Einstein and "spooky actions"

[PolyHedral]

It is something we are familiar with that is why we call it a wave.

That's not how the wave in question works.

 

[idav]

That's not how the wave in question works.

Maybe not but familiar. Thermal light responds appropriately.

 

[LegionOnomaMoi]

Maybe not but familiar. Thermal light responds appropriately.

Why do we still refer to Newton's law of gravity even though it's not a law? Not only that, we know it's wrong? Why do we use equations like F=ma when we know they're wrong too? Mostly convenience. It is convenient to use Newton's gravitational constant because the approximation we end up with is usually good enough and far easier than trying to obtain almost the exact same result using spacetime curvature. It's also much easier to use when teaching physics than having to talk about 4D Minkowski space to a high school physics class.

Likewise, it is more convenient to refer to light as a wave when we are talking about visible light and/or the electromagnetic spectrum, just as it is more convenient to refer to particles when we are talking about photons (which, by definition, are localized).

Newton's law remains wrong even though we use it, and light is neither a wave or a particle even though we refer to light waves and to photons (discrete "particles" of light).

 

[idav]

Why do we still refer to Newton's law of gravity even though it's not a law? Not only that, we know it's wrong? Why do we use equations like F=ma when we know they're wrong too? Mostly convenience. It is convenient to use Newton's gravitational constant because the approximation we end up with is usually good enough and far easier than trying to obtain almost the exact same result using spacetime curvature. It's also much easier to use when teaching physics than having to talk about 4D Minkowski space to a high school physics class.

Likewise, it is more convenient to refer to light as a wave when we are talking about visible light and/or the electromagnetic spectrum, just as it is more convenient to refer to particles when we are talking about photons (which, by definition, are localized).

Newton's law remains wrong even though we use it, and light is neither a wave or a particle even though we refer to light waves and to photons (discrete "particles" of light).

Well what a photon is supposed to really be goes beyond the point. A photon is a packet and it doesn't just suddenly spread out from its packet form but whatever its emitting has a range. At what distance do the slits stop catching the wave function?

 

[LegionOnomaMoi]

A photon is a packet and it doesn't just suddenly spread out from its packet form but whatever its emitting has a range. At what distance do the slits stop catching the wave function?

A packet here means a quantized "portion" of light or a unit of light. How big is this unit? After all, if we're saying it doesn't spread out, we need to know how large it is. if I spread out a tablecloth, I don't say it is nonlocal or something just because it is spread over some area. However, I don't expect that each time I use this tablecloth the size changes. It has only one value for the amount of area it covers. Does a photon? No.


So as you can't tell me, like you could with a tablecloth, how much area on a detection screen a wave packet of light (a photon) has. Which means that when you point to a dots on some detection screen and say "see? Photons aren't spread out, because they cover this specific amount of area" I can show you this isn't true because I can change how localized (how much area is covered when each photon hits the screen) the photons are.

In other words, if you assert that photons aren't spread out, then (just like a tablecloth on a table or a paintball shot at a target) a photon has a certain area it will occupy when we detect it. You can't show this. So how can you show me a wave packet/photon "doesn't just suddenly spread out" when you can't tell me what the spread of the packet is/photon is to begin with?

 

[idav]

A packet here means a quantized "portion" of light or a unit of light. How big is this unit? After all, if we're saying it doesn't spread out, we need to know how large it is. if I spread out a tablecloth, I don't say it is nonlocal or something just because it is spread over some area. However, I don't expect that each time I use this tablecloth the size changes. It has only one value for the amount of area it covers. Does a photon? No.


So as you can't tell me, like you could with a tablecloth, how much area on a detection screen a wave packet of light (a photon) has. Which means that when you point to a dots on some detection screen and say "see? Photons aren't spread out, because they cover this specific amount of area" I can show you this isn't true because I can change how localized (how much area is covered when each photon hits the screen) the photons are.

In other words, if you assert that photons aren't spread out, then (just like a tablecloth on a table or a paintball shot at a target) a photon has a certain area it will occupy when we detect it. You can't show this. So how can you show me a wave packet/photon "doesn't just suddenly spread out" when you can't tell me what the spread of the packet is/photon is to begin with?

This article gives a good description where they are shaping photons but it still exists in a single point in time and space, the shape is the probability, the wave function. Site has a cool picture too it looks like a wave.
Physics - Shaping Single Photons

 

[LegionOnomaMoi]

This article gives a good description where they are shaping photons but it still exists in a single point in time and space, the shape is the probability, the wave function. Site has a cool picture too it looks like a wave.
Physics - Shaping Single Photons

That is a journal published by APS (American Physical Society) intended specifically to summarize and simplify experiments published in full in another APS journal. In this case, the article was talking about the study "Electro-Optic Modulation of Single Photons", published in Phy. Rev. Lett. 101 (if you look at the top right-hand corner of the page you linked to, you'll see the study name and the citation information.

This is the first line from the actual study itself: "This Letter demonstrates how single photons may be modulated so as to produce photon wave functions whose amplitude and phase are functions of time."

The study your article is reporting/talking about absolutely did not show anything about photons being single points. The opposite is true. They are quite clear about what experiment they did and why: "The technique provides the technology for studying the response of atoms to shaped single-photon waveforms on a time scale comparable to the natural linewidth."

A single-photon waveform means that the photon was spread out like a wave. It wasn't localized at a point.

 

[PolyHedral]

...I was just about to post asking Legion if he'd seen a response I made, but apparently I made it in the universe next door. Oops.

Actually, I just missed it. Legion, have you read post #646?

