QFT made Bohmian mechanics a non-starter: missed opportunities?

[WernerQH]

I didn't mean that Ballentine makes no sense but your claim that the standard interpretation of the state within the minimal interpretation were wrong.

I didn't make such a claim. What doesn't make any sense to me is your reference to particular experiments in quantum optics as an argument for your sloppy use of the words "individual" and "system". Not only quantum theory needs interpretation, but the experiments too! What one person perceives as firmly established empirical facts, can be viewed as grounded in deeply engrained habits of thought by another. (Phlogiston, caloric, aether ...) But obviously you can't conceive of such a possibility.

 

[vanhees71]

So what's your interpretation of the quantum state, and how is it consistent with the obvious fact that experiments with ensembles built by preparations of single quantum systems. My photons were an example. There are more examples: Single electrons in a Penning trap, interference experiments with single neutrons, etc. etc.

 

[PeterDonis]

Then I have no idea what do you mean by ensemble interpretation

In the other thread you referenced you criticized Ballentine's ensemble interpretation. His definition (p. 46) is that, given a state preparation procedure, an ensemble is "the conceptual infinite set of all such systems that may potentially result from the state preparation procedure".

 

[PeterDonis]

Obviously it's not the one, Ballentine defines in his book

As far as I can tell, the definition I just posted from Ballentine is consistent with what @martinbn has been saying, and not with what you and @Demystifier have been saying.

I could say the same about what you posted from Ballentine, his definition of the state operator, which is based on the ensemble definition that I posted.

 

[PeterDonis]

what's your interpretation of the quantum state

Ballentine's interpretation is clear; again from p. 46: "...the primary definition of a state is the abstract set of probabilities for the various observables". The ensemble definition I quoted earlier is then introduced with: "...it is also possible to associate a state with an ensemble of similarly prepared systems." He then gives the ensemble definition I quoted. His reason for giving the ensemble definition is (earlier in the same paragraph): "The empirical content of a probability statement is revealed only in the relative frequencies in a sequence of events that result from the same (or equivalent) state preparation procedure."

 

[WernerQH]

So what's your interpretation of the quantum state

I have no doubt at all that QT is a statistical theory. But we seem to disagree on what it is about, what it is that causes the perfect correlations observed in so many experiments. I've tried to explain my view in post #309.

and how is it consistent with the obvious fact that experiments with ensembles built by preparations of single quantum systems.

Sorry, I just can't understand your question. I don't see in which sense there should be an inconsistency.

My photons were an example. There are more examples: Single electrons in a Penning trap, interference experiments with single neutrons, etc. etc.

Also experiments with single particles involve many events, taking a lot of time in the lab.

 

[Structure seeker]

So is the idea that each state in an ensemble is given by a copy of the same stochastic variable? Where the space of possible outcomes of each of these stochastic variables is (perhaps) a subspace of all possible quantum states for these specific quanta?

 

[Quantum Waver]

Freeman Dyson in “THE COLLAPSE OF THE WAVE FUNCTION” in John Brockman’s book “This Idea Must Die: Scientific Theories That Are Blocking Progress (Edge Question Series)” (New York, NY, USA: HarperCollins (2015)):

“Fourscore and eight years ago, Erwin Schrödinger invented wave functions as a way to describe the behavior of atoms and other small objects. According to the rules of quantum mechanics, the motions of objects are unpredictable. The wave function tells us only the probabilities of the possible motions. When an object is observed, the observer sees where it is, and the uncertainty of the motion disappears. Knowledge removes
uncertainty. There is no mystery here.

Unfortunately, people writing about quantum mechanics often use the phrase “collapse of the wave function” to describe what happens when an object is observed. This phrase gives a misleading idea that the wave function itself is a physical object. A physical object can collapse when it bumps into an obstacle. But a wave function cannot be a physical object. A wave function is a description of a probability, and a probability is a statement of ignorance. Ignorance is not a physical object, and neither is a wave function. When new knowledge displaces ignorance, the wave function does not collapse; it merely becomes irrelevant.”

