Quantum physics introduces a new concept of probability which at first glance differs quite a bit from original mathematical concept of probability as defined by Kolmogorov. Sadly it is rarely explained how these frameworks relate to each other and if there may be something missing in mathematics. For a mathematician it feels even more annoying as there is no experiment that is able to challenge classic probability with a result it cannot reproduce.

It turns out that one can take great value to discussing such topics with AI and learn a lot from them, if one has deep enough knowledge of a lot of related topics for a proper discussion.

Now, the AI noted the Kochen-Specker and Bell inequality as the two cases where the fundamental differences become apparent. In either case a deeper investigation showed that both require additional physical assumptions to be made that are not native to probability theory which lead to the corresponding theorems.

For Kochen-Specker that is contextuality - where hidden variables are assumed to have a preexisting values. This assumption breaks down already in classical physics in the case a measurement cannot be done without perturbation of the underlying physical system and therefore possibly changing the state of underlying variables for consequent measurements. Such a behavior of non commuting operators we can reproduce via a classic Markov Decision Process, where each measurement is reflected by a decision (including the decision how to set input variables like the axis along which a detector measures spin) which then can reshuffle all underlying hidden variables / quantum state / wave function.

That leaves only Bell's or for this purpose the more concrete CHSH inequality. Similarly to Kochen-Specker the underlying hidden variables are subject to further physical assumptions and in this case it is locality. of course it is a fair physical assumption that the measurement of one variable far away should not impact another as that would seemingly imply faster then light interaction. however, CHSH experimental results exhibit non locality and so does quantum theory. Classic probability theory does not even have a concept of locality to begin with, so when we do not explicitly enforce it, it is easily able to reproduce quantum behavior. in fact due to the lack of any locality restriction a classic (in the mathematical sense of probability) Markov Decision Process can violate CHSH inequality maximally with CSHS sum = 4.

after a longer discussion the AI then concluded that dropping those two assumptions everything we see in quantum mechanics can be described (or better interpreted) by classic probabilities and doing so does not introduce any assumptions that quantum theory doesn't do itself. The non-locality feature is what distinguishes it notably from all classical theories, but it is just nothing new in terms of probabilities.

3 minutes ago, Killtech said:

Quantum physics introduces a new concept of probability which at first glance differs quite a bit from original mathematical concept of probability as defined by Kolmogorov.

QM predates Kolmogorov’s axioms by a few years, so one might say that Kolmogorov introduced new concepts, if that were the case. But I don’t see what the issues are.

It turns out that one can take great value to discussing such topics with AI and learn a lot from them, if one has deep enough knowledge of a lot of related topics for a proper discussion.

Now, the AI noted as the two cases where the fundamental differences become apparent.

You are free to discuss whatever you like with AI, but you can’t post it here, except as it complies with our rules. We don’t want to waste our time arguing with a possible hallucination.

The Kochen-Specker and Bell theorems are math, used by QM. i.e. they have proofs.

Do you have concrete examples of the alleged failures?

30 minutes ago, Killtech said:

Quantum physics introduces a new concept of probability which at first glance differs quite a bit from original mathematical concept of probability as defined by Kolmogorov. Sadly it is rarely explained how these frameworks relate to each other and if there may be something missing in mathematics. For a mathematician it feels even more annoying as there is no experiment that is able to challenge classic probability with a result it cannot reproduce.

It turns out that one can take great value to discussing such topics with AI and learn a lot from them, if one has deep enough knowledge of a lot of related topics for a proper discussion.

Now, the AI noted the Kochen-Specker and Bell inequality as the two cases where the fundamental differences become apparent. In either case a deeper investigation showed that both require additional physical assumptions to be made that are not native to probability theory which lead to the corresponding theorems.

For Kochen-Specker that is contextuality - where hidden variables are assumed to have a preexisting values. This assumption breaks down already in classical physics in the case a measurement cannot be done without perturbation of the underlying physical system and therefore possibly changing the state of underlying variables for consequent measurements. Such a behavior of non commuting operators we can reproduce via a classic Markov Decision Process, where each measurement is reflected by a decision (including the decision how to set input variables like the axis along which a detector measures spin) which then can reshuffle all underlying hidden variables / quantum state / wave function.

