I Clarification of the postulates of QM

[stevendaryl]

But that doesn't make the assignment of the state $\hat{\rho}_B$ invalid.

It's a perfectly valid description of Bob's knowledge about his particle, but it's not an accurate description of the particle itself. Saying it's spin-down in the z-direction is more accurate.

 

[ShayanJ]

But that doesn't make the assignment of the state $\hat{\rho}_B$ invalid. It still describes Bob's knowledge about the spin state of his particle correctly, no matter what Alice knows about what he will find. Of course, if A tells him what she has measured, B also knows what he will measure, but A can send her message only with at most the speed of light to B. Thus no instantaneous "collapse" happens from B's point of view but just information provided by A separately from the system measured, but I think we just turn in the usual circles here :-(.

Let me see if I get it right. You're saying that in B's view, the state of his particle is $\rho_B= \frac 1 2 \hat 1$ no matter before or after A's measurement. But in A's view, the state of B's particle is $\rho_B=\frac 1 2 \hat 1$ before A's measurement and either $|S_z;+ \rangle$ or $|S_z; - \rangle$(depending on A's result) after A's measurement. But this means there is a collapse, its just subjective!(Which also means you think the state vector only describes the observer's knowledge about the system, and not anything objective.)

 

[stevendaryl]

It's a perfectly valid description of Bob's knowledge about his particle, but it's not an accurate description of the particle itself. Saying it's spin-down in the z-direction is more accurate.

Let me give an analogy. If Bob is handed a deck of 52 cards, and he's asked to describe his probability of getting an ace of spades if he selects one at random, he might say: My chances are 1.9%

But if beforehand, Alice swapped the ace of spades for a Joker, then she knows that Bob has 0% chance of picking the ace of spades.

Bob's prediction was as accurate as he could possibly be, given his knowledge, but Alice's prediction is more accurate.

 

[vanhees71]

Sure, you can have such subjective ideas, but it has nothing to do with the physics of the two particles measured by A and B. I can as well say that for me the state of the German lottery drawing machine instantaneously collapses every Saturday as soon as the numbers are drawn, but that happens for me only if I take notice of the outcome of the experiment. It doesn't do anything on what's going on at the drawing of the actual numbers.

 

[ShayanJ]

Sure, you can have such subjective ideas, but it has nothing to do with the physics of the two particles measured by A and B. I can as well say that for me the state of the German lottery drawing machine instantaneously collapses every Saturday as soon as the numbers are drawn, but that happens for me only if I take notice of the outcome of the experiment. It doesn't do anything on what's going on at the drawing of the actual numbers.

But this inevitably means that you consider the state vector to be a subjective concept only describing the observer's knowledge about the system! This doesn't seem to be minimalistic in that one may ask "what's actually going on down there?", unless by minimalistic, you mean just not asking such questions.

Also, you're a proponent of the ensemble interpretation, so you shouldn't be able to talk about experiments not involving an ensemble. But now you're talking about a single pair of particles!

 

[vanhees71]

Let me give an analogy. If Bob is handed a deck of 52 cards, and he's asked to describe his probability of getting an ace of spades if he selects one at random, he might say: My chances are 1.9%

But if beforehand, Alice swapped the ace of spades for a Joker, then she knows that Bob has 0% chance of picking the ace of spades.

Bob's prediction was as accurate as he could possibly be, given his knowledge, but Alice's prediction is more accurate.

Well, sure. But what does this tell us about the QT example we are discussing? Of course, you'll put different probability distributions on the same situation if you have different knowledge about the system. That's very trivial, but in this case A and B have precisely the same knowledge, namely that each of them measure the spin of one of two spin-entangled particles. So the only thing both know is that each of them just has an unpolarized particle. Only bringing both measurements together in a way that you can compare what was measured for each of the particles prepared in the entangled two-particle state reveals the correlation implied by the entanglement. Of course, you can check it only with some significance at a large enough ensemble of so prepared two-particle states.

 

[stevendaryl]

Well, sure. But what does this tell us about the QT example we are discussing?

It seems perfectly analogous. Bob thinks he has a 50/50 chance of getting spin-up or spin-down, but Alice knows that he has 0% of getting spin-up.

 

[vanhees71]

But this inevitably means that you consider the state vector to be a subjective concept only describing the observer's knowledge about the system! This doesn't seem to be minimalistic in that one may ask "what's actually going on down there?", unless by minimalistic, you mean just not asking such questions.

