I Role of the collapse in the instrumentalist interpretation

[timmdeeg]

TL;DR Summary

Which meaning has the collapse of the wavefunction in the instrumentalist interpretation?

As I understand it, from the instrumentalist perspective the wavefunction is not more than a mathematical tool which predicts probabilities. So he could say a mathematical tool can't collapse because it is not a real physical thing. So to talk about a collapse of the wavefunction is meaningless.

Or is there still something else to consider?

 

[PeterDonis]

he could say a mathematical tool can't collapse because it is not a real physical thing. So to talk about a collapse of the wavefunction is meaningless.

No. In the instrumentalist interpretation, "collapse" means the mathematical process of applying the von Neumann projection postulate when you know the result of a measurement. You have to do that to make correct predictions about the probabilities of future measurements, so it's certainly not meaningless.

 

[Demystifier]

In the instrumentalist interpretation one can think of collapse as an update of information which changes the conditional probability. So the collapse is still there.

 

[timmdeeg]

Thanks for your answers.

So are there two kinds of collapses, a realistic collapse where something real (the wavefunction interpreted to be ontic) contracts instantaneously or abstractly a "mathematical" collapse due to the "von Neumann projection postulate"? Whereby the former is in conflict with SR. Is the von Neumann projection postulate indisputable?

 

[timmdeeg]

In the instrumentalist interpretation one can think of collapse as an update of information which changes the conditional probability. So the collapse is still there.

Is this argumentation roughly the same as to say a not well defined state (spin, polarisation, path,...) reduces to a defined state?

 

[Demystifier]

Is this argumentation roughly the same as to say a not well defined state (spin, polarisation, path,...) reduces to a defined state?

Roughly, yes.

 

[Demystifier]

So are there two kinds of collapses, a realistic collapse where something real (the wavefunction interpreted to be ontic) contracts instantaneously or abstractly a "mathematical" collapse due to the "von Neumann projection postulate"? Whereby the former is in conflict with SR. Is the von Neumann projection postulate indisputable?

Yes and yes.

 

[martinbn]

Thanks for your answers.

So are there two kinds of collapses, a realistic collapse where something real (the wavefunction interpreted to be ontic) contracts instantaneously or abstractly a "mathematical" collapse due to the "von Neumann projection postulate"? Whereby the former is in conflict with SR. Is the von Neumann projection postulate indisputable?

I would say no and yes, because the first comes with a lot of issues on its own.

 

[atyy]

So are there two kinds of collapses, a realistic collapse where something real (the wavefunction interpreted to be ontic) contracts instantaneously or abstractly a "mathematical" collapse due to the "von Neumann projection postulate"? Whereby the former is in conflict with SR. Is the von Neumann projection postulate indisputable?

Conflict with classical SR causality is not a good reason for rejecting realistic collapse. Classical SR causality is not compatible with quantum mechanics. The reason for rejecting collapse is that it depends on the observer, and it seems absurd (but not incoherent) to imagine the physical collapse depends on the subjective assignment of an observer as to when a measurement occurs.

 

[vanhees71]

SR causality is compatible with quantum mechanics as the successful local relativistic QFT (building the Standard Model of elementary particle physics) shows.

I don't know, what you mean by "classical" causality. Physics, including QT, assumes of course causality, because otherwise the study of general physical laws wouldn't make sense to begin with.

The fundamental distinguishing features between classical and quantum physics is rather determinism, i.e., classical physics by assumption is deterministic while quantum physics taught us to give up determinism.

 

[timmdeeg]

The reason for rejecting collapse is that it depends on the observer,

Unfortunately I am not sure how to understand this sentence. What difference does it make whether or not an observer is present during the measurement?
Isn't the reason for rejecting the realistic collapse because this notion seems unphysical (because it occurs instantaneously)?

 

[PeterDonis]

Isn't the reason for rejecting the realistic collapse because this notion seems unphysical (because it occurs instantaneously)?