That leaves only Bell's or for this purpose the more concrete CHSH inequality. Similarly to Kochen-Specker the underlying hidden variables are subject to further physical assumptions and in this case it is locality. of course it is a fair physical assumption that the measurement of one variable far away should not impact another as that would seemingly imply faster then light interaction. however, CHSH experimental results exhibit non locality and so does quantum theory. Classic probability theory does not even have a concept of locality to begin with, so when we do not explicitly enforce it, it is easily able to reproduce quantum behavior. in fact due to the lack of any locality restriction a classic (in the mathematical sense of probability) Markov Decision Process can violate CHSH inequality maximally with CSHS sum = 4.

after a longer discussion the AI then concluded that dropping those two assumptions everything we see in quantum mechanics can be described (or better interpreted) by classic probabilities and doing so does not introduce any assumptions that quantum theory doesn't do itself. The non-locality feature is what distinguishes it notably from all classical theories, but it is just nothing new in terms of probabilities.

I don’t claim to be expert on all this but are you sure what you say concerning non-commuting operators is right? My understanding is the issue of a measurement perturbing the system is known as the “observer effect” , which is quite separate from the non-commutativity of operators for conjugate variables.

I’m also unsure what you mean by hidden variables. My understanding is that no system of hidden variables has been found to work, leading, at least provisionally, to the conclusion they are, to put it bluntly, useless fictions.

5 minutes ago, swansont said:

You are free to discuss whatever you like with AI, but you can’t post it here, except as it complies with our rules. We don’t want to waste our time arguing with a possible hallucination.

The Kochen-Specker and Bell theorems are math, used by QM. i.e. they have proofs.

Do you have concrete examples of the alleged failures?

nothing is failing. i think you misunderstood my statement.

you can look up the proofs of these theorems and see the assumptions they make. of course this discussion is only for those that do know them in detail (or are willing to look them up) and and also have sufficient knowledge of probability theory and Markov Decision Processes to understand and discuss the topic. Indeed that may be a steep requirement which is why i like to double check such discussions with AI first to spot flaws in my logic. Unlike most people, AI is always available, cheap, and does have a vast amount of knowledge over all the areas covered needed to challenge my logic - but of course just like humans it is subject to mistakes. is it against the rules to mention that i double check things with AI before posting?

38 minutes ago, Killtech said:

however, CHSH experimental results exhibit non locality and so does quantum theory.

39 minutes ago, Killtech said:

The non-locality feature is what distinguishes it notably from all classical theories, but it is just nothing new in terms of probabilities.

Not really. What QM and CHSH experimental results display is incompatibility with a somewhat involved hypothesis called "local realism". If I had an alternative life to repeat this discussion again, I think I would be able to convince you that the downfall of local realism is due to the "realism" part of it, not to the "local" part of it. The Schrödinger equation is perfectly local. Field theory is perfectly local. This should be enough of a clue that nothing non-local is going on in QM.

What is formally non-local is the projection postulate (far-away and long-gone subamplitudes of the wave function must "die" immediately). But it has no discernible consequences that would allow to --eg-- send signals, transport energy, etc. Which only reinforces the idea that nothing non-local is actually going on.

2 minutes ago, exchemist said:

I don’t claim to be expert on all this but are you sure what you say concerning non-commuting operators is right? My understanding is the issue of a measurement perturbing the system is known as the “observer effect” , which is quite separate from the non-commutativity of operators for conjugate variables.

question of measurement perturbation is more a question of interpretation - is a change of a quantum state after measurement a perturbation? if you have only observable with operators that do commute, then you cannot construct Heisenberg uncertainty with them and have no issue with Kochen-Specker either because you are in a fully classical maximally boring scenario.

2 minutes ago, exchemist said:

I’m also unsure what you mean by hidden variables. My understanding is that no system of hidden variables has been found to work, leading, at least provisionally, to the conclusion they are, to put it bluntly, useless fictions.

hidden variables are also a thing of interpretation. the Hilbert state space stores a lot of non-observable information that you can just as much name a hidden variable - and quantum mechanics does not work without it. The term however is usually reserved to distinguish quantum mechanics from deterministic hidden variable theories like de Broglie-Bohm theory where these variables take a more pronounced role.

Just now, Killtech said:

For a mathematician it feels even more annoying as there is no experiment that is able to challenge classic probability with a result it cannot reproduce.

