This forum gives conflicting info on the HUP

[Demystifier]

Question: How do you measure the exact pitch for a sound wave in an exact moment in time?

By observing a classical wave, you cannot measure its frequency in an exact moment of time.

Yet, in principle, in QM you can measure any observable in an arbitrarily short time. Since this is valid for any observable, this is valid also for the Hamiltonian. But by measuring Hamiltonian you measure energy, which means that you measure frequency.

So how is it possible that in QM you can measure frequency in an exact moment of time? Essentially, that is because in QM you never really measure frequency. Instead, you really measure something else (the position of some macroscopic pointer) which turns out to be ENTANGLED with frequency.

This is somewhat analogous to the following common sense example. How to make a picture of a very short event (say 1 milisecond) with a very slow camera (with exposition, say, 1 second)? Easy! First make a picture with another sufficiently fast camera, and then use your slow camera to take a picture of the picture made by the fast camera. In this case, by your slow camera you really take a picture of something else (of another picture), but this other picture is "entangled" with the short event, in the sense that it is strongly correlated with it.

 

[Demystifier]

All my posts above can be summarized as follows:
Whenever you measure something in QM, you really measure something else.

 

[DevilsAvocado]

You measure many "exact moments" according a time-resolution of your choice and after you have determined the pitch using information from from all those measurements , you will know precisely what the value was at one of those "exact moments".

Okay Bill, but value and pitch is not the same thing, right? If you pick one off the "exact moments", all you will hear is a very short 'click', right? Or, if you send one "exact moment" to Schrödinger and ask him to calculate/predict which pitch it represents, he will be very mad at you, right?

This is digital signal processing (as in sampling), and in this form of quantization there will always be error/distortion, since the set of possible input values may be infinitely large (thus uncountable).

And even if you could build an extreme super cool quantum clock with a resolution of some yottahertz, you will ultimately run into the wall of Planck time ≈ 5.391 × 10⁻⁴⁴ second (thus HUP) ...

t_P \equiv \sqrt{\frac{\hbar G}{c^5}}

 

[DevilsAvocado]

I didn't use the word "exact". I used the word "sharp". It referred to the value on the macroscopic apparatus. For example, the apparatus may have a digital display which may show a sharp number 7 or a sharp number 8, but it cannot show some unsharp sign which resembles both 7 and 8.
[...]
There is no doubt that both gadgets will show some numbers. As long as you think like an experimentalist without any theoretical prejudices, there is nothing more natural than to interpret these two numbers as simultaneous measurement of position and momentum. That is all.

Okay thanks DM, think I understand now. Simultaneous "sharp" measurement of position and momentum is possible, but arbitrarily precise (including exact) values are not possible, right?

 

[DevilsAvocado]

By observing a classical wave, you cannot measure its frequency in an exact moment of time.

Thanks DM! I knew it!

Yet, in principle, in QM you can measure any observable in an arbitrarily short time. Since this is valid for any observable, this is valid also for the Hamiltonian. But by measuring Hamiltonian you measure energy, which means that you measure frequency.

So how is it possible that in QM you can measure frequency in an exact moment of time? Essentially, that is because in QM you never really measure frequency. Instead, you really measure something else (the position of some macroscopic pointer) which turns out to be ENTANGLED with frequency.

But... what about 'the wall' of Planck time...?

[se post #33]

First make a picture with another sufficiently fast camera, and then use your slow camera to take a picture of the picture made by the fast camera.

But... if "the fast camera" is severely restricted by Mr. Planck, whom is a very close friend to Mr. Heisenberg... doesn't HUP always win in the end...?

 

[DevilsAvocado]

All my posts above can be summarized as follows:
Whenever you measure something in QM, you really measure something else.

Essentially, that is because in QM you never really measure frequency. Instead, you really measure something else (the position of some macroscopic pointer) which turns out to be ENTANGLED with frequency.

