A Assumptions of the Bell theorem

[Nullstein]

Is it simply due to contextuality? Non-commuting observables are modeled by different measurement apparatuses in Bohmian Mechanics, and one cannot measure position and momentum at the same time simply because the measurement setups for each is completely different.

It's contextuality, but at different times, so it is possible to obtain both $A$ and $A(t)$ despite their non-commutativity. As you say, it's impossible to measure non-commuting variables at the same time (independent of the interpretation), but this isn't the situation we're interested in here.

 

[Nullstein]

I meant to formulate it such that all time dependence (in standard QM) is in the time dependence of the state $|\psi(t)\rangle$.

I understand, but in order to compute a commutator, we need to formulate it in terms of operators. But as soon as we agree that BM and QM agree on the predictions of macroscopic measurements, we should be able to perform the computation in the Heisenberg picture anyway.

 

[Kolmo]

I think along the lines that @Demystifier was saying, in the GHZ experiment, the events are not

$A_x = +1$
$A_x = -1$
$B_x = +1$
etc.

The events are:

• Someone measures the spin of particle $A$ in the x-direction, and the result is $+1$
• Someone measures the spin particle $A$ in the x-direction, and the result is $-1$
• etc

These events have well-defined probabilities that obey the Kolmogorov axioms.

I don't think they do as for instance if we take:

• $E =$ "Someone measures the spin of particle $A$ in the x-direction, and the result is $+1$"

can one actually work out $P(E)$?
I don't think one can since one would have to work out the probability of selecting a device measuring a specific spin angle which isn't really something covered by the observable algebra for the particle.

If one just tries to directly take the QM predications it also won't make sense. Consider $X_{+1}$, $X_{-1}$ the obvious $X$-axis events (i.e. somebody measures $X$ and gets $+1$) and $Z_{+1}$, $Z_{-1}$ similar for the $Z$-axis. Then the event:

• $X_{+1} \lor X_{-1} \lor Z_{+1} \lor Z_{-1}$

would have a probability exceeding unity.

 

[Nullstein]

I don't think they do as for instance if we take:

• $E =$ "Someone measures the spin of particle $A$ in the x-direction, and the result is $+1$"

can one actually work out $P(E)$?
I don't think one can since one would have to work out the probability of selecting a device measuring a specific spin angle which isn't really something covered by the observable algebra for the particle.

I don't think one can exclude the possibility of formulating this within a more detailed description completely, but as I said: If your most detailed description is still a quantum theory, the paradoxes won't go away and we would gain no fundamental insight. Maybe we can shift the problem to a realm where it doesn't bother us anymore, but people who care about these things in the first place can hardly be satisfied with such a non-solution.

 

[Demystifier]

I understand, but in order to compute a commutator, we need to formulate it in terms of operators. But as soon as we agree that BM and QM agree on the predictions of macroscopic measurements, we should be able to perform the computation in the Heisenberg picture anyway.

In principle, yes. But when we compare measurements at different times, then we must take into account the effect of "wave function collapse" (or state update, or whatever one likes to call it). This effect is awkward to take into account in the Heisenberg picture.

EDIT: Let me briefly explain an example how collapse can be accounted for in the Heisenberg picture, by first describing it in the Schrodinger picture. Suppose that the state is first prepared in the initial state $|\psi_0\rangle$ at time $t_0$ and then measured at time $t_{\rm m}>t_0$. This means that the state first evolves unitary with the unitary evolution operator $U(t-t_0)$, then at time $t_{\rm m}$ the state gets projected by some randomly chosen projection operator $\pi$. (For POVM measurements it does not need to be a projection operator, but the point is that at $t_{\rm m}$ the unitary evolution is interrupted by a random non-unitary operator.) After measurement we again have unitary evolution described by $U(t-t_{\rm m})$. For $t>t_{\rm m}$ this evolution is described by the non-unitary operator
$$V(t)=U(t-t_{\rm m}) \pi U(t_{\rm m}-t_0)$$
More precisely, the state can be written as
$$|\psi(t)\rangle= \frac{V(t)|\psi_0\rangle}{|V(t)|\psi_0\rangle|}$$
where the denominator accounts for the correct normalization of the state, needed because $V(t)$ is not unitary. Then for an arbitrary observable $A$ we can write
$$\langle\psi(t)|A|\psi(t)\rangle = \langle\psi_0|A(t)|\psi_0\rangle$$
where
$$A(t) \equiv \frac{V^{\dagger}(t)AV(t)}{\langle\psi_0|V^{\dagger}(t)V(t)|\psi_0\rangle}$$
The last equation is naturally interpreted as evolution in the Heisenberg picture, with the effect of measurement taken into account. Due to the measurement, the evolution of the observable is non-unitary (because $V(t)$ is non-unitary), random (because $\pi$ is random) and non-linear (because of the denominator).