Just now, swansont said:

Do you have concrete examples of the alleged failures?

swansont has already asked for examples and so I am asking for them as well.

At the moment you are just piling generality upon generalitywithout the detailed support these claims require.

I would have thought it would have been wise to start with a statement of which of the the three classic types of probability, as taught for UK state school exams, you are referring to and also giving a specific example from QM that runs contrary to it or them.

Edited June 16Jun 16 by studiot
spelling

3 minutes ago, joigus said:

Not really. What QM and CHSH experimental results display is incompatibility with a somewhat involved hypothesis called "local realism". If I had an alternative life to repeat this discussion again, I think I would be able to convince you that the downfall of local realism is due to the "realism" part of it, not to the "local" part of it. The Schrödinger equation is perfectly local. Field theory is perfectly local. This should be enough of a clue that nothing non-local is going on in QM.

What is formally non-local is the projection postulate (far-away and long-gone subamplitudes of the wave function must "die" immediately). But it has no discernible consequences that would allow to --eg-- send signals, transport energy, etc. Which only reinforces the idea that nothing non-local is actually going on.

indeed, QM does not allow to violate CHSM maximally, meaning there is still some restriction on locality but not a full one.

and of course you are right, in that all time evolution remains indeed fully local (deterministic even) and it is just the measurement part that causes the big trouble. you are also referring to the no-communication theorem, which again requires to make some additional assumptions which general validity may be questioned in quantum theory.

that is however not the topic of the discussion. probability theory is neither rooted in realism nor locality. is just offers a framework to describe the outcome frequencies of events given the input information. is is very bare bone.

7 minutes ago, studiot said:

swansont has already asked for examples and so I am asking for them as well.

i can write the CHSH experiment in terms of a classic Markov Decision Process if that is what you want. but it's the same thing you know, just a somewhat different terminology really.

Edited June 16Jun 16 by Killtech

Just now, Killtech said:

i can write the CHSH experiment in terms of a classic Markov Decision Process if that is what you want. but it's the same thing you know, just a somewhat different terminology really.

Thank you, but you don't need need a Markov , or any other, chain to state the probability of a single event.

However you do need to state unequivocally what you mean by the probability of event E is, for the three cases

P(E) = 0

P() = 1

0 < P(E) < 1

since depending upon the type of classic probability you are using these cases, at least, will be different.

37 minutes ago, Killtech said:

question of measurement perturbation is more a question of interpretation - is a change of a quantum state after measurement a perturbation? if you have only observable with operators that do commute, then you cannot construct Heisenberg uncertainty with them and have no issue with Kochen-Specker either because you are in a fully classical maximally boring scenario.

hidden variables are also a thing of interpretation. the Hilbert state space stores a lot of non-observable information that you can just as much name a hidden variable - and quantum mechanics does not work without it. The term however is usually reserved to distinguish quantum mechanics from deterministic hidden variable theories like de Broglie-Bohm theory where these variables take a more pronounced role.

OK so perturbation due to measurement is nothing to do with whether or not pairs of operators commute.

Regarding hidden variables I’m still unclear why you mention them. They are not implied by QM. They are purely an attempted bolt-on extra, designed in the hope of restoring deterministic physics. As they have no observable consequences, science can do without them. So I’m not sure what you mean by saying QM “does not work without it”, or them. Can you explain this further?

1 hour ago, Killtech said:

meaning there is still some restriction on locality but not a full one.

Again, no. There is a restriction on local realism, because there is a restriction on realism, because three perfectly sensible experimental questions, namely "Is A true?", "Is B true?", and "Is C true?" cannot have yes / no as answers simultaneously. In fact, the CHSH state was prepared in one and only one causally connected patch of space-time, so nothing non-local is going on.

Many people misunderstand this. Some of them write books. What can you do...

1 hour ago, Killtech said:

that is however not the topic of the discussion. probability theory is neither rooted in realism nor locality. is just offers a framework to describe the outcome frequencies of events given the input information. is is very bare bone.

Then why bring it up, especially as an incorrect statement?

QM's concept of probability differs from classical probability only in that classical probability doesn't have anything in the way of quantum amplitudes, which as we know give rise to interference phenomena. This gives way for interesting correlations that do not appear classically. Otherwise it's more or less the same concept.

