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OK, ES, you are giving me the same argument you gave me about 10 posts back or so but in an allegedly different context, so we don't seem to get anywhere.

So let me start by making some comments.
1. If you consider just one state |S> you are circumventing the quantum nature of the system. You need a complete set of eigenvectors. A system with only one state is a gross classical system. By gross I mean that you are not even in the statistical regime where to a (classical) macrostate you have an infinye number of (still classical) microstates. Or do you claim that the state |S> is in fact a superposition of elementary states? If yes, mixed or pure?

2. Second of all, you seem to ignore totally the measuring system, or the measurement process. Having observables is not the same thing as measuring them. Having observables only means that you have something that can be consitently measured by different observers in different conditions (i.e. covariance or invariance, as the case may be). And you cannot include the measurement into the observables, because if you incorporate them into the observables, the new self-measuring observables are not abelian anymore, even if unitarity may be preserved. the question is why do you ignore the measurement process?

3. Third of all, averaging on states is not a measurement, according to the postulates of QM. Why did you decide to use the averaging procedure?

Sure, if you have one system with one state (!) and commuting observables, of course the order of the measurements is irrelevant, since you have in fact nothing else than a classical system for any practical purpose. And this is also true if you consider a system of uncorrelated particles where the wavefunction is the tensor product of two pure states.

But then, I fail to see the connection between your argument in the form you presented it now and previously and the issue of entanglement (and its relation to causality). So enlighten me further. And when developing your arguments, try to remember what Murphy said about someone who does twice the same thing and expects different outcomes.

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--- 1. do you claim that the state |S> is in fact a superposition of elementary states? ---
ES: Sure, but we do not care what the "elementary" states are. One can use any of the infinite possible "representations", if you know what it means.

--- 2. why do you ignore the measurement process? ----
ES: I do no such thing.
Measured value in a state | S > is given
by < S | M | S >,
where M is a Unitary operator describing the particular measurement.
The state after measurement is: M | S >

--- 3. Averaging on states is not a measurement, according to the postulates of QM. ----

What you are talking about? There is no such thing as "averaging" in quantum mechanics.

Any physical observable m as represented by operator M, if measured in physical state | S >,
gives < S | M | S > as a result.

Ask your mama :p

ES

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Any physical observable m as represented by operator M, if measured in physical state | S >, gives < S | M | S > as a result
Well, only the eigenvalues of M can be found in experiments. The expectation value is < S | M | S> .

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Quote:
Originally posted by Count Iblis:
Quote:
Any physical observable m as represented by operator M, if measured in physical state | S >, gives < S | M | S > as a result
Well, only the eigenvalues of M can be found in experiments. The expectation value is < S | M | S> .
I would keep this difference aside, for now.

You should agree nevertheless that we have a proof, that "expectation value" of a measurement would not change, regardless whether some other commutting measurement was performed before it or not performed.


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That's pretty trivial, I'm not sure what you're trying to prove. If it is about deterministic theories underlying QM, then you may be interested in doing research on this stuff. Prof 't Hooft is looking for a Ph.D. students and postdocs to work on deterministic theories, see here for a lecture.

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Quote:
Originally posted by Count Iblis:
That's pretty trivial, I'm not sure what you're trying to prove...
It is trivial.
But, people are claiming existence of "quantum entanglement", meaning that a measurent on one "entangled" particle, changes the state of other entangled particle.
It is a patently false claim, do you agree?

ES

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Pasti: ?1. do you claim that the state |S> is in fact a superposition of elementary states??

ES: ?Sure, but we do not care what the "elementary" states are. One can use any of the infinite possible "representations", if you know what it means.?

I am not talking about reps per se, I am just talking about the fact that the Hilbert space of states has a basis, and you can expand any state in such a basis. And since you already claim to have an algebra of observables, you could as well take the basis to be the complete set of the eigenstates of the abelian obsaervables. You want to call this ?representations?, be my guest. But it would be really interesting to see how you associate observables with unitary operators (as you claim you do) without such ?reps?.

So you say that your state |S> is in fact the superposition of eigenstates, so |S>= {Sum} c(n) |S(n)>, with complex coeffs c(n), and with the normalization <S|S>=1

Pasti: ?2. Why do you ignore the measurement process??