 

[Kolmo]

I don't think one can exclude the possibility of formulating this within a more detailed description completely

It can't be eliminated by a more detailed quantum description, you mean some kind of new theory I assume?

My point is more to do with Quantum Theory not being a Kolmogorov probability theory. That would be the standard academic view, i.e. that it breaks Kolmogorov's axioms. "Lost Causes in and Beyond Physics" by R.F. Streater explains this quite well in Chapter 8. This paper by Summers and Rédei explains it well:

https://arxiv.org/abs/quant-ph/0601158

Kolmogorov formulates probability theory as a measure on the sigma algebra of some space. It's Gelfand's famous representation theorem that shows a Kolmogorov probability space is always the representation of some commutative C*-algebra. Thus quantum theory, which uses a non-commutative algebra, isn't a Kolmogorov theory.

 

[stevendaryl]

I don't think they do as for instance if we take:

• $E =$ "Someone measures the spin of particle $A$ in the x-direction, and the result is $+1$"

can one actually work out $P(E)$?
I don't think one can since one would have to work out the probability of selecting a device measuring a specific spin angle which isn't really something covered by the observable algebra for the particle.

What's computed from quantum mechanics is the conditional probability:

"The probability of measuring +1 given that the experimenter chose angle $\alpha$"

The event "Someone measures the spin and the result is +1" can certainly be given an empirical (statistical) probability, but you're right, quantum mechanics by itself doesn't allow us to compute the probability (at least not in practice).

If one just tries to directly take the QM predications it also won't make sense. Consider $X_{+1}$, $X_{-1}$ the obvious $X$-axis events (i.e. somebody measures $X$ and gets $+1$) and $Z_{+1}$, $Z_{-1}$ similar for the $Z$-axis. Then the event:

• $X_{+1} \lor X_{-1} \lor Z_{+1} \lor Z_{-1}$

would have a probability exceeding unity.

Why would that be?

 

[stevendaryl]

Kolmogorov formulates probability theory as a measure on the sigma algebra of some space. It's Gelfand's famous representation theorem that shows a Kolmogorov probability space is always the representation of some commutative C*-algebra. Thus quantum theory, which uses a non-commutative algebra, isn't a Kolmogorov theory.

Okay.

 

[Kolmo]

What's computed from quantum mechanics is the conditional probability:

"The probability of measuring +1 given that the experimenter chose angle α"

Yeah so that's certainly true. The probabilities are intrinsically conditional, Kolmogorov however formulated probability theory with unconditional probabilities as the base and conditionals as derived. So if you fundamentally only have conditionals it's not a Kolmogorov theory.

The event "Someone measures the spin and the result is +1" can certainly be given an empirical (statistical) probability, but you're right, quantum mechanics by itself doesn't allow us to compute the probability (at least not in practice

I would agree that one could give a probability to the event, but this would be a purely subjective credence/guess not something one could work out from quantum theory. Even in principle, I wouldn't say it is an in practice thing.

Why would that be?

The events have no intersection, i.e. $X_{+1} \land Z_{-1}$ has zero probability, so

$P(X_{+1} \lor X_{-1} \lor Z_{+1} \lor Z_{-1}) = P(X_{+1}) + P(X_{-1}) + P(Z_{+1}) + P(Z_{-1})$

and if you put in the predictions a spin-z $+1$ eigenstate for each of these cases then it exceeds $1$.

 

[Demystifier]

So if you fundamentally only have conditionals it's not a Kolmogorov theory.

I disagree. Conditional probabilities satisfy all axioms of Kolmogorov probability, so conditional probabilities are just a special case. Just because Kolmogorov haven't explicitly studied this case does not mean that it's not Kolmogorov probability.

 

[Kolmo]

I disagree. Conditional probabilities satisfy all axioms of Kolmogorov probability, so conditional probabilities are just a special case. Just because Kolmogorov haven't explicitly studied this case does not mean that it's not Kolmogorov probability.

I don't really think there is mathematical scope for disagreement. Kolmogorov's axioms formulate probability theory as a measure function of total weight one assigned to a sigma algebra. By the Gelfand representation such a set up is always equivalent to a commutative C*-algebra, so quantum theory cannot be equivalent to such a system.

The point is that individual conditionals might satisfy Kolmogorov's axioms, but the totality of conditionals does not. So the conditional probabilities for a z-axis spin measurement obey Kolmogorov theory when considered alone, as do x-axis measurements, however both of them together do not. There's no Kolmogorov theory containing the four CHSH measurement contexts for example.