Why do I get the feeling you are saying something very deep that's completely beyond my perception...

In QM we are not measuring reality but an abstract mathematical model that by some amazing 'coupling' turns out to be the most precise description of 'reality' we ever had – it simply works – but nobody really understands why.

* dizzy *

 

[Demystifier]

Okay thanks DM, think I understand now. Simultaneous "sharp" measurement of position and momentum is possible, but arbitrarily precise (including exact) values are not possible, right?

HUP does not prevent arbitrarily precise measurements.

But... what about 'the wall' of Planck time...?

Some physicists believe that time cannot be measured with a precision better than Planck time, but it really depends on which version of quantum gravity you believe in, and there is no any "standard" theory of quantum gravity. After all, we certainly can measure energy with a precision better that Planck energy, so the same could be valid for time as well.

But... if "the fast camera" is severely restricted by Mr. Planck, whom is a very close friend to Mr. Heisenberg... doesn't HUP always win in the end...?

My point is that Mr. Planck and Mr. Heisenberg are not really such close friends.

 

[Demystifier]

Why do I get the feeling you are saying something very deep that's completely beyond my perception...

In QM we are not measuring reality but an abstract mathematical model that by some amazing 'coupling' turns out to be the most precise description of 'reality' we ever had – it simply works – but nobody really understands why.

* dizzy *

To paraphrase Bohr, if you are not confused about QM, then you don't understand it.

But you can deconfuse yourself a lot with a Bohmian way of thinking. With Bohmian thinking, it is quite clear why all measurements reduce to observations of positions of some macroscopic pointers. See e.g.
http://lanl.arxiv.org/abs/1112.2034 [Int. J. Quantum Inf. 10 (2012) 1241016]
Sec. 2.

 

[DevilsAvocado]

Thanks DM, need some time to 'digest' ...

 

[Ken G]

All my posts above can be summarized as follows:
Whenever you measure something in QM, you really measure something else.

But I agree with what you said above, which is that one must be agnostic about QM when doing measurements. This means we first say what a measurement is, and then we use measurements to test theories. We can use indirect measurements if we put our trust in one theory as contingent on the testing of another, but any measurements that involve an element that requires a QM computation to be meaningful is not really a measurement that can be used to test QM.

Also, we can try doing two measurements at once, but we cannot be sure the measurements still mean anything, because the two apparatuses could function at cross purposes. For example, I can have a scale that measures my weight, and a tape measure that measures the circumference of my waist, but if I make both measurements at once then the weight measurement will be altered by the weight of the tape measure. We can assure that we use a light tape measure and get away with those simultaneous measurements, but that only means that we are taking pains to assure the results of both measurements are meaningful. That's not so easy for non-commuting measurements in quantum mechanics, because we describe measurements in quantum mechanics in terms of a superposition of eigenstates of operators, and how we use an apparatus to scramble the coherences within that superposition, but you cannot necessarily scramble the coherences within the eigenstate superpositions of two separate non-commuting operators. So the devices may read something sharp, but it might not qualify as a measurement, just as I would not be measuring my weight if I had a lead tape measure around my waist.

 

[atyy]

What you presented is one particular example of a measuring apparatus which does not allow simultaneous measurement of noncommuting observables. But nothing in your argument shows that this is a general conclusion.

Let me try to describe a variation of SG device which, in principle, would allow simultaneous measurement.

Suppose you want to measure the spin of particle A. For that purpose you first make the particle A interact with another particle B, such that the two particles become entangled. The entanglement may be such that, by measuring spin s_x of particle B, you also know spin s_x of particle A. Then you separate the entangled particles and route particle B in a standard SG device. In this way you measure s_x of particle A without ever routing particle A in a standard SG device.

Similarly, you can measure spin s_y of particle A by entangling it with a third particle C and routing particle C in another standard SG device.

Finally, you can combine these two modified SG devices, such that you can simultaneously measure s_x via particle B and s_y via particle C.

