Does consciousness change the rules of quantum mechanics?

[rational experiences]

If you state human laws. Legal precedence against human behaviour that once decided to do whatever it wanted.

It would involve measurable assertions in life.

Two human adult parents. Everyone's.
Baby girl baby boy.

Babies grew into human adults.

Position mind changes to cause a human to not think in unity.

Unit. One means one purpose.

The measure..

Conscious change to assert agreement with non unit a unit any unit anywhere of any type.

Assertion..family unified who still owned original behaviour.

The changed mind. As behaviour asserts to have what only a changed unit wanted. It was by taught defined force.

For a man baby it was men agreed.

Therefore it would qualify one original father was no longer present.

To assert that murdering and threat of murder and cruelty inflected would cause the larger population to yield. By behavioural differences only.

As humans survival Instinct was not to murder self nor cause harm to self human.

Hence behaviour human consciousness men today assert was a common factor of first scientific belief.

To create was an assertion to cause beyond self body a procreation of one...human species by two to form a baby bio life beyond self.

Men built a machine. How was the machine created? By destructive technique considered removing time by value a burning gas.

So first a mass isn't burning.

Therefore measure in science was falsified.. as practice science involves looking at dead and destroyed things to produce an assessment of what it isn't.

Not what it is.

Ever question why origin civilisation legal human rights said looking at dead things forbidden?

Because they were correct.

 

[Polymath257]

For measurable spaces, you need only a set and your choice of sigma-algebra. Measure theory is the application of measures—which are, and are of necessity, set functions, because measure theory assumes no additional operations or structures—to (sub)sets of measurable spaces. This generality is, again, by design so that measures spaces can be applied to highly abstract spaces having little structure as well as to incredibly rich spaces with level after level of interrelating structures. Finally, the measurable space tuple together with the measure constitute a measure space.

Now, do you agree that the key useful aspect of measure theory is that it allows us to do integration? And that many of the resulting function spaces (L2, L-infinity, Lp) are complete in their appropriate norms?

Do you agree that it is the fact that multiplication of measurable function is commutative that allows the derivation of Bell's inequality in probability spaces?

Do you agree that the measurable functions are analogous to the observables of QM?

Do you agree that the *actual* observables of QM are, in fact, not commuting and that the reason Bell's inequalities fail in QM is, ultimately, the fact that observables do not commute?

Do you agree that L-infinity of a positive measure is a C* algebra that is commutative?

Do you agree that the algebra of operators on a Hilbert space is not commutative?

Do you agree that the states (the positive linear functionals of norm 1) on the C* algebra of continuous complex valued functions correspond exactly the the Borel probability measures on the underlying space?

To you agree that the states on the C* algebra of bounded operators on a Hilbert space correspond exactly to the elements of the underlying Hilbert space of norm 1 up to multiplication by a unit complex number?

So, ultimately, measure theory: the study of integration of measurable functions on some measure space, is the *commutative* version of the study of states on the collection of C* algebras on a Hilbert space?

And, finally, that the reason Bell's inequalities hold in measure spaces is the commutativity of the collection of measurable functions and the violation in QM is due to the non-commutativity of the collection of operators?

Of course, as an expert you already know this. And as an expert you would be familiar with the ways Hilbert spaces can be defined by taking equivalence classes of square-integrable functions with respect to a particular measure, how spectral measures are used for operators on such spaces, how operators have their own integration measures and how e.g., positive operator measures can be used here as well as in quite different contexts.

Yes, I am. And I also know that the reason you get those projection valued operators is that the C* algebra generated by a single, normal operator is commutative. If, instead, you start with two non-commuting operators, you cannot get a simultaneous diagonalization (i.e, no projection valued measure that works for both operators).

More importantly, as an expert you are aware that measure spaces do not come equipped with any operations that are either commutative or non-commutative. You must equip such spaces with additional structure in order to have any commutative properties. It is thus absolutely ludicrous to claim that measure theory is somehow “essentially” commutative (or non-commutative), because to even refer to commutativity requires specifying the additional structure you are equipping your measure space with.

And, just like the definition of a group says nothing about symmetries, but the central aspect of group theory is still all about symmetries, the fact that measures are defined as set functions on some sigma algebra is only the first aspect: that they allow for integration is far more important. And, in this context, multiplication of the measurable functions is a commutative operation.

And, once more, just to be crystal clear, no more attempting to weasel your way out of this by switching from the claim about measure theory being commutative and this being how measure theory fails to your newer attempt to make it about noncommutativity (along with some pointless way of defining a commutative space, which is what? your "proof by it's possible to do something different"?).

So, as an expert, you can start to show that your ridiculous claim is somehow sensible by showing why measure theory is suddenly commutative, although this is just a matter of interest. You've already "described" (largely copied) how one gets a commutative space, but this is wholly irrelevant.

I did not copy. I recalled. And the way you get the collection of measures from a commutative C* algebra and the analogous aspects of non-commutative algebras giving states is exactly the central aspect of why things work in QM (with operators) and not in measure thoery (with measurable functions).

By this method of "proof", C*-algebras fail here because there exist commutative c*-algebras. What you claimed is that measure theory itself here was somehow essentially commutative and thus that's where the difference is, that's where Bell's inequality comes from (and then later, that's where the violations come from), and so forth.

If you follow the proof of Bell's inequalities in a measure space (using integration of measurable functions against a probability measure), and then try to do the analogous proof for operators, you will find that it is the non-commutativity of the operators that leads to the violation of the inequality, while it is the commutativity of the measurable functions that allows the proof to go through in the other context.

Do you disagree with this?

I deleted the rest of your rant. Unless you are willing to stop and read what I am saying and learn why I am saying it, there is little reason to continue. I have explained my meaning multiple times and you have ignored the explanations.

Please read what I wrote above and let me know if you understand it.

And, next time you claim that I am only copying from something, please back up your claim with a source you think I am copying from. Your claims for my dishonesty are completely inappropriate.

 

[Polymath257]

You can’t prove that measure theory is “essentially” commutative. It’s difficult even to make sense out of your claim in a manner that makes it clear how you could prove it. But you are the expert, so there must be some way you can prove that

1) QM contains algebraic structures which are essential to it and are used to do things like calculate probabilities via operators acting on quantum states in e.g., a Hilbert space. To be clear, what I’m really getting at by this condition is that you don’t attempt to sidestep this matter by constructing some non-measurable function or other and hoping I won’t notice this would make QM unusable.

Operators on a Hilbert space. Given an operator, A (which we assume to be normal so it has real valued spectrum, the average value of A when things are in the state x is <Ax,x>.

2) Any choice of measure/measures together with an associated measurable space (i.e., measure spaces) that one attempts to use to work with the noncommutative mathematics required in QM will fail, and this is due to the commutative nature of measure theory itself.