The breaking of Kolmogorov's axioms is the condition in Bell's theorem that quantum theory breaks, as explained in Streater's monograph I mentioned above.

Bell’s inequalities, which hold in any Kolmogorovian theory, do not hold in quantum mechanics

 

[Demystifier]

Kolmogorov formulates probability theory as a measure on the sigma algebra of some space. It's Gelfand's famous representation theorem that shows a Kolmogorov probability space is always the representation of some commutative C*-algebra. Thus quantum theory, which uses a non-commutative algebra, isn't a Kolmogorov theory.

I will not pretend that I fully understand this. However, with my boldings of "some" and "a", it should be clear that there is no contradiction. Just because quantum theory uses a non-commutative algebra does not imply that there is no some other commutative algebra that represents probability. For instance, the probability in the momentum space can be represented with one commutative algebra and the probability in the momentum space can be represented with another one commutative algebra. Of course, those two commutative algebras are different from the non-commutative algebra of quantum observables, but so what? I don't see any contradiction with statements of the theorems.

A real issue is the fact that there is no Kolmogorov probability $P(x,p)$ (with $x$ and $p$ being the position and momentum, respectively) such that the marginals
$$P_1(p)=\int dx \, P(x,p)$$
$$P_2(x)=\int dp \, P(x,p)$$
coincide with quantum probabilities, but again I don't see a contradiction with any Kolmogorov axiom.

 

[Kolmo]

I don't see a contradiction with any Kolmogorov axiom

Each individual context obeys Kolmogorov axioms, but there is no Kolmogorov model of all the variables at once, thus a quantum system as a whole (not focusing on an individual context) does not obey Kolmogorov's axioms.
The fact that the probabilities for each pair of variables in a CHSH test are not marginals of some common Kolmogorov model for all four variables is what allows violation of the Bell inequalities.

The fact that the theory does obey Kolmogorov's axioms within each context is what allows one to rewrite it as a Topos in more advanced treatments.

 

[Demystifier]

Each individual context obeys Kolmogorov axioms, but there is no Kolmogorov model of all the variables at once, thus a quantum system as a whole (not focusing on an individual context) does not obey Kolmogorov's axioms.

OK, but what's wrong with using many Kolmogorov models, one for each context?

Besides, if you add the measuring apparatus into the system as a whole, then all predictions of QM can be reduced to measurement of only one commuting set of observables - the position observables. In this way the quantum system as a whole can be reinterpreted as one Kolmogorov model, which indeed is a basis for the Bohmian interpretation.

Perhaps we can formulate all this as follows. Bell's theorem assumes that there is only one Kolmogorov model and from this one derives nonlocality. Indeed, Bohmian mechanics is an example of this general result. If, on the other hand, we use another Kolmogorov model for each context, then we arrive at consistent histories (which obey locality but non-classical logic when different contexts are discussed at once).

 

[Kolmo]

OK, but what's wrong with using two Kolmogorov models, one for each context?

There's nothing wrong with it. It just means a quantum system as a whole isn't something which obeys the Kolmogorov axioms and that's why we get CHSH violations.

That'd only be a problem if we didn't have some other probability theory to use, but in fact we do as we have quantum theory itself.

 

[Demystifier]

There's nothing wrong with it. It just means a quantum system as a whole isn't something which obeys the Kolmogorov axioms and that's why we get CHSH violations.

That'd only be a problem if we didn't have some other probability theory to use, but in fact we do as we have quantum theory itself.

OK, let me then look at it from a slightly different perspective. I think the Bell's theorem can now be concisely stated as follows: There is no single local Kolmogorov model compatible with all probabilistic predictions of QM.

So it seems that we have 3 options:
(i) abandoning locality (Bohmian mechanics, GRW, ...)
(ii) using many models of Kolmogorov axioms, one for each context (Bohr's complementarity, consistent histories, ...)
(iii) abandoning Kolmogorov axioms at some deeper level (??)

My question for you is this: Is there an explicit proposal of an interpretation in the category (iii)? (Perhaps many worlds?)

 

[Kolmo]

My question for you is this: Is there an explicit proposal of an interpretation in the category (iii)?

Well a quantum system, despite not obeying the Kolmogorov axioms, still obeys certain probabilistic constraints. We could consider giving these up, I'll give an example.

Say the variables $A,B$ have a common Kolmogorov model¹. Then imagine the same is true also for $B, C$ and $A, C$. Quantum Theory implies that since there is a common Kolmogorov model for each pair drawn from $A,B,C$ then for the triplet $A,B,C$ as a whole there is a single Kolmogorov model². In fact you can derive quantum theory as the most general probability theory which permits this structure.

