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

[Demystifier]

How is it a subensemble? In any case it is a different ensemble, so what is collapse exactly?

Suppose that you prepare 100 particles, each in the state
$$|\psi\rangle =|+\rangle +|-\rangle$$
This means that you have an ensemble of 100 particles in the state $|\psi\rangle$. Now suppose that you perform a measurement in the $|\pm\rangle$ basis and find that 48 particles (out of 100) are in the state $|+\rangle$. This means that these 48 particles constitute a subensemble (it's a subensemble because it does not refer to all 100 of them), each in the collapsed state
$$|\psi'\rangle = \pi_+ |\psi\rangle = |+\rangle$$
where $\pi_+=|+\rangle\langle+|$ is the projector.

 

[vanhees71]

I thought POVM descriptions give probabilities as also do the usual idealized von Neumann measurements, which are a special case of a POVM. I still don't understand what either kind of measurements should have to do with the general collapse postulate, which claims (for a von Neuman measurement) that, when you measure an observable $\hat{A}$ with the outcome $a$ (eigenvalue of $\hat{A}$), of a system prepared in the state $\hat{\rho}$ then after the measurement you have to describe the system as being prepared in the state
$$\hat{\rho}'=\frac{\hat{P}_a \hat{\rho} \hat{P}_a}{\mathrm{Tr} \hat{P}_a \hat{\rho} \hat{P}_a}$$
with $\hat{P}_a$ the projector to the eigenspace $\text{Eig}(\hat{A},a)$.
Some similar "collapse postulate" can for sure also be formulated for general POVMS.

In either case, I don't understand, why should this be true for all measurements? For sure it's not true for the drastic example of a photon being absorbed in the measurement process.

 

[Demystifier]

In either case, I don't understand, why should this be true for all measurements? For sure it's not true for the drastic example of a photon being absorbed in the measurement process.

All measurements are POVM measurements. In the linked post above, I explained how it works for photon absorption. You even liked it back then, and it looked as if you understood it.

 

[martinbn]

Suppose that you prepare 100 particles, each in the state
$$|\psi\rangle =|+\rangle +|-\rangle$$

No, absolutely not. If you say each in the state, that is not an ensemble interpretation!!!

This means that you have an ensemble of 100 particles in the state $|\psi\rangle$.

No, I have 100 representatives of that ensemble.

Now suppose that you perform a measurement in the $|\pm\rangle$ basis and find that 48 particles (out of 100) are in the state $|+\rangle$.

Again if the particles are in the state, you are not talking about ensemble interpretation.

This means that these 48 particles constitute a subensemble, each in the collapsed state
$$|\psi'\rangle = \pi_+ |\psi\rangle = |+\rangle$$
where $\pi_+=|+\rangle\langle+|$ is the projector.

I have no problem understang collaps in non ensemble interpretations. I want to know if it.makes sense in the ensemble interpretations.

 

[vanhees71]

But there you explained it as a transition matrix, which is quantum dynamics and not collapse, and that's why I liked it. Maybe I've misunderstood you then, because now you claim it's a collapse, i.e., some "update rule" outside of quantum dynamics.

 

[Demystifier]

But there you explained it as a transition matrix, which is quantum dynamics and not collapse, and that's why I liked it. Maybe I've misunderstood you then, because now you claim it's a collapse, i.e., some "update rule" outside of quantum dynamics.

Yes, you misunderstood me. It's a collapse (though not a projective collapse), it's not quantum dynamics.

 

[Lord Jestocost]

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.”

 

[Demystifier]

No, absolutely not. If you say each in the state, that is not an ensemble interpretation!!!

No, I have 100 representatives of that ensemble.

Again if the particles are in the state, you are not talking about ensemble interpretation.

I have no problem understang collaps in non ensemble interpretations. I want to know if it.makes sense in the ensemble interpretations.

Then I have no idea what do you mean by ensemble interpretation, so I cannot answer your question.

 

[vanhees71]

No, absolutely not. If you say each in the state, that is not an ensemble interpretation!!!

No, I have 100 representatives of that ensemble.

Again if the particles are in the state, you are not talking about ensemble interpretation.

I have no problem understang collaps in non ensemble interpretations. I want to know if it.makes sense in the ensemble interpretations.