That's one interpretation. Ignorance doesn't explain individual particles exhibiting interference patterns when the slit they go through isn't detected. Or other interesting experiments demonstrating wave and particle-like properties depending on measurement.

 

[PeterDonis]

So is the idea that each state in an ensemble is given by a copy of the same stochastic variable?

No. The idea is that the state describes the ensemble, not any individual system. That is because the state describes the probabilities of measurement results, and you can only experimentally measure those probabilities by doing statistics on the results of the same measurement on a large number of identically prepared systems.

 

[Structure seeker]

Identical and independent or identical and perhaps entangled (so that the partial state of one entity in the ensemble is a mixed state)? Usually with such experiments these are by design meant to be independent.

 

[PeterDonis]

Identical and independent

Identically prepared, with each individual preparation being independent.

or identical and perhaps entangled

The preparation procedure can involve preparing an entangled state (for example, parametric down conversion that produces two entangled photons is a valid preparation procedure). But, as above, each individual preparation would still be independent of all the other preparations.

 

[Structure seeker]

Identically prepared, with each individual preparation being independent

So we have $n$ i.i.d. (independent identically distributed) stochastic variables $X_1$, ..., $X_n$ over some space of quantum states, in case of entangled states we just take the density matrix of the whole state as $X_i$ per $i$. However we cannot know the state of any of these, they're quantum. But since they are not correlated, the stochasts that describe their measurement outcomes are also not correlated. Then the actual state $x_i$ (when it were possible to evaluate $X_i$, but the particle must have a state so outside our knowledge $X_i$ has been evaluated) is not entangled with any of the other $x_j$, right? That's what entanglement means.

 

[PeterDonis]

So we have $n$ i.i.d. (independent identically distributed) stochastic variables $X_1$, ..., $X_n$ over some space of quantum states

There are no such "stochastic variables" in standard QM. Are you referring to some particular interpretation? If so, a reference would help since there are no such "stochastic variables" in the usual ensemble interpretation (as described in, for example, Ballentine) either.

If these "stochastic variables" are something you made up yourself, please be aware that personal speculation is off topic here.

 

[Structure seeker]

If the states in an ensemble are partial states, at least they have some state, surely? It's not like we can formulate a quantum that has no state in QFT. So whatever the preparation does, it results in a state $x_i$. If it's not the same for each $i$, while the preparation is identical and independent, the mathematical way to describe this is an i.i.d. stochast set $\{X_i \}_i$ that takes values on the state space.

Let me know where the personal speculation is, if it's in here somewhere.

 

[PeterDonis]

If the states in an ensemble are partial states

I don't know what you mean by this either. In the ensemble interpretation, the state vector $\ket{\psi}$ describes the probabilities for possible measurement results on an ensemble of identically prepared systems. There are no "partial states" or "states in an ensemble".

whatever the preparation does, it results in a state $x_i$.

Not on a single system if we are using the ensemble interpretation, no. The state vector $\ket{\psi}$ describes just what I said above.

Let me know where the personal speculation is, if it's in here somewhere.

The personal speculation is that you appear to be making up a bunch of stuff that doesn't appear anywhere either in the actual math of QM or in the ensemble interpretation.

 

[Structure seeker]

It's not very citable, since it's the basis of most of stochastics: some number of identically prepared independent experiments give as results the outcomes of the same number of i.i.d. stochasts. In this case, I as realist am sure the states exist prior to measurement so they can be seen as outcomes of the preparation.

 

[martinbn]

It's not very citable, since it's the basis of most of stochastics: some number of identically prepared independent experiments give as results the outcomes of the same number of i.i.d. stochasts. In this case, I as realist am sure the states exist prior to measurement so they can be seen as outcomes of the preparation.

The problem is the way you express yourself makes it impossible, at least for me, understand. In fact it looks like nonsense to me. For example you reapetedly wrote "the states in the ensemble", which makes no sense whatsoever. It was pointed out by @PeterDonis and you didnt clarify!