You could imagine probability theories more general than this, i.e. each pair having a Kolmogorov model does not imply the triplet has one. However such theories would violate the Tsirelson bound and thus don't have empirical support.

¹ Conveyed in quantum theory by $[A,B] = 0$
² In QM itself the mathematical expression of this is if each pair has a common eigenbasis, then all three do

 

[Demystifier]

Say the variables $A,B$ have a common Kolmogorov model¹. Then imagine the same is true also for $B, C$ and $A, C$. Quantum Theory implies that since there is a common Kolmogorov model for each pair drawn from $A,B,C$ then for the triplet $A,B,C$ as a whole there is a single Kolmogorov model². In fact you can derive quantum theory as the most general probability theory which permits this structure.

That's axiomatization (which is interesting), but I asked for interpretation. Interpretation is something that is supposed to give an intuitive conceptual picture explaining why is that so. But if thinking in terms of interpretations is not your style, that's OK too.

 

[Kolmo]

That's axiomatization (which is interesting), but I asked for interpretation. Interpretation is something that is supposed to give an intuitive conceptual picture explaining why is that so

I'm not sure I understand what you mean, but the conceptual picture is fairly clear here I would say. In our world being able to perform simultaneous measurements on pairs drawn from some set of variables, implies you can design a single measurement which measures all variables in the set. Quantum Theory is then a probability theory obeying that concept with no further assumptions¹.

¹ No further assumptions because it is the most general probability theory compatible with that assumption.

 

[Demystifier]

I'm not sure I understand what you mean, ... Quantum Theory is then a probability theory ...

I guess I was right that interpretations is not your style. Roughly speaking, in an interpretation quantum theory is supposed to be a physical theory, rather than a probability theory. But of course, it's hard to define precisely what "physical" means, so if you don't get it, forget it!

 

[Kolmo]

I guess I was right that interpretations is not your style. Roughly speaking, in an interpretation quantum theory is supposed to be a physical theory, rather than a probability theory. But of course, it's hard to define precisely what "physical" means, so if you don't get it, forget it!

Well the kinematical core of quantum theory, i.e. Hilbert spaces and operators, is a probability theory. That's just a mathematical fact. I wouldn't get why it's not "supposed to be" that if that is in fact what it is. To me it would be like criticising General Relativity by saying gravity is "supposed to be a physical theory not differential geometry".

 

[PeterDonis]

the kinematical core of quantum theory, i.e. Hilbert spaces and operators, is a probability theory

No, it isn't. Unitary evolution, which is what you are describing, is deterministic. The only way probabilities enter into QM is with measurements, but measurements are not part of the "kinematical core" you describe; they are put in "by hand" in the basic math of QM by assuming the Born Rule and the projection postulate.

The simplest way to see that the above is true is to observe that at least one interpretation of QM, the MWI, has no probabilities; everything is unitary evolution, all the time, and everything is deterministic. Probabilities are appearances only.

 

[Demystifier]

Well the kinematical core of quantum theory, i.e. Hilbert spaces and operators, is a probability theory. That's just a mathematical fact. I wouldn't get why it's not "supposed to be" that if that is in fact what it is. To me it would be like criticising General Relativity by saying gravity is "supposed to be a physical theory not differential geometry".

I understand your point of view and I'm not criticizing it. I think it's illuminating to look at things from different points of view and your point of view enriched my understanding. But perhaps you will also understand my point of view if I offer another analogy: Classical mechanics is supposed to be a physical theory, not simplectic geometry.

 

[Kolmo]

No, it isn't. Unitary evolution, which is what you are describing, is deterministic

Unitary evolution is the dynamics not the kinematics right? I was only describing the kinematics.

The mathematics of quantum theory are usually recognised as a generalised probability theory in the literature. For example in D'Ariano's book "Quantum Theory from First Principles: An Informational Approach", quantum theory is derived as a specific theory within the general space of OPT/GPTs. Hence that it is mathematically a probability theory obeying certain conditions. Doesn't that make it mathematically a probability theory?

Regarding Many Worlds, couldn't we pull this word game with Kolmogorov probability? We have a space $\Omega$, its sigma algebra $\Sigma$ and the measure $\mu$ forming the usual triplet. However because some people think $\mu$ is "physically real" it's wrong to say it is a probability theory?

