot be thought of as subensembles of one ensemble nor alternate course grainings of a common ensemble. They are simply two different ensembles. That is a mathematical fact reflected in the fact that there is no Gelfand homomorphism subsuming all four observables.

However even the whole CHSH set up is a side point. The main point is the KS theorem. Which says either the values don't pre-exist the measurement or they do but are contextual. It's one or the other. If you take the option that they don't pre-exist the measurement, then how do you avoid the measurement creating them?

I think it's a language issue. Perhaps I don't express myself clearly. What I mean is the following:

I consider two spins. The ensemble is described by the spin singlet state, i.e., you prepare photons in this state. In the usual notation not taking into account the Bose nature of the photons, which is sufficient here, we have
$$\hat{\rho}=|\Psi \rangle \langle \Psi|$$
with
$$|\Psi \rangle = \frac{1}{\sqrt{2}} (|HV \rangle-|VH \rangle).$$
now in the first measurement setup A measures here photon's polarization in $\theta_A=0$ and $B$ his in the $\theta_B=\pi/4$ direction. The outcome of the measurement is $+$ or $-$ for each observer, depending on whether the corresponding photon goes through the polarizer or not. Now you count the four possible outcomes $++$, $+-$, $-+$, $--$. In this idealized discussion for each photon in the ensemble with certainty one of these outcomes occurs, and this devides the original ensemble in 4 subensembles.

To check the CHSH relation you only need to count the outcomes and form $E$ as defined in Eq. (3) of

https://en.wikipedia.org/wiki/CHSH_inequality#A_typical_CHSH_experiment

Then you repeat this experiment 3 more times with the other pairs of angles quoted there to evaluate the quantitity in Eq. (2). I haven't checked it explicitly, but if I remember right with the chosen angles that yields the maximal possible violation of the CHSH relation, $S = 2 \sqrt{2}>2$.

Of course, in this case you don't need to argue about "subensembles" as is needed in the quantum eraser or the entanglement-swapping experiments.

 

[DarMM]

I haven't checked it explicitly, but if I remember right with the chosen angles that yields the maximal possible violation of the CHSH relation

This is all correct. The difference is that in the classical case each of the four choices of measurement set ups could be derived as alternate course grainings of a single ensemble. So let's say we have angles $A,B$ for Alice and angles $C,D$ for Bob.

When you do the experiment with $A,C$ say, we get four sub-ensembles for each outcome. Again as you said. I'll just call these subensembles $E_{i}$.
When you do the experiment with $A,D$ say, same thing. Four sub-ensembles $F_{i}$

What makes QM different from classical probability theories is that $E_{i}$ and $F_{i}$ cannot be considered as alternate subensembles/partitions of the same ensemble. This is due to the structure of quantum probability theory being different from Kolomogorovian probability. Only the state and the measurement choice define an ensemble, e.g. the ensemble is given by the triple $\left\{\rho, A, C\right\}$

Streater's monograph explains this quite well. It is odd as it means the experimenters choice of apparatus defines not just the subensembles, but literally the total ensemble itself!

As I said though my original point is with the KS theorem. I think you have to accept that measurements create values in the minimal statistical interpretation and that this is not just sloppy language.

 

[vanhees71]

Sure, that's the whole point of all these Bell ideas, including the CHSH variant. I think I've abused the term "(sub)ensemble(s)".

Now I'm a bit confused about this one sentence:

"Streaters monograph explains this quite well. It is odd as it means the experimenters choice of apparatus defines not just the subensembles, but literally the total ensemble itself!"

Isn't the total ensemble defined by the preparation before the measurements? Of course, the split in subensembles due to a measurement depend on the chosen measurent.

I'm agnostic concerning the meaning of "measurements create values". You measure an observable, and adequately calibrating the apparatus you get the values predicted by QT (i.e., the eigenvalues of the corresponding operators). How it comes that you always get a definite outcome, I think one cannot say. QT describes only the probabilities. The occurance of definite outcomes in measurements is an empirical fact that's described by but not derived from QT. For sure true is that an observable only takes a determined value if the state is accordingly prepared, i.e., in the case that there are no degeneracies the preparation in the eigenstate of the corresponding observable operator. In the case of degeneracy, it should be the projector to the corresponding eigenspace,
$$\hat{\rho}=\frac{1}{d} \sum_{\alpha=1}^d |a,\alpha \rangle \langle a,\alpha|.$$

 

[Mentz114]

Measurements in labs have the structure of POVMs that the KS theorem applies to. Thus if quantum theory is correct, which seems to be the case, the theorem applies to the real world.
They're not "irrelevant and unknowable properties", they are how quantum theory represents actual measurements in labs.

