The interpretation of probability

[Killtech]

I am looking for a way to compare the handling of probability in QT with how it's done in classic PT (probability theory) - and their interpretations. QT does have it's own formalism that works, so there isn't much motivation to bring it into a usual representation which makes it hard to find literature that discusses this in detail.

In QT, the general way to represent probability measures is via DO (density operators). These produce entirely classic real probabilities between 0 and 1 and the sum over all outcomes still needs to be 1. So i want to know how their space differs from a classic measure space. One thing is of course that they are represented via matrices unlike in PT where a usual way to write measures is via real-values operators over the state space - i.e. vectors - using a specific basis for the space. Yet on the other hand the space of operators is itself a linear space so we can think of operators as vectors in that higher dimensional (operator) vector space. Since DOs must produce classic probabilities they form a small simplex within - same as probability measures do. I am looking for some literature that explore that space.

The thing is that when we focus on writing everything by the means of it, we can bring the QT formalism into a familiar form: if we forget the probability normalization of DOs for a second, we get a linear subspace and that has a basis. We are looking for a specific basis though - the barycentric coordinates: i.e. the vertices of the simplex which are special DOs of pure some states. This way we can represent any element of the simplex (any DO) as a positive semidefinite linear combination of the DO basis with total weight 1. That's useful when dealing with probabilities and the way PT does it. But represented in that basis, DOs look just like measures in PT.

Also notice that any transformation like $\tilde U (\rho) := U \rho U^\dagger$ is linear in $\rho$ since
$$\tilde U (a\rho_1 + b\rho_2) = U (a\rho_1 + b\rho_2) U^\dagger= a U \rho_1 U^\dagger+ b U \rho_2 U^\dagger = a \tilde U (\rho_1) + b \tilde U (\rho_2)$$
So in the vector space in which $\rho$ is a vector $\tilde U$ can be represented by a matrix and hence written as $\tilde U \rho$. Note that matrices that maps the simplex onto itself expressed in barycentric coordinates are called stochastic matrices (i.e. operations that don't break probabilities).

The point is that when we represent things in this higher dimensional space everything starts to look very familiar.

 

[A. Neumaier]

The quantum analogue of a finite probability measure is a POVM - not a density matrix. In my paper on quantum tomography I use the name quantum measure for it, to display the analogy most vividly. There you can see how everything is completely analogous, quite close to what you describe.

 

[Killtech]

The quantum analogue of a finite probability measure is a POVM - not a density matrix. In my paper on quantum tomography I use the name quantum measure for it, to display the analogy most vividly. There you can see how everything is completely analogous, quite close to what you describe.

But i am not looking for an analogy but rather want to identify if there are probability measures which produce the same probabilities as calculated in QT. So if QT just uses a different representation of classic measures on some different space.

The thing is that DOs are formulated over the Hilbert space, but the probability simplex they form when it is represented by classic measures is a different space and in particular has a higher dimension then the original Hilbert space.

Admittedly i am not very familiar with POVMs yet. How do they handle systems in mixed states?

Than again, I was just looking in Wikipedia and realized that the example given for the 2 dim Hilbert space does actually the very same i did with the DOs. The optimal POVM elements $F_i$ is exactly the DO barycentric basis i had for the same case. Interestingly they call the motivation for this is unambiguous quantum state discrimination but this is also the exact prerequisite of classic PT handling: you have to separate the pure probability aspects (our knowledge about the state of the system) from the physical state (the actual state of the system). The important thing is that the classical basis for this space must be composed of 3 elements rather then 2, which fixes all the problems.

 

[gentzen]

De-Broglie Bohm gives a model where quantum probabilities arise from classical probabilities. This makes it hard to compare the handling of probability in QT with how it's done in classic PT. (You have to bring in space-time and locality, but then you are no longer talking about probability theory itself.) It seems that DOs don't fit into that De-Broglie Bohm framework, but I recently learned that some Bohmians (Goldstein, Tumulka, et al) attacked that problem head on (https://arxiv.org/abs/quant-ph/0309021):

It is thus not unreasonable to ask, which mu, if any, corresponds to a given thermodynamic ensemble? To answer this question we construct, for any given density matrix rho, a natural measure on the unit sphere in H, denoted GAP(rho). We do this using a suitable projection of the Gaussian measure on H with covariance rho.

Sadly, Jozsa, Robb, and Wootters already constructed the same measure in 1994 (https://journals.aps.org/pra/abstract/10.1103/PhysRevA.49.668), as I learned from a follow up paper by the Bohmians (https://arxiv.org/abs/1104.5482).

I only found their "work", because I recently tried to solve the same problem, and after I "nearly" found the solution, I was in a position to come up with appropriate search terms to locate and identify prior art:

In fact, I now had an idea how to make De-Broglie Bohm work in the quantum computer context, using qubits and their classical states as the ontology.