Not all intepretations reject a realistic collapse, and not all interpretations that do reject it reject it on these grounds. The MWI, for example, rejects a realistic collapse because its interpretation simply doesn't require one--it says that the actual, physical state of the system evolves by unitary evolution all the time, and that's it.

 

[atyy]

Unfortunately I am not sure how to understand this sentence. What difference does it make whether or not an observer is present during the measurement?
Isn't the reason for rejecting the realistic collapse because this notion seems unphysical (because it occurs instantaneously)?

QM has unitary time evolution between measurements, and collapse at the point of measurement. However, QM does not specify when measurements occur. That has to be put in by hand, by the "observer".

 

[timmdeeg]

QM has unitary time evolution between measurements, and collapse at the point of measurement. However, QM does not specify when measurements occur. That has to be put in by hand, by the "observer".

Sorry, perhaps a silly question, but why can't the "observer" be replaced by a machine? Or does this make no difference because it doesn't matter how the occurrence of measurements is specified?

And yet, how about a natural measurement , e.g. the wavefunction of a photon collapses as it hits an atom somewhere at some time?

 

[vanhees71]

Of course, in almost all cases the "observer" is a machine, and "collapse" is nothing else than the adaption of the probabilistic description by an observer given new information.

According to relativistic QFT there cannot be an instantaneous collapse, i.e., a instaneous causal effect over all space, due to a local interaction between the system and a measurement device.

It's also clear from the mathematical formalism. Suppose there is a system in an arbitrary state $\hat{\rho}$ (statistical operator) and now you measure an observable $A$, described by a self-adjoint operator $\hat{A}$ and eigenstates $|a,\alpha \rangle$. Here $a$ runs through the spectrum ("eigenvalues") of the operator $\hat{A}$ and $\alpha$ is some label indicating the orthonormalized basis of the eigenspace $\text{Eig}(\hat{A},a)$ of this eigenvalue.

Now suppose you know you have measured $A$ on the system in some non-destructive way. Now, consider two scenarios:

(a) We know that the measurement of the above kind has happened but we don't have read off the measured value.

(b) We know that the measured value is $a$ (one of the spectral values of $\hat{A}$).

What are the states to be associated in this two scenarios?

For case (a) we know that the system has been measured but we don't know which value has been found. Then according to the orthodox laws we now have to describe the state by the statistical operator
$$\hat{\rho}'=\sum_{a,\alpha} P_{a,\alpha} |a,\alpha \rangle \langle a,\alpha| \quad \text{with} \quad
p_{a,\alpha}=\langle a,\alpha|\hat{\rho} a,\alpha \rangle.$$

For case (b) we know that the measured value is $a$. Then we have to use the statistical operator
$$\hat{\rho}''=\frac{1}{Z} \sum_{\alpha} P_{a,\alpha} |a,\alpha \rangle \langle a,\alpha|, \quad \text{with} \quad Z=\sum_{\alpha} P_{a,\alpha}.$$
In the extreme case that $\mathrm{dim} \mathrm{Eig}(\hat{A},a)=1$ you get a pure state $\hat{\rho}''=|a \rangle \langle a|$.

One should note that the two associtations of the "state after the measurement" result from the same physical manipulations done to take the measurement but also depend on what we know about the system from this measurement, as usual in applied probability theory (statistics).

 

[atyy]

Sorry, perhaps a silly question, but why can't the "observer" be replaced by a machine? Or does this make no difference because it doesn't matter how the occurrence of measurements is specified?

And yet, how about a natural measurement , e.g. the wavefunction of a photon collapses as it hits an atom somewhere at some time?

You can call the observer a "machine" or "pink elephant" or "aksjdakj8" if you like. Apart from "observer" or "measurement apparatus", another traditional term is that it is a "classical apparatus". The important point is that the observer is not included in the quantum state. One needs something outside the quantum state - or - if there is only the quantum state, then one may attempt a Many-Worlds Interpretation.