I have to disagree with the text I've emphasized. No quantum states are measured in labs.
A value which does not appear in the Hamiltonian is irrelevant to the dynamics.
[edit] left out a vital 'not'

 

[Mentz114]

[]

I'm agnostic concerning the meaning of "measurements create values". You measure an observable, and adequately calibrating the apparatus you get the values predicted by QT (i.e., the eigenvalues of the corresponding operators). How it comes that you always get a definite outcome, I think one cannot say. QT describes only the probabilities. The occurance of definite outcomes in measurements is an empirical fact that's described by but not derived from QT. For sure true is that an observable only takes a determined value if the state is accordingly prepared, i.e., in the case that there are no degeneracies the preparation in the eigenstate of the corresponding observable operator. In the case of degeneracy, it should be the projector to the corresponding eigenspace,
$$\hat{\rho}=\frac{1}{d} \sum_{\alpha=1}^d |a,\alpha \rangle \langle a,\alpha|.$$

This cannot be emphasized enough. The actual outcome is created by evolution of dynamic variables not probabilities.

 

[DarMM]

You measure an observable, and adequately calibrating the apparatus you get the values predicted by QT (i.e., the eigenvalues of the corresponding operators). How it comes that you always get a definite outcome, I think one cannot say. QT describes only the probabilities. The occurance of definite outcomes in measurements is an empirical fact that's described by but not derived from QT. For sure true is that an observable only takes a determined value if the state is accordingly prepared

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.

Isn't the total ensemble defined by the preparation before the measurements?

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.

 

[DarMM]

I have to disagree with the text I've emphasized. No quantum states are measured in labs.
A value which does not appear in the Hamiltonian is irrelevant to the dynamics.
[edit] left out a vital 'not'

I never said they were. The KS theorem has nothing to do with quantum states, nor did my post mention quantum states. I don't understand the Hamiltonian remark.

 

[DarMM]

This cannot be emphasized enough. The actual outcome is created by evolution of dynamic variables not probabilities.

Quantum Theory does not seem to describe that evolution as @vanhees71 mentioned. We know such variables if they exist will have to be nonlocal, retro/acausal or involve multiple worlds.

 

[Mentz114]

Quantum Theory does not seem to describe that evolution as @vanhees71 mentioned. We know such variables if they exist will have to be nonlocal, retro/acausal or involve multiple worlds.

Surely if "Quantum Theory does not seem to describe that evolution" then why is this inability considered a problem ? The things I'm talking about are not hidden variables. Pressure, temperature, momentum and energy are and actual stuff are dynamic variables that drive the universe, not probabilities.

 

[DarMM]

The things I'm talking about are not hidden variables. Pressure, temperature, momentum and energy are and actual stuff are dynamic variables that drive the universe, not probabilities.

Fine, but if you try to explain the dynamical evolution with them you will need nonlocality, retro/acausality or multiple worlds.

 

[Mentz114]

Fine, but if you try to explain the dynamical evolution with them you will need nonlocality, retro/acausality or multiple worlds.

A point of agreement ! I believe the non-locality required is present in all physics where spatial translational invariance is present. Unless a finite limit is placed on terms like |\mathbf{x}_1-\mathbf{x}_2|^{2} there is manifest non-locality.

 

[vanhees71]

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.

A value is determined, when an accurate measurement leads to a certain value (in the spectrum of the representing operator) with 100% probability.

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.

Perhaps I miss some subtlety here, but for me the state preparation defines probabilities for the outcome of measurements. As a frequentist I think probabilities describe an ensemble, and in this sense a quantum state represents an ensemble of equally prepared systems.

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.

I guess you talk about two incompatible observables A and B, which usually cannot be measured simultaneously in the sense of ideal (complete) von Neumann measurements. The impossibility of such measurements and corresponding preparations is indeed the main difference between classical and quantum physics.