Interestingly, my efforts were indirectly triggered when Eliezer Yudkowsky asked "I keep wondering if I should take the time to persuade you of MWI." and Scott Aaronson replied:

I still have difficulties with making a straightforward factual belief in MWI “pay rent,” in the same way that my factual beliefs in (e.g.) special relativity, heliocentrism, and the germ theory pay rent for me.

A previous reply by Scott Aaronson had already pointed out the importance of making sense of DOs:

(2) In several posts, Yudkowsky gives indications that he doesn't really understand the concept of mixed states. (...) As I see it, this might be part of the reason why Yudkowsky sees anything besides Many-Worlds as insanity, and can't understand what (besides sheep-like conformity) would drive any knowledgeable physicist to any other point of view. If I didn't know that in real life, people pretty much never encounter pure states, but only more general objects that (to paraphrase Jaynes) scramble together "subjective" probabilities and "objective" amplitudes into a single omelette, ...

What I find intersting is that MWI failed to "pay rent", but De-Broglie Bohm lead some Bohmians to rediscover that measure in 2003, and even lead me to "nearly" (re)discover that measure in 2022. I doubt that Jozsa, Robb, and Wootters were helped by MWI when they initially discovered that measure in 1994.

 

[Killtech]

De-Broglie Bohm gives a model where quantum probabilities arise from classical probabilities. This makes it hard to compare the handling of probability in QT with how it's done in classic PT. (You have to bring in space-time and locality, but then you are no longer talking about probability theory itself.) It seems that DOs don't fit into that De-Broglie Bohm framework, but I recently learned that some Bohmians (Goldstein, Tumulka, et al) attacked that problem head on (https://arxiv.org/abs/quant-ph/0309021):

Lol, i played around with the same measure. There are even some older posts here with a few of my fails to build it properly. I would say this is the canonical approach, which is why all of us stumbled onto it naturally.

But the issue that the density matrix does not uniquely specify the measure, made me think that this isn't the minimal construction. hence i was looking a little bit deeper into DOs - which represent all the possible ensembles over $H$ after all - same as classic probability measures do. Another thing is that QT uses matrix mechanics in just such similar fashion to the Markov processes already. So there has to be a way to make use of the linearity.

And indeed it isn't difficult: we don't need to define the measure on the unit sphere of all of $H$. If we want to remove all redundancy from the model, we first have to realize how incredibly small the space of all possible DOs actually is. So instead of the unit sphere, we can just take a hand full of states - only the ones we can physically distinguish and define that as our state space. So indeed for a 2 state qubit this means our state space consists of merely 3 physical states... while there are other pure states, we don't need them because we won't be able to tell them apart from some mixed states.

 

[Killtech]

This is weird. I don't know what to look for to get some literature on usual decomposition methods for density matrices. You would think that when you have a limited dimensional linear space of objects and everything done with those objects is linear in them, people would jump to apply standard linear algebra concepts.

Everything i can find is always focused on depicting them as matrices over the Hilbert space missing the flexibility their very own linear space gives them. DOs have the complete information of the Hilbert space $H$ and all things can be formulated for DOs rather then for states from $H$. DOs are merely a generalization of the formalism to any mixed state, or am i missing something? But the structure of their space seems way more convenient then the Hilbert space - particularity in terms of probability.

 

[gentzen]

This is weird. I don't know what to look for to get some literature on usual decomposition methods for density matrices.

What sort of decomposition methods are you looking for? Density matrices are Hermitian positive semidefinite, with trace one, so a standard eigenvalue decompositon will work just fine. What more do you want?

So instead of the unit sphere, we can just take a hand full of states - only the ones we can physically distinguish and define that as our state space. So indeed for a 2 state qubit this means our state space consists of merely 3 physical states... while there are other pure states, we don't need them because we won't be able to tell them apart from some mixed states.

Which 3 "physical states" do you have in mind? The classical 4 Stokes parameters, minus the total intensity ($S_0$ which is assumed to be 1)? Or in more mundane terms, the space of Hermitian 2x2 matrixes is 4 dimensional as a real vector space, and if you force the trace to be one, then this reduces to a 3 dimensional space. If you additionally force the matrices to be positive semidefinite, then you get a convex subset of that 3 dimensional space.

Does this descibe more or less, what you have in mind?

 

[A. Neumaier]

Everything i can find is always focused on depicting them as matrices over the Hilbert space missing the flexibility their very own linear space gives them. DOs have the complete information of the Hilbert space $H$ and all things can be formulated for DOs rather then for states from $H$. DOs are merely a generalization of the formalism to any mixed state, or am i missing something? But the structure of their space seems way more convenient then the Hilbert space - particularity in terms of probability.