The only way probabilities enter into QM is with measurements, but measurements are not part of the "kinematical core" you describe; they are put in "by hand" in the basic math of QM by assuming the Born Rule and the projection postulate

The kinematics are the operator algebra and the state space. Same as any theory the kinematics are the DOFs and the states. From Gleason's theorem the algebra of quantum theory has no dispersion-free states so it has probabilities right? The projection postulate is a simple corollary then, defined the same way that one defines conditionals in Kolmogorov probability theory via subalgebras.

 

[RUTA]

Unitary evolution is the dynamics not the kinematics right? I was only describing the kinematics.

The mathematics of quantum theory are usually recognised as a generalised probability theory in the literature. For example in D'Ariano's book "Quantum Theory from First Principles: An Informational Approach", quantum theory is derived as a specific theory within the general space of OPT/GPTs. Hence that it is mathematically a probability theory obeying certain conditions. Doesn't that make it mathematically a probability theory?

Regarding Many Worlds, couldn't we pull this word game with Kolmogorov probability? We have a space $\Omega$, its sigma algebra $\Sigma$ and the measure $\mu$ forming the usual triplet. However because some people think $\mu$ is "physically real" it's wrong to say it is a probability theory?The kinematics are the operator algebra and the state space. Same as any theory the kinematics are the DOFs and the states. From Gleason's theorem the algebra of quantum theory has no dispersion-free states so it has probabilities right? The projection postulate is a simple corollary then, defined the same way that one defines conditionals in Kolmogorov probability theory via subalgebras.

This is the way quantum information theorists think of QM, yes.

 

[Kolmo]

This is the way quantum information theorists think of QM, yes.

I guess I find it confusing as what are quantum stochastic processes then if quantum theory is not a probability theory? They must be misnamed, i.e. they're not really stochastic processes or something. Seems strange to me, not something I've seen in the literature. Quantum Probability is a whole field, is it misnamed or something?

 

[PeterDonis]

Unitary evolution is the dynamics not the kinematics right?

I would say it's both. But I can see how the term "kinematics" could be restricted to just the operator algebra and state space. See further comments below.

One could argue that the full dynamics of QM includes measurements, but exactly what happens, dynamically, in a measurement is interpretation dependent; the basic math of QM does not make any commitment about what "actually happens" in a measurement, only about how we can make predictions about experimental results. But none of that is kinematics.

The mathematics of quantum theory are usually recognised as a generalised probability theory in the literature.

This description includes measurements, which are not kinematics.

Regarding Many Worlds, couldn't we pull this word game with Kolmogorov probability?

One of the major open issues with the MWI is how probabilities can be defined within that interpretation. There is no generally accepted answer to this issue.

The kinematics are the operator algebra and the state space.

And unitary evolution gives a completely deterministic evolution on this space. To bring in any probabilities at all, as I said, you have to include measurements and the projection postulate, which is certainly not kinematics: it's dynamics, and a very weird dynamics at that since it involves discontinuous jumps.

I guess I find it confusing as what are quantum stochastic processes then if quantum theory is not a probability theory?

They're dynamics, not kinematics. Remember that I'm not saying there are no probabilities in QM; I'm just saying that your claim that probabilities are part of the "kinematical core" of QM is not correct.

 

[vanhees71]

OK, let me then look at it from a slightly different perspective. I think the Bell's theorem can now be concisely stated as follows: There is no single local Kolmogorov model compatible with all probabilistic predictions of QM.

So it seems that we have 3 options:
(i) abandoning locality (Bohmian mechanics, GRW, ...)
(ii) using many models of Kolmogorov axioms, one for each context (Bohr's complementarity, consistent histories, ...)
(iii) abandoning Kolmogorov axioms at some deeper level (??)

My question for you is this: Is there an explicit proposal of an interpretation in the category (iii)? (Perhaps many worlds?)

I think that's rather a question of opinion. I would, of course, choose (ii) but of course not Bohr's "complementarity" (because it's vague; I think the quantum theoretical formalism is much clearer by itself without additional philosophical lingo or consistent histories; just the minimal statistical interpretation). QT is indeed simply a theory to predict probabilities for the outcome of measurements ("observables") given a preparation of a system ("state").