Yes, the operators do not commute, while the measurable functions do. The average value of a measurable function is obtained by integrating with respect to the probability measure representing the state of the system.

Of course, you’d be left with the problem of how, in such a pathological space, you could get anything like a physical theory. Then you’d have to explain how every mathematician and scientist in the world whose field concerns or who have ever dealt with quantum mechanics has failed to miss the fact that the operators defined on a measure-theoretic state space with a norm defined via the Lebesgue integral and measure and operators that function in this theory in a manner that crucially depends on spectral measures from measure theory have all somehow failed to see what you have:
That the mathematics underlying quantum theory necessarily fails because measure spaces have a hidden commutative structure you’ve discovered that nobody else knows about.

When you work with only one operator at a time (and a normal or even self-adjoint one at that), the C* algebra generated by that operator is commutative. That is *why* there is a spectral measure that you can use to evaluate <Ax,x>.

But, if you have two non-commuting operators, you do NOT get a single measure that will work for both at the same time.

Until I can see an actual proof, that actually deals with measure theory itself, then you simply aren't worth talking to. You are a disappointment (and really I am disappointed in myself for not realizing it earlier), your blatant misrepresentations reflect poorly on these forums, and I am not entirely sure I would like to be a member of any discussion board that has such charlatan as admin.

Hey, just because you don't really understand the proof of the spectral theorem in, say, Rudin's functional analysis book doesn't mean others fail to understand it.

I'll tell you what. Give me a problem in measure theory. make it one that cannot be easily googled. Let's see if I can show you I can solve it.

Go ahead, suggest any problem in any standard book on measure theory. I'd prefer Rudin, Hewitt&Stromberg, or Cohn, but I can make do with Loeve, for example.

Here's a fun one (although rather trivial) for you: Find a continuous measure on the unit interval (so measures of singletons are zero) that is NOT absolutely continuous with respect to Lebesgue measure (so there is no Radon-Nikodym derivative).

And, yes, it is a Lebesgue-Steiltjes measure, so you only need to describe the corresponding increasing function.

 

[LegionOnomaMoi]

Do you agree that it is the fact that multiplication of measurable function is commutative that allows the derivation of Bell's inequality in probability spaces?

No, it isn't. It is clearly not. This is absolutely basic. But you know NOTHING about measure theory and nothing about QM. You just regurgitate stuff you clearly don't understand. Hence, again, this:

Ultimately, the point is that measure theory is, essentially, commutative operator theory. Under the commutative assumption, you get inequalities that you do not get without it.

I gave you weeks. You're supposed to be an expert in measure theory. I guess that means you're just an expert in the measure theory you made up, this "commutative operator theory" version that nobody but you uses. Because those of us who actually use measure theory, *-algebras, and QM wouldn't make such a ludicrous claim.

Operators, like matrices, are a type of function (matrices are linear maps on vector spaces, operators are more general, but they're both types of functions). The reason that operators, like matrices, do not generally commute with respect to multiplication is because multiplication here is really the composition of functions. And,, regardless of measurability, functions do not commute with respect to the operation of composition. If you ever manage to take some calculus courses (you can find videos for free online), you'll learn about u-substitution and other ways we teach college kids to deal with "noncommutative" functions.

In all your regurgitation of material you lack even a rudimentary understanding of, you seem to have missed (unsurprisingly) the message of the very key points you've mangled here. Noncommutativity isn't something special to QM. It isn't somehow "quantum". It's a basic fact that applies to functions in general, all the way down to those from absolute basic, primary school algebra.

What makes noncommutativity different in QM is the interpretation of the role that the observables (and the operators that represent them) play, including (but not limited to) "compatible" vs. "incompatible" observers. So, for example, since operators play a role similar to random variables in how they are used to calculate expectations, we can try to see what a quantum probability calculus would look like by forcing the observables to both behave as they do in QM while also behave like random variables. We do this in a manner quite standard in noncommutative analysis: we start with a measure space and then define an L^p or (more generally) an L^∞ space over it.

But to actually compute probabilities in experiments or in theoretical work this isn't what we do. We use classical probability. And to the extent that one wishes to drastically oversimplify by stating that probabilities are encoded in the operators representing observables, then we use the spectral theory and some decomposition (not necessarily orthogonal, or even of the Hilbert space vs. the operator) or (better still) POVMs that are designed with the idea in mind of calculating expectation and probabilities more generally (e.g., they are normalized).

So, as an "expert" on measure theory (albeit one you made up), would you care to try again? Because you've shown quite clearly you have no idea what you are talking about, you haven't proved your statement (which you can't, as it is absurdly wrong). In the meantime, I'm going to be going over your previous posts and claims to make sure your complete ignorance doesn't continue to mislead others because you care more about looking knowledgeable than educating yourself.

 

[LegionOnomaMoi]

But, if you have two non-commuting operators, you do NOT get a single measure that will work for both at the same time.

You do, actually. This is fairly basic. Keep in mind that non-commuting operators are just the composition of maps. While you don't know much about the mathematics compared to what you regurgitate, you must have taken algebra or something.

But for the benefit of others, whom I don't want misled by your fundamental ignorance:

The issue of noncommutativity here has to do with the construction of quantum theory in terms of observables that are deemed to be compatible or incompatible. In the "standard" formulation of the measurement process, where measurements are always destructive (i.e., projective-value measures), it is supposed to be impossible to simultaneously measure two incompatible observables to the degree of precision one has in the case of compatible observables. Whether this is due to an ontological indeterminacy is still debated. The EPR paper shows that QM predicts in certain cases that supposedly indeterminate quantities can be known exactly using a single measurement (see below).

These days, we know much, much more about how to perform careful, precision measurements, much more about the physics of quantum systems, and have a much more fully developed formalism to treat quantum measurement schemes systematically and mathematically. So the old notion of incompatibility is no longer adequate in general. If one uses the more general formalism of density operators to represent the states of quantum systems and POVMs to model measurements, than unsharp joint measurements can be made on incompatible observables. One sacrifices instead the knowledge of the state of the system after measurement (which one would, in many cases, not have in the textbook case as e.g., photons are typically destroyed after measurement and it is not generally true that even in the special case of an eigenstate a measurement of a system yielding the associated eigenvalue means the system is or was actually in the eigenstate as in the textbook case).

Also for the benefit of others, here's how measure theory actually works in quantum mechanics (Hint: Commutativity is not only not required, but frequently wholly irrelevant).

In the standard Hilbert state space formalism, one associates the state of a quantum system with an element of a Hilbert space.

Textbook QM typically focuses on the Schrödinger picture and wavefunctions. Here, quantum states are represented by waves in Hilbert space. What Hilbert space? Well, for a single particle the associated state space in wave mechanics is called L^2(R). This is both notation as well as shorthand (omitting parts of the notation that would otherwise be required) for a space consisting of square-integrable functions.