 

[atyy]

According to relativistic QFT there cannot be an instantaneous collapse, i.e., a instaneous causal effect over all space, due to a local interaction between the system and a measurement device.

This is not true. Relativistic QFT is consisent with a physical, instantaneous collapse.

 

[vanhees71]

How? In relativistic QFT the Hamilton density commutes with any local observables at spacelike distances. So how should a local measurement have causal effects at spacelike separated events?

 

[atyy]

How? In relativistic QFT the Hamilton density commutes with any local observables at spacelike distances. So how should a local measurement have causal effects at spacelike separated events?

If we say the collapse is physical, no predictions of the theory change. So a physical collapse is consistent with relativistic QFT.

If we say the collapse is not physical, no predictions of the theory change. So a non-physical collapse is also consistent with relativistic QFT.

 

[Morbert]

You can call the observer a "machine" or "pink elephant" or "aksjdakj8" if you like. Apart from "observer" or "measurement apparatus", another traditional term is that it is a "classical apparatus". The important point is that the observer is not included in the quantum state. One needs something outside the quantum state - or - if there is only the quantum state, then one may attempt a Many-Worlds Interpretation.

We can include the observer/apparatus in the quantum state while still maintaining an instrumentalist interpretation, although it is not typically useful to do so. E.g. Say we are interested in measuring the observable $A = \sum_i a_i \Pi_{a_i}$ of system $s$ at time $t$ using apparatus $m$ (where $a_i$ and $\Pi_{a_i}$ are eigenvalues and corresponding projectors respectively). As instrumentalists, we are interested in computing the relative frequencies of the possible measurement results. So we compute the probabilities $$p(a_i) = \mathrm{Tr}\left[\rho_s\Pi_{a_i,t}\right]$$But we could also include the apparatus explicitly in our description and compute $$p(a_i) = \mathrm{Tr}\left[\rho_s\otimes\rho_m\Pi_{a_i,t}\right]$$Or we could define some observable of the apparatus (like a monitor reading or dial position) $E = \sum_i\epsilon_i\Pi_{\epsilon_i}$ that registers the result and compute the probabilities for a particular outcome $$p(\epsilon_i) = \mathrm{Tr}\left[\rho_s\otimes\rho_m\Pi_{\epsilon_i,t}\right]$$All three calculations will generate the correct relative frequencies, though of course the first is the simplest.

 

[atyy]

We can include the observer/apparatus in the quantum state while still maintaining an instrumentalist interpretation, although it is not typically useful to do so.

If you do that, you need one more observer to observe the "observer".

 

[Morbert]

If you do that, you need one more observer to observe the "observer".

Could you expand on this? I don't see how it follows.

 

[atyy]

Could you expand on this? I don't see how it follows.

In the indirect measurement framework, you put the apparatus in the quantum state. But there are no outcomes unless a measurement is made, so one still needs an observer to measure the apparatus.

 

[Demystifier]

How? In relativistic QFT the Hamilton density commutes with any local observables at spacelike distances. So how should a local measurement have causal effects at spacelike separated events?

This is like asking how the QFT Hamiltonian, which gives a deterministic evolution of the state, be compatible with the probabilistic interpretation? The point is that the probabilistic outcomes are not created by the Hamiltonian. Those probabilistic outcomes can be described by a collapse postulate, or a projection postulate, or an update postulate, or whatever you want to call it. Of course, philosophically, a collapse is not the same thing as an update. But there is no measurable difference between a collapse and an update, so whatever argument you use that update is good and collapse is not, your argument is philosophical and in that sense not strictly scientific.

 

[vanhees71]

I find it very scientific to choose the interpretation of a theory which doesn't contradict the mathematical formulation it is based on. If this is philosophy, for the first time, I see a merit of philosophy for science ;-).

 

[Morbert]

In the indirect measurement framework, you put the apparatus in the quantum state. But there are no outcomes unless a measurement is made, so one still needs an observer to measure the apparatus.