For violating Bell's inequality you have to do different measurements which cannot be done on one system. In the example treated in Sakurai's textbook you prepare two spins 1/2 in the $S=0$ state
$$|\Psi \rangle=1/\sqrt{2}(|\hat{z} 1/2,\hat{z}-1/2 \rangle - |\hat{z} -1/2,\hat{z} 1/2 \rangle,$$
where $|\hat{a} \pm 1/2 \rangle$ denotes the eigenstate of the spin component in direction of the unit vector $\hat{a}$ (I use Sakurai's notation). Then you have to consider the probalities for the outcome of three measurements
$$P_1=P(\hat{a} 1/2,\hat{b} 1/2), \quad P_2=P(\hat{a} 1/2,\hat{c} 1/2), \quad P_3=P(\hat{b} 1/2,\hat{c} 1/2),$$
where the angles between the unit vectors are $\theta_{ab}=2 \theta$ and $\theta_{ac}=\theta_{bc}=\theta$. Bell's inequality says
$$P_1 \leq P_2+P_3 \quad (\text{any local HV theory a la Bell}).$$
The QT probabilities are
$$P_1=\frac{1}{2} \sin^2(\theta_{ab}/2)$$
etc. and thus Bell's inequality would be
$$\sin^2 \theta \leq 2 \sin^2(\theta/2)?,$$
but
$$\sin^2 \theta-2 \sin^2(\theta/2)=2 \cos \theta \sin^2(\theta/2),$$
which is $>0$ for $0<\theta<\pi/2$.

For the measurement of each of the $P_j$ ($j \in \{1,2,3\}$) you have to prepare an ensemble of systems in the state $|\Psi \rangle \langle \Psi|$ and perform the specific measurement for each case. There's no way to do all three measurements at one particle and use one big ensemble to get the probabilities needed to check Bell's inequality. You can only measure the spin component in one direction for each of the two particles but never spin components in different directions on one particle.

Just think how to do that measurement in the lab: You have to use a Stern-Gerlach apparatus with a magnetic field in the direction of the spin component you want to measure. This excludes the accurate determination of any other component of the spin in another direction. Note that this is true already for the classical picture of the description in terms of a classical magnetic dipole moment, which precesses rapidly around the direction given by the magnetic field, and thus only the component in diretion of the magnetic field is determined when defined as a time average of the rapidly precessing dipole (of course this is not the accurate quantum picture but it's close to make the SGE plausible!).

 

[RUTA]

They're dynamics, not kinematics. Remember that I'm not saying there are no probabilities in QM; I'm just saying that your claim that probabilities are part of the "kinematical core" of QM is not correct.

In quantum information theory, the kinematics of QM reside in its probability structure. Here is a quote from Bub (https://arxiv.org/abs/1210.6371):

The quantum theory is a nonlocal box theory, i.e., it is a no-signaling theory with counter-intuitive probabilistic features like those of a PR-box. Hilbert space as a projective geometry (i.e., the subspace structure of Hilbert space) represents the structure of the space of possibilities and determines the kinematic part of quantum mechanics. This includes the association of Hermitian operators with observables, the Born probabilities, the von Neumann-Lu ̈ders conditionalization rule, and the unitarity constraint on the dynamics, which is related to the possibility structure via a theorem of Wigner [29],[27].

 

[Kolmo]

I would say it's both

I've never seen unitary evolution described as part of the kinematics honestly. Since it describes dynamical evolution I've always seen it put with the dynamics.

I'm just saying that your claim that probabilities are part of the "kinematical core" of QM is not correct

I don't see the difference with classical probability theory though.

The states themselves obey all the theorems probabilities do, for example a de Finetti theorem.

For state reduction:
In both cases we have a state $\omega$ and some algebra of random variables $\mathcal{M}$ for the system. If we measure some variable $N$ which forms its own subalgebra $\mathcal{N}$, $\mathcal{N} \subset \mathcal{M}$. Then a conditional expectation is a map:
$E_{\mathcal{N}}: \mathcal{M} \rightarrow \mathcal{N}$
with norm one whose restriction to $\mathcal{N}$ is the identity.

In classical probability this map gives the Bayesian update rule, in quantum theory it gives the projection postulate. What are we putting in by hand here and why is it wrong that the kinematical core, i.e. states and observables, are a probability theory?

 

[PeterDonis]

For state reduction

Which is dynamics, not kinematics.

In classical probability this map gives the Bayesian update rule

Which is dynamics, not kinematics.

If you are going to exclude unitary evolution from kinematics (which, as I said, I can see would make sense), then you have to also exclude any other kind of "evolution" of the state from kinematics.

 

[PeterDonis]

In quantum information theory, the kinematics of QM reside in its probability structure.

If one adopts an interpretation of QM in which the state is not ontic, then I can see how the probability structure can become primary as opposed to the states. I'm just not sure I would call that "kinematics".

 

[vanhees71]

Which is dynamics, not kinematics.Which is dynamics, not kinematics.

If you are going to exclude unitary evolution from kinematics (which, as I said, I can see would make sense), then you have to also exclude any other kind of "evolution" of the state from kinematics.