We haven’t yet come to operators and measure theory is absolutely essential already. First, because in order for a space to actually be a Hilbert space it must not only be a normed vector space, but also complete with respect to the norm induced by the inner product. That’s a double requirement. One can consider other norms, but what distinguishes a Hilbert space from a Banach space is exactly the norm that one gets from the inner product. And for L^2(R), the inner product is given by integration with respect to the Lebesgue measure (generally, with the Lebesgue or Lebesgue-Stieltjes integral). The notational bit suppressed above is the measure µ: L^2(R, µ). So measure theory gives us our norm and our inner product.

How about completeness? Here again we need measure theory, and here again commutativity is irrelevant. We need measure theory because even if we considered Riemann-type integration as non-measure-theoretic, Riemann integrals are inadequate here. The space contains non-Riemann integrable functions that are Lebesgue integrable.

Then there’s the actual nature of quantum states in these spaces. One typically hears/sees them defined in terms of rays (or, worse, “vectors” or something similar) in Hilbert space, and for non-normalized states this is almost true. In fact, though, we quickly run into a problem of non-uniqueness. That is, a given state may have multiple “realizations” in the Hilbert space. Of course, if the states are equivalent in the appropriate sense, then this doesn’t matter physically. And since the appropriate sense is given mathematically through measure theory, it doesn’t matter theoretically either. We generally ignore the distinction between [ᴪ], the equivalence class of functions that differ on sets of measure 0 (or “agree” almost everywhere, denoted a.e. in the literature as this is a term from measure theory), and a specific function ᴪ in the space. But we can only do this thanks again to measure theory.


Note that so far there isn’t anything about operator algebras at all, still less an issue with commutativity or noncommutativity.

 

[LegionOnomaMoi]

(continued from above)

How about observables? This is, after all, where commutativity comes into play. In the standard formulation of QM, observables are what give us a connection between theory and experimental outcomes. They are akin to “random variables” and encode information about the expectation values associated with particular systems. So if it makes any sense to think of measure theory itself as somehow commutative (it doesn’t), we would definitely see evidence here.

We don’t, of course. It’s certainly true that noncommutativity plays a central role in the formulation of QM and our understanding of its theoretical structure. It’s even played a foundational role in developing the textbook “orthodox” interpretation of QM. To better understand this, we can bring in Einstein, EPR, and two fundamental observables: position and momentum.

In classical mechanics, specification of these two “observables” yields basically all that you need. This is one reason why Einstein so desperately wanted to show that there must, at least in principle, be a way to show that these quantities were well-defined for quantum systems even if they couldn’t be experimentally determined to any arbitrary degree of precision. He first did this in direct debates with Bohr in the 20s using thought experiments. His arguments were generally regarded as failing. EPR was a continuation of these same arguments in the sense that Einstein was still trying to show that a system’s position and momentum must be well-defined.

His approach, however, was far subtler. He used quantum theory itself to show that these properties must exist. The idea is the heart of what came to be called (initially by Schrödinger) entanglement. He showed that one could use the formalism of QM to describe an experiment in which the measurement of one subsystem would yield a definite value a noncommuting observable in the same “system” (different “subsystem”). To oversimplify enormously, consider two “particles” A and B described by a single wavefunction (in an entangled state, such that e.g., the state space is not given by the product of the Hilbert spaces of A and B, or better yet that the wavefunction isn’t a composite system describing two seperable systems A and B but rather is nonseparable). EPR sets up a situation in which a single wavefunction describes a situation in which the (linear) momentum of A is exactly anti-correlated with B. This means that knowing the momentum of A is q tells you that the momentum of B is -q. Likewise, the position (along e.g., the x-axis) is such that again the measurement of position p for one observable means that the other observable has position -p.

So, argued EPR (and again, I’m horribly oversimplifying), a measurement on A’s position would tell us what B’s position was, and likewise a measurement on B’s momentum would tell us exactly what A’s momentum was. In short, quantum mechanics predicts that these observables have no determinate values prior to measurement, yet also tells us precisely the determinate values they have without any measurement whatsoever. More importantly, these observables are not supposed to have any definite “reality” because they are represented by noncommuting operators. This was (and still often is) interpreted to mean that the more accurately one knows the value given by one observable the more “indefinite” any noncommuting observable becomes.

Back to measure theory. How does it come into play here (and in general for observables), and what role does commutativity play? The answer to the first question is that, in the standard formalism one extracts predictions from observables using the spectral theory. In finite dimensions, where one represents observables with matrices, the spectrum of the operator corresponds to the set of eigenvalues for a chosen basis and a given state (represented by a vector the matrix “acts on”). Measure theory gives us a way to treat the spectrum of an operator as a probability measure, and also to perform the functional analysis equivalent of diagonalizing a matrix (spectral decomposition). More fundamentally, it is measure theory that is behind the “collapse of the wavefunction” for observables such as position and momentum. This collapse corresponds mathematically to a projection-valued measure.

Now, how about commutativity? Well, this is irrelevant. Spectral measures/PVMs, and even POVMs for a given operator representing some observable are found by dealing with that operator, not with some algebra of observables (still less the commutativity or noncommutativity of the elements of the algebra). It is again akin to what one does with elementary linear algebra in finite-dimensional cases. If I have a state represented by a vector, and I want to measure some property represented by a matrix, I don’t need to think about other observables’ matrix representations and whether these commute with the one of interest to me in order to find the eigenvalues for that one, i.e., the one representing the property I intend to measure. In fact, I can find the spectrum for any and all of them regardless of whether they commute or not.

So, independently of how ridiculous it is to assert that measure theory somehow is or isn’t commutative, and even ignoring that we need measure theory before we can even talk about the operators because its required to define the domain, we’re still left with the fact that noncommutativity of observables has absolutely no relevance here. It can be studied, sure. Various formulations of QM (quantum probability, algebraic QM, etc.) can be and have been formulated with one goal being the clear elucidation in a mathematically rigorous way the nature (and compatibility) of observables for a given system using operator algebras (or defining quantum random variables in terms of these). But in these cases, the more central role played by noncommutativity and its analytic treatment is by design, not forced on us by experiment. And the fact that, experimentally, none of these treatments can do better than standard formulations (indeed, rarely have they proved viable for use in experiments at all), gives us no reason to suppose that somehow noncommutativity is more central that e.g., nonseparability or better yet noncontextuality.

However, even if noncommutativity alone were enough to explain all there is that makes QM “quantum” vs. what makes classical physics “classical”, that still wouldn’t alter what measure theory actually is. And it is not, at all, in any way, in any context, somehow “essentially” commutative anything. We need measure theory to extract predictions from observables, but since we do this for a given observable using projection-valued measures or operator-valued measures, commutativity or noncommutativity don’t matter.

Finally, a little look behind the scenes at the ludicrous claim that “ultimately, measure theory is, essentially” somehow something to do with commutative operators.