But can this work on an arbitrary state? When you entangle you may alter the state of the particle you are measuring, and when you measure the state of the probe, you will collapse the state of both the probe and the particle.

But I agree with what you said above, which is that one must be agnostic about QM when doing measurements.

But it is the theory that says what we measure, in particular, whether a measurement is "sharp" or not is defined by QM.

 

[Demystifier]

Ken G, I fully agree with you, and I like your analogy with weight-and-circumference measurement.

 

[Demystifier]

But can this work on an arbitrary state?

Yes.

When you entangle you may alter the state of the particle you are measuring, and when you measure the state of the probe, you will collapse the state of both the probe and the particle.

True, but this can also be said for a measurement of a single observable.

But it is the theory that say what we measure, in particular, whether a measurement is "sharp" or not is defined by QM.

I both agree and disagree with this. I agree probably for the same reason as you do, so let me only explain why I disagree.

If theory determines what we measure, that it can be argued that the same experiments measure different quantities, depending on which interpretation of QM you adopt. Hence, to avoid interpretation-dependence, it is wise to remain partially agnostic about the theory, and to define measurement in terms of concepts which do not depend much on the theory. One way to reduce the amount of assumed theory is to define all measurements in terms of classical concepts only. Of course, there are other ways too, so in reality it is important to define precisely what do you mean by a "measurement" when you claim that you have performed one.

 

[Fredrik]

That is actually quite easy. If you know
1) how A entangles with B when C is not present, and
2) how A entangles with C when B is not present,
then linearity alone is sufficient to determine how A entangles with both B and C when both are present.

I still don't understand what you have in mind. Are you referring to the linearity of the time evolution operator? We're not talking about having one time evolution operator act on two different states. We would have to deal with two different (and non-commuting) time evolution operators acting on one state. So linearity doesn't help.

I also don't see an obvious way to combine something like
$$\frac{1}{\sqrt 2}\big(|z+\rangle|z-\rangle|\sigma\rangle - |z-\rangle|z+\rangle|\sigma\rangle\big)$$ and
$$\frac{1}{\sqrt 2}\big(|x+\rangle|\rho\rangle|x-\rangle - |x-\rangle|\rho\rangle|x+\rangle\big)$$ into one state. Do you just add them and normalize? What about $|\sigma\rangle$ and $|\rho\rangle$? How do you choose them?

 

[atyy]

@Demystifier, I meant: can it work for accurate simultaneous joint measurements on arbitrary states? I agree this works for accurate measurements of a single observable. The entangling with a probe should be equivalent to the usual measurements which collapse the wave function. So if we measure position accurately it should collapse to a position eigenstate, and if we measure momentum it should collapse to a momentum eigenstate. If you try to measure both, it seems that you will measure something else.

 

[Ken G]

That's my issue as well, I think Demystifier's point is well taken that all measurements involve something that QM would regard as entanglement, so invoking EPR-type entanglement is not something completely unheard of. But we have to be careful about what we mean by a simultaneous measurement, and what we mean by simultaneous knowledge of two observables. I wonder if we can agree on the following things:

1) To be able to claim we have "simultaneous knowledge" that observables X and Y take on values x and y, to precision sigma(X) and sigma(Y), we are making the claim that if we next choose to do a very precise measurement of either X or Y, then we should get results within sigma(X) of x and within sigma(Y) of y. Is that not a reasonable meaning of simultaneous knowledge?

2) To be able to claim we have done a "simultaneous measurement" of X and Y, we must claim we are doing something that conveys simultaneous knowledge of X and Y.