This is pretty much semantics. The description of "states and observables" in Hilbert space are "kinematics" (as in classical mechanics the Euclidean space and the 1D time, the trajectories of point particles etc.) and the equations defining unitary time evolution (e.g., in the wave-mechanical formulation the time-dependent Schrödinger equation) is "dynamics" (as in classical mechanics is Newton's 2nd Law).

 

[Kolmo]

If you are going to exclude unitary evolution from kinematics (which, as I said, I can see would make sense), then you have to also exclude any other kind of "evolution" of the state from kinematics

Unitary evolution involves postulating the actual Hamiltonian and thus the interaction terms and coupling constants and how systems interact with each other over time. Bayesian updating is just a common statistical procedure whose form is completely fixed by the kinematical structure.
The view of unitary evolution as kinematical must be a very uncommon one, are there worked examples where kinematics fixes the Hamiltonian in a non-trivial manner?

 

[vanhees71]

Indeed, as in Newtonian mechanics the forces, in QT the Hamiltonians are not fixed by the "kinematics" of the theory.

 

[PeterDonis]

This is pretty much semantics.

If it is (and I'm not saying it isn't), then so are the words "kinematics" and "dynamics". They add nothing to the actual physics; they're just labels that some people like to put on certain parts of the physics.

 

[PeterDonis]

Unitary evolution involves postulating the actual Hamiltonian and thus the interaction terms and coupling constants and how systems interact with each other over time.

Yes. But this is more than just the "form" of the procedure. See below.

Bayesian updating is just a common statistical procedure whose form is completely fixed by the kinematical structure.

The "form" of Bayesian updating may be fixed by the "kinematical structure", but so is the "form" of unitary evolution: that "form" is just $i \hbar \partial_t \psi = \hat{H} \psi$, which is the same no matter what $\hat{H}$ actually is. So if the "form" is what "kinematics" is, then the "form" of unitary evolution is just as much kinematics as the "form" of Bayesian updating.

Of course, to actually do a Bayesian update, you need to know more than just the "form"; you need to know the actual priors and the actual conditional probabilities for your particular problem. Just as you need to know the actual Hamiltonian in order to actually do unitary evolution. So I do not see any difference between the two cases.

The view of unitary evolution as kinematical must be a very uncommon one

As above, one can consider just the "form" of unitary evolution to be "kinematical", in exact analogy to your statement about Bayesian updating.

 

[PeterDonis]

as in Newtonian mechanics the forces, in QT the Hamiltonians are not fixed by the "kinematics" of the theory.

But the equations those things appear in are fixed. See my post #287 just now.

 

[Kolmo]

This is a really strange view to me and I've never really heard views like yours. The kinematic/dynamic division is a very common one, calling it just semantics doesn't match anything I've read, but I'll just leave it at that.

The original point was that quantum theory is a probability theory. That seems to be borne out by the fact that:

• It belongs to a braided monoidal category like all probability theories
• It's commutative version is literally Kolmogorov probability theory
• The quantum states obeys all the usual theorems of classical probability theory such a de Finetti theorem, form of conditioning, embedding of subalgebras and hundreds of others
• It can be derived from conditions placed on families of random variables.

I don't see what suggests that the mathematical structure of quantum theory is not a probability theory.

The "form" of Bayesian updating may be fixed by the "kinematical structure", but so is the "form" of unitary evolution

How would the kinematics tell you if the evolution was Unitary or CPTP?

 

[PeterDonis]

The original point was that quantum theory is a probability theory.

The full theory of QM is, yes, because, as I've already said a couple of times now, the full theory includes measurements, which is where probabilities and discontinuities in the state enter in.

I have no problem with the above. My only issue is with using the term "kinematics" to describe the above. To me, the way QM treats measurement is dynamics, not kinematics, and, as I said, it's a weird dynamics since it involves discontinuous jumps. The only QM interpretation in which there are no such jumps in the dynamics is the MWI, and the MWI is pure unitary evolution, all the time. And you have already insisted that unitary evolution is dynamics, not kinematics. It seems very strange to me to say that and yet to still say that the full theory of QM, which is unitary evolution plus all the weird stuff happening with measurements, is "kinematics".

How would the kinematics tell you if the evolution was Unitary or CPTP?

What is CPTP?

 

[Kolmo]

My only issue is with using the term "kinematics" to describe the above

The details of the dynamics involve physical postulates that move beyond just generalized probability theory, so usually one cannot say that part is just a probability theory.

And you have already insisted that unitary evolution is dynamics, not kinematics. It seems very strange to me to say that and yet to still say that the full theory of QM, which is unitary evolution plus all the weird stuff happening with measurements, is "kinematics"

I never said that. I said the main kinematical part, i.e. not including the details of dynamical evolution, is just a generalized probability theory mathematically. Since I was claiming there was a kinematic/dynamic distinction I certainly wasn't saying "all of QM is kinematics" or anything like it.