13 minutes ago, studiot said:

However you do need to state unequivocally what you mean by the probability of event E is, for the three cases

in the CHSH proof you have a hidden variable {\lambda\} which describe the underlying state. if we dropped locality restriction we can simple take lambda to be the quantum state as the hidden variable and the quantum state space as our probability space, hence we start with deterministic distribution \{\mu_0\} starting in the state \{\lambda\}. in quantum mechanics we would instead write it as the corresponding density operator for the state. each measurement we do model as a decision in a markov process - which means that at each decision the probability space is transformed by a Markov transition matrix which depends on the decision (measurement configuration) taken, i.e. \{\mu_t = \mu_{t-1} p(\theta_a,\theta_b)_{i j} \}. coincidentally this is the same the quantum mechanic calculus prescribes for the density matrix after each measurement with the corresponding observable operator.

Since this merely reframes the calculation from quantum mechanics in the CHSH case into the classical framework without any change to the calculation or assumption, all probabilities stay the same and thus the result as well.

34 minutes ago, exchemist said:

Regarding hidden variables I’m still unclear why you mention them. They are not implied by QM. They are purely an attempted bolt-on extra, designed in the hope of restoring deterministic physics. As they have no observable consequences, science can do without them. So I’m not sure what you mean by saying QM “does not work without it”, or them. Can you explain this further?

because both Bell and Kochen-Specker deal with them and they are needed to even understand what these theorems are meant to say.
but i used the term more general to challenge a simple fact about quantum mechanics: consider the information a wave function stores (i..e. an infinite series of complex number determining its value at each point in spacetime). this information is not directly observable, yet it is crucial for quantum mechanics to make all the predictions of all observables. you cannot do QM without any such hidden variables. But: in distinction to hidden variable theories which consider them real physical quantities, QM does not leaves the Interpretation for them open - or makes them into something even more obscure as in many worlds interpretations.

Edited June 16Jun 16 by Killtech

1 hour ago, Killtech said:

is it against the rules to mention that i double check things with AI before posting?

It’s pointless to do so, arguably off-topic (against the rules), and possibly invoking argument from authority (which as a fallacy, is also against the rules)

5 minutes ago, joigus said:

gain, no. There is a restriction on local realism, because there is a restriction on realism, because three perfectly sensible experimental questions, namely "Is A true?", "Is B true?", and "Is C true?" cannot have yes / no as answers simultaneously

I am not an expert on realism to discuss that nor is it a topic of my thread. but i am unsure why you would even insist 3 experiments being able to have unique answers simultaneously?

that does not even work in simple real life: like i could want to make an experiment to measure how happy i would have been at the age of 30 if i took job offer A. another experiment would be to measure the same with me taking on job opportunity B. obviously conducting one experiment disqualifies the other from being possible, hence those two cannot have unique answers at the same time (except in a multiverse i guess) - so measuring one leave the other at an large uncertainty. Markov Decision Processes are the model for these scenarios where the answers to some questions become inaccessible due to the decisions taken along the path.

16 minutes ago, joigus said:

Then why bring it up, especially as an incorrect statement?

QM's concept of probability differs from classical probability only in that classical probability doesn't have anything in the way of quantum amplitudes, which as we know give rise to interference phenomena. This gives way for interesting correlations that do not appear classically. Otherwise it's more or less the same concept.

that is an incorrect statement. of course you can handle quantum amplitudes and interferences. these correlations do not appear in classical physics but that has nothing to do with probability.

Edited June 16Jun 16 by Killtech

47 minutes ago, Killtech said:

in the CHSH proof you have a hidden variable {\lambda\} which describe the underlying state. if we dropped locality restriction

Bell’s theorem is a restriction on local hidden variables, so if you drop the local requirement you’re talking about something else.

In the words of Bell, "If [a hidden-variable theory] is local it will not agree with quantum mechanics, and if it agrees with quantum mechanics it will not be local."

https://en.m.wikipedia.org/wiki/Bell%27s_theorem

46 minutes ago, Killtech said:

that does not even work in simple real life

But we’re talking about entangled quantum states, where you can have superposition. “real life” is classical, and you don’t get such behavior. That’s kinda the point - QM is not secretly classical

47 minutes ago, Killtech said:

I am not an expert on realism to discuss that nor is it a topic of my thread. but i am unsure why you would even insist 3 experiments being able to have unique answers simultaneously?