 

[DarMM]

A value is determined, when an accurate measurement leads to a certain value (in the spectrum of the representing operator) with 100% probability

Certainly, but I assume you don't think the system actually possesses such a value prior to measurement. It's just certain to produce a given outcome upon measurement.

Perhaps I miss some subtlety here, but for me the state preparation defines probabilities for the outcome of measurements. As a frequentist I think probabilities describe an ensemble, and in this sense a quantum state represents an ensemble of equally prepared systems

The state preparations define the probabilities for future measurements, but these probabilities do not satisfy the conditions of a statistical model/ensemble unless one selects a context. A quantum state cannot be thought of as an ensemble.

The impossibility of such measurements and corresponding preparations is indeed the main difference between classical and quantum physics.

It's not, you can have classical theories with non-commutativity and observables that are impossible to measure simultaneously.

The main difference is that in such a classical theory all observables and their outcomes can be thought of as alternate partitions of a single ensemble and thus the state can be thought of as an ensemble.

 

[A. Neumaier]

Perhaps I miss some subtlety here, but for me the state preparation defines probabilities for the outcome of measurements. As a frequentist I think probabilities describe an ensemble, and in this sense a quantum state represents an ensemble of equally prepared systems.

The sublety is more apparent in a simpler experimental setting:

Suppose you prepare completely unpolarized light.

It in the first experiment you pass it through polarizers polarizing it horizontally and vertically. According to Born's rule, the subsequently observed photon counting probabilities add up to 1, hence you would conclude that the original ensemble has been split into two subensembles of linearly polarized photons.

It in the second experiment you pass it through polarizers polarizing it left circularly and right circularly. According to Born's rule, the subsequently observed photon counting probabilities add up to 1, hence you would conclude that the original ensemble has been split into two subensembles of circularly polarized photons.

But the original ensemble cannot simultaneously be an ensemble consisting of linearly polarized photons and an ensemble consisting of circularly polarized photons. Thus a polarizer cannot be said to select a subensemble but must be said to create a new ensemble!

This is meant by having disjoint samle spaces, depending on the experimental setup.

 

[Auto-Didact]

A point of agreement ! I believe the non-locality required is present in all physics where spatial translational invariance is present. Unless a finite limit is placed on terms like |\mathbf{x}_1-\mathbf{x}_2|^{2} there is manifest non-locality.

Fully agree. To use another example from physics, the Navier-Stokes equation is manifestly nonlocal in almost exactly the same way as QT; the only proven methodology to understand the causes and effects of this type of manifest nonlocality so far is by approaching Navier-Stokes from a dynamical systems theorist perspective, i.e. not as a dynamical system per se but studying the equation using the most sophisticated technical non-perturbative tools available in the practice of applied mathematics, i.e. the tools of the modern dynamical systems theorist. This methodology, from a structuralist history of physics perspective, is the natural structural evolution of the portfolio of pure mathematical tools of the theoretical physicist. Experience has taught us that the conceptual problems of nonlocality and nonlinearity seem to be quite deeply intertwined at a pure mathematics level.

Your arguments are also full of assumptions solely based on your faith, none of them verifiable.

To answer you more directly, @A. Neumaier: the successfulness of applying dynamical systems methodology onto both complicated empirical phenomena and pure mathematics - even subjects which were deemed 'clearly' non-dynamical - is where my 'unfounded faith' is based upon; I am not the originator of such ideas for even Heisenberg said it himself [Heisenberg 1967]. The naive perspective of something being 'clearly non-dynamical and therefore should not be approached by dynamical methods' has more often than not in STEM and beyond turned out to be about as true as Euclid's parallel postulate is; once there is sufficient recognition that the naive perspective is actually completely contingent it gets reduced to an axiom which can and will be removed once deemed necessary, creating an entirely new form of mathematics in the process - fully analogous to the creation of non-Euclidean geometry by ditching Euclid's parallel postulate.

 

[DarMM]

A point of agreement ! I believe the non-locality required is present in all physics where spatial translational invariance is present. Unless a finite limit is placed on terms like |\mathbf{x}_1-\mathbf{x}_2|^{2} there is manifest non-locality.

Bear in mind then that the Kochen-Specker theorem says that such a nonlocal account must be contextual and Hardy's theorem demonstrates that in such an account a given system must have an infinite number of dynamical degrees of freedom.