Operators can be thought of as generalizations of matrices. Matrices have the advantage (given by the finite-dimensionality) of allowing one to write them down. But all row and column entries for a matrix A are just basis specific representations of a more general linear map. And linear maps are functions that take elements of one vector space to another, not necessarily distinct, vector space. Why, then, is matrix multiplication not commutative in general while (pointwise) multiplication of functions is?

Because they are very different operations. Matrix multiplication is actually the composition of linear maps, and therefore of functions. Likewise with operator algebras. Both commute with respect to the operation addition and not (in general) with multiplication because (again) it is actually a composition of maps.

But even in basic, elementary calculus one learns that composition of functions in noncommutative. So, for example, the cosine of the square of x is not generally equal to the square of the cosine of x. Things get more complicated when one seeks to integrate. In elementary calculus, measure theory is swept under the rug because it isn’t needed (rather, it's implicit and the technicalities are too advanced for such students). But when it is needed for more general spaces, more general functions, and in general more complicated and abstract mathematics we find the same noncommutativity: composition isn’t generally commutative.

Yet measure theory was built up specifically to better handle the partitioning of spaces and the consistent assignment of measures as one requires when integrating. And one still has the kinds of tools one learns in first year calculus to integrate composite functions. Yet these functions, like matrices and operators, don’t commute in general.

Luckily, again, measure theory isn’t commutative. It can’t be, as there are no operations in measure spaces with respect to which the elements of the space could commute. And if it relied on the kinds of commutativity that do not hold with matrices and operators, it would be useless as this is the same kind of noncommutativity we find for matrices.

 

[LegionOnomaMoi]

Hey, just because you don't really understand the proof of the spectral theorem in, say, Rudin's functional analysis book doesn't mean others fail to understand it.

Excellent! Great point at which you can start your proof. Because those of us who actually use this mathematics and need to understand it and apply it here would use spectral theory to obtain PVMs from an operator. You're the only "genius" who somehow thinks that we need an algebra of observables usually reserved for algebraic QFT (oh, by the way, now that we all know you're bluffing about your mathematical "knowledge", you really should be more careful when doing so as you claims about unboundedness not being an issue kind of stop working when you are asserting nonsense about c*-algebras, as elements of these are bounded by definition; hence, to use a c*-algebra means to exclude most operator representations in QM, and indeed abandon the algebra generated by the canonical commutation relations).

But by all means, as an "expert" in measure theory, show how the spectral theorem (or spectral theory more generally), which is one central use of measure theory in QM, shows us that

measure theory is, essentially, commutative operator theory.

Because if it were true that measure theory is, essentially or in any other sense actually "commutative operator theory", then you'd have to show how we can only really use spectral measures on operator algebras and only then on elements that commute with respect to the "multiplication" operation of the algebra.

Keep digging this hole. It's making me feel better about ever taking anything you said seriously (no wonder you kept using Riesz representation theorem incorrectly, asserted that the Dirac notation actual mathematicians despise because it hides the distinctions mathematicians require and which many physicists like to forget or ignore, and have in this thread asserted continually confused the algebraic approach to quantum theory with the more standard Hilbert space version while asserting that c*-algebras which are bounded by definition somehow work when one is dealing with operators that are generally unbounded).

In the meantime, I'll be going back over the thread to correct your errors on quantum theory specifically so that others need not be misled by someone pretending to have knowledge they utterly lack.

But by all means, please prove me wrong. Show me how I've misunderstood Rudin's non-existent proof that

measure theory is, essentially, commutative operator theory.

And show how we can't spectral decomposition or PVMs on an operator, but must instead consider the algebra itself and thus it is somehow true (in some sense), in QM at least, that "measure theory is, essentially, commutative operator theory".

I won't hold my breath or anything, but as you are supposed to be an expert in measure theory and you've so fundamentally misunderstood the basics you could have gotten from the introductory paragraphs of an elementary introduction to measure theory you've conflated an entire branch of mathematics with "commutative operator theory". Worse still, rather than admit a mistake, the closest you've gotten to clarifying is to claim that you meant the statement to be understood only in the context of quantum theory. Of course, it's still absolute nonsense, as I've shown over and over again, and you've implicitly acknowledge my more regurgitated statements about function spaces and Lebesgue integrals you should have looked up before you made such a ridiculous claim.

 

[LegionOnomaMoi]

Do you agree that L-infinity of a positive measure is a C* algebra that is commutative?

A positive measure is a real number greater than equal to 0. You're asking if I agree that a function space L^∞ "of a positive measure" (i.e., a non-negative number) is a C*-algebra?

Do you agree that the algebra of operators on a Hilbert space is not commutative?

There is no "the algebra of operators on a Hilbert space". There are infinitely many possible algebras. The reals are a Hilbert space and if you had the knowledge and were so inclined you could take a few nice functions like trig functions on appropriate domains and construct a C*-algebra out of them and a field (just use the reals) over the real numbers. In addition to showing you that your "commutative measure theory" would fail when it comes to handling trivial cases, this would show you that there is no "the algebra of operators on a Hilbert space".

Do you agree that the states (the positive linear functionals of norm 1) on the C* algebra of continuous complex valued functions correspond exactly the the Borel probability measures on the underlying space?

No. For several reasons, firstly because you're again confusing algebraic quantum theory (in which the states are secondary and are defined not using Hilbert space but an abstract algebra), secondly because you're regurgitating a simple example you've misunderstood (the word "state" in the literature sometimes corresponds to the state of a system in algebraic quantum theory or other applications in physics but also, both in the mathematics and physics literature, is a technical term that is defined as the "positive linear functionals" and the algebra need not be unital)., thirdly because the entire point of the algebraic approach is to rid the theoretical structure of any underlying (Hilbert or otherwise) space, fourthly because there are infinitely many Borel probability measures for most spaces even if there were an underlying space, and fifthly because you think measure theory is commutative operator theory.

Which means you know nothing about measure theory.

To you agree that the states on the C* algebra of bounded operators on a Hilbert space correspond exactly to the elements of the underlying Hilbert space of norm 1 up to multiplication by a unit complex number?

No. Yes, there's an isomorphism and you can use the GNS construction to show how there is a correspondence, but
1) There is no such thing as a C*-algebra of unbounded operators so why say "bounded operators"? You know those symbols || || or | | you must come across in the sources you keep regurgitating? They're called norms. They're required by definition for c*-algebras. And they mean the operators are bounded by definition. So, for example, the operators corresponding to momentum and position in QM can't be in a C*-algebra, nor can the algebra of the CCRs be a C*-algebra (but these algebras of operators defined over Hilbert spaces."

So, ultimately, measure theory: the study of integration of measurable functions on some measure space, is the *commutative* version of the study of states on the collection of C* algebras on a Hilbert space?