 

[atyy]

That's my issue as well, I think Demystifier's point is well taken that all measurements involve something that QM would regard as entanglement, so invoking EPR-type entanglement is not something completely unheard of. But we have to be careful about what we mean by a simultaneous measurement, and what we mean by simultaneous knowledge of two observables. I wonder if we can agree on the following things:

1) To be able to claim we have "simultaneous knowledge" that observables X and Y take on values x and y, to precision sigma(X) and sigma(Y), we are making the claim that if we next choose to do a very precise measurement of either X or Y, then we should get results within sigma(X) of x and within sigma(Y) of y. Is that not a reasonable meaning of simultaneous knowledge?

2) To be able to claim we have done a "simultaneous measurement" of X and Y, we must claim we are doing something that conveys simultaneous knowledge of X and Y.

From my understanding, Ozawa's definition of an accurate measurement of A is simply that the distribution of measurement outcomes is the same as that when a projective measurement of A is performed on an ensemble of identically prepared particles in that state. So it depends on the textbook definition of accurate measurement that collapses the state into an eigenstate of the observable.

It seems that if we have some knowledge of the state, then we can tailor the measurement procedure for that state. However, these special procedures will not work on arbitrary states, so they cannot work for an an ensemble of particles in an unknown state. I think a special case is where one already knows the exact state, then a special procedure that accurately measures all observables simultaneously is to toss the state into the garbage, and just output measurement outcomes calculated using knowledge of the state and quantum mechanics.

 

[Demystifier]

We would have to deal with two different (and non-commuting) time evolution operators acting on one state. So linearity doesn't help.

You are right and I have to withdraw the statement that linearity is enough. In reality, you need to do the following. If the first measurement is achieved with the Hamiltonian H_1 and the second measurement with the Hamiltonian H_2, then the combined measurement is described by the evolution with the Hamiltonian H_12=H_1+H_2. The corresponding evolution operator U_12 is well defined, but different from both U_1 U_2 and U_2 U_1.

 

[Demystifier]

So if we measure position accurately it should collapse to a position eigenstate, and if we measure momentum it should collapse to a momentum eigenstate. If you try to measure both, it seems that you will measure something else.

You are right. As I already said, if you try to measure both it will collapse to a coherent state. That's why the repeated measurement will not give the same value of either position and momentum.

But if you insist that the only meaningful measurement is the one in which repeated measurement gives the same value, then what about a photon measurement which typically destroys the photon? Does it mean that photon measurements are not meaningful?

 

[Demystifier]

I wonder if we can agree on the following things:

1) To be able to claim we have "simultaneous knowledge" that observables X and Y take on values x and y, to precision sigma(X) and sigma(Y), we are making the claim that if we next choose to do a very precise measurement of either X or Y, then we should get results within sigma(X) of x and within sigma(Y) of y. Is that not a reasonable meaning of simultaneous knowledge?

2) To be able to claim we have done a "simultaneous measurement" of X and Y, we must claim we are doing something that conveys simultaneous knowledge of X and Y.

I cannot agree on 1) because, as I explained in the post above, it would imply that we cannot have a reasonable knowledge about photons.

Concerning 2), you tacitly assume that X or Y is real when we measure it. But the example of Bohmian mechanics teachs us that spins may never be real, even when we "measure" them.

This, indeed, is why Bell insisted that we should not talk about measurement (which, as we see, is a misleading concept) but about experiment.

 

[atyy]

You are right. As I already said, if you try to measure both it will collapse to a coherent state. That's why the repeated measurement will not give the same value of either position and momentum.

But if you insist that the only meaningful measurement is the one in which repeated measurement gives the same value, then what about a photon measurement which typically destroys the photon? Does it mean that photon measurements are not meaningful?

Yes, that's why repeated measurement is not the only accurate measurement, which opens the door to the acccurate measurements on a specific state that Ozawa (and others) talk about.

How's this:

An accurate measurement of A on any state is a procedure that gives measurement outcomes according to the Born rule for any state. If the state survives, it will be in an eigenstate of A with probability given by the Born rule.