Also, you're a proponent of the ensemble interpretation, so you shouldn't be able to talk about a experiments not involving an ensemble. But now you're talking about a single pair of particles!

No. The state is an objective description of the system, which is prepared in this state. Of course, any probabilistic statement can be checked only at an ensemble.

 

[vanhees71]

It seems perfectly analogous. Bob thinks he has a 50/50 chance of getting spin-up or spin-down, but Alice knows that he has 0% of getting spin-up.

So?

 

[ShayanJ]

No. The state is an objective description of the system, which is prepared in this state. Of course, any probabilistic statement can be checked only at an ensemble.

So before A's measurement, her spin is in the state $\frac 1 2 \hat 1$. Then she measures her spin and gets the result +, so she knows that her spin is pointing upward. But she's actually wrong and her measurement is meaningless because her spin is still in the state $\frac 1 2 \hat 1$ and not $|S_z;+\rangle$? This doesn't make sense!

 

[atyy]

So before A's measurement, her spin is in the state $\frac 1 2 \hat 1$. Then she measures her spin and gets the result +, so she knows that her spin is pointing upward. But she's actually wrong and her measurement is meaningless because her spin is still in the state $\frac 1 2 \hat 1$ and not $|S_z;+\rangle$? This doesn't make sense!

vanhees71 is a secret many-worlder, since that is the interpretation in which the state is objective, and neither hidden variables nor collapse :)

In fact, it is because he thinks the state is objective that he objects to collapse. If the state were subjective or just FAPP, there would be no problems with collapse.

Of course, in a sense, if MWI works it is certainly the minimal interpretation!

 

[martinbn]

To say that measurement "prepares" the state is the same thing as collapse.

Not the way I see it. Collapse asumes that the state decribes the system and that before and after the measurement you have the same system but in a different state. But if the state describes the set of equivalently prepatered systems (not an individual one), then after the measurement you have a diffrent system, not the same with collapsed state. You may say that this is just words, but that's what interpretations are.

 

[atyy]

Not the way I see it. Collapse asumes that the state decribes the system and that before and after the measurement you have the same system but in a different state. But if the state describes the set of equivalently prepatered systems (not an individual one), then after the measurement you have a diffrent system, not the same with collapsed state. You may say that this is just words, but that's what interpretations are.

The point is that collapse allows you to calculate the conditional probability of the second measurement outcome conditioned on a sub-ensemble of outcomes from the the first measurement.

In calculating the conditional probability, you need the state of a sub-ensemble.

The collapse rule assigns the state of a sub-ensemble. So collapse is still needed within the ensemble interpretation.

 

[martinbn]

The point is that collapse allows you to calculate the conditional probability of the second measurement outcome conditioned on a sub-ensemble of outcomes from the the first measurement.

In calculating the conditional probability, you need the state of a sub-ensemble.

The collapse rule assigns the state of a sub-ensemble. So collapse is still needed within the ensemble interpretation.

I know all this, but my point is that what you call the second measurment of the sub-ensemble is in fact the first measurment of a new ensemble. Before the measurment you didn't have that ensemble i.e. you cannot talk about a sub-ensemble. So what allows me to calculate conditional probability is the preparation of a new ensemble.

 

[atyy]

I know all this, but my point is that what you call the second measurment of the sub-ensemble is in fact the first measurment of a new ensemble. Before the measurment you didn't have that ensemble i.e. you cannot talk about a sub-ensemble. So what allows me to calculate conditional probability is the preparation of a new ensemble.

Sure, collapse is a form of state preparation. The point of collapse is that it links measurement and state preparation. Without the collapse, you do not have that link.

 

[vanhees71]

So before A's measurement, her spin is in the state $\frac 1 2 \hat 1$. Then she measures her spin and gets the result +, so she knows that her spin is pointing upward. But she's actually wrong and her measurement is meaningless because her spin is still in the state $\frac 1 2 \hat 1$ and not $|S_z;+\rangle$? This doesn't make sense!

If you have a usual polarization filter and the detection of the particle doesn't destroy the corresponding spin state then of course A will associate the state represented by $|\sigma_z=1/2 \rangle$ with it. What else? I only say that this is an association of the state that Alice does to her particle and I don't think that anything instantaneous happens to Bob's particle, because this contradicts the dynamics of relativistic quantum fields, which by definition is microcausal, i.e., local field operators that represent observables commute at space-like distances and thus the local interaction of the particle with A's measurement apparatus doesn't change immediately B's particle in any way. That's math and not subject to any interpretational issue!