Even if we were simply dealing with measure-theoretic "integration" and "integrals" such as in e.g., probability theory or quantum theory or any number of fields, we'd have to include e.g., determinants, sums, discrete distributions, and other things one generally doesn't call integrals. More importantly, if measure theory is limited to commutative operator theory, then we'd have no way of defining commutation.
This: [AB-BA]= (something)
i.e., the commutator, requires that we can calculate both AB and BA regardless of whether or not they commute. In fact, we can only show they don't commute if we can calculate the composition of both. Which means they're measurable. If they're weren't measurable then you couldn't evaluate the product you need to show that they don't commute.

By your logic, we can't integrate elementary functions, because composition of functions and maps is generally noncommutative and therefore outside the realm of measure theory (well, for you anyway).

And, finally, that the reason Bell's inequalities hold in measure spaces is the commutativity of the collection of measurable functions and the violation in QM is due to the non-commutativity of the collection of operators?

Unravelling most of this nonsense, the answer is still NO. In algebraic QFT, one can show (using actual C*-algebras and not your drivel) that, again, the vacuum state maximally violates Bell inequalities.
More generally, since it is the entangled state that is crucial, the only reason that needs to deal with the observables for two systems in Bell tests is because Bell tests are not tests of QM and these days are so removed from that approach that much of the work here goes under the name "device independent physics", indicating that not only are the results independent of quantum theory (or any theory) but any particular implementation via a particular device.
Also, you can't violate Bell inequalities using a collection of operators, regardless of commutativity or noncommutativity. This is, again, basic stuff. Google LOCC, PR- (Popescu-Rohrlich) box, nonlocal games, and maybe read up a little on Bell's theorem and basic probability too.

Every set can be made into a measure space (indeed, a probability space). You've confused the basics again. It isn't that noncommutativity is a problem for classical probability. It isn't even that, in quantum probability (which, again, isn't probability used in quantum mechanics but a seperate formulation) one treats the observable operators as random variables and forces them not to commute, it's that states are nonseperable. Taking many of the most common formulations of a quantum system violating Bell inequalities, we prepare a single state that cannot be factored which we call a "singlet" state. Treating it as such is no problem, except that such states have multiple "parts" that correspond in measurements to e.g., two different particles.
It's the states, not the observables.

Yes, I am. And I also know that the reason you get those projection valued operators is that the C* algebra generated by a single, normal operator is commutative.

Wrong. Because apart from anything else you can't generate C*-algebras from operators in QM like this. It's not how it works, especially because C*-algebras are bounded by definition, and operators in QM are generally unbounded. Also, the PVMs are generally obtained from decomposing the Hilbert space, making the operators somewhat secondary and unnecessary. They're convenient, that's all (maybe even elegant). Of course, if you weren't endlessly conflating and confusing the algebraic approach with the Hilbert space approach, you might start regurgitating stuff that made more sense.

If, instead, you start with two non-commuting operators, you cannot get a simultaneous diagonalization (i.e, no projection valued measure that works for both operators).

Luckily for those of us using measure theory in QM, this is about as meaningful as saying that we can't use PVMs on birds or use orthogonal decomposition of Dilbert and Dogbert spaces with spectral theory to get the necessary PVMs.
The operators can commute or not commute. It's irrelevant.
If you know a little matrix algebra, you can see this clearly in the finite-dimensional case (which, after all, is all one needs to perform Bell tests). I have a state and some operators representing observables. I can perform the calculations and operations needed to diagonalize each matrix representation and calculate the expectation values using the trace. Commutativity is irrelevant.

 

[Polymath257]

Operators, like matrices, are a type of function (matrices are linear maps on vector spaces, operators are more general, but they're both types of functions). The reason that operators, like matrices, do not generally commute with respect to multiplication is because multiplication here is really the composition of functions. And,, regardless of measurability, functions do not commute with respect to the operation of composition. If you ever manage to take some calculus courses (you can find videos for free online), you'll learn about u-substitution and other ways we teach college kids to deal with "noncommutative" functions.

Yes, the multiplication for operators is composition and that is why it is non-commutative. But the operation of multiplication for random variables is ordinary multiplication, which *is* commutative.

In all your regurgitation of material you lack even a rudimentary understanding of, you seem to have missed (unsurprisingly) the message of the very key points you've mangled here. Noncommutativity isn't something special to QM. It isn't somehow "quantum". It's a basic fact that applies to functions in general, all the way down to those from absolute basic, primary school algebra.

The point is that the relevant C* algebra (that of bounded operators on a Hilbert space) is non-commutaitve while the relevant C* algebra for measure theory (Linfty of some measure) is commutative.

What makes noncommutativity different in QM is the interpretation of the role that the observables (and the operators that represent them) play, including (but not limited to) "compatible" vs. "incompatible" observers. So, for example, since operators play a role similar to random variables in how they are used to calculate expectations, we can try to see what a quantum probability calculus would look like by forcing the observables to both behave as they do in QM while also behave like random variables. We do this in a manner quite standard in noncommutative analysis: we start with a measure space and then define an L^p or (more generally) an L^∞ space over it.

Um, Linfty is not more general than Lp. The essentially bounded functions don't even need the integral to define them. the Lp spaces do.

So, as an "expert" on measure theory (albeit one you made up), would you care to try again? Because you've shown quite clearly you have no idea what you are talking about, you haven't proved your statement (which you can't, as it is absurdly wrong). In the meantime, I'm going to be going over your previous posts and claims to make sure your complete ignorance doesn't continue to mislead others because you care more about looking knowledgeable than educating yourself.

Once again, that you missed my point doesn't mean that the point is invalid.

 

[Polymath257]

A positive measure is a real number greater than equal to 0. You're asking if I agree that a function space L^∞ "of a positive measure" (i.e., a non-negative number) is a C*-algebra?

I really can't believe you just wrote that. Did someone hack your account?

You have already talked about having a measure space. In other words, a set with a sigma algebra and a *positive valued measure* on that sigma algebra. It is standard to just say that you have a positive measure (the measure is what assigns a non-negative real number to each measurable set).

So, yes, we are talking about Linfty of some measure space (in other words, Linfty(mu) where mu is a positive measure).

Do you agree that Linfty is a commutative C* algebra of (equivalence classes of )functions?

Just as an aside, there are measures that are not positive valued. It is also standard to have real or complex valued measures. These turn out to be linear combinations of positive measures, but they have many interesting properties including important representation theorems.

There is no "the algebra of operators on a Hilbert space". There are infinitely many possible algebras. The reals are a Hilbert space and if you had the knowledge and were so inclined you could take a few nice functions like trig functions on appropriate domains and construct a C*-algebra out of them and a field (just use the reals) over the real numbers. In addition to showing you that your "commutative measure theory" would fail when it comes to handling trivial cases, this would show you that there is no "the algebra of operators on a Hilbert space".

Once again, I find it incredible you actually wrote this. it is standard terminology to say 'the algebra of operators on a Hilbert space) when what is meant is the collection of all bounded operators, B(H), on that Hilbert space.