Ok. So let's leave it alone, as long as you agree that Bell's theorem experimental violation implies there can be no local hidden variables (a local reality that underlies quantum mechanics).

I agree with @studiot that you don't need Markovian probabilities to accomodate quantum mechanics. You only need a concept of probability in the terms that he defined.

What's different is the existence of these "potentialities" if you will, that we call "probability amplitudes".

5 minutes ago, swansont said:

Bell’s theorem is a restriction on local hidden variables, so if you drop the local requirement you’re talking about something else.

indeed, my topic is about probabilities.

the point is that classical probability theory knows no restriction on locality and hence is no in conflict with Bell's theorem on that ground.

6 minutes ago, joigus said:

What's different is the existence of these "potentialities" if you will, that we call "probability amplitudes".

that is neither a restriction for classic probability theory.

in fact look at the density matrix which is the equivalent object to a probability distribution in classic probability. apparently there is a difference that it is a matrix while in math it would be a vector. so let's dig deeper to work out what the core difference boils down to. if we assume the Hilbert space is our classic probability space then a probability density \(\mu\) is a measurable function from the Hilbert space to the positive real numbers - while a density matrix is much smaller as it dimensions reduce to the basis o the Hilbert space rather then the total Hilbert space. Now, it is an interesting feature of Markov theory that while it is entirely linear, it is perfectly able to model any non-linear processes - and does so by blowing up in the dimension of the probability vectors.

So QM makes the novel assumption that the underlying state space itself is a linear space which description therefore we can reduce via a basis of pure states. so we have two linearities: the one is fundamental to the definition of probability vectors and another for the states (superpositions). if we introduce this additional restriction/simplification to probability theory, we can then utilize the very same density matrix formalism. note that this simplification also has to be applied to all the Markov transition matrices - which we can now identify by the very same operators quantum theory uses.

So it would seem QM is a special case of probability theory by adding special assumption about the state space \(\Omega\) with a somewhat novel technique to handle it.

Edited June 16Jun 16 by Killtech

1 hour ago, Killtech said:

in the CHSH proof you have a hidden variable {\lambda\} which describe the underlying state. if we dropped locality restriction we can simple take lambda to be the quantum state as the hidden variable and the quantum state space as our probability space, hence we start with deterministic distribution \{\mu_0\} starting in the state \{\lambda\}. in quantum mechanics we would instead write it as the corresponding density operator for the state. each measurement we do model as a decision in a markov process - which means that at each decision the probability space is transformed by a Markov transition matrix which depends on the decision (measurement configuration) taken, i.e. \{\mu_t = \mu_{t-1} p(\theta_a,\theta_b)_{i j} \}. coincidentally this is the same the quantum mechanic calculus prescribes for the density matrix after each measurement with the corresponding observable operator.

Since this merely reframes the calculation from quantum mechanics in the CHSH case into the classical framework without any change to the calculation or assumption, all probabilities stay the same and thus the result as well.

because both Bell and Kochen-Specker deal with them and they are needed to even understand what these theorems are meant to say.
but i used the term more general to challenge a simple fact about quantum mechanics: consider the information a wave function stores (i..e. an infinite series of complex number determining its value at each point in spacetime). this information is not directly observable, yet it is crucial for quantum mechanics to make all the predictions of all observables. you cannot do QM without any such hidden variables. But: in distinction to hidden variable theories which consider them real physical quantities, QM does not leaves the Interpretation for them open - or makes them into something even more obscure as in many worlds interpretations.

Regarding Bell, surely the point is his theorem rules out hidden variables, unless they are non-local.

I don’t see how can you say hidden variables are required by QM, when it makes no reference to them anywhere. I don’t recall the algebraic expressions for the wave function being an infinite series of complex numbers. They are just algebra, surely, expressing a probability amplitude at each point in space.