To me, the way QM treats measurement is dynamics, not kinematics, and, as I said, it's a weird dynamics since it involves discontinuous jumps

That's the way any stochastic theory works though. For example the Black-Scholes or any stochastic process has continuous evolution and then Bayesian updating "jumps". I don't see what is weird about it.

What is CPTP?

I don't see the point in discussing QM without knowing such basic terminology. It means Completely Positive Trace Preserving. It's the general form of time evolution in QM, unitary evolution being a special case.

 

[PeterDonis]

I don't see the point in discussing QM without knowing such basic terminology.

If it's "such basic terminology", then you should be able to point me to where in all of the standard QM textbooks this term appears. For example, where is it in Ballentine?

 

[PeterDonis]

I said the main kinematical part, i.e. not including the details of dynamical evolution, is just a generalized probability theory mathematically.

But the only place probabilities arise in QM is in the context of measurement, so it seems strange to me to ignore the fundamentally dynamic nature of measurement when talking about probabilities in QM.

That's the way any stochastic theory works though.

I know that. But other stochastic theories do not claim that the system being described is actually making discontinuous jumps; the discontinuities are only in our knowledge of the system, which discontinuously changes when we obtain new data and make a Bayesian update. The system itself is assumed to have an underlying dynamics which is continuous; we just aren't able to track it precisely.

In QM, however, at least under certain interpretations, the system itself is claimed to actually discontinuously change its state when a measurement happens. That is the "weird dynamics" I am talking about.

Possibly you are implicitly using an interpretation of QM where this issue does not arise, such as the statistical or ensemble interpretation.

 

[Kolmo]

But the only place probabilities arise in QM is in the context of measurement, so it seems strange to me to ignore the fundamentally dynamic nature of measurement when talking about probabilities in QM

Probabilities arise entirely at the kinematic level in QM. It's a fact that given the algebra of observables there are no non-dispersive states. It might seem weird to you, but you don't need the dynamics to develop the probabilistic side of the theory. It's directly induced from the kinematical side.

Take classical probability theory. There we have a triplet $(\Omega, \Sigma, \mu)$, the sample space, the sigma algebra and the probability measure. We also have random variables $X: \Omega \rightarrow \mathbb{R}$. This already has all the probability structure in it. We don't need to understand the detailed dynamics of how an individual $X$ is measured to make this claim.

 

[Kolmo]

Classical mechanics is supposed to be a physical theory, not simplectic geometry

I don't really see the opposition. The fact that the systems are described by symplectic geometry is just a say very terse encoding of certain physical insights. I would see the fact that a quantum system as a whole does not obey Kolmogorov's axioms in a similar light.
Similarly axiomatisations improve my understanding, but you see them as separate to proper conceptual understanding.

 

[PeterDonis]

It's a fact that given the algebra of observables there are no non-dispersive states.

Can you give a reference that develops this in more detail? If one has already been given earlier in the thread, just point me at the post.

Also, I'd be interested in your response to this from me:

Possibly you are implicitly using an interpretation of QM where this issue does not arise, such as the statistical or ensemble interpretation.

 

[Kolmo]

Can you give a reference that develops this in more detail?

It's a corollary of Gleason's theorem. Most textbook proofs of Gleason's theorem will mention it.

 

[PeterDonis]

It's a corollary of Gleason's theorem.

Ok. I have a todo item on my list to refresh my understanding of Gleason's Theorem anyway.

 

[Kolmo]

Ok. I have a todo item on my list to refresh my understanding of Gleason's Theorem anyway.

I'd definitely have a look at the POVM based proof first:
https://arxiv.org/abs/quant-ph/9909073

It's the extension to PVMs where the real difficulty arises and requires most of the heavy duty mathematics.

 

[Kolmo]

Possibly you are implicitly using an interpretation of QM where this issue does not arise, such as the statistical or ensemble interpretation

I would say I don't see why it matters since all of this stuff has a direct analogue in classical probability theory that doesn't prevent us from using the phrase "probability theory" there.

When we set up a Stochastic model in classical probability having the usual $(\Omega, \Sigma, \mu)$ and apply it to say a dice. We wouldn't usually consider the fact that somebody could consider $\mu$ to be a real physical wave (MWI analogue) or that accounts of how a given $X: \Omega \rightarrow \mathbb{R}$, such as "Is the dice result even?", is measured would require detailed dynamics to be reasons to not call the structure $(\Omega, \Sigma, \mu)$ a probability model.