A Measurement problem in the Ensemble interpretation

[vanhees71]

Well, according to QT there's not more knowable about your particle than the probabilities for the outcome of measurements. The probabilities evolve from a given (prepared) initial to the state at the time of detection, and what we get (repeating the experiment with equal preparations) the distribution by measuring the observable we are interested in. According to the standard model of elementary particles the interactions are local. So your point (1-3) are all well included in the standard model: (1) is ensured by the conservation laws: If I prepare, e.g., some particle with a given charge, then at least this charge must exist all the time; (2) is implemented by construction in any local and microcausal relativistic QFT, (3) is just the only way to answer reasonable the "when" question. How else would you define the "time of detection"?

 

[Demystifier]

Well, according to QT there's not more knowable about your particle than the probabilities for the outcome of measurements. The probabilities evolve from a given (prepared) initial to the state at the time of detection, and what we get (repeating the experiment with equal preparations) the distribution by measuring the observable we are interested in. According to the standard model of elementary particles the interactions are local. So your point (1-3) are all well included in the standard model: (1) is ensured by the conservation laws: If I prepare, e.g., some particle with a given charge, then at least this charge must exist all the time; (2) is implemented by construction in any local and microcausal relativistic QFT, (3) is just the only way to answer reasonable the "when" question. How else would you define the "time of detection"?

I don't agree that standard QT ensures 1). The standard QT talks only about probabilities of measurement outcomes. In particular, the conservation of charge in standard QT says that if you measure charge at two times, then you will get the same number at both times. But it does not claim that charge will exist between the two measurements. It is an extra assumption which may or may not be true, but cannot be tested by experiment. It is a reasonable assumption indeed, but it cannot be strictly derived from principles of standard QT.

So, if you want to think in lines of standard QT, I think you should give up 1).

 

[RockyMarciano]

Symmetry principles aren't the goal. The goal is modeling the world. If the world obeys certain symmetry principles, then of course, we should find out what they are, but finding out symmetry principles isn't the goal.
That is the heart of the measurement problem. Why do macroscopic systems have well-defined macroscopic values?

Sorry to get back to this but aren't precisely the symmetry principles-conservation laws what makes macro systems have (approximately) well defined values by preserving measurements(i.e. measurability)?

 

[Demystifier]

(1) is ensured by the conservation laws: If I prepare, e.g., some particle with a given charge, then at least this charge must exist all the time

Another counterargument:
Suppose that we talk about barion charge (not electric charge) and suppose that, due to some GUT effects, there is a very small probability that the charge will not be conserved. The probability can be arbitrarily small, say $10^{-100}$, but it is not zero. For all practical purposes the charge can be considered conserved. In this case, would you say that charge must exist all the time? Or almost all the time? Or would you say that it comes to existence only when the detector clicks?

 

[vanhees71]

Well, then you could measure that baryon number is not conserved and then you could indeed only give a probability for still finding the same baryon number as you started with.

 

[Demystifier]

Well, then you could measure that baryon number is not conserved and then you could indeed only give a probability for still finding the same baryon number as you started with.

So you (seem to) introduce a step function. If probability of conservation is smaller than 1, then charge does not exist between measurements. If probability of conservation is strictly 1, then charge exists between measurements.

Don't you have a feeling that there should be a continuous transition, that the difference between 1 and 0.99999999 should not be so radical?

 

[Demystifier]

Well, then you could measure that baryon number is not conserved and then you could indeed only give a probability for still finding the same baryon number as you started with.

Another reason why your reasoning doesn't make sense (to me).

You essentially say that
1) Only conserved quantities exist between measurements.
2) Dynamics is local.
But the goal of dynamics is to describe the change, i.e. to describe the behavior of quantities which are not conserved. By 1), this means that the goal of dynamics is to describe the things which do not exist between measurements. So the fact that dynamics is local really means that dynamics of non-existing entities is local. It says nothing about locality or non-locality of existing entities. Therefore the fact that QFT has local dynamics is not an argument that existing entities obey local laws.

Indeed, the Bell theorem says essentially that if there are changing entities that exist between measurements, then their dynamics must be non-local. It is compatible with the conclusion above that local QFT dynamics is only dynamics on non-existing entities.

 

[Demystifier]

But the goal of dynamics is to describe the change, i.e. to describe the behavior of quantities which are not conserved.

Let me put this point forward.

As an extreme case, consider the Lagrangian
$$L(q,\dot{q})=0$$
This Lagrangian is invariant under any conceivable transformation, so by Noether theorem everything is conserved in this theory. In other words, in this theory, there is no dynamics at all.