There are other procedures that give accurate measurements of A on specific states in the sense that the distribution of outcomes is the same as that given by the Born rule. These procedures require some knowledge of the state, and don't work for any state. In these cases, if the state survives, it need not be in an eigenstate of A with probability given by the Born rule. For example, if the state is known, a state-dependent measurement is to toss the state into the garbage or leave it unchanged and simply write down measurement outcomes according to the Born rule.

 

[Ken G]

From my understanding, Ozawa's definition of an accurate measurement of A is simply that the distribution of measurement outcomes is the same as that when a projective measurement of A is performed on an ensemble of identically prepared particles in that state. So it depends on the textbook definition of accurate measurement that collapses the state into an eigenstate of the observable.

True, he uses measurement in the sense of a predictable outcome, not in the sense of conveying knowledge of an observable. We may have to recognize different types of measurement-- getting away from our classical prejudices that any measurement that represents a testable outcome also conveys knowledge of the observable!

It seems that if we have some knowledge of the state, then we can tailor the measurement procedure for that state. However, these special procedures will not work on arbitrary states, so they cannot work for an an ensemble of particles in an unknown state.

That's an important issue as well, but is a separate feature in the landscape. So it seems we really have three levels of measurement here:
1)general measurements: they convey knowledge of the observable on any state, without any prior knowledge,
2)specific measurements: they convey knowledge of the observable, but only if you already know something about the state, and
3)non-repeatable measurements: they do not convey knowledge of the observable in the sense defined above, but they do register a result on a pointer. Destructive measurements are of this type, but so are non-destructive measurements involving EPR entanglements that are broken by the measurement.

It seems to me that not only are Ozawa's theorems not about simultaneous measurements of type (1), they are not even of type (2)-- they invoke type (3)! Since the measurements are non-repeatable, they make it impossible to test if the two measurements are interfering with each other.

As as example, consider a device that sends out two photons in opposite directions with the same known wave packet, and in a momentum-conserving way, so the photons are entangled. We do a precise p measurement on one, and a precise x measurement on the other. The wave packet determines the distributions we get, so we have a precise measurement in Ozawa's sense. Do we have simultaneous knowledge of x and p of both photons, in the sense I defined above? No, we have a form of the HUP. Yet doesn't the entanglement say that we are doing a simultaneous x and p measurement for both photons? In the sense Ozawa means, apparently yes, but not in the sense of conveying knowledge of both those observables on both photons, even given our knowledge of the entangled state.

 

[Ken G]

I cannot agree on 1) because, as I explained in the post above, it would imply that we cannot have a reasonable knowledge about photons.

We can have reasonable knowledge of photons, but we have to recognize different types of knowledge. Usually we only use measurements to see if our predictions were right, so we don't care if the photon is destroyed or not. But we are only claiming "my theory worked", we are not claiming "I have knowledge that the photon is in state X." The latter is a very different kind of claim, and so we might have measurements that can support a prediction, yet not convey knowledge of the state of the photon. A state is a kind of preparation, so knowledge of a state must be knowledge about a preparation, not knowledge of a destruction.

Concerning 2), you tacitly assume that X or Y is real when we measure it. But the example of Bohmian mechanics teachs us that spins may never be real, even when we "measure" them.

To me, the only thing that is real is the knowledge, not the thing-in-itself.

This, indeed, is why Bell insisted that we should not talk about measurement (which, as we see, is a misleading concept) but about experiment.

Perhaps we can talk about both-- as long as we are clear what we mean.

 

[Ken G]

I think I've had an insight that will help with this. The problem seems to be that if we have two particles, we are wondering if there are really 4 things to precisely know there (two x and two p), or just 2 things to know there (an x or p from one, and an x or p from the other)? The HUP suggests the latter, but EPR entanglement might be used to make it seem like there actually are 4 things there we can know. This relates to the issue of what kind of experiment can impart "simultaneous knowledge" of x and p for both particles, if they are entangled.