Probably because density operators do not form a liner space but a cone in a linear space.

 

[Killtech]

Probably because density operators do not form a liner space but a cons in a linear space.

Of course the Hermitian operators form a linear space and the subspace the DOs span within it is called a simplex. This structure is generated by the probability condition that the sum over all possibilities must be one. There are a lot of well known methods for this particular subspace - specifically because it ends up being the same space structure classic probability measures span. So the way DOs relate to hermitian operators is the same as probability measures relate to measures.

Density operators and probability measures are designed to provide a description for the very same thing: ensembles / mixed states. They also end up using the very same concept of probability so i guess this is what makes their spaces isomorph.

What sort of decomposition methods are you looking for? Density matrices Hermitian positive semidefinite, with trace one, so a standard eigenvalue decompositon will work just fine. What more do you want?

Oh, i am not looking for a decomposition into the inner structure of a given density matrix. Instead, given any density operator i am looking for a decomposition into a standardized set of operators - i.e. a basis of that operator space. I am looking for a decomposition in term of the outer structure i.e. relating it to all other possible density operators / the space of DOs. Practically a decomposition into the coordinate system of the DO space.

Which 3 "physical states" do you have in mind? The classical 4 Stokes parameters, minus the total intensity (S0 which is assumed to be 1)? Or in more mundane terms, the space of Hermitian 2x2 matrixes is 4 dimensional as a real vector space, and if you force the trace to be one, then this reduces to a 3 dimensional space. If you additionally force the matrices to be positive semidefinite, then you get a convex subset of that 3 dimensional space.

This sort of thing: https://en.wikipedia.org/wiki/POVM#An_example:_unambiguous_quantum_state_discrimination
As you see the $F_i$ matrices are hermitian but do not suffice to all conditions to render them valid DOs. You could however adapt it by $F_i = p_i D_i$ to be usable and always produce DOs like $D = \sum_{i} p_i D_i$ such that $D$ and $D_i$ are all DOs. By $\exists p_i : id = \sum_{i} p_i D_i, \sum_{i} p_i = 1$ you guarantee that for any state the whole space is covered such that the sum over all possibilities is always 1 ($p_i$ are barycentric coordinates of the simplex).
This won't work for any DOs $D_i$ though and requires a restriction, so instead of taking any two states you have to start with an orthonormal state basis and follow the remaining construction to get the specialized decomposition. This construct should cover the whole DO space and for the qubit case needs exactly 3 $D_i$.

 

[A. Neumaier]

the subspace the DOs span within it is called a simplex.

The manifold of (unnormalized) density operators is a convex cone, not a simplex. A simplex is spanned by n+1 vertices if n is the dimension of the affine space.

So the way DOs relate to hermitian operators is the same as probability measures relate to measures.

No. Already the cone of density operators in the 4-dimensional real vector space of complex Hermitian 2x2 matrices has infinitely many extreme points, while a simplex in a 4-dimensional space has exactly 5 extreme points.

given any density operator i am looking for a decomposition into a standardized set of operators

The best you can do is to decompose it into a linear combination of extreme states. The extreme states (i.e., the ones which cannot be decomposed) are the pure states (density operators of rank 1). The decomposition is not unique, as known since density operators were introduced about 90 years ago.

 

[Killtech]

The manifold of density operators is a convex cone, not a simplex. A simplex is spanned by n+1 vertices if n is the dimension of the affine space.

Since convex condition is a "$\geq$" a simplex ends up being both convex and concave. It's also sometimes described as the convex hull of its k + 1 vertices.

No. Already the cone of density operators in the 4-dimensional real vector space of complex Hermitian 2x2 matrices has infinitely many extreme points, while a simplex in a 4-dimensional space has exactly 5 extreme points.

$\rho = \sum_i p_i |\phi_i\rangle\langle \phi_i|$ assuming the $\phi_i$ are orthogonal is a valid density operator if and only if all $p_i$ are taken from a simplex (i.e. probabilities) . In particular $1 |\phi_i\rangle\langle \phi_i|$ is each a DO and a vertex of a simplex (there are many difference simplexes that can be formed that way).

(Edit: Oh, stupid me, this standard way to write a density operator is just exactly a simplex decomposition; the thing i was looking for. Yeah, i guess the most obvious things end up easily overlooked)

However the above sum does not span the entire DO space, so it must be supplemented with further states. Those states cannot be orthogonal in the Hilbert space anymore as it's dimension is fully spanned already but there is still space in the DO dimensions to make them orthogonal there. UQSD does just that. The whole thing remains therefore a simplex.

The decomposition isn't unique indeed and that's actually a very interesting feature as it means the state space cannot be uniquely identified like that.

 

[A. Neumaier]

The whole thing remains therefore a simplex.