This demonstrates my more general point that conservation laws show that some quantities are not dynamical. But the point of dynamics is to describe quantities which are dynamical. So conservation laws, although very useful, do not describe dynamics. Conservation laws describe non-dynamics; they tell us which of the potentially interesting quantities are not dynamical, so can be excluded from further dynamical considerations.

 

[RockyMarciano]

Let me put this point forward.

As an extreme case, consider the Lagrangian
$$L(q,\dot{q})=0$$
This Lagrangian is invariant under any conceivable transformation, so by Noether theorem everything is conserved in this theory. In other words, in this theory, there is no dynamics at all.

This demonstrates my more general point that conservation laws show that some quantities are not dynamical. But the point of dynamics is to describe quantities which are dynamical. So conservation laws, although very useful, do not describe dynamics. Conservation laws describe non-dynamics; they tell us which of the potentially interesting quantities are not dynamical, so can be excluded from further dynamical considerations.

How would you reconcile this with the fact that measurements must be possible in a dynamical world for science to make sense(for different local measurements to be coherent with one another) which seems to imply that at least there must be conservation laws for dynamical measuring tools?

 

[vanhees71]

So you (seem to) introduce a step function. If probability of conservation is smaller than 1, then charge does not exist between measurements. If probability of conservation is strictly 1, then charge exists between measurements.

Don't you have a feeling that there should be a continuous transition, that the difference between 1 and 0.99999999 should not be so radical?

I've not said what you seem to have understood. All I said was what holds for any unstable particle: You prepare it, and then with some probability it's decayed after a given time. That's all you can know in such cases. I still don't get, what should be a problem with that. To the contrary thanks to Q(F)T we have a theory to describe such decays very well.

If there's a conserved charge, at least you know that it will be there forever in the one or the other form. To be sure that a once prepared particle is always there, of course it must be stable, because if there is only the tiniest probability for its decay, then you can never be sure that it is still there after some time. That's why it's called unstable.

 

[Lord Jestocost]

Here is why it is problematic. You simultaneously assume that
1) The measured system (particle) exists even before measurement.
2) The dynamics is local.
3) The random decision happens when the detector clicks (not before).
Indeed, each assumption by itself seems reasonable. But the problem is that they cannot all be simultaneously true. At least one must be wrong. You must give up at least one of them.

Let me explain why they cannot all be true. From 3) and 1) it follows that, immediately before the click, the system exists not only near one detector, but near both of them. But then, puff, at the time of click, the system suddenly ceases to exist near the detector that didn't click. How did this part of the system knew that the click happened near the other part? Since the two parts are spatially separated, there must have been some non-local (even if random) mechanism, which contradicts 2). Hence assumptions 1) and 3) contradict 2), which implies that it is not possible that all three assumptions are true.

And yet, you seem not be ready to give up any of the three assumptions. That's the problem.

Note that the argument above is even simpler than the Bell theorem, because the system studied above does not involve entanglement. The Bell theorem derives a contradiction by assuming 1), 2) and entanglement. The argument above derives a contradiction by assuming 1), 2) and 3).

To my mind, assumptions 1) and 2) are misleading when thinking about quantum phenomena. These assumptions are based on classical conceptions.

Regarding assumption 3), I follow J . Marburger: „We can only measure detector clicks. But when we hear the click we say “there’s an electron!” We cannot help but think of the clicks as caused by little localized pieces of stuff that we might as well call particles. This is where the particle language comes from. It does not come from the underlying stuff, but from our psychological predisposition to associate localized phenomena with particles.“ (J. Marburger, “On the Copenhagen interpretation of quantum mechanics” in Symposium on The Copenhagen Interpretation: Science and History on Stage, National Museum of Natural History of the Smithsonian Institution, 2 March 2002)

 

[Demystifier]

How would you reconcile this with the fact that measurements must be possible in a dynamical world for science to make sense(for different local measurements to be coherent with one another) which seems to imply that at least there must be conservation laws for dynamical measuring tools?

I don't see how this implies a need for conservation laws.

 

[RockyMarciano]

I don't see how this implies a need for conservation laws.

I mean that something must be conserved for measurements being valid regardless where and when they are performed and how(at which energy, etc), i.e. for the physics not to depend on any special factor in a dynamical or changing context.

 

[Demystifier]

I mean that something must be conserved for measurements being valid regardless where and when they are performed and how(at which energy, etc), i.e. for the physics not to depend on any special factor.

