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

  • #251
Re/ the QZE: An ideal physicist would write down a suitable model Hamiltonian for the unstable atom + detector + EM field + environment, and then compute the relevant decay rates/probabilities for some time of interest. This "shut up and calculate" procedure isn't contingent on the acceptance or rejection of any interpretation, statistical, minimalist, or otherwise.

What makes QZE interesting is the challenge of shutting up and calculating: the construction of the model Hamiltonian and the computation of decay rates etc. From https://arxiv.org/pdf/quant-ph/0307075.pdf:
The latter condition explains why the QZE was not obtained for exponentially decaying systems by theories based on the projection hypothesis. By applying the projection operator, the quantum coherences between |x, 0〉 and |g, k〉’s are destroyed regardlessly of the energy of the emitted photon. Therefore, the projection-based theory corresponds to ∆ → ∞ [20]. In such a limit, however, inequality (10) cannot be satisfied and the QZE never occurs. Since ∆ of any real detector is finite, such a limit is rather unphysical.
Naively projecting here and there is not a feature of an interpretation. It's a feature of poor application of a theory.
 
Last edited:
  • Like
Likes mattt, gentzen and vanhees71
Physics news on Phys.org
  • #252
Demystifier said:
The purpose of interpretation is not the truth, the purpose is intuition. Intuition is a thinking tool.
I agree with this, but I'm not sure all physicists who are proponents of particular interpretations do. Many of them appear to think that their preferred interpretations are "true", not just useful thinking tools.
 
  • Like
Likes nnunn, Lord Jestocost, Demystifier and 1 other person
  • #253
PeterDonis said:
I agree with this, but I'm not sure all physicists who are proponents of particular interpretations do. Many of them appear to think that their preferred interpretations are "true", not just useful thinking tools.
Yes, and I believe I can explain why is that. At the beginning, when one hears about a particular interpretation for the first time, nobody thinks it's "true". At best, it looks like "OK, maybe there is some truth in it". But with this state of mind, the tool is not yet fully efficient, one must create a stronger bond with it to exploit its full potential. For many people, the thinking tool may work the best when one imagines that it is "really true". But when one comes to that psychological state of imagination, the problem is then to later step out from this state of mind. So after a while the imagination may start to look like reality. The map may start to look like the territory.

My own way to cope with this is changing the roles. For example, when I speak with very strong proponents of the Bohmian interpretation, I often present them the counterarguments, some of which I've learned in this forum.
 
  • #254
vanhees71 said:
And what do you think is wrong with this?
Let me put it this way. Do you know a reference where the real quantum Zeno effect (which is a measured effect) is explained theoretically without reference to any kind of collapse? I would really like to see how one can avoid it, in actual calculation.
 
  • #255
Can you refer to a reference, where "the real quantum Zeno effect" is demonstrated? I could imagine it's some atom in a cavity with a laser kicking it somehow to stay (maybe in principle for ever) in an excited/metastable state. I'm not sure, whether one can treat the full quantum-time evolution of this exactly.

The main objection against the hand-waving arguments using 1st-order perturbation theory for the transition rate for "infinitesimal times" and then treat the process as if it were Markovian is at least misleading.

The more we think about open quantum systems in our research work the more we come to the conclusion that Markovian approximations are making more trouble than good. Particularly it's not clear, how to ensure proper thermalization in the long-time limit, but that's a different topic.
 
  • #257
Demystifier said:
Let me put it this way. Do you know a reference where the real quantum Zeno effect (which is a measured effect) is explained theoretically without reference to any kind of collapse? I would really like to see how one can avoid it, in actual calculation.
Does collapse make sense in an ensemble interpretation?
 
  • #260
martinbn said:
Does collapse make sense in an ensemble interpretation?
Collapse doesn't make sense to me, because it claims that there is dynamics not described by QT and then you ad hoc assume some magic that violates relativistic causality. I have no clue, what the collapse postulate is needed for.
 
  • Like
Likes Lord Jestocost
  • #261
vanhees71 said:
Collapse doesn't make sense to me, because it claims that there is dynamics not described by QT and then you ad hoc assume some magic that violates relativistic causality. I have no clue, what the collapse postulate is needed for.
That is not what i am asking. In a non ensemble interpretation it is clear what collapse is, whether it is needed or not is a separate question. But my question is what is it in the ensemble case?
 
