Classical and Quantum probabilities

[seratend]

Can we say that the classical (Kolgomorov axiomatization) and quantum probabilities results are different, when we consider only the statistical results (i.e. the probability of events)?
Can we really distinguish a classical statistical result from a quantum one (i.e. the born rules)?

What are the real differences between these 2 mathematical tools when we consider only the experimental results (the statistics or if you prefer the frequencies of the events)?

Seratend.

 

[bw]

I don't know if I understand your question correctly, but I'll try.

According to usual probability rules, if A and B are two mutually exclusive events, then P(AUB) = P(A)+P(B). In words, if an event can occur via two mutually exclusive ways, then the probablity that the event occurs is the sum of the probability of each alternative.

Now consider the two slit experiment in quantum mechanics.

Let's say we see an electron on a certain spot on the screen. A is the event that the electron has arrived through the top hole, B is the event that it has passed through the bottom hole. A and B are two mutally exclusive alternatives. The meanings of P(A) and P(B) are self evident.

So as above P(AUB) =P(A)+P(B). AUB is the event that the electron has passed through exactly one hole.

Mathematically this is perfectly sound.

But in order to identify the left hand side of the equation as the probability of observing an electron we made a crucial PHYSICAL assumption informed by classical mechanics, which turns out to be wrong.

The assumption is that the electron must have gone through a hole and one hole only in order to arrive at the screen. That is, we mistakenly assumed the detection of the electron is equivalent to the event AUB, which is not!

The probability P(AUB) is STILL CORRECTLY given by the sum rule.Only that AUB does not represent the physical situation. P(AUB) = P(A)+P(B) still holds. But P(AUB) is NOT the probability of detecting the electron because the event that you detect an electron's is not given by AUB!

On the other hand, if you put detectors around the holes to track the electrons, then, indeed, the arrival of an electron at the screen implies it must have gone through exactly one hole. In this case the event AUB and the detection of the electron are equivalent.Hence the probability of observing an electron is P(A)+P(B)

The often made but incorrect claim that quantum mechanics invalidates the logical rule of "exclusion of the middle" basically stems from the same fallacy.
The rule always holds. Quantum mechanics merely says that the "either -or" assumption may not apply.

IMO there is no such thing as "quantum statistics" different from "classical" statistics. There is only one statistics and it applies even in quantum mechanics (OK, two if you consider Baysian v.s frequentist statistics). You only need to be careful with your physical assumptions when you use your
mathematical tools.

 

[kleinwolf]

Another difference between both is expressed by Bell's inequalities.

 

[SpaceTiger]

Can we say that the classical (Kolgomorov axiomatization) and quantum probabilities results are different, when we consider only the statistical results (i.e. the probability of events)?

I would say that the difference, as we think we understand it, is that in classical systems the probabilities arise because we lack information. For example, in thermodynamics, we don't know the properties of each individual particle. If we did, we wouldn't need to talk about probability. In the popular interpretation of QM, however, the probabilities are a fundamental characteristic of nature and do not arise from a lack of information.

 

[seratend]

I don't know if I understand your question correctly, but I'll try. .

I just want to see if we are able to distinguish the quantum and classical probability results (I would like to have a good feedback from this community).
I just separate the problem of the time evolution or simply the evolution of a probabilistic state from the probability results/statistics (i.e. the number of times we get the same event in the frequentist view).

By classical probability, I just mean the mathematical tool. I am not assuming that the system under study evolves under classical or quantum mechanical laws (i.e “the time evolution of the probability lax).

In order to avoid extra complications/confusions (as long as it is not needed), we may try to focus on the frequentist approach in this thread (used in both classical and quantum experiments to compute the probabilities).

Now consider the two slit experiment in quantum mechanics.

Let's say we see an electron on a certain spot on the screen. A is the event that the electron has arrived through the top hole, B is the event that it has passed through the bottom hole. A and B are two mutally exclusive alternatives. The meanings of P(A) and P(B) are self evident.
So as above P(AUB) =P(A)+P(B). AUB is the event that the electron has passed through exactly one hole.
Mathematically this is perfectly sound.

It is a good example to start understanding the initial questions. You have already made some implicit assumptions, where I think you may mix the time evolution (the path of electron) with the results of the experiment: the spot of the electron on the screen.
In classical probability we have formally the event SPOT, the result of the experiment (the electron has been detected on the screen at a given position labelled by the event SPOT). We thus have the probability P(SPOT). If we introduce the classical random variable X_screen (the position of the electron on the screen, we have P(SPOT)=P(X_screen=SPOT).
By saying that, I am not saying that X_screen is the path of the electron, just the detected impact of the electron on the screen.
Therefore P(X_screen=spot_x) is the interference probability result of the experiment expressed in a classical probability form (where spot_x are the different possible locations on the screen).

Now, you have the legitimate right to ask for the conditional probability P(A) and P(B). But why do you want that P(AUB)=P(A)+ P(B)? Classical probabilities deal only with the results of experiments and not with the [classical/quantum] path of particle (implicit assumption to get P(AUB)=P(A)+P(B)).
Moreover, Classical probability does not forbid that, if you measure the “which slit the electron passes” event, the probability law is changed as in QM so that the interferences disappear. (We have an evolution of the probability law due to an active measurement at the slits as in QM: it is the freedom of classical probability as well as QM probability).

If I connect this classical approach to the QM approach, I may also, formally, compute the statistics at the slits *without* doing any measurement on the slits. I will therefore have:
* P(A)= |<slitA|psi>|^2
* P(A)=|<slitB|psi>|^2
* P(X_screen=SPOT)=|<SPOT|psi>|^2 where |SPOT> is the position of the spot on the screen
* P(X_screen=spot_x)= |<spot_x|psi>|^2 is the interference pattern (where spot_x is the position on the screen).

In both cases case if we do an active measurement on the slits, we loose the interference pattern on the screen (evolution natively included in the Qm formalism but external to the classical probability formalism).