Your comments show that you don't know much about C* algebras. For one they are all Banach algebras (hence, Banach spaces, with complete norms) with involutions. But not every Banach algebra with involution is a C* algebra. In addition, the C* algebra property (that the norm of xx* is the square of the norm of x for all x in the algebra) is required. And this is a very restrictive property. Oh, and in almost every case, the underlying vector space is over the complex numbers (not just the real numbers).

In particular, if you 'took a few trig functions' and attempted to create a C* algebra out of them, you would end up with all continuous functions on your domain.

As an aside, the collection of all complex measures on R has an interesting Banach algebra property where the multiplication is convolution of measures. This is a commutative algebra with involution but is NOT a C* algebra. There are still many research level questions about this algebra. For example, the Shilov boundary of the structure space is unknown.

No. For several reasons, firstly because you're again confusing algebraic quantum theory (in which the states are secondary and are defined not using Hilbert space but an abstract algebra),

Actually, I am using C* algebras, which are Banach algebras with involution and some nice properties. This is enough to define positive linear operators and thereby states.

secondly because you're regurgitating a simple example you've misunderstood (the word "state" in the literature sometimes corresponds to the state of a system in algebraic quantum theory or other applications in physics but also, both in the mathematics and physics literature, is a technical term that is defined as the "positive linear functionals" and the algebra need not be unital).

And, in the specific example, of B(H), the bounded operators on a Hilbert space, how can such positive linear operators be constructed? By taking a vector v in the Hilbert space and looking at the map A--><Av,v> from B(H) to C. For each vector v, this will be a positive linear operator. Furthermore, if you multiply v by a complex number of absolute value 1, you will get the same functional. In other words, the *ray* is what gets identified with the state, just as is required in the other formulation.

, thirdly because the entire point of the algebraic approach is to rid the theoretical structure of any underlying (Hilbert or otherwise) space,

And yet, B(H) is a central example.

fourthly because there are infinitely many Borel probability measures for most spaces

First, your underlying set need to have a topology to even discuss Borel sets (the elements of the smallest sigma algebra containing the open sets) and thereby Borel measures.

But yes, for any infinite set where singletons are measurable, there will be infinitely many probability measures (the point masses will be examples, but also any convex combination).

Some care is required because there are compact spaces where the only Borel measures are linear combinations of point masses (infinite sums allowed). To get beyond the 'discrete measures' requires some additional assumptions on the topology. To get a 'continuous measure' (where the measure of all single tons is 0) is non-trivial in general. of course, Lebesgue measure on [0,1] is a typical example of a continuous measure.

even if there were an underlying space, and fifthly because you think measure theory is commutative operator theory.

Which means you know nothing about measure theory.

Nope, nothing at all. You seem to only know about Lebesgue and Lebesgue-Steiltjes measures on R, but I am the one that knows nothing about measure theory. You seem to think that every measure is a Borel measure, but don't seem to understand the topology necessary for that concept.

No. Yes, there's an isomorphism and you can use the GNS construction to show how there is a correspondence, but
1) There is no such thing as a C*-algebra of unbounded operators so why say "bounded operators"? You know those symbols || || or | | you must come across in the sources you keep regurgitating? They're called norms. They're required by definition for c*-algebras. And they mean the operators are bounded by definition. So, for example, the operators corresponding to momentum and position in QM can't be in a C*-algebra, nor can the algebra of the CCRs be a C*-algebra (but these algebras of operators defined over Hilbert spaces."

Finally a reasonable criticism! But, of course, the usual way to deal with unbounded operators is to consider densely defined operators that have closed graphs. Since most linear differential operators are of this sort, this isn't a huge restriction. Also, by doing a Cauchy transformation, it is possible to make a bounded operator out of such and do the relevant spectral theory. Rudin does this.

Even if we were simply dealing with measure-theoretic "integration" and "integrals" such as in e.g., probability theory or quantum theory or any number of fields, we'd have to include e.g., determinants, sums, discrete distributions, and other things one generally doesn't call integrals.

What are you talking about? You won't get a determinant in an infinite dimensional setting.

I'm guessing you are using the term 'discrete distribution' for the distribution function of a discrete measure. The underlying measure is then just a countable sum of point masses. And yes, integration against these measures would be called integrals. They are also infinite sums.

But do you realize that not all measures (even on R) are either discrete or absolutely continuous with respect to Lebesgue measure? Is it possible the distribution function is continuous but has derivative 0 almost everywhere with respect to Lebesgue measure. An example is given by the Cantor staircase function (and associated Lebesgue-Steiltjes measure/integral).

More importantly, if measure theory is limited to commutative operator theory, then we'd have no way of defining commutation.
This: [AB-BA]= (something)
i.e., the commutator, requires that we can calculate both AB and BA regardless of whether or not they commute. In fact, we can only show they don't commute if we can calculate the composition of both. Which imeans they're measurable. If they're weren't measurable then you couldn't evaluate the product you need to show that they don't commute.

Huh? All I am saying is that the C* algebra of Linfty functions is commutative (pointwise multiplication) while the C* algebra of bounded operators on A Hilbert space, B(H) s not commutative. The first is the 'operator theoretic' version of measure theory.

The operators can commute or not commute. It's irrelevant.
If you know a little matrix algebra, you can see this clearly in the finite-dimensional case (which, after all, is all one needs to perform Bell tests). I have a state and some operators representing observables. I can perform the calculations and operations needed to diagonalize each matrix representation and calculate the expectation values using the trace. Commutativity is irrelevant.

And you better be using non-commuting observables. If the observables commute, you can do a simultaneous diagonalization and the violations do not occur.

 

[Polymath257]

Every set can be made into a measure space (indeed, a probability space). You've confused the basics again. It isn't that noncommutativity is a problem for classical probability. It isn't even that, in quantum probability (which, again, isn't probability used in quantum mechanics but a seperate formulation) one treats the observable operators as random variables and forces them not to commute, it's that states are nonseperable. Taking many of the most common formulations of a quantum system violating Bell inequalities, we prepare a single state that cannot be factored which we call a "singlet" state. Treating it as such is no problem, except that such states have multiple "parts" that correspond in measurements to e.g., two different particles.
It's the states, not the observables.

That depends on which representation you use. If you use a representation where the states are fixed and the operators change over time, then it is the observables are relevant.

And you can always use that 'nonseparable state' as one of your basis vectors, which would mean the individual 'part's are non-separable. Whether a state can be 'factored' depends on what basis is being used for the tensor product.

 

[PureX]

Does consciousness change the rules of quantum mechanics?

"Rules" are phenomenological aspects of conscious awareness. No consciousness, ... no "rules".

Does this change physical behavior on the quantum level?