 

[Auto-Didact]

Bear in mind then that the Kochen-Specker theorem says that such a nonlocal account must be contextual and Hardy's theorem demonstrates that in such an account a given system must have an infinite number of dynamical degrees of freedom.

I am hearing two necessary requirements:
1) contextuality
2) an infinite dimensional state space

Abramsky already demonstrated that the former can be treated using sheaf theory, while the latter requires functional analysis.

A naive reply from my part would be that a pure mathematical cross between Morse theory - a methodology from nonlinear functional analysis - and Grothendieck's sheaf cohomology should be able to achieve what is needed.

 

[vanhees71]

The sublety is more apparent in a simpler experimental setting:

Suppose you prepare completely unpolarized light.

It in the first experiment you pass it through polarizers polarizing it horizontally and vertically. According to Born's rule, the subsequently observed photon counting probabilities add up to 1, hence you would conclude that the original ensemble has been split into two subensembles of linearly polarized photons.

It in the second experiment you pass it through polarizers polarizing it left circularly and right circularly. According to Born's rule, the subsequently observed photon counting probabilities add up to 1, hence you would conclude that the original ensemble has been split into two subensembles of circularly polarized photons.

But the original ensemble cannot simultaneously be an ensemble consisting of linearly polarized photons and an ensemble consisting of circularly polarized photons. Thus a polarizer cannot be said to select a subensemble but must be said to create a new ensemble!

This is meant by having disjoint samle spaces, depending on the experimental setup.

Sure, I've never claimed something else. It depends on the selection, i.e. the measurement used to select, into which subensembles you sort a given ensemble. Of course, I don't claim that before the measurement the photons had the properties measured. The only properties a system has is described by the state it is prepared in, and these properties are probabilities for outcomes of measurements.

 

[DarMM]

It depends on the selection, i.e. the measurement used to select, into which subensembles you sort a given ensemble

Indeed the measurement outcomes sort the ensemble into subensembles.

However it's the state and the measurement choice together that define the ensemble in the quantum case. The state alone does not define an ensemble unlike in classical probability. Quantum observables are not alternate partitions of a common ensemble, but define alternate ensembles.

 

[atyy]

Indeed the measurement outcomes sort the ensemble into subensembles.

However it's the state and the measurement choice together that define the ensemble in the quantum case. The state alone does not define an ensemble unlike in classical probability. Quantum observables are not alternate partitions of a common ensemble, but define alternate ensembles.

But the Bohmian case shows that the ensemble and subensembles can be defined before the measurement.

 

[vanhees71]

This formulation I find very confusing. For me a state represents operationally a preparation procedure. With this preparation procedure you create a well-defined ensemble. On this ensemble you can perform whatever measurement is possible. Then you can define subensembles by sorting them out dependent on the outcomes of these measurements. Of course, these subensembles are not predetermined as for a deterministic classical system. The QT randomness is not like classical randomness due to incomplete knowledge about the system (like the initial conditions of a die to predict the outcome when throwing it) but is a property of the system itself.

 

[DarMM]

But the Bohmian case shows that the ensemble and subensembles can be defined before the measurement.

In Bohmian Mechanics there is a single sample space, so indeed you can.

 

[DarMM]

With this preparation procedure you create a well-defined ensemble

You don't. That's a mathematical fact of the formalism. A preparation and a context give a well-defined ensemble, not a preparation alone. In a classical theory a preparation alone creates a well defined ensemble.

 

[vanhees71]

But the preparation does not predetermine what I measure. I can still choose any physically possible measurement. Which mathematical fact are you referring to?

In other words, why doesn't the preparation of a proton beam in the LHC define an ensemble of protons?

 

[atyy]

In Bohmian Mechanics there is a single sample space, so indeed you can.

But if BM is consistent with the minimal interpretation, then it would seem that one can also have subensembles defined before the measurement in the minimal interpretation (ie. the minimal interpretation doesn't force us to accept that the subensembles are chosen by the measurement).

 

[vanhees71]

You cannot define subensembles before the measurement. This doesn't make sense at all, not even in classical physics: To create subensembles you must somehow choose them, i.e., you must sort the subensembles using some criterion, i.e., you have to observe something to choose the subensembles.