You are misusing standard mathematical terminology in your attempt to hunt a chimera.

The decomposition isn't unique indeed

whereas the decomposition of a point of a simplex into a convex combination of its vertices is unique.

 

[Killtech]

You are misusing standard mathematical terminology in your attempt to hunt a chimera.

Oh, sorry, i missed that you claimed that the DO manifold is a cone before. I am not sure if it's the decomposition into a Bloch sphere you meant by it since for the vector we have $|a|\leq 1$, hence it's a cone.

But note that this decomposition always require the adding of a $id$ operator. So this mapping to a Bloch vector $a(\rho)$ is not linear since if we have DOs $\rho_1$, $\rho_2$ and $\rho_{1+2} = \frac 1 2 \rho_1 + \frac 1 2 \rho_1$ then $\frac 1 2 a(\rho_1) + \frac 1 2 a(\rho_2) \neq a(\rho_{1+2})$. Therefore the vector $a$ represents a non-linear coordinate of the DO space. I don't think you can make that mapping isometric, hence it will always reshape the manifold.

 

[A. Neumaier]

Oh, sorry, i missed that you claimed that the DO manifold is a cone before. I am not sure if it's the decomposition into a Bloch sphere you meant by it since for the vector we have $|a|\leq 1$, hence it's a cone.

The unnormalized density matrices (i.e., the positive semidefinite Hermitian operators) from a convex cone. For qubits, this cone is isomorphic to the Lorentz cone, and the isomorphism is given by the Pauli representation in terms of the 4-dimensional Stokes vector.

The normalized density matrices form a convex subset of this cone, but not a simplex. For qubits, the normalized states form the 3-dimensional Bloch ball, and the pure states (its extreme points) form its boundary, the 2-dimensional Bloch sphere. Every point in the ball can be written in infinitely many ways as a convex combinations of the infinitely many extremen points, while an n-dimensional simplex has only n+1 extreme points exist.

 

[Killtech]

The normalized density matrices form a convex subset of this cone, but not a simplex. For qubits, the normalized states form the 3-dimensional Bloch ball, and the pure states (its extreme points) form its boundary, the 2-dimensional Bloch sphere. Every point in the ball can be written in infinitely many ways as a convex combinations of the infinitely many extremen points, while an n-dimensional simplex has only n+1 extreme points exist.

I understand that. But you are aware that the mapping of DOs onto a Bloch ball is not an isomorphism? it is bijective but not a linear mapping. While it covers the whole space, It does not preserve the structure of its origin (that's what isomorph literally means: preserve the structure).

Same as spherical coordinates map from a plane onto the sphere and hereby reshape the manifold. Things that appear linear in Cartesian coordinates will look entirely different in spherical coordinates. The Bloch mapping seems to do the the very same thing. The Bloch mapping does not even generally preserve first order differtiability around the balls origin.

But there are infinitely many different coordinate maps for a given manifold. You cannot view the manifold purely though one specific coordinates system.

 

[A. Neumaier]

I understand that. But you are aware that the mapping of DOs onto a Bloch ball is not an isomorphism? it is bijective but not a linear mapping. While it covers the whole space, It does not preserve the structure of its origin (that's what isomorph literally means: preserve the structure).

This is irrelevant. Since the Bloch ball is not a linear space, linearity need not be conserved.

But convex combination (i.e., the structure that the set of density matrices has) is preserved, otherwise the representation would not be really useful.

 

[Killtech]

This is irrelevant. Since the Bloch ball is not a linear space, linearity need not be conserved.

But convex combination (i.e., the structure that the set of density matrices has) is preserved, otherwise the representation would not be really useful.

Hmm, i figured that $\rho_1 = \frac 1 2\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, $\rho_2 = \frac 1 2\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, $\rho_3 = \frac 1 2\begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}$ are all valid DOs and all other DOs lie within their linear span. However, i realize now that while every DO is a linear combination of those, it doesn't guarantee to have positive coefficients i.e. not all states lie within the simplex of those vertices. So this cannot be interpreted directly as a classic probability space.

On the other hand, i do think that preserving the linearity is very much relevant: every DO can be written as a vector in this basis. Since time evolution is linear in $\rho$ we can now use this decomposition to represent it via a matrix in the DO basis. Nothing alike is available for the Bloch ball, since it breaks linearity.

Adding a slightly more complicated representation as $\rho = R \sum p_i \rho_i$ where $R$ is a rotation matrix and $p_i$ from a simplex (representation isn't uniquely determined) we can relate it to a classical Markov chain: A process that has a transition matrix that instead of being stochastic is instead a product of a rotation and a stochastic matrix. Hence we can split it into a simple Markov chain and a rotation process.