In the scenario I am considering, it is both the case that the apparatus can be described within a quantum theoretic framework, and that the apparatus is measuring the observable $A$ of $s$. The data generated by the instrument constitute measurement outcomes if i) for a given datum $\epsilon_i$ and measurement result $a_i$ $$p(a_i\cap\epsilon_i) = p(\epsilon_i) = p(a_i)$$ ie $$\mathrm{Tr}\left[\rho_s\otimes\rho_m\Pi_{a_i,t}\Pi_{\epsilon_i,t}\right] = \mathrm{Tr}\left[\rho_s\otimes\rho_m\Pi_{\epsilon_i,t}\right] = \mathrm{Tr}\left[\rho_s\Pi_{a_i,t}\right]$$ and ii) $\{\epsilon_i\}$ are sufficiently approximated by classical physics. You don't need some secondary apparatus to justify explicit computation of the relative frequencies in the dataset.

 

[kith]

@Morbert: From an instrumentalist viewpoint, the identification of $\text{tr}\{\rho \Pi\}$ with a probability is an axiom (the Born rule) which applies whenever a measurement of the quantum system is made. So if you include the first apparatus in the quantum system, the Born rule only applies when a measurement of this combined quantum system is made. I don't see a way to both take an instrumentalist perspective and use the Born rule without a reference to an external observer who performs this measurement of the combined system.

 

[Morbert]

@Morbert: From an instrumentalist viewpoint, the identification of $\text{tr}\{\rho \Pi\}$ with a probability is an axiom (the Born rule) which applies whenever a measurement of the quantum system is made. So if you include the first apparatus in the quantum system, the Born rule only applies when a measurement of this combined quantum system is made. I don't see a way to both take an instrumentalist perspective and use the Born rule without a reference to an external observer who performs this measurement of the combined system.

If this is how instrumentalism is understood then ok. My last few posts were written under a broader but perhaps nonstandard notion of instrumentalism: quantum mechanics offers us procedures for predicting relative frequencies in our experimental data, and does not concern itself with a realist account of physics.

 

[atyy]

If this is how instrumentalism is understood then ok. My last few posts were written under a broader but perhaps nonstandard notion of instrumentalism: quantum mechanics offers us procedures for predicting relative frequencies in our experimental data, and does not concern itself with a realist account of physics.

In your broader but perhaps nonstandard notion of instrumentalism, are the experimental data real?

 

[Morbert]

In your broader but perhaps nonstandard notion of instrumentalism, are the experimental data real?

Quantum mechanics is an instrument used by the physicist to compute predictions regarding the data. Whether or not the physicist believes the data are real probably depends on the metaphysical beliefs of the physicist.

 

[atyy]

Quantum mechanics is an instrument used by the physicist to compute predictions regarding the data. Whether or not the physicist believes the data are real probably depends on the metaphysical beliefs of the physicist.

Do I dare ask whether the physicist believes himself/herself/itself to be real?

 

[Morbert]

Do I dare ask whether the physicist believes himself/herself/itself to be real?

I'd say most physicists think they're real.

 

[vanhees71]

Some are imaginary (e.g., specialists on lattice QCD).

 

[kith]

My last few posts were written under a broader but perhaps nonstandard notion of instrumentalism: quantum mechanics offers us procedures for predicting relative frequencies in our experimental data, and does not concern itself with a realist account of physics.

Note that you have references to an external observer in your phrasing as well ("us", "our"). Instrumentalism has the observer baked in because there's always the question of instrumental to whom. A stick may exist without something external but it becomes an instrument only when there's an ape that uses it to fish for termites.

 

[Morbert]

Note that you have references to an external observer in your phrasing as well ("us", "our"). Instrumentalism has the observer baked in because there's always the question of instrumental to whom. A stick may exist without something external but it becomes an instrument only when there's an ape that uses it to fish for termites.