 

[ShayanJ]

If you have a usual polarization filter and the detection of the particle doesn't destroy the corresponding spin state then of course A will associate the state represented by $|\sigma_z=1/2 \rangle$ with it. What else? I only say that this is an association of the state that Alice does to her particle and I don't think that anything instantaneous happens to Bob's particle, because this contradicts the dynamics of relativistic quantum fields, which by definition is microcausal, i.e., local field operators that represent observables commute at space-like distances and thus the local interaction of the particle with A's measurement apparatus doesn't change immediately B's particle in any way. That's math and not subject to any interpretational issue!

But saying "this is an association of the state that Alice does to her particle", means you think the wave-function is subjective and only describes the observer's knowledge about the system!

 

[vanhees71]

Sure, what else should it describe?

 

[ShayanJ]

Sure, what else should it describe?

Its just that in post #68, you insisted that "the state is an objective description of the system".

 

[stevendaryl]

Sure, what else should it describe?

The wave function of an electron might describe facts about an electron, maybe?

 

[vanhees71]

Sure, it describes objective facts about the electron; objective facts the physicist knows due to the (equivalence class of) preparation procedures defining the state. I think, all this is a pretty empty debate about semantics, as usual in such discussions about "interpretation".

I think it's pretty clear that A in our example knows something different than B, because she has done a "preparation procedure" with her particle by determining its spin component. Due to the $\sigma_z$-entangled state the two-particle system was prepared before, she knows also B's $\sigma_z$, but B doesn't know it, before he has measured it and just finds with 50% probability the one or the other outcome. So everything is consistent without any necessity to envoke "spooky action at a distance", which is indeed not implemented in the dynamics of the theory by construction, since we use a relativistic, local, microcausal QFT.

There'd be only a problem with this "interpretation" (which is just what the formalism, particularly Born's rule tells us, nothing else, and in this sense it's "minimal") if it would make a difference for B whether or note A measures her particle's $\sigma_z$ first or not, but it doesn't. So it's all consistent.

 

[stevendaryl]

Sure, it describes objective facts about the electron; objective facts the physicist knows due to the (equivalence class of) preparation procedures defining the state. I think, all this is a pretty empty debate about semantics, as usual in such discussions about "interpretation".

It's an empty debate because there is no real distinction between your position and those who believe that observation collapses the wave function.

What is a "preparation procedure"? Can you define it without invoking either the macroscopic/microscopic distinction, or the observer/observed distinction?

 

[stevendaryl]

So everything is consistent without any necessity to envoke "spooky action at a distance", which is indeed not implemented in the dynamics of the theory by construction, since we use a relativistic, local, microcausal QFT.

I don't see how the dynamics of QFT is relevant. There are two aspects to quantum theory: (1) Smooth evolution of the wave function (or smooth evolution of the field operators, in QFT), and (2) the Born interpretation of quantum amplitudes.

The issue is whether the combination is local, not whether the smooth evolution part is local.

 

[atyy]

Sure, it describes objective facts about the electron; objective facts the physicist knows due to the (equivalence class of) preparation procedures defining the state. I think, all this is a pretty empty debate about semantics, as usual in such discussions about "interpretation".

I think it's pretty clear that A in our example knows something different than B, because she has done a "preparation procedure" with her particle by determining its spin component. Due to the $\sigma_z$-entangled state the two-particle system was prepared before, she knows also B's $\sigma_z$, but B doesn't know it, before he has measured it and just finds with 50% probability the one or the other outcome. So everything is consistent without any necessity to envoke "spooky action at a distance", which is indeed not implemented in the dynamics of the theory by construction, since we use a relativistic, local, microcausal QFT.

There'd be only a problem with this "interpretation" (which is just what the formalism, particularly Born's rule tells us, nothing else, and in this sense it's "minimal") if it would make a difference for B whether or note A measures her particle's $\sigma_z$ first or not, but it doesn't. So it's all consistent.

But if collapse is a preparation, and the preparation prepares an objective state, then collapse would seem to be "objective" in your terminology.

 

[stevendaryl]

But if collapse is a preparation, and the preparation prepares an objective state, then collapse would seem to be "objective" in your terminology.

It occurs to me that there is (perhaps) a subtle distinction between measurement and preparation. For example, when you send a stream of electrons through a Stern-Gerlach device, those with spin-up are deflected one way and those with spin-down are deflected the other way. You haven't measured the spin of any electron, though.