ES: ?I do no such thing. Measured value in a state | S > is given
by < S | M | S >, where M is a Unitary operator describing the particular measurement.
The state after measurement is: M | S >?

Sure you do. You are basically ignoring the measurement process even in its most elementary form. From the Postulate #? (I can look up the number if you so desire) of quantum mechanics you know that if you have a state |S> as you say, which is a superpostion of eigenstates, the results of the measurement of say observable M is

M|S>= m(i)c(i)|S(i)> (no summation)

i.e. you leave the system in the state |S(i)> after the measurement, and the state |S(i)> occurs with the probability c(i)c(i)*.So in your argument, M|S> is an eigenstate of M, which you have omitted to mention. So let's now elaborate along the lines of your reasoning.

A.If you measure first M1, you have

M1|S>= m1(n)c(n)|S(n)> (no sum)

and if you now measure M2 after the measurement on M1 has been performed, you have

M2M1|S>= M2(M1|S>)= m1(n)c(n)M2|S(n)>= m1(n)m2(n)c(n)|S(n)> (no sum)

which gives you the average value (not that it makes sense in this context)

<S|M2M1|S>= m1(n)m2(n)c(n)c(n)* (no sum)

B. If you measure now M2 first, you have

M2|S>= m1(p)c(p)|S(p)> (no sum)

where in general p is different from n. Along the same lines you have

M1M2|S>= M1(M2|S>)= m2(p)c(p)M1|S(p)>= m1(p)m2(p)c(p)|S(p)> (no sum)

And once again, the ?average?
<S|M1M2|S>= m1(p)m2(p)c(p)c(p)* (no sum)

I believe that it is clear now that the order of measurements for individual measurements matters in terms of the final result unless the very special case where n=p. This case can of course occur, but only as a concidence, if you know what I mean.

Your argument holds true only if you have ?prepared? your system before measurement, so to speak, in the state |S> which is an eigenstate of both operators M1 and M2. But then, the averaging (expectation value) makes no sense anymore since you are doing deterministic measurements, and there is no entanglement involved by construction (since you have already projected your state onto some convenient eigenstate).


Pasti: ?3. Averaging on states is not a measurement, according to the postulates of QM.?

ES: ?What you are talking about? There is no such thing as "averaging" in quantum mechanics. Any physical observable m as represented by operator M, if measured in physical state | S >, gives < S | M | S > as a result.?

You have no clue what you are talking about. A measurement of an observable implies collapse of a wavefunction into an eigenstate, hence

M|S>= m(i)c(i)|S(i) (no summation)

while calculation of the expectation value, or of the average value of the observable involves no such collapse, i.e.

M|S>={Sum} m(i)c(i)|S(i)>

such that in the end you get:

<S|M|S>={Sum} m(i)c(i)c(i)*

The latter expression can be interpreted in two different ways (and once again, I am not using any fancy concepts, just elementary quantum mechanics), and usually is considered to be the link between quantum mechanics and the classical theory:
(i) for a very large number of individual measurements of the observable M, <S|M|S> is nothing else than the average of the quantity m(I) that occurs with probability c(i)c(i)*, showing the statistical nature of quantum mechanics (here I have considered a discrete spectrum, but the same conclusion holds if the spectrum of the operators is continuous or mixed).
(ii) For any individual measurement process, the probability to obtain the value m(i) is of course c(i)c(i)*.

I think that it has also become clear why applying the average/calculating the expectation value of an observable on an eigenstate does not make too much sense.

So, going back to your model and your reasoning, your argument is nothing else than a particular case of more general argument. Furthermore, the conclusions of your argument do not hold for the general case, as you can see from the above, but you seem to have incorrectly generalized them.

Even worse, I still don?t see the relation of your argument to entanglement and to subsequent violations of causality.

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Quote:
Originally posted by Pasti:
...
Read Dirac's "Principles of Quantum Mechanics".

States are not needed to describe anything of substance.
If you need a notion of state to describe your concept or theory, they are wrong.