But in order to identify the left hand side of the equation as the probability of observing an electron we made a crucial PHYSICAL assumption informed by classical mechanics, which turns out to be wrong.

Why do you want to use classical mechanics with classical probabilities. You are free to allow other evolutions (e.g. bohmian mechanics or whatever else): we are trying to see if we can differentiate the results of the laws.
I use the word classical probability, just to differentiate the way we compute the probabilities results in the QM or in a Kolgomorov axiomatic basis. If you prefer, we can use the word “Kolgomorov probability” instead of the word “classical probability” in order not to confuse the evolution of a system with the probability results.

IMO there is no such thing as "quantum statistics" different from "classical" statistics. There is only one statistics and it applies even in quantum mechanics (OK, two if you consider Baysian v.s frequentist statistics). You only need to be careful with your physical assumptions when you use your mathematical tools.

May you develop? It is the centre of my question. Can we really distinguish quantum and classical statistics.
Moreover, Classical probability doe not intrinsically include the evolution of the probability law, while the QM formalism includes it natively. Because in classical probability the evolution of the law is external, can we say that it allows in fact a wider set of possibilities compared with the QM formalism and its forced law evolution (the measurement, not the SE time evolution)?


Seratend

 

[seratend]

Another difference between both is expressed by Bell's inequalities.

I think your are confusing the locality problem and the statistics by themselves. Bell's inequalities may be reproduced by the classical probability formalism:
E(a,b)= Int dPab(w)Aa(w)Bb(w) instead of the formula E(a,b)= Int dP(w)Aa(w)Bb(w).

Again: do not confuse "classical probabilty" with a classical system following the Newtonian time evolution on an unkonw state. (I must admit that my terminology is not the best one, sorry :). As I already said in a previous post, we can rename classical probability into Kolgomorovian probability to avoid confusions.

Seratend.

 

[seratend]

I would say that the difference, as we think we understand it, is that in classical systems the probabilities arise because we lack information.

However, we must not confuse the lack of information with the evolution of the knowledge of the system. Once again, by classical probability, I just mean Kolgomorov probability. The increase/decrease of information (the evolution of the knowledge) of the system may be controlled by SE or a Newton equation.

For example, in thermodynamics, we don't know the properties of each individual particle. If we did, we wouldn't need to talk about probability.

In thermodynamics, we just can say that the observed statistics of some systems may be described by a Newtonian type time evolution equation. However, for thoses systems, I would like to say that we cannot distinguish the obtained statistics (E, P,T, etc ..) from a quantum time evolution (i.e therefore my questions).

Seratend

 

[SpaceTiger]

However, we must not confuse the lack of information with the evolution of the knowledge of the system. Once again, by classical probability, I just mean Kolgomorov probability. The increase/decrease of information (the evolution of the knowledge) of the system may be controlled by SE or a Newton equation.

Yes, but with Newton's equations, there is no limit to amount of information you can potentially obtain. There is in QM. This is important because:

In thermodynamics, we just can say that the observed statistics of some systems may be described by a Newtonian type time evolution equation. However, for thoses systems, I would like to say that we cannot distinguish the obtained statistics (E, P,T, etc ..) from a quantum time evolution (i.e therefore my questions).

In traditional Kolmogorov probabilities, we can make measurements of any of the set of multiple quantities at one time, but in QM that is not always possible. The statement, "What is the probability that this particle will have position x and velocity v?" is meaningful with classical probabilities, but not in QM because you can't measure both simultaneously.

 

[seratend]

In traditional Kolgomorov probabilities, we can make measurements of any of the set of multiple quantities at one time, but in QM that is not always possible. The statement, "What is the probability that this particle will have position x and velocity v?" is meaningful with classical probabilities, but not in QM because you can't measure both simultaneously.

You are right but you must take into account that QM includes the law evolution in the measurements (the "projection postulate") while the Kolgomorov probabilities do not (external to the axiomatic formulation). Therefore, in QM, because of the projection postulate (evolution of the law), you have 2 kind of observables, the ones that does not change the law upon a measure, and the others that change the law (e.g. your x and v).
However your are free to define equivalent random variables under the kolgomorov formalism if you add an add hoc probability law under a measurement: this is almost what is done with the bohmian mechanics formal formulation (i.e when you remove the interpretation phylosophical aspects ; ).

If you look more precisely at your statement “you can't measure both simultaneously”, you must see that it is only significant, if you have a law evolution of the statistics (“the projection postulate”). Therefore, If I add an external law evolution to the Kolgomorov probabilities (formally, I have the right to do that) that behaves as the projection postulate of QM, do I retrieve the same statistics?
i.e. “(Kolgomorov probability + add hoc projection law evolution) ~ (QM probabilities)?”

I think so, but I am not sure (existence problem). It is why I appreciate your and the others feedback in PF.


Seratend.

 

[seratend]

I must add some precisions to my previous post. Instead of law evolution, it is more correct to say the evolution of the Kolgomorov probability space: after a measurement, we may formally consider the probability space updates into a new one through an ad hoc equivalent projection postulate (freedom of the Kolgomorov axiomatisation) to match the QM probability statistic updates (state evolution).
Note that we are also free (formally) to update the random variables instead of the probability space to match the state evolution of QM (analogue to the schroedinger and Heisenberg QM state representation).

Seratend.

 

[bw]

By classical probability, I just mean the mathematical tool. I am not assuming that the system under study evolves under classical or quantum mechanical laws (i.e “the time evolution of the probability lax).
Seratend

This is exactly my point. I am not sure why you're bringing in evolutionary law in the first place.You can ask whether a certain mathematical/statistical model is appropiate in solving a physical problem. But that is the problem of the physicists, not that of the statistician or the mathematician. Operator theory is "correct" regardless whether the rules of quantim mechanics are or not.

