Role of the collapse in the instrumentalist interpretation

[timmdeeg]

TL;DR

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.