 

[Killtech]

What sort of decomposition methods are you looking for? Density matrices are Hermitian positive semidefinite, with trace one, so a standard eigenvalue decompositon will work just fine. What more do you want?

Which 3 "physical states" do you have in mind? The classical 4 Stokes parameters, minus the total intensity (S0 which is assumed to be 1)? Or in more mundane terms, the space of Hermitian 2x2 matrixes is 4 dimensional as a real vector space, and if you force the trace to be one, then this reduces to a 3 dimensional space. If you additionally force the matrices to be positive semidefinite, then you get a convex subset of that 3 dimensional space.

Does this descibe more or less, what you have in mind?

The decomposition i wrote in my post above is what i had in mind. For the qubit case the basis has 3 elements.

Indeed the Stokes parameter or Bloch spheres are a proper full map of the DO space which i was looking for, but both aren't linear. The decomposition i had in mind is linear but the spanned space contains more then just DOs. Anyhow, looking at both approaches i realized that they both are very closely related.

We have a very similar situation in probability theory: the probability measures forming a simplex are part of an affine surface, hence not a linear space itself. But we still use the linear span anyway that allows us to represent everything in terms of matrices and vectors instead of more abstract measures. The crucial part is finding the class of matrices that maps the subspace of p.measures onto itself which in case of classic probability are stochastic matrices.

So i have to correct my question to looking for an embedding of DOs into a minimal linear space and a proper linear decomposition within that space.

It would seem that for DOs this seems to work almost in the same way. Their subspace is a different one though, hence the class of matrices that preserves it differs. But for as far as i can see the product of stochastic and rotation matrices seems to do the job instead.

This is what i had in mind. I was hoping someone already explored what such a representation entails - even though i see now that it differs a bit from the classic case in probability.

 

[Son Goku]

Density matrices can't be represented as probability measures on a single sample space. That's what lies behind no-go theorems like GHZ or Tsirelson's version of the CHSH inequalities. It's why quantum probability is more general than classical/Kolmogorov probability, since a single sample space would entail the assumption that all variables can take well-defined values at the same time.

 

[Killtech]

Density matrices can't be represented as probability measures on a single sample space. That's what lies behind no-go theorems like GHZ or Tsirelson's version of the CHSH inequalities. It's why quantum probability is more general than classical/Kolmogorov probability, since a single sample space would entail the assumption that all variables can take well-defined values at the same time.

That's a crude misinterpretation of those no-go theorems. All those are based on Bell's idea and derived from Bell's factorization assumption representing the classical concept of locality. Now the issue is classic probability doesn't have a concept like that, so you can just take the truth table of the material conditional and build two random variables that produce a CHSH sum of 4, much higher then Tsirelson's. 4 is the limit for these kind of inequalities in classic probability theory (you can exceed that if you accept negative or above 1 probabilities). Now these kind of truth tables are very important for computers in real live and we can check that their physical implementation in reality strictly requires locality to achieve that - so no contradiction to Bell. These no-go theorems are restrictions for physical models that have to model locality but are irrelevant for Kolmogorovs probability which doesn't. The no-go is that slapping a probability model over a classical physical model has no chance of succeeding.

Kolmogorovs axioms could also be summarized into two statement: the chance for an event is a number between 0 and 1 and the sum over all possibilities has to be one (the second one is technically so challenging in the most general case of infinite spaces that we need a whole theory to handle it). No quantum experiment contests either statement.

Also, here is an example of a classical measure for QM:

https://arxiv.org/abs/quant-ph/0309021

The canonical construction: take the whole of normalized states from the Hilbert space as the state space. Doing so practically means you can take all of the calculus of QM and port it one to one into a classic probability framework. It's just that you create a huge infinitely dimensional probability space even for the most simple of problems which also suffers from a high redundancy given how many different probability measures will end up representing the very same mixed state / DO. Hence why i was looking if that can be compacted to much fewer dimensions.

 

[Son Goku]

Kolmogorovs axioms could also be summarized into two statement: the chance for an event is a number between 0 and 1 and the sum over all possibilities has to be one (the second one is technically so challenging in the most general case of infinite spaces that we need a whole theory to handle it). No quantum experiment contests either statement.

Kolmogorov's axioms involve more than those conditions, they explicitly define probability models as measures on the Sigma algebra of a single sample space.

build two random variables that produce a CHSH sum of 4, much higher then Tsirelson's. 4 is the limit for these kind of inequalities in classic probability theory

Show me such a classical model. It is impossible per Tsirelson's formulation of CHSH.

 

[Son Goku]

Hence why i was looking if that can be compacted to much fewer dimensions.

No. That's the content of Hardy's infinite baggage theorem.

 

[Killtech]

Kolmogorov's axioms involve more than those conditions, they explicitly define probability models as measures on the Sigma algebra of a single sample space.