It is a good example to start understanding the initial questions. You have already made some implicit assumptions, where I think you may mix the time evolution (the path of electron) with the results of the experiment: the spot of the electron on the screen.
In classical probability we have formally the event SPOT, the result of the experiment (the electron has been detected on the screen at a given position labelled by the event SPOT). We thus have the probability P(SPOT). If we introduce the classical random variable X_screen (the position of the electron on the screen, we have P(SPOT)=P(X_screen=SPOT).
By saying that, I am not saying that X_screen is the path of the electron, just the detected impact of the electron on the screen.
Therefore P(X_screen=spot_x) is the interference probability result of the experiment expressed in a classical probability form (where spot_x are the different possible locations on the screen).
Seratend

I don't see the relevance of this comment. Of course I was talking about the EXPERIMENTAL OUTCOME (the classical event) of "detecting an electron". Probability can only be referred to the OUTCOMES of measurement. Not to the evolution(dynamics), which is completely deterministic(so I am not quite sure what do you mean by "the evolution of probability law")

I wasn't mentioning evolution(though I did later, implicitly in pointing out the inappropiateness of the sum rule) because it was irrelevant at that stage.

Now, you have the legitimate right to ask for the conditional probability P(A) and P(B). But why do you want that P(AUB)=P(A)+ P(B)?
Classical probabilities deal only with the results of experiments and not with the [classical/quantum] path of particle (implicit assumption to get P(AUB)=P(A)+P(B)).
Moreover, Classical probability does not forbid that, if you measure the “which slit the electron passes” event, the probability law is changed as in QM so that the interferences disappear. (We have an evolution of the probability law due to an active measurement at the slits as in QM: it is the freedom of classical probability as well as QM probability).
Seratend

But I explicitly said that P(AUB)= P(A)+ P(B) gives the correct answer in our example ONLY IN SO FAR AS classical assumptions apply, which allows us to identify the event "electron detected" with AUB. The fact that it is NOT a justifiable assumption(which I also explicitly stated) is a physics problem. This was exactly the point I was making.

The equation P(AUB)=P(A)+ P(B)
is always correct. You just can't use it in this instance. So this is not a question of probability theory not giving you the right answer. It is because you use it inappropiately.

"Classical" probability does not tell you how to assign probabilities to elementary events. It only tells you how to compute the probabilities of certain types of composite events(those that can be expressed through Boolean operations) when the probabilities of elementary events are given.

The questions of what should be considered elementary events and how to assign probability to such events are specific to the field you study. In QM measuring a system in pure state (described by a wave function) is an elementary event (hence in the two slit experiment you can't just add the probabilities, the event that you detect an electron is elementary). Born's rule tell you how to assign probabilities when mesauring such a system.

Economists, sociologists and gamblers have other definitions for "elementary events" and other rules to assign probabilities for these events.
Whether these rules are justified or not is not the problem of statistics.

Why do you want to use classical mechanics with classical probabilities. You are free to allow other evolutions (e.g. bohmian mechanics or whatever else): we are trying to see if we can differentiate the results of the laws.
I use the word classical probability, just to differentiate the way we compute the probabilities results in the QM or in a Kolgomorov axiomatic basis. If you prefer, we can use the word “Kolgomorov probability” instead of the word “classical probability” in order not to confuse the evolution of a system with the probability results.
Seratend

You use the same probability theory independent of what theory of physics you assume. Only that you have to be careful not to use the wrong formulae. Perhaps I wasn't making my pont clear enough. I was saying the sum rule is ALWAYS correct. In classical physics (if you track the electron) you get the correct answer because in this case "detecting an electron" is the same as the event AUB. But in QM you get the wrong answer with the sum rule, NOT because the rule itself is wrong, but because it is used inappropiatelty. In this case "detecting an electron" is an elementary event, whose probability has to be computed/assigned OUTSIDE of statistics(sum the amplitude and take mod square)

Moreover, Classical probability doe not intrinsically include the evolution of the probability law, while the QM formalism includes it natively. Because in classical probability the evolution of the law is external, can we say that it allows in fact a wider set of possibilities compared with the QM formalism and its forced law evolution (the measurement, not the SE time evolution)?
Seratend

I am not completely sure what do you mean by "evolution of probability law"(not evolution of probability, which is the study of stocastic processes)

If you suggest QM may offer some exciting new angles for statisticians and mathematicians to develope new theories in statistics, I would tend to agree(no doubt someone is already working on that) But I don't know in what direction one should explore.

 

[bw]

One more thing.

Statistics can't tell you whether a system is "truly" random or just behave as though it is(say, because of lack of information) It only offeres tests for sequences that appear to be "sufficiently random", but,a good pseudo random number generator(a completely deterministic machine) can fool a lot of statistics tests. There are many proposals for what may serve as a definition of "true randomness", but none is satisfactory for all concievable occasions as far as I know. There are some interesting attempts to define radomness in terms of complexity theory. But I don't know how they can be actually applied in, say a test.

Therefore we cannot actually say QM is "truly random". We don't know what randomness is!(I am not only saying that non local hidden variables may be at work, the very concept of "randomness" is not well understood, even theoretically!)

 

[bw]

In traditional Kolmogorov probabilities, we can make measurements of any of the set of multiple quantities at one time, but in QM that is not always possible. The statement, "What is the probability that this particle will have position x and velocity v?" is meaningful with classical probabilities, but not in QM because you can't measure both simultaneously.

No. "Traditional" probability theory has nothing to say about whether you can
measure quantities simultaneously or not. Again that is a physics question.

Probability theory only says IF the values of these observables are such and such, THEN blah blah...

 

[SpaceTiger]

No. "Traditional" probability theory has nothing to say about whether you can
measure quantities simultaneously or not. Again that is a physics question.

It has to do with whether the statements of the probability theory make sense for the physical model. Any other interpretation of his question leads to the trivial "of course math works" response. I think he's asking for something deeper than that.