Of course, you must be able to say (to some accuracy at least) that each single realization of the system is prepared in this state. Otherwise you cannot build the ensembles we are talking about to begin with. The probabilistic meaning refers to the outcome of measurements, the state itself refers to a preparation procedure.

E.g., if you want to do a double-slit experiment with, e.g., an electron, you must prepare the electron with a sufficiently accurate momentum running towards the double slit, i.e., then it's prepared in a state $\hat{\rho}=|\psi_{\vec{P}} \rangle \langle \psi_{\vec{P}}|$ with
$$|\psi_{\vec{P}} \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} g(\vec{p}-\vec{P}) |\vec{p} \rangle,$$
where $g$ is some normalized $\mathrm{L}^2(\mathbb{R}^3)$ function, sharply peaked around $\vec{0}$.

Then you ask, where the electron will be detected behind the two slits, and of course QT gives you only probability-density distributions and not a determined spot, where the electron will be registered. To test this, you have to prepare an ensemble of equally prepared electrons, justifying that this ensemble is described by the function $g$. This is realized (with sufficient accuracy) by the "electron source".

 

[martinbn]

Then I have no idea what do you mean by ensemble interpretation, so I cannot answer your question.

If you dont know what an ensemble interpretation is, then you shouldnt make statments about it.

 

[martinbn]

Of course, you must be able to say (to some accuracy at least) that each single realization of the system is prepared in this state.

May be as a sloppy way of phrasing it. But strictly you cannot. That is the point of being an ensemble interpretation.

 

[Morbert]

Of course, you must be able to say (to some accuracy at least) that each single realization of the system is prepared in this state.

This is fine to say loosely since everyone will know what you mean. But strictly, according to a minimalist ensemble interpretation, we don't assign any significance to a single system. $|\psi\rangle$ only refers to an ensemble of identically prepared systems, not the state of each member of the ensemble, though the distinction is not always important.

This is also why collapse is not real under this interpretation. Arguably, even time-evolution is not real. All that is physical is the preparation and the dataset once the measurements or sequences of measurements are concluded.

 

[Demystifier]

If you dont know what an ensemble interpretation is, then you shouldnt make statments about it.

I think I know what it is (I think @vanhees71 agrees with me on that), I just don't know what you mean by it. Perhaps you would like to explain it to us?

 

[vanhees71]

Obviously it's not clear what "the ensemble interpretation" is since you also deny that it is possible to prepare single representants of an ensemble in a given state. If this were true, you couldn't use QT at all to describe real-world experiments. That's obviously wrong.

For me the ensemble interpretation means that the only meaning of the quantum state are to provide probabilities for the outcome of measurements on equally prepared systems. The preparation procedure refers to single instances that constitute the ensemble.

 

[vanhees71]

This is fine to say loosely since everyone will know what you mean. But strictly, according to a minimalist ensemble interpretation, we don't assign any significance to a single system. $|\psi\rangle$ only refers to an ensemble of identically prepared systems, not the state of each member of the ensemble, though the distinction is not always important.

This is self-contradictory. How can you prepare each single system "identically" such that it is described by some state $\hat{\rho}$? To build an ensemble you must be able to prepare each single system in such a way to build this ensemble, right?

This is also why collapse is not real under this interpretation. Arguably, even time-evolution is not real. All that is physical is the preparation and the dataset once the measurements or sequences of measurements are concluded.

Of course time-evolution is real. How else do you explain that standard S-matrix theory works in HEP?

 

[martinbn]

I think I know what it is (I think @vanhees71 agrees with me on that), I just don't know what you mean by it. Perhaps you would like to explain it to us?

You say that a single particle has the givne state. That contradicts the main idea of the ensemble interpretation.

 

[vanhees71]

No, it doesn't. How do you come to this idea?

 

[martinbn]

Obviously it's not clear what "the ensemble interpretation" is since you also deny that it is possible to prepare single representants of an ensemble in a given state. If this were true, you couldn't use QT at all to describe real-world experiments. That's obviously wrong.

You can prepare representatives, but they are not in the state. Only ensembles can have states.

For me the ensemble interpretation means that the only meaning of the quantum state are to provide probabilities for the outcome of measurements on equally prepared systems. The preparation procedure refers to single instances that constitute the ensemble.

This sounds more like Copenhagen.

 

[Demystifier]

This is one of the happiest moments in my life, @vanhees71 and me agree against @martinbn .