Fun fact: they actually are just those conditions. But it turns out that if you formulate them as naively as i did you run into weird Banach-Tarsky issues in infinitely enough spaces. which is why mathematicians came up with sigma algebras to make that simple concept work in every case.

And the most of Bell related examples always just look at a discrete or even finite number of possible measurement outcomes where all of the sigma algebra stuff is completely irrelevant and you could just as much use my simplistic formulation.

Show me such a classical model. It is impossible per Tsirelson's formulation of CHSH.

Sure, let's take Bells idea an deprive it from its central assumption and see what happens. Let $a, a'$ be two random variables with values in $\{-1,1\}$. Let $d$ be a random variable that is $1$ when it's decided to measure $a$ and $-1$ otherwise. Let $b = a$ and $b' = 1_{\{d=-1\}}a' - 1_{\{d=1\}}a$ be another two random variables. A classic simple decision process. I leave it to the interested reader to show CHSH sum $S=4$.

 

[Son Goku]

Fun fact: they actually are just those conditions.

Kolmogorov's axioms assume a single sample space, which is an assumption beyond the two you gave. This is true in the finite and infinite cases. Taking the finite case, note in the QM case the sum over the probability of atomic events exceeds one, unlike in finite Kolmogorov probability. Kolmogorov probability assumes the underlying set of events forms a Boolean algebra, in QM this is not the case as the set of events is an orthomodular lattice.

Sure, let's take Bells idea an deprive it from its central assumption and see what happens. Let be two random variables with values in . Let be a random variable that is when it's decided to measure and otherwise. Let and be another two random variables. A classic simple decision process. I leave it to the interested reader to show CHSH sum .

That's not a classical probability model of CHSH, just consider writing down the sample space and write a and d as random variables on it and you'll see.

 

[A. Neumaier]

Indeed the Stokes parameter or Bloch spheres are a proper full map of the DO space which i was looking for, but both aren't linear. The decomposition i had in mind is linear but the spanned space contains more then just DOs. Anyhow, looking at both approaches i realized that they both are very closely related.

We have a very similar situation in probability theory: the probability measures forming a simplex are part of an affine surface, hence not a linear space itself. But we still use the linear span anyway that allows us to represent everything in terms of matrices and vectors instead of more abstract measures. The crucial part is finding the class of matrices that maps the subspace of p.measures onto itself which in case of classic probability are stochastic matrices.

The vector space spanned by the density matrices in d dimensions is simply the $d^2$-dimensional space of all Hermitian d by d matrices. For d=2 one can represent them as linear combinations of the identity and the Pauli matrices; the coefficients are the components of the Stokes vector. That's why the latter is relevant in optics. For $d>2$ a vector representation is much less useful than the matrix representation.

But all this is quite unrelated to the d-dimensional simplex in d+1 dimensional vector space, which characterizes classical probability.

Note that every finite-dimensional manifold may be regarded as part of an affine space, but the latter plays only rarely an important role in the analysis of the former.

 

[Son Goku]

Also, here is an example of a classical measure for QM

https://arxiv.org/abs/quant-ph/0309021

I had a look at this and it still doesn't look like an actual classical measure for QM. It seems to classicalize part of the probability structure, at the cost of severally altering the geometry of the state space to the point where I'd doubt its ability to replicate dynamics.

How does it classically replicate a pure state for example?

 

[gentzen]

I had a look at this and it still doesn't look like an actual classical measure for QM. It seems to classicalize part of the probability structure, at the cost of severally altering the geometry of the state space to the point where I'd doubt its ability to replicate dynamics.

How does it classically replicate a pure state for example?

Great that you had a look. That paper defines a canonical measure (from a given density matrix) on the space of pure states. So it doesn't even try to replicate a pure state.

Your confusion seems to arise from the distinction between classical probability theory, and classical physics. The measure from that paper is a perfectly classical probability measure on the space of pure states. But this has no connection to classical physics, because that was never the goal.

The goal instead is to improve the handling of density operators (DOs) in pure wavefunction based formalisms (or interpretations, like De-Broglie Bohm or MWI), and especially provide an alternative to that (in)famous "church of the larger Hilbert space" treatment:

De-Broglie Bohm gives a model where quantum probabilities arise from classical probabilities. This makes it hard to compare the handling of probability in QT with how it's done in classic PT. (You have to bring in space-time and locality, but then you are no longer talking about probability theory itself.) It seems that DOs don't fit into that De-Broglie Bohm framework, but I recently learned that some Bohmians (Goldstein, Tumulka, et al) attacked that problem head on (https://arxiv.org/abs/quant-ph/0309021):

 

[Son Goku]

Your confusion seems to arise from the distinction between classical probability theory, and classical physics

Thanks for your response.
Honestly I'm not confusing classical physics and classical probability. This construction seems parasitic on quantum probabilities for the pure case. How are the pure case probabilities replicated?