 

[A. Neumaier]

It depends on the selection, i.e. the measurement used to select, into which subensembles you sort a given ensemble.

To call it subensembles is misleading terminology.

The only properties a system has is described by the state it is prepared in, and these properties are probabilities for outcomes of measurements.

By this definition it is a transformation of the original ensemble, of the same kind as when you apply a rotator in place of a polarizer.

 

[A. Neumaier]

For me a state represents operationally a preparation procedure. With this preparation procedure you create a well-defined ensemble.

why doesn't the preparation of a proton beam in the LHC define an ensemble of protons?

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!

You cannot define subensembles before the measurement. This doesn't make sense at all, not even in classical physics: To create subensembles you must somehow choose them, i.e., you must sort the subensembles using some criterion, i.e., you have to observe something to choose the subensembles.

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

 

[A. Neumaier]

But the Bohmian case shows that the ensemble and subensembles can be defined before the measurement.

But if BM is consistent with the minimal interpretation, then it would seem that one can also have subensembles defined before the measurement in the minimal interpretation (ie. the minimal interpretation doesn't force us to accept that the subensembles are chosen by the measurement).

This is because additional hidden variables are posited. BM is not a minimal interpretation. The ensemble is determined in @vanhees71 version of the minimal interpretation by the state alone, but in BM by the state plus the position assignments. The latter provide the additional preferred basis mentioned by DarMM.

 

[DarMM]

But if BM is consistent with the minimal interpretation, then it would seem that one can also have subensembles defined before the measurement in the minimal interpretation (ie. the minimal interpretation doesn't force us to accept that the subensembles are chosen by the measurement).

@A. Neumaier has stated this already but Bohmian Mechanics is adding additional variables which restore the ability to speak of a single ensemble. This makes its probability theory quite different from that of the quantum formalism.

So if Bohmian Mechanics is correct and its additional variables exist then the ensemble (note: not subensemble) is defined prior to a choice of measurement.

As an analogy in a similar "minimal" interpretation of Newtonian Mechanics one could say that you aren't forced to say spacetime isn't Lorentzian. However clearly in the theory itself spacetimes are not Lorentzian. Ultimately one could minimally interpret any theory in a sense to leave open completely different structures and truths that hold in a yet undiscovered completion.

Similarly the actual mathematical structure of QM literally does not have ensembles well-defined prior to a measurement choice, that's part of its actual mathematical structure and the statistics of real experiments do not act in accord with the notion of an ensemble defined from preparation alone.

Bohmian Mechanics as a different theory which would constitute a completion of QM would allow such a pre-defined ensemble notion. However the formalism and experimental results we currently have do not.

 

[DarMM]

But the preparation does not predetermine what I measure. I can still choose any physically possible measurement

Correct and the quantum formalism does not give a well-defined ensemble until this choice is specified.

Which mathematical fact are you referring to?

The deviation of QM from Kolomogorov's probability theory as shown in for example the violations of the total law of probability.

 

[atyy]

@A. Neumaier has stated this already but Bohmian Mechanics is adding additional variables which restore the ability to speak of a single ensemble. This makes its probability theory quite different from that of the quantum formalism.

So if Bohmian Mechanics is correct and its additional variables exist then the ensemble (note: not subensemble) is defined prior to a choice of measurement.

As an analogy in a similar "minimal" interpretation of Newtonian Mechanics one could say that you aren't forced to say spacetime isn't Lorentzian. However clearly in the theory itself spacetimes are not Lorentzian. Ultimately one could minimally interpret any theory in a sense to leave open completely different structures and truths that hold in a yet undiscovered completion.

Similarly the actual mathematical structure of QM literally does not have ensembles well-defined prior to a measurement choice, that's part of its actual mathematical structure and the statistics of real experiments do not act in accord with the notion of an ensemble defined from preparation alone.

Bohmian Mechanics as a different theory which would constitute a completion of QM would allow such a pre-defined ensemble notion. However the formalism and experimental results we currently have do not.

Yes, what I mean is that the minimal interpretation does not say that ensembles and subsensembles cannot be defined, rather that it does not define a unique ensemble.

 

[vanhees71]

To call it subensembles is misleading terminology.

By this definition it is a transformation of the original ensemble, of the same kind as when you apply a rotator in place of a polarizer.