I would argue that entanglement does not work to extend what we can simultaneously know about the state of two particles, for the simple reason that entanglement means that there are still only 2 things to know about those two particles. As when the particles are unentangled, we can choose between 4 precise measurements to do, but we can only do 2 measurements that can be reproduced immediately, so our ongoing knowledge of the system is restricted to 2 things. When the particles are entangled, this continues to be true-- so there are still only two things about that system we can obtain "simultaneous knowledge" of.

In other words, if we have two entangled particles that conserve total p=0, and we do a p measurement on one, we can predict precisely the p of the other. And if we instead do an x measurement on the other, we can say we know x and p of that particle, but actually what we measured was an x and p of the entangled system, so we still only know two things about a system that gives us only 2 things to know. If we want to say we know 4 things, then we have to treat the system as no longer entangled, which is true because we broke the entanglement, but now we cannot use the other p to get the p of a particle, we'd have to do a new p measurement on that no-longer-entangled particle. Bottom line: if one interprets one form of the HUP as saying that two particles only contain two pieces of precise x,p information that we could be in a position to predict, this continues to be true whether the two particles are entangled or not.

 

[atyy]

I think I've had an insight that will help with this. The problem seems to be that if we have two particles, we are wondering if there are really 4 things to precisely know there (two x and two p), or just 2 things to know there (an x or p from one, and an x or p from the other)? The HUP suggests the latter, but EPR entanglement might be used to make it seem like there actually are 4 things there we can know. This relates to the issue of what kind of experiment can impart "simultaneous knowledge" of x and p for both particles, if they are entangled.

I would argue that entanglement does not work to extend what we can simultaneously know about the state of two particles, for the simple reason that entanglement means that there are still only 2 things to know about those two particles. As when the particles are unentangled, we can choose between 4 precise measurements to do, but we can only do 2 measurements that can be reproduced immediately, so our ongoing knowledge of the system is restricted to 2 things. When the particles are entangled, this continues to be true-- so there are still only two things about that system we can obtain "simultaneous knowledge" of.

In other words, if we have two entangled particles that conserve total p=0, and we do a p measurement on one, we can predict precisely the p of the other. And if we instead do an x measurement on the other, we can say we know x and p of that particle, but actually what we measured was an x and p of the entangled system, so we still only know two things about a system that gives us only 2 things to know. If we want to say we know 4 things, then we have to treat the system as no longer entangled, which is true because we broke the entanglement, but now we cannot use the other p to get the p of a particle, we'd have to do a new p measurement on that no-longer-entangled particle. Bottom line: if one interprets one form of the HUP as saying that two particles only contain two pieces of precise x,p information that we could be in a position to predict, this continues to be true whether the two particles are entangled or not.

The way I've been thinking about measurements above doesn't have anything to do with knowledge or information. It's just a procedure to produce a distribution of outcomes. That's why I said, well if you know the state you can measure all observables simultaneously, but obviously you get no information.

But if one wants to use measurements to get information, then there is this notion of "no information without distrubance". I googled it, and this paper seems interesting: http://arxiv.org/abs/quant-ph/9512023v1 (10.1103/PhysRevA.53.2038).

 

[Fredrik]

http://arxiv.org/abs/0911.1147

And a claimed proof of "Theorem 14. In any Hilbert space with dimension more than 3, there are nowhere commuting observables that are simultaneously measurable in a state that is not an eigenstate of either observable." is provided on p10.

There's one detail in theorem 14 (and elsewhere) that stands out. The theorem you're quoting isn't saying that a simultaneous measurement of two non-commuting observables is possible, period. It's saying that there's a state such that if the particle is in that state, then a simultaneous measurement is possible.

I still haven't read enough to see the significance of this. The author has a very specific definition of terms like simultaneous measurement. I would have to study those definitions to know what impact his theorems have on my conjecture that "simultaneous measurements are possible if and only if the measuring devices can exist in the same place without interfering with each other".