ES

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Quote:
Originally posted by extrasense:
Quote:
Originally posted by Count Iblis:
That's pretty trivial, I'm not sure what you're trying to prove...
It is trivial.
But, people are claiming existence of "quantum entanglement", meaning that a measurent on one "entangled" particle, changes the state of other entangled particle.
It is a patently false claim, do you agree?

ES
From Pasti's reply I see that you either don't fully understand QM or that you have some deterministic agenda. I can understand why people don't agree with the projection postulate, because that seems to imply that if one particle of a two particle entangled state is measured, the other particle instantaneously collapses.

However, you can't just get rid of the projection postulate in the way you suggest, because that would lead to a theory that doesn't agree with experiment (it is an experimental fact that only eigenvalues of operators can be the result of experiments).

You can get rid of the projection postulate by including the observer in the QM description, which will lead to a many worlds theory, or by trying to explain QM as some effective statistical theory, like e.g. 't Hooft has suggested.

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Quote:
Originally posted by Count Iblis:
you can't just get rid of the projection postulate in the way you suggest, because that would lead to a theory that doesn't agree with experiment (it is an experimental fact that only eigenvalues of operators can be the result of experiments).
The measurement operators normally have continuous specter. In fact, I do not know of any example that it is not the case.
Therefore, it can not be "experimental fact" that "only eigenvalues of operators can be the result of experiments".

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ES: "Read Dirac's "Principles of Quantum Mechanics"."

How about you read von Neumann's book, the "Mathematical foundations of quantum mechanics"?

"States are not needed to describe anything of substance.If you need a notion of state to describe your concept or theory, they are wrong."

Huh? You sure you read Dirac's book? You cannot have quantum mechanics without a Hilbertian space of states.

But let's assume for a moment that indeed you don't need states, as you claim. Then please be so kind and describe the quantization procedure for a Hamiltonian system, and to make it simpler, a Hamitonian system without any type of constraints.

As for your answer to Iblis' comment: ES:"The measurement operators normally have continuous specter. In fact, I do not know of any example that it is not the case."

Let me see, angular momentum and spin have a discrete spectrum (not spectres), particle in a box has a discretre spectrum of states, energy in the H atom has a discrete spectrum of states, a lattice has a discrete spectrum of states (although they appear as quasicontinuous in the bands, unless hopping and percolation phenomena are taken nto account), the harmonic oscillator has a discrete spectrum of states, and I can give you plenty more examples.

You seem to once again confuse continuity of the state function in the configuration variables with the discreteness of the spectrum for an operator.
Let me spell it for you: say in the position representation for the H atom the wavefunction is roughly a function F(n,x), where n is the discrete principal quantum numbr and x is the configuration variable. F is discrete in n, but continuous in x. Which has nothing to do whatsoever with the fact that the spectrum of the postion operator is continuous. You should have known that.

ES: "Therefore, it can not be "experimental fact" that "only eigenvalues of operators can be the result of experiments"."

Hm,really? Then exactly how do you explain say, atomic spectra? You will agree with me that the spectra of atoms, molecules, etc. is discrete, and measurement of the energy of an atom involves discrete values of energy. Not to mention photon counting techniques.

And don't even dare to go into the issue of "measurement is not instantaneous, but an integral process" because if you do, your argument is not worth a dime.

So, cut the baloney and let's get back to the arguments above. Within the conceptual bounds that you yourself have imposed.

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Quote:
Originally posted by Pasti:
...
Read this, it is all you need to know about "projection postulate":

Von Neumann's projection postulate:
Sometimes also von Neumann's projection (or reduction) postulate,
stating that during a measurement of standard observable A the state vector |y>
undergoes a discontinuous transition |y> ---> |am>
if the measurement result is am,
is taken as part of the formalism of quantum mechanics.
I do not consider von Neumann's projection (or reduction) postulate
to be either a necessary or a useful property of a quantum mechanical measurement.
It is not necessary, because it is a consequence of a certain interpretation
of the quantum mechanical state vector
(viz. a realist version of the individual-particle interpretation)
that is dispensable.
It is not useful because it is not satisfied by many practical experimental measurement procedures.
For instance, an ideal photon counter detects photons by absorbing them.
Hence, ideally the final state of the electromagnetic field is the vacuum state
rather than the eigenvector of the photon number observable corresponding
to the detected number of photons.
Since photons that have survived the detection process are not registered at all,
it follows that the functioning of the photon counter even depends crucially
on not satisfying von Neumann's projection postulate:
it is operating better to the extent it violates the projection rule.
Consistency problems of quantum mechanical measurement, arising because of von Neumann projection,
could better be dealt with by abandoning the postulate as a measurement principle,
rather than by ignoring the existence of such well-tried measuring instruments like photon counters.