1 minute ago, exchemist said:

I don’t recall the algebraic expressions for the wave function being an infinite series of complex numbers

i was trying to make you aware how much information you actually need to uniquely define a wave function (or any function of a continuous space really). to define the position of an object you need merely 3 numbers. to uniquely define a function, how much do you need? let's say your wave function is mathematically a nice one and can be written in terms of Furier like series, that is written as a linear combination of a basis of pure states. this series has a countable infinite amount of coefficients \(a_i\). every distinct combination of coefficients produces a distinct wave function, right? so you really need each coefficient to uniquely specify your function. and each of the coefficients is a variable in your theory.

you cannot do QM without this information. but the wave function is not measurable itself. hence, it and all variables that describe it are technically hidden - but they are only called that (i think) when you interpret them to be real (whatever that actually means).

Edited June 16Jun 16 by Killtech

7 minutes ago, Killtech said:

i was trying to make you aware how much information you actually need to uniquely define a wave function (or any function of a continuous space really). to define the position of an object you need merely 3 numbers. to uniquely define a function, how much do you need? let's say your wave function is mathematically a nice one and can be written in terms of Furier like series, that is written as a linear combination of a basis of pure states. this series has a countable infinite amount of coefficients ai. every distinct combination of coefficients produces a distinct wave function, right? so you really need each coefficient to uniquely specify your function. and each of the coefficients is a variable in your theory.

you cannot do QM without this information. but the wave function is not measurable itself. hence, it and all variables that describe it are technically hidden - but they are only called that (i think) when you interpret them to be real (whatever that actually means).

No. The wave function for the electron in the hydrogen atom is just a rather messy algebraic expression, involving an exponent in the radial part and spherical harmonics for the angular part. Nothing at all about infinite numbers of coefficients.

11 minutes ago, Killtech said:

apparently there is a difference that it is a matrix while in math it would be a vector.

The key to the concept of a density matrix is that, whenever your quantum state is not maximally determined, the most natural thing to assume mathematically is that the collectivity you're handling is a statistical mixture of different so-called "pure" vectors with statistical weights p₁ , p_2, etc. It is a mathematical convenience to define it as a matrix: p₁|1><1|+p₂|2><2|+... Nothing more. Anything you can define with a matrix you could define equally well with a series of scalars (bilinears, in the case of QM).

20 minutes ago, Killtech said:

while a density matrix is much smaller as it dimensions reduce to the basis o the Hilbert space rather then the total Hilbert space.

This doesn't make much sense, as a basis of the Hilbert space spans the whole set of possible states. The total is the span of the basis states... Or I'm missing your point completely.

Quantum mechanics is non-Markovian. Present states States at any given time depend on their past histories...

Errr... more stuff...

Non-linearity could be relevant at many levels. Evolution could be non-linear, observables could be non-linear functionals defined on elements of a linear vector space, etc. These things have been tried to death, of course.

And sub-quantum Markovian processes have and are being tried. Cellular automata, for example. I'm almost sure of that, although I'm no expert. Look up Gerard 't Hooft interpretation of quantum mechanics...

Edited June 16Jun 16 by joigus
minor syntax correction+conceptual correction

Just now, Killtech said:

in fact look at the density matrix which is the equivalent object to a probability distribution in classic probability.

Actually you are correct there is a difference between probability and probability density that is often forgotten and folks just use the word probability.

As I understand matters that is because of the way QM is formulated so we have

probability is a point function which is a map from the sample space to an interval between zero and 1 of the reals.

Locality is not involved.

On the other had there is no such thing as a point function in QM.

probability density is the (classical) probability of the

Event E lying between say x and (x + dx) for one axis, suitably adjusted for the normalisation condition.

Edited June 16Jun 16 by studiot

3 minutes ago, exchemist said:

No. The wave function for the electron in the hydrogen atom is just a rather messy algebraic expression, involving an exponent in the radial part and spherical harmonics for the angular part. Nothing at all about infinite numbers of coefficients.

and what do you think the spherical harmonics are? these are the basis functions of your quantum mechanical Hilbert space. you use them to expand any given wave function into an infinite series of them (the spherical harmonics consist of infinitely many functions). Note that in general, a hydrogen atom may be in any state superposition, not just in an energy eigenstate. in the general case it is hence \( \psi (x) = \Sigma_i a_i \psi_i(x) \) where \(\psi_i(x)\) are the wave functions of energy eigenstates.

3 minutes ago, joigus said:

This doesn't make much sense, as a basis of the Hilbert space spans the whole set of possible states. The total is the span of the basis states... Or I'm missing your point completely.