Well, to measure a distance with a meter, the length of meter should not change. But there is no law of conservation of length. What we need here is stability, not conservation laws.

 

[RockyMarciano]

Well, to measure a distance with a meter, the length of meter should not change. But there is no law of conservation of length. What we need here is stability, not conservation laws.

how do you maintain this stability without conservation laws in a dynamical context?

 

[Demystifier]

how do you maintain this stability without conservation laws in a dynamical context?

The meter can exchange energy with the environment (e.g. it can absorb heat), so energy of the meter is not conserved. Yet, the meter as a solid object is stable. If the meter was made from liquid it would not be stable (and not useful as a meter) even when it's energy is constant. This demonstrates that stability and energy conservation are not directly related to each other.

More formally, consider a particle in a potential of the form
$$V(x,t)=\frac{kx^2}{2}+U(t)$$
where $U(t)$ is a positive non-constant function of time $t$. Clearly this potential does not conserve energy. Yet, the particle position $x=0$ is stable, provided that $k$ is positive. If $k$ were negative the position $x=0$ would not be stable, even if $U(t)$ were zero. That's another demonstration that stability and energy conservation are not related.

 

[Demystifier]

I still don't get, what should be a problem with that. To the contrary thanks to Q(F)T we have a theory to describe such decays very well.

If there's a conserved charge, at least you know that it will be there forever in the one or the other form. To be sure that a once prepared particle is always there, of course it must be stable, because if there is only the tiniest probability for its decay, then you can never be sure that it is still there after some time. That's why it's called unstable.

Let me try to explain the problem once again. The conserved quantities (charge, energy, ...) do not change. But the role of dynamics is to describe the change. There are two types of change in QFT:
1) Changes of probabilities between detections. Those are described by local deterministic laws.
2) Clicks of detectors. Those are described by non-deterministic laws.
The 2) clearly does not reflect all the properties of 1), because 1) is deterministic and 2) is not. So, given that 2) is so fundamentally different from 1), what makes you think that 2) must be local?

 

[vanhees71]

Now it's totally confusing. All that QT gives me is, given the preparation of the state initially, the probabilities for finding a certain value for any observable possible for the system. The click of a detector is such a measurement (e.g., for the presence of a particle in the detector). Of course, I don't know more than the probability for it to click. The click is due to interactions of the particle with the particles in the detector, governed by the same laws, so it's local.

 

[Demystifier]

The click is due to interactions of the particle with the particles in the detector, governed by the same laws, so it's local.

If it's due to the same laws, how can it be one of them are deterministic and other non-deterministic?

Or do you deny that unitary evolution of probability between measurements is deterministic?

Or do you claim that even between measurements something changes in a non-deterministic way? If so, what is it?

 

[vanhees71]

I don't understand what you mean by deterministic then. The state implies only probabilities. So there's no deterministic content in it (except for the case of a precisely determined value of an observable). The (ideal) detector clicks with the probability given by the state according to Born's rule.

 

[Demystifier]

I don't understand what you mean by deterministic then. The state implies only probabilities. So there's no deterministic content in it (except for the case of a precisely determined value of an observable). The (ideal) detector clicks with the probability given by the state according to Born's rule.

By unitary evolution, if you know probability $P(t)$ for some $t$, then you can calculate the probability $P(t+\Delta t)$. That's deterministic evolution of probability. Between the clicks, probability evolves deterministically. At the moment of click, it is not so obvious whether it does or not.

Nevertheless, try to answer my last question in the post above:
Do you claim that even between measurements something changes in a non-deterministic way? If so, what is it?

 

[vanhees71]

Of course between the clicks you have unitary time evolution. It's misleading to call it "deterministic", but I know what you mean. I don't know what you mean with the question whether the click is deterministic of not. It's the measurement, and about the measurement I only know probabilities. That's the whole point of saying QT is a probabilistic description, and what's probabilistic is not determined (except that the probability for the outcome of the measured observable is 100%).

 

[Demystifier]

Of course between the clicks you have unitary time evolution.

And how about the clicks themselves? Are they unitary too?

It's the measurement, and about the measurement I only know probabilities.

And how about non-measurements? Can you say anything about probabilities of non-measured quantities?

 

[vanhees71]

What do you mean by the "clicks are unitary" or "clicks are not unitary"?

As a physicist I don't need to talk about non-measured quantities since if I don't measure them, what should I be able to say about them?

 

[Demystifier]

What do you mean by the "clicks are unitary" or "clicks are not unitary"?