 

[gentzen]

Honestly I'm not confusing classical physics and classical probability.

You have a confusion in your communication with Killtech, and in your interpretation of that paper. The goal of Killtech is not to reduce quantum probabilities to classical physics. He wants to understand the structure of the "quantum probability space" given by the density operators. Then A. Neumaier suggested that a more promising approach would be to look at the structure of POVMs instead.

This construction seems parasitic on quantum probabilities for the pure case. How are the pure case probabilities replicated?

For De-Broglie Bohm, a pure state defines a measure on the configuration space of particle positions. Those reproduce the quantum probabilities by the usual interpretation of De-Broglie Bohm. For MWI, again you would use the usual interpretation of how probabilites arise in MWI. And for a textbook interpretation of QM focusing on pure states, the quantum probabilities are simply given by the Born rule.

That you call the construction from that paper "parasitic" still gives me the impression that you misinterpret its goals. That construction sort of gives you the barycentric coordinates that Killtech was looking for:

We are looking for a specific basis though - the barycentric coordinates: i.e. the vertices of the simplex which are special DOs of pure some states.

Note that matrices that maps the simplex onto itself expressed in barycentric coordinates are called stochastic matrices (i.e. operations that don't break probabilities).

Lol, i played around with the same measure. There are even some older posts here with a few of my fails to build it properly. I would say this is the canonical approach, which is why all of us stumbled onto it naturally.

(I thought about generalized barycentric coordiantes while trying to construct such a canonical measure, or rather when trying to prove that my construction was canonical. That made me realize the flaws of my construction(s), so I came closer to the correct construction, and in fact tried each ingredient of it on my way, but found that reference before I managed myself to put those ingredients together in the right way.)

 

[Son Goku]

Okay I see now. My understanding was that a purely deterministic formalism cannot model the quantum probabilities due to the presence of Kolmogorov–Levin–Cháitín randomness.
I understood the goals of the paper itself, but Killtech seemed to be saying it gave a classical model of quantum probability. The paper itself didn't seem to say so itself, the "parasitic" remark was more to do with trying take it as doing so, but I can see the authors don't mean that.

If somebody wants to understand the quantum state space I'd use
"Geometry of Quantum States: An Introduction to Quantum Entanglement" by Ingemar Bengtsson and Karol Życzkowski.

Naivly Killtech's goal seems to involve understanding a density matrix as ignorance of pure states, which we know to be an incorrect approach.

 

[Son Goku]

Those reproduce the quantum probabilities by the usual interpretation of De-Broglie Bohm

Sorry thinking about this again, how is this done? The underlying equations are deterministic so how do you ensure you get 1-random frequencies like QM.

The goal of Killtech is not to reduce quantum probabilities to classical physics

Genuinely I understand this isn't about classical physics. I'm solely talking about how quantum probability cannot be replicated by classical probability.
I fully get that that paper is defining a classical probability measure over pure states, but I don't see how that recovers quantum probabilities without simply interpreting pure states as probabilistic constructs already as opposed to "real waves". Once again I understand the distinction between classical probabilities and classical physics.

 

[gentzen]

Sorry thinking about this again, how is this done? The underlying equations are deterministic so how do you ensure you get 1-random frequencies like QM.

Good question, even so maybe this is not the best thread for asking it, because the resident Bohmian experts will probably not answer it here. The probability measure on the configuration space of particle positions, and drawing a configuration of particle positions based on that measure is not the problem. That part is just assumed to be given.

One problem is how to get randomness from one particular configuration of particle positions drawn according to that measure. This is achieved by the notion of typicality. This means that the particular drawn configuration of particle positions has an overwhelmingly huge probability of being typical, at least if the configuration space is big enough. How big is big enough? Not sure, but at least $10^{23}$ is definitively big enough.

The other problem is how to get actual measurement results from particle positions that are in a certain sense not directly observable either. But at least they exist, according to the ontology of Bohmian mechanics. And so you end-up with a measurement theory which tries to explain this.

 

[Killtech]

Work & life is keeping me busy, so sorry i am unable to respond in a timely manner.

Genuinely I understand this isn't about classical physics. I'm solely talking about how quantum probability cannot be replicated by classical probability.

I have tried to find any indication that quantum probabilities are different from classical ones, but i couldn't find anything that would show that. All the no-go theorems are very interesting but all of them ultimately are focused on using some classical physical assumptions and showing exposing that it cannot work. Unfortunately they have to ebbet it into some probability model due to the nature of probabilistic results in real experiments.