 

[PeterDonis]

It's not very citable, since it's the basis of most of stochastics

We're not talking about "stochastics" here, we're talking about quantum mechanics. Either what you're saying has some basis in either the standard math of QM or some recognized interpretation, or it doesn't.

some number of identically prepared independent experiments give as results the outcomes of the same number of i.i.d. stochasts

Again, either this has some basis in the actual math of QM, or it doesn't. What you're describing does not look like QM to me; it looks like classical statistical mechanics. That is off topic in this thread and this forum.

I as realist

Then you're not using the ensemble interpretation and nothing you're saying is relevant to the posts of mine you were responding to, which were specifically about the ensemble interpretation. Realist interpretations make different claims, which are often inconsistent with claims made by the ensemble interpretation. That's not something that is resolvable by discussion. (The guidelines for this subforum talk about this.)

 

[Structure seeker]

So it appears that "realism of quantum states" predicts that ensemble states are never entangled.

 

[PeterDonis]

So it appears that "realism of quantum states" predicts that ensemble states are never entangled.

No, it appears that you are trying to mix two different QM interpretations which say inconsistent things. The correct thing is to not do that. You can't take statements that are valid for an ensemble interpretation, such as the ones I made, and then try to interpret them using a realist interpretation. That just leads to nonsense.

 

[vanhees71]

I have no doubt at all that QT is a statistical theory. But we seem to disagree on what it is about, what it is that causes the perfect correlations observed in so many experiments. I've tried to explain my view in post #309.

I don't know, which correlations you are referring to. Is it about entanglement? This is not so relevant for the very basic question what the state means for an individual quantum system. The standard answer within the statistical interpretation is that the quantum state, represented by a statistical operator $\hat{\rho}$, is the description of (an equivalence class of) preparation procedures, as in the example with the single electron prepared with pretty sharp momentum (as done in accelerators) in Ballentine's RMP article. The quantum state must have this operational meaning within the statistical interpretation, because otherwise you couldn't associate ensembles, which must be formed by some preparation procedures, with the formal definition of the state in the theory.

Now the association of the state with a real-world situation does imply and only imply the statistical properties for the outcomes of measurements, and in this sense the quantum state describes not the properties of an individual quantum system but of ensembles of "equally prepared" quantum systems. This also implies that observables only take determined values if the system is prepared in a corresponding state, i.e., that with 100% probability you find a specific value. It's something like
$$\hat{\rho}=\sum_{\alpha} p_{\alpha} |a,\alpha \rangle \langle a,\alpha|,$$
where the $|a,\alpha \rangle$ are a orthonormal system spanning the eigenspace $\text{Eig}(\hat{A},a)$ of the self-adjoint operator $\hat{A}$, representing the observable $A$, of the eigenvalue, $a$, and $\sum_{\alpha} p_{\alpha}=1$, $\p_{\alpha} \geq 0$.

Sorry, I just can't understand your question. I don't see in which sense there should be an inconsistency.

If you deny that the state describes a preparation procedure on a single system, you can't say, how the well-defined ensembles, described by a statistical operator $\hat{\rho}$ are formed. I also don't understand, why the standard definition as describing a preparation procedure on a single system should be problematic. It's just reflecting what's done by experimentalists: They prepare large ensembles of equaly prepared individual quantum systems and perform measurements on them. The probabilistic predictions of QT are confirmed by these experiments. So there must be some truth in the standard association of quantum states with the ability to form ensembles with the corresponding specific, statistical properties.

Also experiments with single particles involve many events, taking a lot of time in the lab.

Exactly. To be able to do so, it must be possible to prepare ensembles in a well-defined quantum state, and the preparation procedure refers to the single members of these ensembles.

 

[vanhees71]

So is the idea that each state in an ensemble is given by a copy of the same stochastic variable? Where the space of possible outcomes of each of these stochastic variables is (perhaps) a subspace of all possible quantum states for these specific quanta?

Standard QT is not described by stochastic differential equations. I don't see, where you find "stochastic variables" in the theory.

 

[martinbn]

@vanhees71 I think your favorite interpretation, which you call the minimal statistical interpretation, is not the ensemble interpretation. It is the minimal interpretation of the formalism of QM including the statistical interpretation of the state vector (via the Born rule). But it is not a statistical interpretation.