 

[bw]

I must add some precisions to my previous post. Instead of law evolution, it is more correct to say the evolution of the Kolgomorov probability space: after a measurement, we may formally consider the probability space updates into a new one through an ad hoc equivalent projection postulate (freedom of the Kolgomorov axiomatisation) to match the QM probability statistic updates (state evolution).
Note that we are also free (formally) to update the random variables instead of the probability space to match the state evolution of QM (analogue to the schroedinger and Heisenberg QM state representation).

Seratend.

You mean like a classical phase space but instead of the Liouville flow you have discontinuous flow corresponding to state reduction(?) What information can you gain from that, as you can only "update" after you make a measurement(of course)?

 

[bw]

Actually, now that I think of it I did go to a seminar talk on quantum probability by a mathematician. He was basically trying to incoporate some sort of uncertainty principle type constriants to his system of axioms.I don't know if that led to any interesting result, he didn't seem to claim that he had any new discovery other just setting up the system, But I can't really member the details.

The talk was really bad. The guy got his slides and notes mixed up with a different talk and ended up confusing everyone and the most confused person appeared to be himself. The talk ended in a Sienfeld moment when the guy standing there red faced, hair all messed up with slides all over the bench. We joked that that was the uncertainty principle at work.

 

[SpaceTiger]

If you look more precisely at your statement “you can't measure both simultaneously”, you must see that it is only significant, if you have a law evolution of the statistics

Perhaps you could be more specific, as I'm not sure how you intend to describe the combined probability distribution of x and v by including a law for time evolution. Attempts have been made to describe it classically by including non-local "hidden" variables (perhaps these are the extra parameters you're describing), but this is rather ad hoc. On the other hand, that seems ok with you, so I suspect the answer to your question is yes, it can be done. That doesn't mean, however, that it should be done.

 

[seratend]

This is exactly my point. I am not sure why you're bringing in evolutionary law in the first place.You can ask whether a certain mathematical/statistical model is appropiate in solving a physical problem. But that is the problem of the physicists, not that of the statistician or the mathematician. Operator theory is "correct" regardless whether the rules of quantim mechanics are or not.

In the beginning of this thread, even if not correctly expressed (or somewhat fuzzy :), I would like to know if we can distinguish the statistical results of quantum probability (born rules + projection postulate) from the one derived from the Kolgomorovian probability.
The example you gave is correct, but why do you want to mix Kolgomorov probability results with the implicit behaviour of a classical (“Newtonian”) electron? You have implicitly connected the probability slits results at the slits with the probability results at the screen. You have 2 probability laws of 2 random variables. The way you link them is what I call your implicit formal assumption.
Formally, these statistics are not logically connected if you do not add an external assumption (e.g. formally projection postulate, conditional probability, time evolution, etc …).
As you said, we are not considering the time evolution of the system, but rather the evolution of (or the link between) statistics knowing certain results. The projection postulate of QM as well as the conditional probability in Kolgomorov probability include both an implicit formal “evolution” (you can choose a better word) of the statistics (formal view). In Kolgomorv probability you have the basic link P(A=a|B=b) to connect the new probability law to a given result. However, we are free (formally) to build objects that modify the probability space under some experiment results in order to reflect the QM projection postulate of non commuting observables events.

But I explicitly said that P(AUB)= P(A)+ P(B) gives the correct answer in our example ONLY IN SO FAR AS classical assumptions apply, which allows us to identify the event "electron detected" with AUB. The fact that it is NOT a justifiable assumption(which I also explicitly stated) is a physics problem. This was exactly the point I was making.

"Classical" probability does not tell you how to assign probabilities to elementary events. It only tells you how to compute the probabilities of certain types of composite events(those that can be expressed through Boolean operations) when the probabilities of elementary events are given.
In classical physics (if you track the electron) you get the correct answer because in this case "detecting an electron" is the same as the event AUB. But in QM you get the wrong answer with the sum rule, NOT because the rule itself is wrong, but because it is used inappropiatelty. In this case "detecting an electron" is an elementary event, whose probability has to be computed/assigned OUTSIDE of statistics(sum the amplitude and take mod square).

I do not understand completely your statement (as well as I think I am not clear with my statements ; ). A classical probability space is a set of outcomes, a sigma algebra and a probability law. On this space, I may, formally, define any random variable. These random variables (that are not necessarily independent) define the elementary events of experiments based on these random variables (i.e. elementary events are defined respectively to a given random variable). As in QM probability (for a given observable), these elementary events are mutually exclusive for a given random variable (observable). A random variable induces a probability law on its own space as in QM probability. Only the state resulting from the assumption a given event is true seems to be different in QM and Kolgomorov probabilities (what i call the evolution of the law, which is not the best designation).
As an example, I recall the previous post with additional comments:
A is the elementary event “electron in slit A” and B is the elementary event “electron in slit B” for the random variable “electron position at slits” (we may call X_slit)
* (Kolgomorov) P(A)=P(X_slit=xa)= |<slitA|psi>|^2 (QM)
* (Kolgomorov) P(B)=P(X_slit=xb)=|<slitB|psi>|^2 (QM)

We have A inter B = empty set (because we associate them with the random variable X_slit) that is the same as <slitA|slibB>=0 when we consider the observable X_slit (|SlitA> and |SlitB> are eigen values of the observable X_slit).

In this case we have P(AUB)=P(X_slit={xa,xb})=|<slitA|psi>|^2+|<slitB|psi>|^2= P(A)+P(B).

It is the random variable that defines the elementary events in Kolgomorov probability and the observable in QM probability. Formally we have:
* in QM probability (A,|psi>) defines the probability law of the observable A events
* In Kolgomorov probability (A, probability space) defines the probability law of the random variables A events.

Therefore, choosing a probability space (~the law) or a state seems, formally, almost identical to get the same statistical results in an experiment.