Ok, I'll cancel "subensemble" in this sense from my vocabulary.

I found it useful to describe situations like the delayed-choice experiments, like the entanglement-swapping experiments.

There you prepare two polarization entangled photon pairs, which themselves are unentangled and then by performing a local Bell measurement with two photons with both photons from the two different pairs. One just needs to make a measurement protocol and then after the measurement (in this case literaly "after the fact") you can choose each of the four possible Bell states the two other photons are in without having ever been in common local interactions with any device. I called these four selections "subensembles", but it's perhaps indeed misleading, because the original ensemble is of course destroyed completely by the measurement process.

 

[vanhees71]

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

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

On the physical operational level an observable is (an equivalence class of) a measurement procedure. In my example you can define any kind of observable on "colliding proton beams". At the LHC these are given by the various detectors. An observable would, e.g., be the $p_T$ spectrum of charged particles or an invariant-mass spectrum of electron-positron pairs (dileptons) etc. etc.

It is important to note that the measurement is independent of the preparation, i.e., no matter how you prepare your system you can always perform any sensible measurement on it, e.g., preparing a proton beam with well-defined momentum you can always measure very precisely the position of the proton (or the proton distribution in the proton bunch). This is important to interpret, e.g., the uncertainty relation right (Heisenberg got it wrong at first and was corrected by Bohr): It's not describing the impossibility to measure position and momentum accurately at the same time but it describes the impossibility to prepare a particle with well-defined position and momentum simultaneously.

The mathematical formulation we have discussed very often, and I don't think that we differ much concerning it. Of course, I refer to the idealized measurements. In mathematical language, these are the projective measurements, which are clearly a subset of the more general POVMs.

Concerning the latter, in all the nice mathematical sources you quoted not a single one gives a clear physical description of a POVM measurement of position or the "fuzzy common measurement of position and momentum" (as I'd translate what seems to be intended by the very abstract formulations of the POVM formalism I've seen so far).

 

[vanhees71]

Correct and the quantum formalism does not give a well-defined ensemble until this choice is specified.The deviation of QM from Kolomogorov's probability theory as shown in for example the violations of the total law of probability.

Well, obviously we just have a different understanding of "ensemble" (see my previous posting of #235). For me it's sufficient to specify the state by some preparation procedure. The state provides probability distributions for the outcome of any sensible experiment you can do on the so prepared system. Of course the complete description of the random experiment you need to specify also the measurement (if I understand it right nowadays in the most general sense formally by a POVM).

What do you mean by "the total law of probability"? To my understanding for any completely defined measurement the probabilities fulfill the Kolmogorov axioms (at least for the standard projector valued measurement formalism in textbooks).

 

[DarMM]

Yes, what I mean is that the minimal interpretation does not say that ensembles and subsensembles cannot be defined, rather that it does not define a unique ensemble.

I assume the "it" is refers to the state. Well a state $\rho$ and a context $\left\{A_{i}\right\}$ define an ensemble under some Gelfand map $G$. So in this sense an ensemble is not defined until you have a state and a context.

However this has nothing to do with Bohmian Mechanics, I'm not sure how it changes this statement.

 

[DarMM]

Well, obviously we just have a different understanding of "ensemble" (see my previous posting of #235). For me it's sufficient to specify the state by some preparation procedure. The state provides probability distributions for the outcome of any sensible experiment you can do on the so prepared system. Of course the complete description of the random experiment you need to specify also the measurement (if I understand it right nowadays in the most general sense formally by a POVM).

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.

For me the preparation procedure delivers a well-defined ensemble of colliding proton beams

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.

What do you mean by "the total law of probability"?

I gave it in #207

To my understanding for any completely defined measurement the probabilities fulfill the Kolmogorov axioms

Yes 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.

 

[atyy]

I assume the "it" is refers to the state. Well a state $\rho$ and a context $\left\{A_{i}\right\}$ define an ensemble under some Gelfand map $G$. So in this sense an ensemble is not defined until you have a state and a context.

However this has nothing to do with Bohmian Mechanics, I'm not sure how it changes this statement.

Well, if BM = QM then do the hidden variables define a context?

Or is BM not equivalent to QM?

 

[DarMM]

Well, if BM = QM then do the hidden variables define a context?

Or is BM not equivalent to QM?

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.

 

[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.