The person using QM to make predictions constitutes an observer in the sense described above, but is the idea of an observer limited to this sense? Observer can refer more generally to a classical apparatus, correlated with the quantum system, that renders a measurement outcome. With the double slit experiment, most people would consider the interference-destroying measurement to be made by the detector at the slits, rather than the scientist reviewing the detector data at a later time.

What I'm challenging is the insistence that an observer in this sense must be excluded from the quantum state, since the classical properties necessary to establish a measurement outcome can be identified in a quantum framework.

 

[atyy]

The person using QM to make predictions constitutes an observer in the sense described above, but is the idea of an observer limited to this sense? Observer can refer more generally to a classical apparatus, correlated with the quantum system, that renders a measurement outcome. With the double slit experiment, most people would consider the interference-destroying measurement to be made by the detector at the slits, rather than the scientist reviewing the detector data at a later time.

What I'm challenging is the insistence that an observer in this sense must be excluded from the quantum state, since the classical properties necessary to establish a measurement outcome can be identified in a quantum framework.

The point is that ultimately, one needs an observer that is excluded from the quantum state. Whether the measurement apparatus is considered part of the observer is subjective. If the physicist considers its "self" to be real, it must still define "self". For example, does a physicist's "self" include its fingernails, or clothes? If that is subjective, then whether the measurement apparatus is part of the "self" is also subjective.

 

[Morbert]

The point is that ultimately, one needs an observer that is excluded from the quantum state. Whether the measurement apparatus is considered part of the observer is subjective. If the physicist considers its "self" to be real, it must still define "self". For example, does a physicist's "self" include its fingernails, or clothes? If that is subjective, then whether the measurement apparatus is part of the "self" is also subjective.

Then I wouldn't object to this convention, provided we acknowledge that the emergence of classicality necessary for a measurement scenario is not similarly subjective. Whether we use $\rho_s$ or $\rho_s\otimes\rho_m$, the data $\epsilon_i$ can be included in a classical (boolean) logic long before the scientist or their fingernails need to be considered.

Also, it seems this notion of an observer as "the entity using the theory to make predictions about an experiment they choose to conduct" would also be excluded from the state even if we use a classical theory.

 

[kith]

The person using QM to make predictions constitutes an observer in the sense described above, but is the idea of an observer limited to this sense? Observer can refer more generally to a classical apparatus, correlated with the quantum system, that renders a measurement outcome. With the double slit experiment, most people would consider the interference-destroying measurement to be made by the detector at the slits, rather than the scientist reviewing the detector data at a later time.

Things are simple, if we use the word "observer" to refer to the actual person who performs the measurement in union with her means of investigation. Statements like "the measurement is made by the detector" make things unnecessarily complicated. They replace the clear meaning of the term "measurement" as an intentional process of an entity to gather data for building models with something unclear.

If we want to talk about the physical processes happening at the detector, we have a precise technical language at our disposal (the language of decoherence) which we should use.

 

[kith]

Also, it seems this notion of an observer as "the entity using the theory to make predictions about an experiment they choose to conduct" would also be excluded from the state even if we use a classical theory.

Yes and we could give up the realism of classical mechanics and instead use a similar ontology as in instrumental QM to reflect this. But we can't go the other route: in QM we can't simply take the state of the observer as determined but unknown without getting into problems. In classical mechanics, we can, so we usually do.

 

[Morbert]

Things are simple, if we use the word "observer" to refer to the actual person who performs the measurement in union with her means of investigation. Statements like "the measurement is made by the detector" make things unnecessarily complicated. They replace the clear meaning of the term "measurement" as an intentional process of an entity to gather data for building models with something unclear.

The convention you describe above seems reasonable, but perhaps we can obtain further specificity. Von Neumann, in his theory of measurement[1], makes a distinction between measurement as a physical process and observation as perception. He uses an example of a thermometer measuring the temperature of a system, and an observer observing the length of the mercury in the thermometer. I.e. Perhaps it is useful to make a distinction between the generation of data by an apparatus (measurement), and the subjective experience of/inference from this data by the user (observation).