"Behavior" is pattern recognition and the "rules" codify the patterns that we recognize. Both are phenomenological aspects of conscious awareness. So, ... no conscious awareness; no pattern recognition ("behavior") and no behavioral "rules".

Whatever is left is of no cognitive consequence.

 

[Polymath257]

How about observables? This is, after all, where commutativity comes into play. In the standard formulation of QM, observables are what give us a connection between theory and experimental outcomes. They are akin to “random variables” and encode information about the expectation values associated with particular systems. So if it makes any sense to think of measure theory itself as somehow commutative (it doesn’t), we would definitely see evidence here.

Yes. The operators do not commute. The random variables of measure theory do. That is the crucial distinction.

So, argued EPR (and again, I’m horribly oversimplifying), a measurement on A’s position would tell us what B’s position was, and likewise a measurement on B’s momentum would tell us exactly what A’s momentum was. In short, quantum mechanics predicts that these observables have no determinate values prior to measurement, yet also tells us precisely the determinate values they have without any measurement whatsoever. More importantly, these observables are not supposed to have any definite “reality” because they are represented by noncommuting operators. This was (and still often is) interpreted to mean that the more accurately one knows the value given by one observable the more “indefinite” any noncommuting observable becomes.

But it is NOT 'without any measurement whatsoever'. The measurement was done for the values for A's position and B's momentum.

Back to measure theory. How does it come into play here (and in general for observables), and what role does commutativity play? The answer to the first question is that, in the standard formalism one extracts predictions from observables using the spectral theory. In finite dimensions, where one represents observables with matrices, the spectrum of the operator corresponds to the set of eigenvalues for a chosen basis and a given state (represented by a vector the matrix “acts on”).

Good so far.

Measure theory gives us a way to treat the spectrum of an operator as a probability measure,

No. The spectrum is still a subset of the plane (or the real line if the operator is self-adjoint). The projection valued measure corresponds to the eigenspaces and the integral of the identity function corresponds to recovering the operator as a sum of the projections multiplied by the eigenvalues.

and also to perform the functional analysis equivalent of diagonalizing a matrix (spectral decomposition).

Yes. And in order to be able to do this, the operator itself must be normal (commute with its adjoint). If this fails, then the diagonalization isn't even possible in the finite dimensional setting.

More fundamentally, it is measure theory that is behind the “collapse of the wavefunction” for observables such as position and momentum. This collapse corresponds mathematically to a projection-valued measure.

No, the collapse corresponds to the projection itself. The projection valued measure corresponds to the possible (eigen)subspaces corresponding to different values of the observable.

Now, how about commutativity? Well, this is irrelevant. Spectral measures/PVMs, and even POVMs for a given operator representing some observable are found by dealing with that operator, not with some algebra of observables (still less the commutativity or noncommutativity of the elements of the algebra).

Wrong. Even in the finite dimensional situation, you need that the operator commute with its adjoint in order to be diagonalizable. In the infinite dimensional setting, the invariant subspace problem is still open for separable Hilbert spaces. Such invariant subspeaces would correspond, roughly, to eigenspaces.

It is again akin to what one does with elementary linear algebra in finite-dimensional cases. If I have a state represented by a vector, and I want to measure some property represented by a matrix, I don’t need to think about other observables’ matrix representations and whether these commute with the one of interest to me in order to find the eigenvalues for that one, i.e., the one representing the property I intend to measure. In fact, I can find the spectrum for any and all of them regardless of whether they commute or not.

Yes, but the fact that you cannot find simultaneous diagonalizations is due to the non-commutativity. And the violations of Bell's inequalities is directly tied to this.

So, independently of how ridiculous it is to assert that measure theory somehow is or isn’t commutative, and even ignoring that we need measure theory before we can even talk about the operators because its required to define the domain, we’re still left with the fact that noncommutativity of observables has absolutely no relevance here.

And that is nonsense. it is, for example, directly tied to the uncertainty principle. it is directly tied to the fact that there cannot be simultaneous diagonalization. It is directly tied to the fact that violations of Bells' inequalities are possible (yes, even in the finite dimensional setting).

However, even if noncommutativity alone were enough to explain all there is that makes QM “quantum” vs. what makes classical physics “classical”, that still wouldn’t alter what measure theory actually is. And it is not, at all, in any way, in any context, somehow “essentially” commutative anything. We need measure theory to extract predictions from observables, but since we do this for a given observable using projection-valued measures or operator-valued measures, commutativity or noncommutativity don’t matter.

I didn't say that non-commutativity *alone* is all that is going on. There is also the little matter that we are talking operators and functional analysis.

Measurable functions (random variables) commute. Operators do not.

Bell's inequalities are specific to measurable functions because those functions commute. They do not apply in QM because the relevant analogs of random variables (the operators) to not commute.

Finally, a little look behind the scenes at the ludicrous claim that “ultimately, measure theory is, essentially” somehow something to do with commutative operators.

Operators can be thought of as generalizations of matrices. Matrices have the advantage (given by the finite-dimensionality) of allowing one to write them down. But all row and column entries for a matrix A are just basis specific representations of a more general linear map. And linear maps are functions that take elements of one vector space to another, not necessarily distinct, vector space. Why, then, is matrix multiplication not commutative in general while (pointwise) multiplication of functions is?

Because they are very different operations. Matrix multiplication is actually the composition of linear maps, and therefore of functions. Likewise with operator algebras. Both commute with respect to the operation addition and not (in general) with multiplication because (again) it is actually a composition of maps.

But even in basic, elementary calculus one learns that composition of functions in noncommutative. So, for example, the cosine of the square of x is not generally equal to the square of the cosine of x. Things get more complicated when one seeks to integrate. In elementary calculus, measure theory is swept under the rug because it isn’t needed (rather, it's implicit and the technicalities are too advanced for such students). But when it is needed for more general spaces, more general functions, and in general more complicated and abstract mathematics we find the same noncommutativity: composition isn’t generally commutative.

Yes, and it is crucial that the appropriate multiplication of operators is composition and the relevant multiplication of random variables is, well, multiplication.

Yet measure theory was built up specifically to better handle the partitioning of spaces and the consistent assignment of measures as one requires when integrating. And one still has the kinds of tools one learns in first year calculus to integrate composite functions. Yet these functions, like matrices and operators, don’t commute in general.

Composition isn't the appropriate multiplication of measurable functions (in fact the composition of Lebesgue measurable functions need not even be measurable--nice exercise for you).

Luckily, again, measure theory isn’t commutative. It can’t be, as there are no operations in measure spaces with respect to which the elements of the space could commute.

The relevant objects in this case are the measurable functions (random variables) and yes, those do commute with their appropriate multiplication.

And if it relied on the kinds of commutativity that do not hold with matrices and operators, it would be useless as this is the same kind of noncommutativity we find for matrices.

You are, once again, completely missing the point.