Dr Willem M. de Muynck

Department of Applied Physics
Eindhoven University of Technology
P.O.B. 513, 5600 MB Eindhoven
the Netherlands
Phone: +31 40 2474351
Fax: +31 40 2445253
E-mail: W.M.d.Muynck@tue.nl


Well said, doctor Muynck!

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ES: ''Therefore, it can not be "experimental fact" that "only eigenvalues of operators can be the result of experiments".''

You seem to have forgotten that the set of eigenvalues can be a continuum. Also, you must remember that the projection postulate assumes an ''ideal measurement''. The fact that the real world is more complicated cannot be used to argue that the basics of QM are wrong.

In the real world (assuming that QM remains valid), there is no collapse, but rather an interaction with the measurement apparatus. This causes the apparatus plus observer and rest of the universe to go into some entangled state with the measured system. The final state consists of a superposition of different states in which the observer has observed something different.

For each observer only the projection of the full state onto his subspace is relevant. So, ultimately the projection postulate is valid.

Dr Willem M. de Muynck is of course right, but that doesn't mean that you were too!


Similarly, in equilibrium statistical physics you implicitely assume properties like ergodicity. But I wouldn't mark a student who fails to calculate the partition function of a simple system because he says it's not valid because in real systems the assumptions are not 100% realistic.

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Quote:
Originally posted by Count Iblis:
Dr Willem M. de Muynck is of course right
Thank you.
He is right, that projection postulate must be dropped, as unnecessary and misleading.
This is all I am claiming.
In the quantum theory there are two sorts of rules.
1. The actual quantum rules, that are an essense of the theory and are absolute in it.
2. The euristics rules, that are used to come up with quantum solutions, starting from classic systems and concepts.

The euristics rules are helpful,but are not absolute and serve just as hints.
The result of their use has to be checked against experiment, and has to be accepted or rejected depending on outcome of said check.

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OK, so let's review. ES, you do not approve of the projection postulate, since it contradicts experimental evidence (maybe I should remind you that there is also experimental evidence to support it, but you seem to ignore these facts), and in support of your claim, instead of presenting your argument, you provide something that might resemble a reference.

First of all, let?s clear a few things, to avoid any further confusion.
a)The projection postulate as I presented and used it
M|S>= m(n)|S(n)> (no sum)
is the simplest(idealized) version of the measuring process, involves the loathed collapse concept, does not consider the measuring system as an integral part of the experiment, and IS NOT a von Newmann process.
b) The von Newmann process considers the measurement system as a quantum system and as a part of the measurement process, involves entanglement, and involves (once again) collapse on a final state of the entangled system but IT IS NOT the projection postulate as presented in traditional quantum mechanics.

Both of them are nothing else than more or less idealized models describing the reality of measurement, and as such of course they have limited applicability. And I urge you to read Iblis' comments regarding the conceptual use of such models.

So, now let?s review van Muynck?s argument. While the projection postulate does not apply indeed to his photon counter example, the von Neumann measurement process does, and from the bit you posted from his argument, I have to conclude that somehow he either does not distinguish between the ?Projection postulate? and the von Neumann measurement process, or he is simply wrong.
If he confuses the postulate with the von Neumann process, he is right, but this is of little comfort since he treats the measuring system pretty much classically, so the overall resulting picture is confusing at best. von Neumann himself pointed out this aspect somewhere in the late forties.
If he does not confuse the two processes, then he is simply wrong, since a von Newmann measurement process includes the measuring system as a quantum system, and as such will measure the final state of the total system as the state in which you have an extra M electrons present in the measurement system according to the quantum absorption process, the remaining non-absorbed photons (in accord to the statistical character/ behaviour of the total system). Of course, this is the correct outcome of the experiment since by knowing this state one ?observes? that there is a correlation between the number of photons in the system and the number of electrons that are measured. So the von Neumann measuring process indeed describes correctly the photon counter example. Don?t bother to tell me that that the argument is wrong, just go and read some elementary quantum mechanics, say Liboff?s book. You have in there all the details you need. And as I said before, it would not hurt you to read the papers by Adami.