I am not completely sure what do you mean by "evolution of probability law"(not evolution of probability, which is the study of stocastic processes)

If you suggest QM may offer some exciting new angles for statisticians and mathematicians to develope new theories in statistics, I would tend to agree(no doubt someone is already working on that) But I don't know in what direction one should explore.

I am just trying to see (for one aspect) if QM probabilities may be formally equivalent to Kolgomorov probabilities plus some additional evolution on the probability space (i.e. to match the projection postulate). If this is true, we can view QM probabilities as just a specific case of Kolgomorov probabilities (i.e. the projection postulate may be mapped into an update of the probability space).
Note that the statistics rules are given before the projection of the state in QM probability.

Seratend.

 

[seratend]

One more thing.

Statistics can't tell you whether a system is "truly" random or just behave as though it is(say, because of lack of information) It only offeres tests for sequences that appear to be "sufficiently random", but,a good pseudo random number generator(a completely deterministic machine) can fool a lot of statistics tests. There are many proposals for what may serve as a definition of "true randomness", but none is satisfactory for all concievable occasions as far as I know. There are some interesting attempts to define radomness in terms of complexity theory. But I don't know how they can be actually applied in, say a test.

Therefore we cannot actually say QM is "truly random". We don't know what randomness is!(I am not only saying that non local hidden variables may be at work, the very concept of "randomness" is not well understood, even theoretically!)

Well, I must admit that I do not understand exactly the term random alone.
The only results I know are the results of the strong law of large numbers that allow one to measure the probability law of events on a given experiment (frequentist view of probability).
However, the problem of the strong law of large numbers is that you get a convergence in law. That means, formally when you do an experiment, the peculiar sequence of results you obtain may belong to the events of null probability (i.e. they do not converge to the probability law).
Now, you can say that it is because you have not realized fully "random" trials to explain this result (i.e. if your sequence of trials is not fully random, therefore you may have a sequence of events that does not converge to the probability law). But it seems somewhat artificial.

Seratend.

 

[selfAdjoint]

It is at least arguable that there does not exist any truly random process in nature. Radioactive disintegration is presumed to be random, but AFAIK there is no evidence that it is not at some level deterministic. In a sense this is like the problem of deciding if a given physical length is irrational or rational. Physical measurements just aren't up to determining that.

 

[seratend]

I think he's asking for something deeper than that.

You are right even I am not clear. I just want to remove the interpretation problems (philosophy) form the discussion and keep logical statements when it is possible.

There is already a good trial to transpose QM probabilities into Kolgomorov probabilities. However, it is embedded with what seems to be unuseful interpretational problems and a complicated mathematical expression (mix of time evolution with probabilies and radom variables):bohmian mechanics (1952).

I am looking for a more basic view at the statistics without the time evolution.
In QM probability, when we remove the schroedinger equation, we just have the born rules and the projection postulate to compute probabilities. However, when we try to look at the differences between Kolgomorov and QM probabilities, we may view the projection postulate as an update of the probability space (or of the random variables). But may be, there is other possibilities.

One advantage of the possible formal identity between QM and Koglomorov probabilities may be a simpler explanation of the projection postulate (interpretation independent) as well as a simpler view into the decoherence results.

Seratend.

 

[seratend]

Actually, now that I think of it I did go to a seminar talk on quantum probability by a mathematician. He was basically trying to incoporate some sort of uncertainty principle type constriants to his system of axioms.I don't know if that led to any interesting result, he didn't seem to claim that he had any new discovery other just setting up the system, But I can't really member the details.

The talk was really bad. The guy got his slides and notes mixed up with a different talk and ended up confusing everyone and the most confused person appeared to be himself. The talk ended in a Sienfeld moment when the guy standing there red faced, hair all messed up with slides all over the bench. We joked that that was the uncertainty principle at work.

I was not that man ; ). My intention is not to bring any confusion. Just to see some logical statements about possible connections between 2 abstract theories.

Seratend.

 

[seratend]

You mean like a classical phase space but instead of the Liouville flow you have discontinuous flow corresponding to state reduction(?) What information can you gain from that, as you can only "update" after you make a measurement(of course)?

Yes you can view it, in most of the case, as the discontinuity on the liouville flow (as in QM you have a discontinuity in the density probability). However, we must not forget that the liouville flow assumes a peculiar decomposition of the probability law: we assume that the probability of a system is given by the density probability f(p,q)dpdq. In other words, the associated Hilbert space is |p>(x)|q> while in QM, we are only dealing with the Hilbert space |q> (note: it is why in Bohmian mechanics, it is added the path of a particle, in order to get a (q,p) phase space system).

We are dealing with a probabilistic view (choice). Both in Kolgomorov and QM probabilities, all we can do is an update of the state (or the probability space) after a measurement.

As said in previous posts, this approach can be viewed, under this peculiar aspect, as a way to explain, formally, the projection postulate (transposition into the Kolgomorov probability axiomatization).

Seratend.

 

[seratend]

Perhaps you could be more specific, as I'm not sure how you intend to describe the combined probability distribution of x and v by including a law for time evolution. Attempts have been made to describe it classically by including non-local "hidden" variables (perhaps these are the extra parameters you're describing), but this is rather ad hoc. On the other hand, that seems ok with you, so I suspect the answer to your question is yes, it can be done. That doesn't mean, however, that it should be done.

I am not trying (I think) to make the description of the joint probability of a certain (x, v) random variables. An attempt is already done with bohmian mechanics and I think it is not an interesting one (creation of a "virtual path" of a virtual particle).
Therefore, a good explanation should not need the addition of a new variable (not very efficient).

Seratend.

 

[slyboy]

OK, I have 2 contributions to this discussion.

1. The way the state is updated after a measurement is NOT fixed by the quantum formalism. The projection postulate is one possibility, just like updating by Bayes rule (conditioning) is one possibility in the Kolmogorov theory. However, generally there is the same sort of freedom in how quantum states can be updated as there is in the Kolmogorov theory (I can go into more detail if you wish).