Then you can send those electrons deflected in one direction to go on to a second Stern-Gerlach device, with a different orientation. Those electrons split into two groups, as well.

Eventually, you will (presumably) do a real measurement, by checking for the presence/absence of an electron. But all the selection prior to this doesn't involve measurement, and presumably doesn't collapse the wave function.

So I think that vanhees71 might be saying that it is possible to do experiments so that there is only one real measurement/observation, at the very end. So you don't need collapse (because you don't do any further experiments on the electron after the actual observation).

 

[atyy]

It occurs to me that there is (perhaps) a subtle distinction between measurement and preparation. For example, when you send a stream of electrons through a Stern-Gerlach device, those with spin-up are deflected one way and those with spin-down are deflected the other way. You haven't measured the spin of any electron, though.

Then you can send those electrons deflected in one direction to go on to a second Stern-Gerlach device, with a different orientation. Those electrons split into two groups, as well.

Eventually, you will (presumably) do a real measurement, by checking for the presence/absence of an electron. But all the selection prior to this doesn't involve measurement, and presumably doesn't collapse the wave function.

So I think that vanhees71 might be saying that it is possible to do experiments so that there is only one real measurement/observation, at the very end. So you don't need collapse (because you don't do any further experiments on the electron after the actual observation).

Not all preparations involve measurements, but some measurements are preparations.

 

[vanhees71]

I don't see how the dynamics of QFT is relevant. There are two aspects to quantum theory: (1) Smooth evolution of the wave function (or smooth evolution of the field operators, in QFT), and (2) the Born interpretation of quantum amplitudes.

The issue is whether the combination is local, not whether the smooth evolution part is local.

Well, if you only consider non-relativistic QT, there's no problem with a collapse concerning causality. You an assume an instantaneous action at a distance without any problem, and then there's no debate to begin with, i.e., you can use collapse arguments without contradictions.

The final sentence doesn't make sense to me. What do you mean by "local" here. Of course, in relativistic QFT by construction all interactions are local in space and time (you write down a Lagrangian with field operators multiplied at the same space-time point only). What's in some sense "non-local" in QT is related to our debate and entanglement, but that I'd not call "non-local" but long-range correlations between parts of a quantum system. One must not misunderstand long-range correlations with non-local interactions at a distance!

 

[stevendaryl]

Not all preparations involve measurements, but some measurements are preparations.

Yes. The tricky thing about applying the collapse hypothesis is that it's rare that you can do nothing more than measure an observable. To measure the position of an electron, you might have the electron collide with a photographic plate and see where the dot is. But that's a destructive measurement. The electron is gone afterward (absorbed by the photographic material).

That's what's special about entangled systems: You can perform a destructive measurement on one subsystem and that counts as a non-destructive measurement of the other subsystem.

 

[vanhees71]

Yes, and this is the truly interesting feature. One thing you can do nowadays, which sounds trivial first, is that you can prepare heralded single photons, i.e., a true single-photon Fock state by creating an entangled photon pair by parametric downconversion and measure one of the photons (the "trigger photon"), and then you now with certainty that you also have another photon (the "idler photon"), even with a specific polarization when the trigger photon's polarization state was determined by the measurement (putting a usual polarizer or other "optical elements" like quarter-wave plates before the detector). As usual, here you have preparation by filtering, i.e., you consider only a subensemble of many prepared photons selecting the wanted ones by a measurement. The point here is that you sort out the photons you want by this preparation procedure, but the measurement of the idler is not the cause of the idlers state, but the cause is that in the very beginning the two photons were prepared in the entangled two-photon state via parametric downconversion, and then you just sort out what's unwanted!

 

[atyy]

Well, if you only consider non-relativistic QT, there's no problem with a collapse concerning causality. You an assume an instantaneous action at a distance without any problem, and then there's no debate to begin with, i.e., you can use collapse arguments without contradictions.

The final sentence doesn't make sense to me. What do you mean by "local" here. Of course, in relativistic QFT by construction all interactions are local in space and time (you write down a Lagrangian with field operators multiplied at the same space-time point only). What's in some sense "non-local" in QT is related to our debate and entanglement, but that I'd not call "non-local" but long-range correlations between parts of a quantum system. One must not misunderstand long-range correlations with non-local interactions at a distance!

But the interesting point is that there is no problem with collapse in relativistic QFT.

If you think there is a problem, then you are using the wrong definition of causality.