Issue is that people seem to have quite a lot of misconceptions about classic probability - like for example where do you get the idea from that

Kolmogorov's axioms assume a single sample space, which is an assumption beyond the two you gave

Except from people here on the physics forums claiming that, i have not ever seen such an assumption in probability theory. Please look up the axioms, you won't find any mention of it. Kolmogorov definition outlines the minimum conditions to have a probability measure with its technical requirements (taken together they are define a probability space). But there is no limit on how many of those you many have (why would there be?). In fact every time you define a random variable, you define a new probability space on the possible values that variable can take (the state space of said variable) with its own measure and sigma algebra.

In case of CHSH we work with random variables, so we are just discussing such a situation. Maybe we should first try to find out where you got these strange idea from?

Kolmogorov's axioms involve more than those conditions, they explicitly define probability models as measures on the Sigma algebra of a single sample space.

Sure, Kolmogorov also needs set theory as a basis to define functions and sets, but Hilbert spaces need the same. I indeed didn't count the technical assumptions that QM also requires. As for the sigma algebras, you are aware that those are purely needed to ensure integrals are always well defined? It's particularly relevant for QM which Hilbert spaces is some form of $L^2$ space - A measurable space (named after Lebesque, a measure theory mathematician) that only works due to the choice Borel's sigma algebra. So QM has to make use of the very same technical framework to begin with - and quite a lot more. That may not be exclusively mentioned in physics lectures, but it doesn't change the facts.

Honestly I'm not confusing classical physics and classical probability. This construction seems parasitic on quantum probabilities for the pure case. How are the pure case probabilities replicated?

A pure state translates into a Dirac probability measure in probability theory - i.e. an (almost) certain state. So if you calculate $\rho_\mu = \int \mu(d_\psi) |\psi\rangle \langle \psi|$ with $\mu$ being a Dirac delta distribution you simply get the DO for that pure state. i.e. you get the exact same description you have in QM.

Hence this measure is really nothing else than a inconvenient way to denote DOs but in a classic probability formalism. And as @gentzen said, it is indeed:

That construction sort of gives you the barycentric coordinates that Killtech was looking for

where each $|\psi\rangle \langle \psi|$ becomes a vertex of an infinitely dimensional simplex.

I knew this probability measure and one intention of this thread was to understand if this is the most "minimal" probability measure that works.

 

[Killtech]

The vector space spanned by the density matrices in d dimensions is simply the $d^2$-dimensional space of all Hermitian d by d matrices. For d=2 one can represent them as linear combinations of the identity and the Pauli matrices; the coefficients are the components of the Stokes vector. That's why the latter is relevant in optics. For $d>2$ a vector representation is much less useful than the matrix representation.

But all this is quite unrelated to the d-dimensional simplex in d+1 dimensional vector space, which characterizes classical probability.

Note that every finite-dimensional manifold may be regarded as part of an affine space, but the latter plays only rarely an important role in the analysis of the former.

Hmm, indeed you are right a basis of Hermitian matrices should do the trick. With them we can make use that an operation like $S \rho S^\dagger$ is linear in $\rho$. For the qubit case $d=2$ we can write $\rho$ as a 4 dimensional vector of that basis and use $\tilde S \rho$ to express the same operation via a 4x4 matrix. Since the commutator is also linear in $\rho$ this also works with $[H, \rho]$ within that space which implies that there must exist a 4x4-matrix $\tilde H$ such that $\rho(t) = e^{t \tilde H}\rho(0)$.

Now that formulation looks almost the same as a continuous time Markov chain with the exception that the DOs form a different subspace. So we have to ask what class of matrices $\tilde S$ maps the DO space onto itself? For a probability space those are the stochastic matrices, but here we need a slightly larger class.
What is it?

But all this is quite unrelated to the d-dimensional simplex in d+1 dimensional vector space, which characterizes classical probability.

Well, after this discussion i have come to the conclusion that the smallest simplex that can correctly encompass all pure states must indeed have infinite dimension. That is the probability space of the classic probability measure gentzen posted links to.

I never liked it because the infinite dimensionality makes it very tedious to work with. And on the other hand there is a lot of linearity in the Hilbert space that measure cannot make any use of. Given that we are dealing with such very special probability spaces i was trying to understand if its construction might not have been minimal. But now i think it is and i understand the challenge.

With something like the Hermitian basis however i come to realize that we can get around needing to use the continuous state space probability formalism. Because instead it seems we can use a helper measure that while not being a proper probability measure, it still fully determines the experimentally distinguishable part of the actual probability measure. That way a classic probabilistic description won't have to deal too much with Markov kernel formalism.

 

[A. Neumaier]

it seems we can use a helper measure that while not being a proper probability measure, it still fully determines the experimentally distinguishable part of the actual probability measure.

Quite likely what you are looking for is the Wigner quasi-probability representation...