2. Do quantum statistics conflict with Kolmogorov theory? The answer is a definite NO. Of course, it entirely depends on how you choose to set up your sample space. For example, if you choose a separate sample space for every possible observable and then just take a Cartesian product of them, you will end up with a space on which the quantum probabilities can be represented by a Kolmogorov measure.

However, any way you choose to set up such a space, you will end up with some strange features. For example, you can prove that there is no classical sample space rich enough to include the probability assignments from any quantum state (of a system with a fixed Hilbert space) such that all measures on that space correspond to probability assignments of quantum states. To put it another way, there will always be classical measures on the space that are not valid in quantum mechanics.

For this reason, amongst others, some people prefer to work with modified notions of sample spaces that do not obey the Kolmogorov axioms. This is called quantum logic and quantum probability. The lattice of projectors onto the closed subspaces of a Hilbert space is the simplest example of a quantum logic. The measures on this structure all correspond to probability assignments of quantum states, as was proven by Andrew Gleason.

In summary - you can't use QM to rule out the Kolmogorov formalism, but you can formulate the theory using an alternative axiomatization of probability if you choose to do so. It's a matter of taste. Personally, I quite like the quantum logic - quantum probability approach, but it has not had a big impact on physics as of yet,

 

[seratend]

1. The way the state is updated after a measurement is NOT fixed by the quantum formalism. The projection postulate is one possibility, just like updating by Bayes rule (conditioning) is one possibility in the Kolmogorov theory. However, generally there is the same sort of freedom in how quantum states can be updated as there is in the Kolmogorov theory (I can go into more detail if you wish).

Thanks for your contribution. And I would welcome more details.
I think you are speaking about POVM and generalized measurements, aren’t you? If this is the case, you always have (my knowledge) the projection of the state embedded into a more general model of measurement with the following principle (simplified formula: pure state etc ..., there are a lot of good papers on the more general form of POVM and generalized measurements):
|in> -> projection -> |m> -> unitary evolution ->|out> (state after the generalized measurement), where |m> is the state associated with the observed measured value.
i.e. |out>=UoP|in> where U is a unitary operator and P a projector.
Therefore, we always have natively the projection postulate (we can always assume, formally, that the unitary evolution U is part of a kind of dynamical time evolution, e.g. a dirac evolution, if not time is elapsed during the measurement model).
If you say that you have, in QM, a way to build formally other types of measurements that do not include natively a projection, I am interested on. I would welcome any precisions (currently I, personally, do not know any QM “measurement” that does not involve a projection in one step of the measurement).

2. Do quantum statistics conflict with Kolmogorov theory? The answer is a definite NO. Of course, it entirely depends on how you choose to set up your sample space. For example, if you choose a separate sample space for every possible observable and then just take a Cartesian product of them, you will end up with a space on which the quantum probabilities can be represented by a Kolmogorov measure. However, any way you choose to set up such a space, you will end up with some strange features.

That’s one of my questions. The projection of the QM state “after” a measurement (note that I am just speaking about the formal result, not the interpretations of this QM postulate) seems to require the update of the probability space (in the Kolgomorov axiomatization formulation) or the update of random variables.
Therefore, as you say, we may try, formally to make a Cartesian product of all these individual probability spaces (the sample spaces), however we obtain a non countable Cartesian product. The problem with non countable Cartesian products of sets is the object you obtain. This object may be very singular. I do not know the type of sigma algebra we obtain nor the probability law if it is indeed a probability law (problem of the sigma additivity on this sigma algebra). May be there is known theorems on that but I do not know them.

For example, you can prove that there is no classical sample space rich enough to include the probability assignments from any quantum state (of a system with a fixed Hilbert space) such that all measures on that space correspond to probability assignments of quantum states.

Can you tell us the theorem or the result you are dealing with? This seems to underline perfectly the problem of non-countable Cartesian products of sample spaces. You obtain an object where I do not know if it is a probability space, if the individual spaces follow the probabilities of the states of a given Hilbert space (problem of the sigma additivity on this new space).
However, this does not preclude the possibility to update externally the Kolgomorov probability space. I.e. we may not require the infinite non-countable Cartesian product to be a probability space (no reason).

To put it another way, there will always be classical measures on the space that are not valid in quantum mechanics.

You mean the inverse? Or may be I do not understand what you are saying.

For this reason, amongst others, some people prefer to work with modified notions of sample spaces that do not obey the Kolmogorov axioms. This is called quantum logic and quantum probability. The lattice of projectors onto the closed subspaces of a Hilbert space is the simplest example of a quantum logic. The measures on this structure all correspond to probability assignments of quantum states, as was proven by Andrew Gleason.

Yes, I have read some papers on it. I really appreciate the “formal” formalism (even if I am far from understanding the whole subject). However, we still have 2 parts in this formalism:

1) how to construct formally general measures P(B,rho)=tr(M(B).rho). Where B is an event (of a given observable, M(B) is a sigma additive function into the sets of orthogonal projectors and rho the state of the QM system.
2) the new state after the measurement.

The main point that bothers me is the state after the measurement. As P(B,rho) is a probability law ([0,1], sigma additivity, etc ...), it seems that the state after the measurement is only the [possibly external] feature added by the QM probability formalism when compared with Kolgomorov formulation. In other words, can we say that QM probability minus the state after the measurement is equivalent to the Kolgomorov probability?

In summary - you can't use QM to rule out the Kolmogorov formalism, but you can formulate the theory using an alternative axiomatization of probability if you choose to do so. It's a matter of taste. Personally, I quite like the quantum logic - quantum probability approach, but it has not had a big impact on physics as of yet.