For specificity, and to make it easier on you, take the measure space ([0,1], M, lambda) where lambda is Lebesgue measure on the interval [0,1] and M is the sigma algebra of Lebesgue measurable sets. Construct the Hilbert space of square integrable functions, H=L2([0,1],lambda).

I know you can understand at least this much because you have already written about this much.

Now, take an L-infty ([0,1]) function g. We can create an operator on the Hilbert space H via multiplication: T_g (f)=fg. This will be a bounded operator which is even a normal operator, so it has a spectral decomposition.

And, in fact, the spectrum of the operator T_g is precisely the essential range of g. The projection valued measure is also easy to describe: if E is a Borel subset of [0,1], then P(E) is simply multiplication by the characteristic function of E.

Notice that the same projection valued measure works for every possible L-infty function g. The operators T_g are simultaneously 'diagonalized' by this projection valued measure.

Related to this is the fact that if g and h are two L-infty functions, then T_g and T_h are commuting operators.

Now a crucial fact:

If you limit yourself to the operators/observables of the form T_g, then Bell's inequalities will always hold. the commutativity of the operators is what makes the inequalities hold.

In order to get violations of Bell's inequalities for operators, you need to use operators that do not commute.

 

[Polymath257]

Keep digging this hole. It's making me feel better about ever taking anything you said seriously (no wonder you kept using Riesz representation theorem incorrectly,

Where did I use RRT incorrectly? Do you realize that there are several different results all called the Reisz Representation Theorem?

One has to do with bounded linear operators on a Hilbert space and identifies them as all inner products with elements of that Hilbert space. In essence, this shows that a Hilbert space and its dual are conjugate isomorphic.

The other results called RRT have to do with measure theory. The basic result is that a positive linear functional on the space of continuous functions (on a locally compact hausdorff space) with compact support is given as integration with respect to some positive regular Borel measure on the underlying space.

The deeper results give a similar representation of the collection of bounded linear functions on the collection of continuous functions that vanish at infinity as integration with respect to either a signed measure (for real valued functions) or a complex measure (for complex valued functions) and in such a way that the norm of the functional is the same as the full measure of the total variation of the complex measure.

Of course, to discuss those requires first understanding a bit about signed and complex measures as opposed to just the positive measures that you seem to be familiar with.

There are generalizations of this result to either positive or bounded functionals on the space of all bounded real valued functions on a Tychonoff space, but the measures need to be Baire measures and not simply Borel. The distinction is important in this cases where the underlying space isn't a complete metric space (like R is).

I'd also point out that there are many subtleties involved when doing product measures. For something easy like the Lebesgue measure, most things work out, although the distinction between Borel and Lebesgue measurability is important (and neglected by Rudin).

Things get even more subtle if products of infinitely many factors are investigated. Here, there is more than one natural sigma algebra and the RRT result is necessary to even get the 'diagonal' to be a measurable set.

asserted that the Dirac notation actual mathematicians despise because it hides the distinctions mathematicians require and which many physicists like to forget or ignore, and have in this thread asserted continually confused the algebraic approach to quantum theory with the more standard Hilbert space version while asserting that c*-algebras which are bounded by definition somehow work when one is dealing with operators that are generally unbounded).

Well, there are several different contexts where a Dirac 'measure' is relevant. In measure theory, it is simply the point mass at some point. The measure of a set is 1 or 0 depending on whether the set has or does not have the point defining the point mass.

A more subtle context is in distribution theory where, instead of all continuous functions, the test functions are either those that are infinitely differentiable with compact support OR those that are infinite differentiable and going to zero quickly at infinity . This good thing about these contexts is that the derivative of the point mass can also be defined. This can also allow the use of harmonic analysis techniques for solving differential equations.

And yes, physicists like to ignore the subtleties of distribution theory. It is made a bit more subtle because the relevant test functions no longer even form a Banach space. They do form a topological vector space, though.

 

[wellwisher]

We live in space-time, where space and time are connected and dependent on each other. Photons of light, for example, display both wavelength and frequency which are connected in a way that multiplies to the speed light. This is the same in all inertial references.

The quantum world appears to behaves more like space and time are not connected. In this case, we have separated distance and separated time interacting with space-time, to alters the expected results within space-time; uncertainty. Acceleration due to force is d/t/t which is two parts time and one part distance; space-time plus time. Quantum coupling moves forward in time, in a way that is not dependent on space; synchronized in time by extra time potential beyond just space-time.

If we could move in distance without the constraint of time, we could be everywhere in the universe at the same time; omnipresent. If we could move in time, independent of distance, we would know what was happening everywhere, at any point in time; omniscience. The gravitational force, which acts to infinity, and integrates the universe, sort follows these rules. The speed of light is the same in all references could be induced with time potential superimposed on space-time.

Two classic attributes of God are omniscience and omnipresence. This appears to imply that God is more connected to the quantum world, where and space and time are not connected; different realm.

If space and time were not connected, we would be in a state of infinite entropy; complexity, since there would be no space-time limits as to what combinations would be possible. This state of infinite entropy, would set a potential with space-time, which has finite entropy; the basis for the second law where entropy has to increase in space-time. The quantum world is sort of in the middle; bridge between, with the universe working its way back toward infinite entropy.

Consciousness has a connection to entropy and entropic potential. The brain expends 90% of its metabolic energy pumping and exchanging sodium and potassium cations. This energy intensive process lowers the entropy of cations, against the direction of the 2nd law; segregates two materials that would prefer form a uniform solution in water. This sets an enhanced entropic potential at the neuron membrane. The flow of entropic potential, throughout the brain is moving in the direction of increasing the entropy, which is part of the consciousness effect; short term and long term memory meet.

If you look at the frontal lobe of the human brain. it is connected to the consciousness process called the imagination. The imagination, in terms of information, is not bound by space-time. The material of the brain is bound by space-time, but not the flow of information via the entropic potential.

For example, I can imagine flying to the sun in one second and then burrowing to the solar core to get a nice nuke tan. This is not for possible for material objects in space-time; human body. It breaks a bunch of physical laws of space-time. However, it is possible in terms of our imagination as it processes information in imaginary ways detached from space-time. The frontal lobe and imagination process uses a quantum affect where space and time are not connected. This is proof of concept and connects consciousness to the quantum world where space and time can be separate.

Innovation can organize matter in ways that would not naturally form; iPhone. The matter of the universe is limited to space-time and its laws, unless we add time and/or distance potential via the imagination. Then we can shape materials in unnatural ways, to get unnatural results that can enhance. This brings us back to the topic but not in the way most people expected.

 

[shunyadragon]

"If you think you understand quantum mechanics, you don't understand quantum mechanics"
Richard P Feynman

This philosophical quip does not make sense in modern science of Quantum Mechanics, which is understood to the point of applications in technology and industry.

Quantum entanglement has been described generated artificially, which has lead to techology and industrial applications,

To try to include in matters of consciousness muddles the issue.