 

[vanhees71]

You can prepare representatives, but they are not in the state. Only ensembles can have states.

They are representatives of the state you use to describe the ensemble, or how else do you explain the tremendous success of QT in application to real-world observations of Nature?

This sounds more like Copenhagen.

That may well be. The Copenhagen interpretation cover such a wide range of unsharp defined interpretations that nearly all interpretation fall into that category. For me even the ensemble interpretation is quite in the Copenhagen spirit.

 

[Morbert]

This is self-contradictory. How can you prepare each single system "identically" such that it is described by some state $\hat{\rho}$? To build an ensemble you must be able to prepare each single system in such a way to build this ensemble, right?

Given some equivalence class of preparations, you can of course prepare an individual system, but it has no significance. All that is significant is an ensemble of such systems.

Of course time-evolution is real. How else do you explain that standard S-matrix theory works in HEP?

My wording here was probably too strong. What I mean is the significance of time-evolution is the establishment of the frequencies and correlations we expect to see in an experimental dataset, based on the preparation, as opposed to a fine-grained account of the history of a system.

 

[martinbn]

They are representatives of the state you use to describe the ensemble, or how else do you explain the tremendous success of QT in application to real-world observations of Nature?

Yes, this is true. But the individual particles are not in the state. The state does not describe induviduals, just ensembles.

That may well be. The Copenhagen interpretation cover such a wide range of unsharp defined interpretations that nearly all interpretation fall into that category. For me even the ensemble interpretation is quite in the Copenhagen spirit.

 

[martinbn]

This is one of the happiest moments in my life, @vanhees71 and me agree against @martinbn .

If being wrong makes you happy, congrats.

 

[Demystifier]

Yes, this is true. But the individual particles are not in the state. The state does not describe induviduals, just ensembles.

The state describes what we know about the individuals, this is common to Copenhagen, ensemble and Bohmian interpretation. The three interpretations only differ on whether this knowledge is complete or not.

 

[vanhees71]

Yes, this is true. But the individual particles are not in the state. The state does not describe induviduals, just ensembles.

That doesn't make sense. Just take the definition by Ballentine in his textbook:

Postulate 2. To each state there corresponds a unique state operator. The
average value of a dynamical variable $\hat{R}, represented by the operator$\hat{R}, in
the virtual ensemble of events that may result from a preparation procedure
for the state, represented by the operator $\hat{\rho}$, is
$$\langle R \rangle = \frac{1}{\mathrm{Tr} \hat{\rho}} \mathrm{Tr} (\hat{R} \hat{\rho}).$$

Of course he goes on to make the usual definition for the state operator (or statistical operator): It must be a positive semidefinite self-adjoint operator with a finite trace, and it's customary to properly normalize it right away such that $\mathrm{Tr} \hat{\rho}=1$.

 

[vanhees71]

The state describes what we know about the individuals, this is common to Copenhagen, ensemble and Bohmian interpretation. The three interpretations only differ on whether this knowledge is complete or not.

But isn't it complete in any of these interpretations? The Bohmian trajectories are just deriable from the state, i.e., they don't provide any additional "knowledge" not already contained in the states, right?

 

[martinbn]

The state describes what we know about the individuals, this is common to Copenhagen, ensemble and Bohmian interpretation. The three interpretations only differ on whether this knowledge is complete or not.

Well, no, it is not in the ensemble interpretation.

 

[martinbn]

That doesn't make sense. Just take the definition by Ballentine in his textbook:Of course he goes on to make the usual definition for the state operator (or statistical operator): It must be a positive semidefinite self-adjoint operator with a finite trace, and it's customary to properly normalize it right away such that $\mathrm{Tr} \hat{\rho}=1$.

Where does he say that the state describes the individual?!

 

[Demystifier]

But isn't it complete in any of these interpretations? The Bohmian trajectories are just deriable from the state, i.e., they don't provide any additional "knowledge" not already contained in the states, right?

The Bohmian trajectory (singular, not plural) for a single particle is derivable from the state, given its initial position. But the initial position is not derivable, so the initial position is the additional information not present in Copenhagen and ensemble interpretations.

 

[vanhees71]

Then please give a reference, where your flavor of the ensemble representation is clearly stated. Obviously it's not the one, Ballentine defines in his book (and already in his RMP).