Using this language, an observer is excluded from the system modeled by quantum mechanics, but a measurement process can be included.

If we want to talk about the physical processes happening at the detector, we have a precise technical language at our disposal (the language of decoherence) which we should use.

Just a point of clarification: Decoherence is important in this process I was calling measurement, but it isn't sufficient. Decoherence establishes a logic for classical properties in the apparatus so that it has the ability to express data, but it is the establishment of a logical equivalence between classical properties of the apparatus and quantum properties of the measured system that is key to a measurement process.

[1] https://press.princeton.edu/books/h...mathematical-foundations-of-quantum-mechanics

 

[kith]

Von Neumann, in his theory of measurement[1], makes a distinction between measurement as a physical process and observation as perception.

I don't agree with this reading of von Neumann.

In his presentation, $I$ is the system of interest, $II$ are the means the observer uses to perform the measurement and $III$ is the "actual" observer (which does not necessarily include her whole body). What he does is to prove that the boundary between what is considered to be the quantum system and what is considered to belong to the observer can be shifted, i.e. the divisions $I \,\, | \,\, II\!+\!III$ and $I\!+\!II \,\, | \,\, III$ give the same predictions.

He doesn't say that the physical process involving $I + II$ generates data which are simply perceived by the observer, which is false (see below).

I.e. Perhaps it is useful to make a distinction between the generation of data by an apparatus (measurement), and the subjective experience of/inference from this data by the user (observation).

The quantum interaction between the system and the apparatus doesn't generate data, it generates possible data. The observer is not just there to take note of previously generated data, he is necessary to objectify one of the possibilities. I disagree with your notion of "measurement" because it doesn't reflect this.

 

[Morbert]

In his presentation, $I$ is the system of interest, $II$ are the means the observer uses to perform the measurement and $III$ is the "actual" observer (which does not necessarily include her whole body). What he does is to prove that the boundary between what is considered to be the quantum system and what is considered to belong to the observer can be shifted, i.e. the divisions $I \,\, | \,\, II\!+\!III$ and $I\!+\!II \,\, | \,\, III$ give the same predictions.

Von Neumann discusses two divides: The "observer/observed" divide and the "actually observed/measuring instrument/actual observer" divide. The bit I've put in bold implies you consider the quantum/classical divide to be the same as the observer/observed divide.

The division $I\!+\!II \,\, | \,\, III$ means the interaction between the measuring instrument and the actually observed system is explicitly modeled, and $II$ (whether it is a thermometer, or the light + observer eye, or the retina + brain) must have classical properties if they are to be perceived by the actual observer. So this would be an example of the classical/quantum divide differing from the observer/observed divide, since the classical properties needed to express the measurement result are on the "observed" side of the observer/observed divide. A useful distinction can therefore be made between the classical apparatus serving as the measuring instrument and the actually observed system, independent from what we implicitly or explicitly model.

You might object to a description of these measuring instrument properties as classical under the division $I\!+\!II \,\, | \,\, III$ because they are handled within a quantum framework. But we can identify classical properties in a quantum theoretic framework[1] and it is useful to do so when understanding the measuring process.

He doesn't say that the physical process involving $I + II$ generates data which are simply perceived by the observer, which is false (see below).

The quantum interaction between the system and the apparatus doesn't generate data, it generates possible data. The observer is not just there to take note of previously generated data, he is necessary to objectify one of the possibilities. I disagree with your notion of "measurement" because it doesn't reflect this.

Instrumentalism frames QM as a theory that makes statistical claims. It returns relative frequencies and expectation values that the user can compare against the data produced by the experiment. If this is what you mean by generates possible data, I agree, but this doesn't contradict the common sense understanding of a measuring instrument interacting with $I$ and producing data, to be studied by the actual observer. We just have to not expect QM to describe the actualisation of the data. It instead describes the statistics the actualised data will follow.