A Hilbert space as a linear space and hence has a basis, right? if you drop that assumption that is is linear, you no longer have a basis to work with and must now handle every possible state in that space separately. Note that in probability theory we do not assume any kind of structure for \(\Omega\) automatically, so that complicates things.

9 minutes ago, joigus said:

Quantum mechanics is non-Markovian. Present states depend on their past histories...

the question of Markov-Property depends on what you consider your underlying state space to be. if you take the entire Hilbert space as your state space, then it contains all necessary information because all previous history is memorized by the quantum state or more generally the density matrix. it is entirely irrelevant how the density matrix came to be - it is the only thing determining observable results. It's history is irrelevant, right? Therefore the process specifying the time evolution of the density matrix is Markovian.

if you however choose a much smaller state space for your probabilities, like for example a naive classically inspired description of point particles characterizing them by only 6 values (3 position, 3 for momentum), these cannot incorporate all information required to produce the probabilities of quantum experiments, hence will require to include the information what previous measurements occurred and in what order. In that case your model is non-Markovian as you need to know the entire history of your state to reproduce the needed probabilities.

19 minutes ago, studiot said:

On the other had there is no such thing as a point function in QM.

to prevent a misunderstanding, can you specify what you mean by a point function? did you mean a quantile function? note that this only exist for state spaces compatible with numeric operations because in order to write \(Pr(X \geq x)\), one requires that \(x\) is something that understand what \( \geq \) means. if \(x \epsilon \{red, green, blue\}\) this does not work. quantiles do not exist in general case of probability theory but are specific to numeric valued random variables usually.

Edited June 16Jun 16 by Killtech

5 minutes ago, Killtech said:

A Hilbert space as a linear space and hence has a basis, right? if you drop that assumption that is is linear, you no longer have a basis to work with and must now handle every possible state in that space separately. Note that in probability theory we assume no kind of structure for Ω. that complicates things.

Sure, but you said,

1 hour ago, Killtech said:

if we assume the Hilbert space is our classic probability space then a probability density μ is a measurable function from the Hilbert space to the positive real numbers - while a density matrix is much smaller as it dimensions reduce to the basis o the Hilbert space rather then the total Hilbert space.

It is that what I argued not to make much sense. For starters, typical quantum mechanical space states are infinite-dimensional. But even in the case of relevant sub-spaces (eg, spin), the space of pure states has dimension 2S+1, while the space of density matrices has dimension (2S+1)². So, again, what do you mean a density matrix is "much smaller"?

10 minutes ago, joigus said:

It is that what I argued not to make much sense. For starters, typical quantum mechanical space states are infinite-dimensional. But even in the case of relevant sub-spaces (eg, spin), the space of pure states has dimension 2S+1, while the space of density matrices has dimension (2S+1)². So, again, what do you mean a density matrix is "much smaller"?

Let's assume a simple quantum space with only two states \( |0\rangle \) and \( |1\rangle \). density matrices are then 2x2. however, the Hilbert space itself has still infinitely many elements, even uncountable many, because there are so many possible superpositions. If i naively take the Hilbert space as my state space \(\Omega\) of some classical Markov process without assuming anything about its structure, then how would my probability vectors look for that scenario? Basically i have to model it as a Markov chain on a continuous state space - which adds quite a bit of complexity because my 'probability vector' has uncountable infinite dimension, that is, it is a density function on a continuous space. as you can see a probability vector in that case requires far more information to be uniquely specified then a density matrix.

But if i am allowed to use the additional assumptions of state linearity from quantum mechanics, then i can do the same tricky reduction of complexity and reduce the space of my process to the size of the density matrix. that is any probability vector is fully specified by only 4 values - which are enough to reproduce all probabilities of this quantum system. however, this reduction comes at a cost. if i want to represent it as a vector, then there exist no linear time evolution of it, unlike for my much more dimensional original probability distribution. but QM still found a nice way to calculate it, and that is the von Neumann equation which allows states to interfere with each other, something probabilities cannot do.

I mean these are all great techniques to reduce these special continuous space Markov process to a seemingly discrete space. you swap the overhead of a continuous state space for a much lower dimension non-linear matrix evolution. as cool as that may be, nothing of it conflicts with classic probability theory in any way.