 

[vanhees71]

Then, what do you think differs in what you call "my interpretation" to that one given by Ballentine. I don't see any difference.

 

[martinbn]

Then, what do you think differs in what you call "my interpretation" to that one given by Ballentine. I don't see any difference.

For you the state vector is a description of the individual system. If you have many equally prepared ones, they will have the same state vector. But that is not the same as the state vector describes the whole ensemble and not the individuals.

 

[WernerQH]

If you deny that the state describes a preparation procedure on a single system, you can't say, how the well-defined ensembles, described by a statistical operator are formed.

The problem is your imprecise use of the term "single system". Of course a polarization filter will produce a beam of photons in some definite polarization state. Subsequent measurements with a polarizer in the same direction will confirm that the photons are in this same state. But the word "state" refers to an abstraction, a "typical" or "average" photon in that beam. It is an assumption that the state is a complete description of each photon in that beam. I'm not proposing hidden variables, but we cannot prove that the photons in the beam are identical. So the second best option is to think of the polarization state as a stable statistical distribution that doesn't change if we add more polarization filters with the same orientation. It still describes an ensemble. Classically circularly polarized light is characterized by a correlation of $E_x$ with $E_y$ at some point in space a quarter period later. This carries over to the microscopic, quantum picture. (See, for example, my post in another thread.)

 

[vanhees71]

For you the state vector is a description of the individual system. If you have many equally prepared ones, they will have the same state vector. But that is not the same as the state vector describes the whole ensemble and not the individuals.

No! It's not that superficial. It's precisely as you say, and that's what I say the whole time: The state (not the state vector but the statistical operator $\hat{\rho}$ btw.) describes on the one hand for the single system a preparation procedure but on the other concerning the properties of the so prepared system probabilities for the outcome of measurements and nothing else, which implies that it describes properties of the ensemble an not the individual system.

Of course you need both meanings of the state, i.e., it must refer to the single system as a description of the preparation procedure, because otherwise you couldn't make the connection between the formalism (statistical operator) with the system under consideration. At the same time the preparation in a state only provides probabilistic properties and as such make only sense for an ensemble of so (sic!) prepared systems.

 

[Morbert]

No! It's not that superficial. It's precisely as you say, and that's what I say the whole time: The state (not the state vector but the statistical operator $\hat{\rho}$ btw.) describes on the one hand for the single system a preparation procedure but on the other concerning the properties of the so prepared system probabilities for the outcome of measurements and nothing else, which implies that it describes properties of the ensemble an not the individual system.

Of course you need both meanings of the state, i.e., it must refer to the single system as a description of the preparation procedure, because otherwise you couldn't make the connection between the formalism (statistical operator) with the system under consideration. At the same time the preparation in a state only provides probabilistic properties and as such make only sense for an ensemble of so (sic!) prepared systems.

While I won't try to give labels to these positions, I will at least point out that these two positions are distinct:

i) Given a single microscopic system, a state represents an equivalence class of preparations the system was subject to.

ii) Given a single microscopic system, a state represents a fictitious infinite ensemble of equivalently prepared systems, of which the system is a member.

 

[vanhees71]

I think both statements are correct and both are necessary to make physical sense of the quantum formalism, and both are part of the minimal statistical interpretation a la Balentine.

 

[Morbert]

I think both statements are correct and both are necessary to make physical sense of the quantum formalism, and both are part of the minimal statistical interpretation a la Balentine.

i) Is an account given by Asher Peres, and might be too minimalist even for Balentine.

 

[Simple question]

these two positions are distinct:

But in both cases, the "state as an ensemble" will be tested by a series of individual measurements events, that cannot be reduced nor averaged.

"equivalence class of preparations" is quite vague, as it cannot be equivalent as defined by QM itself (no cloning). Either way that ensemble is the most non-local thing there is in physics.

 

[Morbert]

@Simple question I've responded to your post here

 

[vanhees71]

But in both cases, the "state as an ensemble" will be tested by a series of individual measurements events, that cannot be reduced nor averaged.