Yes, but what I think interesting to know is the extra axioms added to the Kolgomorov probability to get the full compatibility with QM axioms (formal separation of the problems: probability calculation and state evolution). QM formalism or QM logic is efficient to compute many results. However, it has some drawbacks especially when we deal with the connection of discrete and continuous measures (i.e. the Hilbert space is not large enough, it requires less simple objects such as generalized Hilbert spaces to handle correctly the formulation). While in Kolgomorov formalism, it is formally evident and consistent (i.e. the probability law on discrete or continuous random variable).

Seratend.

 

[slyboy]

OK, I don't have time to respond to all your questions right now, so let me just deal with the first one.

I think you are speaking about POVM and generalized measurements, aren’t you?

Yes, that was what I was referring to - and you are quite correct to say that every generalized measurement update has a representation as a unitary operator acting on an extended system, followed by a projection, followed by a tracing out of the extra degrees of freedom that were added. However, you are quite wrong to say that this gives the projection postulate any special significance. There is an almost exact analog of this in classical probability theory. Any measurement of a random variable can be modeled by introducing a second random variable, performing a joint stochastic operation on the pair of variables, looking at the second random variable and doing a Bayesian update of the pair, and finally taking the marginal state of just the original variable. Try it for yourself - it is quite instructive to see how closely analogous the proofs are in the classical and quantum case.

However, if we are not going to use this classical result to argue that Bayes rule should be an axiom of Kolmogorov probability, and I presume from your earlier comments that you see no reason to do so, then there is no reason to do so for the projection postulate either. In fact, there is a perfectly adequate mathematical formalism of generalized measurements that does not refer to an extended system and replaces the unitary and projection with an arbitrary completely positive map (these are sometimes called quantum instruments in the more mathematical literature). This is just about the most general way you could hope to transform one density matrix into another, and is just the analog of applying an arbitrary stochastic map to a classical probability distribution.

 

[slyboy]

OK, time to respond to a couple more of your comments.

That’s one of my questions. The projection of the QM state “after” a measurement (note that I am just speaking about the formal result, not the interpretations of this QM postulate) seems to require the update of the probability space (in the Kolgomorov axiomatization formulation) or the update of random variables. Therefore, as you say, we may try, formally to make a Cartesian product of all these individual probability spaces (the sample spaces), however we obtain a non countable Cartesian product. The problem with non countable Cartesian products of sets is the object you obtain. This object may be very singular. I do not know the type of sigma algebra we obtain nor the probability law if it is indeed a probability law (problem of the sigma additivity on this sigma algebra). May be there is known theorems on that but I do not know them.

I'm not sure it is all that big a problem. Of course, I was only talking informally about such a sample space and it needs to be made more rigourous. One way would be to treat the measurement directions, or choices of observable, in the same way we treat any other continuous parameters in rigourous probability theory, e.g. we only treat the statistics of Borel sets of possible directions for example. Quantum measurement theorists do not usually bother with this, because most problems deal with a single "optimal" measurement or some finite collection of measurements. However, I don't think there would be an issue with setting this up formally. In fact, I wouldn't be surprised if one of the Japanese guys who worked on this stuff in the 80s had already done it, or some subset of Busch, Lahti and Mittlestadt. If not, it would make a good thesis project for a rigorous math inclined grad student.

Another thing is that we haven't proved that you need this uncountable Cartesian product - that was just one way of doing it. There could be some countable sample space, such that all quantum statistics can be derived from a Boolean sigma-algebra. I have never seen any theorems proving this one way or the other.

Another way out is to use the finite precision loophole. Remeber, we only ever control the observable we are measuring to finite precision and we only ever measure probabilities as relative frequencies. Therefore, we could replace the continuum of observables with some countable set without affecting the emprical adequacy of the theory. Admittedly, this would be somewhat inelegant, but it can never be ruled out, so it is an option for someone committed to preserving classical logic and probability.

Quote:
Originally Posted by slyboy
For example, you can prove that there is no classical sample space rich enough to include the probability assignments from any quantum state (of a system with a fixed Hilbert space) such that all measures on that space correspond to probability assignments of quantum states.

Can you tell us the theorem or the result you are dealing with? This seems to underline perfectly the problem of non-countable Cartesian products of sample spaces. You obtain an object where I do not know if it is a probability space, if the individual spaces follow the probabilities of the states of a given Hilbert space (problem of the sigma additivity on this new space).
However, this does not preclude the possibility to update externally the Kolgomorov probability space. I.e. we may not require the infinite non-countable Cartesian product to be a probability space (no reason).

Quote:
Originally Posted by slyboy
To put it another way, there will always be classical measures on the space that are not valid in quantum mechanics.

You mean the inverse? Or may be I do not understand what you are saying.

Well, it's basically a consequence of the Bell-Kochen-Specker theorem. It's actually a theorem about hidden-variable theories, but we can also discuss it more abstractly. You want tot set up a classical sample-space (i.e. Boolean sigma-algebra) such that quantum states are represented by classical probability measures on that space. Furthermore, the outcomes of sharp measurements are to be represented by characteristic functions on that space. Also, any outcomes that are always assigned the same probability by QM for any state are to be identified by the same characteristic function. Well, the BKS theorem says that this is impossible.

As a corollary, any Boolean sample-space that is adequate for QM must represent at least some outcomes that always have the same probability in QM by distinct characteristic functions. Then, there must exist measures on that space that assign the two outcomes different probabilities and we know that these do not arise in QM.

A simple example of this would be representing position and momentum measurements by a Cartesian product of a sample space for position and another one for momentum. This space has delta-function distributions that assign a definite position and a definite momentum, but we know that these do not arise in QM due to the uncertainty principle.

In other words, can we say that QM probability minus the state after the measurement is equivalent to the Kolgomorov probability?

It should be clear by now that the answer to this is not straightfoward. I think we can preserve the Kolmogorov axioms, but only by sacrificing considerable mathematical elegance.