You said: "Of course between the clicks you have unitary time evolution."
In the same sense you meant that, I ask you: Do I also have unitary time evolution at the time of clicks?

As a physicist I don't need to talk about non-measured quantities since if I don't measure them, what should I be able to say about them?

If so, then why do you keep saying that there is conserved charge in the absence of measurement? Why don't you say that you don't need to talk about conserved charge in the absence of measurement? It looks as if you use double standards.

 

[RockyMarciano]

the meter as a solid object is stable.

Are you referring only to classical theory? Because this doesn't seem to be a valid assertion in the quantum realm, at least if we go by its theoretical principles. A solid meter is most likely made up of atoms joined by chemical bonds that act as springs with a ground state energy that fluctuates, the corresponding uncertainty in the length of the spring makes the separation between atoms at each step not well defined so that they shouldn't add up to a fixed and stable expected distance between marks on the meter and therefore it can't justify a robust measure remaining stable independently of how and when it is used as a measuring tool.

Of course in practice these shortcomings are overcome by obtaining a measurement that gives a defined distance that allows to introduce an idealized meter and the atomic fluctuations only produce a minor blurring for the position of each atom(for instance in x-ray scattering).

This demonstrates that stability and energy conservation are not directly related to each other.

You would have to show from first principles how the meter is stable taking into account the ground state energy fluctuations.

More formally, consider a particle in a potential of the form
$$V(x,t)=\frac{kx^2}{2}+U(t)$$
where $U(t)$ is a positive non-constant function of time $t$. Clearly this potential does not conserve energy. Yet, the particle position $x=0$ is stable, provided that $k$ is positive. If $k$ were negative the position $x=0$ would not be stable, even if $U(t)$ were zero. That's another demonstration that stability and energy conservation are not related.

See above.

 

[vanhees71]

You said: "Of course between the clicks you have unitary time evolution."
In the same sense you meant that, I ask you: Do I also have unitary time evolution at the time of clicks?If so, then why do you keep saying that there is conserved charge in the absence of measurement? Why don't you say that you don't need to talk about conserved charge in the absence of measurement? It looks as if you use double standards.

Sure, but you cannot evaluate it in practice since the detector is a macroscopic device. All you are interested in is a macroscopic very coarse-grained obsevable (in this case simply "click" or "no click").

The argument with conserved charge was to the question, why a particle is with certainty there. As I argued that's of course the case only for stable particles, and that usually conservation laws forbid its decay. If the particle is unstable, of course it decays with some probability and you cannot with certainty say whether it's still there but only give the probability of its survival. I don't know, why all of this is a "measurement problem". If you just stick to the minimal interpretation, there's never a contradiction. QT seems to be a pretty logically consistent probabilistic description of nature. It's also very successful, i.e., it's tested very well against observations, and the loopholes concerning the possibility of some local deterministic description are more and more closed too. So if you want to get back to a deterministic theory, you'd have to invent something non-local, and that seems to be very difficult, because so far nobody has come up with a convincing model. Maybe Bohmian mechanics is the most convincing, but on the other hand there seem to be predictions of "trajectories" that cannot be verified by experiment.

 

[Demystifier]

So if you want to get back to a deterministic theory, you'd have to invent something non-local

My main objection concerns the claim above. If by "deterministic" you mean the opposite of probabilistic, then, I claim, even without determinism you need something non-local. That's what I am repeatedly trying to explain to you in various ways, and that's what even Ballentine in his book explains in his own way. But somehow you fail to grasp any argument in that direction, because you always and up with an argument of the form: "The QFT dynamics is local" (which is true) "and hence we don't need anything non-local" (which is at least doubtful).

 

[vanhees71]

This I don't understand. Classical electrodynamics (with a classical continuum description for the charged matter) is a local deterministic theory par excellance. Why, in your opnion, do I need to get non-local even in the non-quantum context?

The other argument is related to standard relativistic QFT, which is indeed local and probabilistic. So far we don't need anything non-local, because QFT (even the Standard Model) is very successful in describing all observed facts.

Here we discuss something else, namely possible theories going beyond standard Q(F)T, maybe deterministic ones. In the latter case, imho it's pretty clear that we'd need a non-local formulation if you want to have a deterministic theory that can describe what's described by entanglement in QFT.

 

[Demystifier]

Why, in your opnion, do I need to get non-local even in the non-quantum context?

I didn't say that we need it in non-quantum context. Why do you think I did?

So far we don't need anything non-local, because QFT...

And I disagree. I claim that even QFT has something implicit non-local in it. But I cannot explain it to you without repeating my arguments which you failed to grasp.