"equivalence class of preparations" is quite vague, as it cannot be equivalent as defined by QM itself (no cloning). Either way that ensemble is the most non-local thing there is in physics.

The point is you must be able to build ensembles in the lab to begin with and indeed this is done by some "preparation procedure" on each single system, and the quantum state, $\hat{\rho}$, thus refers to the preparation procedure and in this respect refers to the single system. On the other hand what it also describes are the properties of the system due to this preparation procedure, i.e., what this preparation implies concerning the outcome of measurements/observations on each of the so prepared systems. Now in the development of the theory it turned out that the only concistent interpretation is Born's probabilistic meaning of the state, and in this sense the state refers only to an "ensemble of equally prepared systems", i.e., you need to be able to prepare the system in this state in a reproducible way.

 

[WernerQH]

The point is you must be able to build ensembles in the lab to begin with and indeed this is done by some "preparation procedure" on each single system, ...

What does it mean to "build ensembles"? I think you've got this backwards: it is Nature providing the ensembles, and we struggle to adjust our statistical descriptions to what we observe. A preparation procedure is empirically known to have produced specific statistical patterns in the past, and it is reasonable to assume that it will produce the same patterns in the future. Of course it is possible to apply probability theory to a "single" system, e.g. the throw of a die. But it's still statistical reasoning. The definiteness of the "state" of the "system" is more grounded in psychology than in the quantum formalism: your desire to describe it in a "Newtonian" way as a system evolving continuously with time, rather than a sequence of events.

Quantum theory has a vastly broader scope than experiments in the laboratory. (Consider for example the nuclear reactions in the sun.) Isn't it ridiculous to base the formulation of this microscopic theory on undefined (primitive) notions like state preparation and measurement? On the grounds that there is no other way to describe the real world?

 

[vanhees71]

What does it mean to "build ensembles"? I think you've got this backwards: it is Nature providing the ensembles, and we struggle to adjust our statistical descriptions to what we observe.

Well, yes. Nature provides also some kind of ensembles. In our everyday life it's many-body systems close to thermal equilibrium. Here, of course, we have to deal with the question, how to describe systems with incomplete, coarse-grained knowledge about the state, and that's why there thermodynamics plays such an important role.

Otherwise physics deals with quite artificial idealized "preparations" in order to investigate aspects of Nature under controlled conditions. E.g., in Bell experiments one prepares entangled photon pairs by parametric down conversion with help of a laser shot at BBO crystals and filtering out precisely these entangled pairs.

A preparation procedure is empirically known to have produced specific statistical patterns in the past, and it is reasonable to assume that it will produce the same patterns in the future. Of course it is possible to apply probability theory to a "single" system, e.g. the throw of a die. But it's still statistical reasoning. The definiteness of the "state" of the "system" is more grounded in psychology than in the quantum formalism: your desire to describe it in a "Newtonian" way as a system evolving continuously with time, rather than a sequence of events.

With a single throw of a die you cannot say anything about the statistics. You have to throw the die many times in the same way to get statistics to empirically check predicted probability distributions. E.g., just knowing nothing about the die you simply assume that's a fair die and you assume P=1/6 for each of the possible outcomes. You can test this hypothesis of course only by repeating the experiment very often, i.e., probabilistic statements are about ensembles.

I know there's all this philosophical sophistication about some "Bayesian approach" to probabilities, but this is completely, well, philosophical, i.e., it doesn't say, how to figure out whether the predicted probabilities of some theory are correct for the given situation/preparation procedure.

Quantum theory has a vastly broader scope than experiments in the laboratory. (Consider for example the nuclear reactions in the sun.) Isn't it ridiculous to base the formulation of this microscopic theory on undefined (primitive) notions like state preparation and measurement? On the grounds that there is no other way to describe the real world?

There's no problem whatsoever to apply the formalism to the fusion processes in the Sun. This is done since Bethe et al in the 30ies. To confirm the "solar standard model", of course needed till the mid to late 1990ies, when neutrino mass and oscillation was discovered, but that's another story.