Yes, but what I think interesting to know is the extra axioms added to the Kolgomorov probability to get the full compatibility with QM axioms (formal separation of the problems: probability calculation and state evolution). QM formalism or QM logic is efficient to compute many results. However, it has some drawbacks especially when we deal with the connection of discrete and continuous measures (i.e. the Hilbert space is not large enough, it requires less simple objects such as generalized Hilbert spaces to handle correctly the formulation). While in Kolgomorov formalism, it is formally evident and consistent (i.e. the probability law on discrete or continuous random variable).

Yes, I admit that current axiomatizations are far from adequate and there are still a number of challenging open questions. But, it is still a viable area for new research and we don't have to close the book on it yet. Even if we cannot remove the dependency on the Hilbert space formalism, which is the main point of these axiomatizations, I still think it is more elegant than trying to introduce Kolmogorv probability spaces.

 

[vanesch]

Oops, I now see that this has already been discussed...

So as above P(AUB) =P(A)+P(B). AUB is the event that the electron has passed through exactly one hole.

Mathematically this is perfectly sound.

But in order to identify the left hand side of the equation as the probability of observing an electron we made a crucial PHYSICAL assumption informed by classical mechanics, which turns out to be wrong.

The assumption is that the electron must have gone through a hole and one hole only in order to arrive at the screen. That is, we mistakenly assumed the detection of the electron is equivalent to the event AUB, which is not!

The probability P(AUB) is STILL CORRECTLY given by the sum rule.Only that AUB does not represent the physical situation. P(AUB) = P(A)+P(B) still holds. But P(AUB) is NOT the probability of detecting the electron because the event that you detect an electron's is not given by AUB!

On the other hand, if you put detectors around the holes to track the electrons, then, indeed, the arrival of an electron at the screen implies it must have gone through exactly one hole. In this case the event AUB and the detection of the electron are equivalent.Hence the probability of observing an electron is P(A)+P(B)

The often made but incorrect claim that quantum mechanics invalidates the logical rule of "exclusion of the middle" basically stems from the same fallacy.
The rule always holds. Quantum mechanics merely says that the "either -or" assumption may not apply.

IMO there is no such thing as "quantum statistics" different from "classical" statistics. There is only one statistics and it applies even in quantum mechanics (OK, two if you consider Baysian v.s frequentist statistics). You only need to be careful with your physical assumptions when you use your
mathematical tools.

I think that what is written here is very true: there is often a confusion between the hypothesis (a physical hypothesis) of some underlying classical mechanism, which can sometimes clash with QM predictions, and the fact that QM "violates Kolmogorov probabilities" (the last statement, I think, is wrong).

In fact, there IS already a completely classical model of quantum mechanics which gives identical probability results as QM: it is Bohmian mechanics !

The thing that Bohmian mechanics does not respect is *locality*. But that's physics, that's not probability theory. So I think that the very existence of Bohmian mechanics disproves the statement that QM probabilities "are not Kolmogorov probabilities".

cheers,
Patrick.

 

[slyboy]

In fact, there IS already a completely classical model of quantum mechanics which gives identical probability results as QM: it is Bohmian mechanics !

Bohmian mechanics is a good example of the kind of thing I am talking about. There your ontic state (state of reality) is just the particle positions and the quantum state vector. You just have an ordinary Kolmogorv distribution over particle positions. However, there are perfectly good Kolmogorov distributions over position that are not allowed in this theory, since they have to obey the ``equilibrium hypothesis'' in order to agree with the predictions of QM, i.e. the prob. distribution must be the one coming from the Born rule applied to the state vector.

One can, and I think one should, ask why the equilibrium hypothesis holds. We can either just state that it is a law of nature, which I think is inadequate as an explanation, or one can try to argue that nonequilibrium states will relax rather quickly to the equilibrium state, like in statistical mechanics (see http://www.arxiv.org/abs/quant-ph/0403034 for some evidence of this). This would be a better explanation, but if it is true then it would mean that Bohmian mechanics without the equilibrium hypothesis is not entirely equivalent to standard QM because it would predict that we ought to see nonequilibrium states somewhere in the universe (see http://www.arxiv.org/abs/astro-ph/0412503, http://www.arxiv.org/abs/hep-th/0407032, http://www.arxiv.org/abs/quant-ph/0309107, http://www.arxiv.org/abs/quant-ph/0104067).

 

[vanesch]

Bohmian mechanics is a good example of the kind of thing I am talking about. There your ontic state (state of reality) is just the particle positions and the quantum state vector. You just have an ordinary Kolmogorv distribution over particle positions.

Yes, this is the part of Bohmian mechanics that is sufficient to show that you do not NEED any "quantum probability rules" that go beyond Kolmogorov probability ; the very fact that it is possible to construct a model that gives you the same predictions as QM using Kolmogorov probability is sufficient to show that. Don't understand me wrong: I didn't want to say that Bohmian mechanics must therefore be "right" or whatever ; it is just that the very existence of that model, giving QM probabilities, and obeying Kolmogorov probabilities, disproves the statement that you cannot have QM probabilities satisfying Kolmogorov's axioms.

However, there are perfectly good Kolmogorov distributions over position that are not allowed in this theory, since they have to obey the ``equilibrium hypothesis'' in order to agree with the predictions of QM, i.e. the prob. distribution must be the one coming from the Born rule applied to the state vector.

Yes, this is correct ; however, it doesn't invalidate the claim that QM probabities CAN be generated by a system obeying Kolmogorov's axioms. If that system can also generate other probabilities (not those of QM) doesn't matter. The very fact that with the right distributions, you DO get out the QM predictions, is enough.

Just as a side note: this doesn't mean I endorse Bohmian mechanics. I have mixed feelings towards it, my main complaint being that the guiding equations do not obey the imposed symmetries on the wave dynamics (like Lorentz invariance). But I do think that it is very useful to study Bohmian mechanics because it is a model that disproves a lot of claims about QM, like the one we're discussing here, namely that "quantum probabilities" are somehow not "normal probabilities as we know them" (Kolmogorov axioms).

cheers,
Patrick.