Now, if you agree with Muynck and his argument (i.e. you advocate the elimination of the projection postulate) you have to deal with the much more complicated problem that while you can drop the postulate, you have to replace it at least with the von Newmann measurment procedure or with something better. And I am pretty sure that you don?t want to replace the postulate the postulate with von Neumann?s procedure, since in an earlier post I also tried to use the latter and you were displeased, to put it very mildly.That's the reason why I swithched back to the postulate.

If you disapprove of both these concepts/postulates/procedures, which is fine with me, what do you replace them with? Since you do consider these models unrealistic, and want to drop them, then you must replace them with a more realistic, in your opinion, model of measurement in order to carry out your argument(s). So, cutting down to the chase, how do you define the measurement of an observable of a system in a certain state? And be specific, since the goal is ultimately to return to your argument regarding the inexistence of entanglement, and how (the inexisting) entanglement violates causality.

Furthermore, you still need to explain the quantization procedure for a classical system that does not involve states (!), as you claimed earlier.

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Quote:
Originally posted by Pasti:
... you still need to explain the quantization procedure for a classical system that does not involve states (!), as you claimed earlier.
Sure,
You take classical Hamiltonian, which is the energy expressed as function of coordinates Xi, and momentums Pi.
You substitute Pi ==> i*h*d/dXi.
i*i=-1; h is Plank constant
This gives you quantum Hamiltonian in the coordinate representation. It acts on the Psy function of coordinates.

This is an example of euristic rules of quantum mechanics.
They only can be used as a hint, and are not a part of the quantum theory/normal quantum theory./
The "projection postulate" is an another example of euristic rules of quantum mechanics.
It can be used or dropped, and the quantum mechanics itself is not affected by that.

The problem arises,
when people are trying to use euristic rules of quantum mechanics,
as components of proofs or theories.
This has happen to the "quantum computing" apologists. Their "proofs" are a bunch of crap, since they do not understand the difference between euristic and actual rules of quantum mechanics.

ES

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OK, so from your statements above I will take it that we agree with Dirac's canonical quantization program. But in this case, you still need to explain the statement you made earlier that in a quantum theory you don't need a state(!). From your description above, you seem to include all the necessay ingredients, Hilbert space and so on and so forth, including the states of a quantum system.

And it sems you have missed entirely the other question I was asking you. If you drop the prjection postulate and the von Newmann process, what do you replace them with?You need to replace them with something, so please specify in as much detail what is the concept that you introduce in their place, so that we can continue the discussion about quantum entanglement.

ES:"This is an example of euristic rules of quantum mechanics.They only can be used as a hint, and are not a part of the quantum theory/normal quantum theory. The "projection postulate" is an another example of euristic rules of quantum mechanics.It can be used or dropped, and the quantum mechanics itself is not affected by that."

We seem to have a syntax problem now, but its fine, we can overcome this too.
I agree with you that we are dealing on a daily bases with more or less idealized models, which can never describe nature in totality. At least up to this moment, this seems to be the situation, and it is mostly a matter of philosophy if we will ever have a model describing totally and accurately nature. As such, this could be the subject of another thread, I would be very happy to discuss it, time permitting.

But I was under the impression that it was self-evident that we confine ourselves to those models describing the best the phenomena under consideration, and that are also in reasonable agreement with the "previous" physics (by this I agreement with concepts and agreement with observational evidence). So as Ibliss said, you cannot just use the blindly the difference between "real" reality and "model" reality to totally disregard/invalidate all conceptual models. I mean, of ceourse, you could, but then you are only left with a form of religion instead of science, with what mysticism calls the "awe", which does not explain anything per se.