[1] https://aip.scitation.org/doi/10.1063/1.531886

 

[kith]

Instrumentalism frames QM as a theory that makes statistical claims. It returns relative frequencies and expectation values that the user can compare against the data produced by the experiment.

Textbook instrumentalism goes beyond probabilities and expectation values and talks about the state of the individual system after a single measurement. I didn't realize that you might have an ensemble view of instrumentalism in mind. When you talk about "data" are you always referring to many runs?

If this is what you mean by generates possible data, I agree [...]

Sorry, I should have been clearer here. By "possible data" I meant that if we include $II$ in the quantum description, the final state of the combined system is an entangled state which corresponds to a multitude of possible measurement outcomes and not a state which corresponds to a single, actualized measurement outcome.

[...]but this doesn't contradict the common sense understanding of a measuring instrument interacting with $I$ and producing data, to be studied by the actual observer.

If we take $II$ to be the thermometer we don't get a definite temperature after a single measurement but a superposition which corresponds to different temperatures. When the actual observer determines the temperature by looking at the thermometer, she doesn't simply perceive an external fact but is herself integral to its establishment. We shouldn't call the interaction between $I$ and $II$ a "measurement" because it isn't sufficient to establish the fact.

 

[Morbert]

Textbook instrumentalism goes beyond probabilities and expectation values and talks about the state of the individual system after a single measurement. I didn't realize that you might have an ensemble view of instrumentalism in mind. When you talk about "data" are you always referring to many runs?

I had a frequentist unserstanding of probabilities in mind, but a Bayesian understanding would be compatible as well. But more to your point: I was understanding the initial quantum state as an input, and only probabilities (however they are interpreted) as output. As opposed to both the final quantum state and probabilities as output. The physicist can of course incorporate her knowledge of a measurement result from a previous experiment in her preparation of a subsequent experiment. E.g. If she learns of a measurement result $a_i$, she can write down the quantum state for the next experiment $$\rho'_s = \Pi_{a_i,t_0}\rho_s\Pi_{a_i,t_0}$$
that is if we use the convention $I \,\, | \,\, II\!+\!III$. If we instead use $I\!+\!II \,\, | \,\, III$ then she constructs $$\rho'_{s,m}\otimes\rho_{m'} = (\Pi_{\epsilon_i,t_0}\rho_s\otimes\rho_m\Pi_{\epsilon_i,t_0})\otimes\rho_{m'}$$ where $m'$ is the 2nd measuring instrument

Sorry, I should have been clearer here. By "possible data" I meant that if we include $II$ in the quantum description, the final state of the combined system is an entangled state which corresponds to a multitude of possible measurement outcomes and not a state which corresponds to a single, actualized measurement outcome.

If we take $II$ to be the thermometer we don't get a definite temperature after a single measurement but a superposition which corresponds to different temperatures. When the actual observer determines the temperature by looking at the thermometer, she doesn't simply perceive an external fact but is herself integral to its establishment. We shouldn't call the interaction between $I$ and $II$ a "measurement" because it isn't sufficient to establish the fact.

We both agree that in the real world, a thermometer reports a definite temperature. We both agree that we can use QM to construct a final state that is a superposition*. Where we seem to disagree is you say "this final state is what QM reports as the result of $I$ and $II$ interacting, but instead a definite temperature is recorded , so this interaction is not a measurement in any meaningful sense". I say "QM does not report the final state as the result of $I$ and $II$ interacting. Instead QM reports a probability for each possible result of $I$ and $II$ interacting, and the final state is merely an ingredient in the computation of these probabilities". I don't know how substantive this disagreement is, and it might just come down to language convention.

*Ignoring for the time being things like information loss irreversibly and mixed states, which are admittedly important factors if we want to explore what it means to establish a fact.

 

[timmdeeg]

Thanks for answering my question, very much appreciated.