  • Like
Likes gentzen and vanhees71
  • #262
martinbn said:
Then what is collapse in an ensemble interprwtatoon?
Information updwatee. :oldbiggrin:
 
  • #263
But your information update should relate to what you look at. You cannot say that whenever you measure an observable the correct information update is that then the system is in an eigenstate of the measured observable with the eigenvalue given by the measured value. It depends on, how the system interacts with the measurement device, how to describe its state after the measurement. Sometimes the very object you were measuring is destroyed in the measurement process (like a photon hitting a photoplate). What sense then does the collapse postulate make?
 
  • #264
Demystifier said:
Information updwatee. :oldbiggrin:
No, i mean how is it formulated if we have ensembles and the wave function describes them, not the idividual system?
 
  • #265
vanhees71 said:
But your information update should relate to what you look at. You cannot say that whenever you measure an observable the correct information update is that then the system is in an eigenstate of the measured observable with the eigenvalue given by the measured value. It depends on, how the system interacts with the measurement device, how to describe its state after the measurement. Sometimes the very object you were measuring is destroyed in the measurement process (like a photon hitting a photoplate). What sense then does the collapse postulate make?
https://www.physicsforums.com/threads/difference-between-collapse-and-projection.998545/post-6445809
 
  • #266
martinbn said:
That is not what i am asking. In a non ensemble interpretation it is clear what collapse is, whether it is needed or not is a separate question. But my question is what is it in the ensemble case?
If you have some experiment involving a sequence of measurements, then an example of a "collapse" would be a partitioning of the ensemble of experimental runs into subensembles in accordance with the outcome of one of the measurements.
 
  • #267
What has this to do with the collapse postulate?
 
  • #268
martinbn said:
No, i mean how is it formulated if we have ensembles and the wave function describes them, not the idividual system?
Then the collapsed wave function describes a subensemble, in which all members show the same measurement outcome.
 
  • Like
Likes mattt, PeterDonis and gentzen
  • #269
Demystifier said:
Then the collapsed wave function describes a subensemble, in which all members show the same measurement outcome.
How is it a subensemble? In any case it is a different ensemble, so what is collapse exactly?
 
  • #270
vanhees71 said:
What has this to do with the collapse postulate?
If you ask me, the last formula describes a generalized collapse rule for general POVM measurements.
 
  • #271
martinbn said:
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.
 
Last edited:
  • #272
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.
 
  • #273
vanhees71 said:
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.
 
  • #274
Demystifier said:
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!!!
Demystifier said:
This means that you have an ensemble of 100 particles in the state ##|\psi\rangle##.
No, I have 100 representatives of that ensemble.
Demystifier said:
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.
Demystifier said:
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.
 
  • #275
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.
 
  • #276
vanhees71 said:
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.
 
  • #277
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.
 
  • Like
  • Skeptical
Likes gentzen, Quantum Waver and WernerQH
  • #278
martinbn said:
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.
 
  • #279
martinbn said:
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".
 
  • Like
Likes mattt, gentzen and Demystifier
  • #280
Demystifier said:
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.
 
  • #281
vanhees71 said:
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.
 
  • #282
vanhees71 said:
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.
 
  • Like
Likes mattt, WernerQH and martinbn
  • #283
martinbn said:
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?
 
  • #284
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.
 
  • Like
Likes mattt, gentzen and Demystifier
  • #285
Morbert said:
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?
Morbert said:
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?
 
  • #286
Demystifier said:
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.
 
  • #287
No, it doesn't. How do you come to this idea?
 
  • Like
Likes Demystifier
  • #288
vanhees71 said:
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.
vanhees71 said:
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.
 
  • #290
martinbn said:
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?
martinbn said:
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.
 
  • Like
Likes Demystifier
  • #291
vanhees71 said:
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.
 
  • Like
Likes mattt and martinbn
  • #292
vanhees71 said:
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.
vanhees71 said:
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.
 
  • #293
Demystifier said:
This is one of the happiest moments in my life, @vanhees71 and me agree against @martinbn . :partytime:
If being wrong makes you happy, congrats.
 
  • #294
martinbn said:
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.
 
  • #295
martinbn said:
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##.
 
  • #296
Demystifier said:
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?
 
  • #297
Demystifier said:
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.
 
  • #298
vanhees71 said:
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?!
 
  • #299
vanhees71 said:
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.
 
  • #300
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).
 
  • Like
Likes Demystifier
Back
Top