As for the projection postulate, yes, I agree with you that in can be dropped, but as I said previously, you cannot just simply drop it and say that quantum mechanics can survive without it.It cannot, because the postulate, or von Neumann's measuring process introduce/describe the concept of measurement in the theory, and by simply dropping this concept off, without replacing it you end up with a "PRINCIPIALLY UNTESTABLE" theory, which is useless. You must replace the old measurement concepts with new ones.SO ONCE AGAIN, WHAT IS THE NEW CONCEPT OF MEASUREMENT THAT YOU INTRODUCE IF YOU DROP THE PROJECTION POSTULATE?

ES: "The problem arises, when people are trying to use euristic rules of quantum mechanics, as components of proofs or theories. This has happen to the "quantum computing" apologists. Their "proofs" are a bunch of crap, since they do not understand the difference between euristic and actual rules of quantum mechanics."

I agree that one can use stupidly a model out of its context, and get "odd" results. But I hope that by the argument you present above, your argument regarding entanglement is also a bunch of crap, so is mine, so is Iblis's, so is Bohr's, Einstein's, and so on and so forth. Furthermore, since we cannot actually know in detail the reality, this entire discussion about entanglement, is a bunch of crap. And incidentally (and believe me that I am not saying this because I want to be nasty or similar, it is just a direct extension of your arguments), so are your other arguments regarding the existence of life on Mars, for the same reasons.

Once again, let's cut down to the chase, and return to the argument regarding entanglement. You don't like the projection postulate. Fine. What do you replace it with?

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Quote:
Originally posted by Pasti:
You don't like the projection postulate. Fine. What do you replace it with?
If you follow my line of thinking, the "projection postulate" can not be a part of the quantum mechanics itself. So it must be dropped from the pure quantum side.
But, it still can be renamed as a "projection euristic rule", and remain as such in the quantum realm. Until something better invented :p .
The only change will happen to the validity of any proof that relies on the projection rule being be a "postulate". Such a proof is invalid.

ES

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Damn, I lost all I wrote. I don't have time anymore to rewrite everything so I will be rather brief.

No, you cannot apriorically discard heuristic proofs, because of something called "the predictive power" of a theory. When a theory is "designed" it is designed such that heuristic proofs agree with known observational data, and that by the means of these heuristic proofs you can make predictions. You or I or anyone else cannot just discard such proofs just because they are based on more or less idealized heuristic arguments. It is only observation that can decide whether a prediction is true or not. Remember the laws of planetary motion, the discovery of Pluto the precession of Mercury perihelion? They were all (at the time) predictions of certain theories.

But let's try now your argument without any heuristic rules. You have a system in a state |0> which we consider to be reproducibe, i.e. we can somehow prepare the system to be in such a state whenever we need it (it might take some trials to do that, but in the end we can find this state).
And we measure two quantities, M1,M2 which are just QUANTITIES, not operators, and I don't care if they commute or not since I don't measure them simultaneously anyway.
a) I measure first M1, and by measuring M1, the system is left in a state |0,1>. All this ket notation is absolutely meaningless,no eigenstates, no left and right products, nothing of the sort. I just use it as a convenient notation for the states. Now I measure M2, and after the measurement, the system is left in the state |0,1,2>
b) I measure first M2, which brings my sysytem into the state |0,2'>, and if afterwards I measure M1, the sytem will get ino the state |0,2',1'>

Now, the states |0,1>,|0,1,2>, |0,2'> |0,2',1'> are in the general case not the same by themselves or combinations, except for the particular case of a coincidence. And if you associate in an univalued manner states with the values of the observations/measurement, the result of the two measurements will be different.

Sure, if you make statistical measurements (i.e. if you measure both M1M2 and M2M1 a large number of times), you will obtain the same distribution of the values m1m2 and m2m1, but then, what does this tell you? Absolutely nothing because you are ignoring the "correlations" between the individual measurements, you understand what I mean. You are now making correlations on the entire enasmbles, which erases basically the individual measurement "correlations".

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Joined: Oct 2004
Posts: 427
Quote:
Originally posted by Pasti:
But let's try now your argument without any heuristic rules. .... we measure two quantities, M1,M2 which are just QUANTITIES, not operators, and I don't care if they commute or not since I don't measure them simultaneously anyway...
This is childish smile
The quantum theory represents ALL measureable characteristics as operators.

Relax,

es

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