 

[atyy]

There's one detail in theorem 14 (and elsewhere) that stands out. The theorem you're quoting isn't saying that a simultaneous measurement of two non-commuting observables is possible, period. It's saying that there's a state such that if the particle is in that state, then a simultaneous measurement is possible.

I still haven't read enough to see the significance of this. The author has a very specific definition of terms like simultaneous measurement. I would have to study those definitions to know what impact his theorems have on my conjecture that "simultaneous measurements are possible if and only if the measuring devices can exist in the same place without interfering with each other".

I think quite general relations for simultaneous measurements are given in http://arxiv.org/abs/1312.1857. Some of those results are shown graphically in Fig. 1 of http://arxiv.org/abs/1304.2071, which also has a useful appendix A1 that uses the basic idea you mentioned that if you attempt to measure jointly you really measure something else. Branciard's papers follow work by Arthurs and Kelly, and by Ozawa http://arxiv.org/abs/quant-ph/0310070. This paper by Ozawa shows in section III that an accurate simultaneous measurement of two observables is possible only if the two observables commute.

There are somewhat different relations if one uses different definitions of error http://arxiv.org/abs/1306.1565. This paper is about sequential measurements, but it's related since the Ozawa relation for sequential measurements http://arxiv.org/abs/quant-ph/0207121 is pretty similar for to the above-mentioned Ozawa relation for simultaneous measurements.

Apparently there is some controversy over whose definitions of error are better http://physicsworld.com/cws/article...y-reigns-over-heisenbergs-measurement-analogy.

If the probes in sequential measurements are correlated, it seems that one can do better than Ozawa's relation for sequential measurements http://arxiv.org/abs/1212.2815. There's a comment just after Eq 21: "Notice, that in the EPR thought-experiment [37] the measurements can be carried out simultaneously, violating the joint uncertainty principle. This is a consequence of the probes and the measured system being initially correlated, contrary to what is assumed herein ...".

 

[Demystifier]

So it seems we really have three levels of measurement here:
1)general measurements: they convey knowledge of the observable on any state, without any prior knowledge,
2)specific measurements: they convey knowledge of the observable, but only if you already know something about the state, and
3)non-repeatable measurements: they do not convey knowledge of the observable in the sense defined above, but they do register a result on a pointer. Destructive measurements are of this type, but so are non-destructive measurements involving EPR entanglements that are broken by the measurement.

Well summarized!

 

[Demystifier]

How's this:

An accurate measurement of A on any state is a procedure that gives measurement outcomes according to the Born rule for any state. If the state survives, it will be in an eigenstate of A with probability given by the Born rule.

The notion of state survival is not well defined. For example, even when the photon is destroyed, you can say that the state survives because you still have some state of quantum electrodynamics. A state with zero number of photons is still a state.

If we accept only the first requirement above (that outcomes need to be given by the Born rule), then "my" simultaneous measurement of non-commuting observables is "accurate".

 

[Fredrik]

If the state survives, it will be in an eigenstate of A with probability given by the Born rule.

The article by Ozawa that you linked to in #24 says this:

In fact, it is widely accepted nowadays that any observable can be measured correctly without leaving the object in an eigenstate of the measured observable; for instance, a projection $E$ can be correctly measured in a state $\psi$ with the outcome being 1 leaving the object in the state $M\psi/\|M\psi\|$, where the operator $M$ depends on the apparatus and satisfies $E=M^\dagger M$ (see, for example, a widely accepted textbook by Nielsen and Chuang [10]).
​

I'm going to have to read up on this, because I have no idea what he's talking about.

I see that you're also saying this:

There are other procedures that give accurate measurements of A on specific states in the sense that the distribution of outcomes is the same as that given by the Born rule. These procedures require some knowledge of the state, and don't work for any state. In these cases, if the state survives, it need not be in an eigenstate of A with probability given by the Born rule.

Maybe he's just talking about that. (He doesn't say if his $\psi$ is arbitrary). I will think about it.