 

[idav]

That is a journal published by APS (American Physical Society) intended specifically to summarize and simplify experiments published in full in another APS journal. In this case, the article was talking about the study "Electro-Optic Modulation of Single Photons", published in Phy. Rev. Lett. 101 (if you look at the top right-hand corner of the page you linked to, you'll see the study name and the citation information.

This is the first line from the actual study itself: "This Letter demonstrates how single photons may be modulated so as to produce photon wave functions whose amplitude and phase are functions of time."

The study your article is reporting/talking about absolutely did not show anything about photons being single points. The opposite is true. They are quite clear about what experiment they did and why: "The technique provides the technology for studying the response of atoms to shaped single-photon waveforms on a time scale comparable to the natural linewidth."

A single-photon waveform means that the photon was spread out like a wave. It wasn't localized at a point.

That's great when you were saying I couldn't show it was spread out. I showed it.

It is also a specific point as the article noted about the photon being in a particular space and time. Thing is they shoot a packet that remains a packet despite having to go through slits to land on the screen. When it is one slit no interference, with two slits the photon does the same thing except interference is coming from somewhere. The photon doesn't just suddenly spread out cause we add a slit, it is still a single photon effeted by its own wave amplitude.

 

[LegionOnomaMoi]

Actually, I just missed it. Legion, have you read post #646?

I have. I started to respond a few times, but gave up, because some extremely basic aspects of linear algebra, function spaces, fields, etc., seemed to be unfamiliar to you (perhaps you don't use them regularly, and like everyone else knowledge tends to fade). You stated that you couldn't imagine functions being defined differently. That is so basic high school students are taught some of the differences. And I outlined in detail problems with your posts here.

When you asked about a vector with 3 entries spanning a space it cannot even exist on, the basic knowledge of linear algebra you had seems to have decreased over time, and as this is far more basic than fields, it seemed unproductive to get into that at all. However, as I still have my drafts, I'll attempt a quick amalgamation for you.

For instance: I say that a vector space, V<F> over some field F is not itself a field. You respond with this section. This is very informative, but it does not actually answer my point, AFAIK - it does not tell me if V is a field or not.

What is a field? A field is a set of rules that creates a mathematical structure on some arbitrary set. Some sets, like the natural numbers, do not have the properties necessary to create this structure. However, a subset of the natural numbers does. Simply construct a set that consists of all multiples of 2 (including 2 * 1). Now you have both the necessary additive properties and inverse properties.

In other words, a field isn't one specific thing. By that I mean I can refer to other structures using the same set, or build fields out of subsets of sets that cannot be fields, etc.

When you ask, "Is V a field?" then your question is basically meaningless. A field takes an arbitrary set and imposes a mathematical structure via particular conditions. The most fundamental of these is an additive and multiplicative operation. A vector space can be a field because scalar multiplication and vector addition satisfy the necessary operations over the set of all vectors in that space. Things like the multiplicative inverse become easy. All we need is an arbitrary scalar that maps a vector to the coordinate set of an arbitrary point multiplied by -1. And vector addition satisfies the other condition for a field. The more minor conditions are defined by a vector space itself.

There's your field.

 

[PolyHedral]

So why are your responses not matching what I've said?

 

[LegionOnomaMoi]

So why are your responses not matching what I've said?

Because you are wrong.

I don't see how you can possibly define addition, subtraction, multiplication and division differently for real/complex-valued functions in any useful way.

Such operators on trig functions are different even though they are real-valued. Basic single-variable calculus requires specific rules for multiplication and division of functions. What you can't "possibly see" is high school mathematics or beginning college mathematics.

But that is not an approximation at all, unlike if we were working with an arbitary function. In the case of an n-degree polynomial, the equality is exact, and I can prove that to you if I have to. The above equation is (the definition of!) a polynomial, not an approximation or converging series.

So prove it.

However,

Wrong.

Any set of vectors of any dimension which contains precisely 2 linearly independent members will span R^2

Completely wrong. In fact, it is so off target that you cannot use linear algebra, complex analysis, or anything that refers to dimensions and/or vectors if you think that statement is true.

You are smart guy. Probably a lot smarter than I am. But whether you had bad teachers or you are just out of practice, you have fundamentally misrepresented the basics.

 

[idav]

Quantum Principles
Kinda what I've been trying to say.

Making sense out of the double slit and other mysterious things:
Most physicists say that it went through both slits as a wave, and its particle nature is not there until it is detected. All you need to make sense out of it is the tunneling-like magic. You do not need the additional magic of the particle nature being absent as the wave passes through the double slit. You can say that a particle going toward a double slit may go through one or the other, but its wave goes through both. The particle can jump around anywhere in its wave, and the square of the wave is the probability of finding it in any location. Is it doing all this jumping around as it travels through space? It doesn't hurt to picture it that way, but all we can say from the evidence is that the probability of its location where it will be detected is governed by its wave. (And of course its wave nature immediately changes upon detection- the "collapse of the wave function." The wave is nature's expression of the randomness, and the randomness of its location necessarily disappears when it is detected.)

 

[PolyHedral]

Oh dear, I think I get to the source-quotign thing now.

What is a field?

Intuitively, a field is a set F that is a commutative group with respect to two compatible operations, addition and multiplication, with "compatible" being formalized by distributivity, and the caveat that the additive identity (0) has no multiplicative inverse (one cannot divide by 0).

However, a subset of the natural numbers does. Simply construct a set that consists of all multiples of 2 (including 2 * 1). Now you have both the necessary additive properties and inverse properties.

So which multiple of two is equal to the multiplicative inverse?

When you ask, "Is V a field?" then your question is basically meaningless.

It asks if the set V has specific properties and functions.

A vector space can be a field because scalar multiplication and vector addition satisfy the necessary operations over the set of all vectors in that space.

Closure of F under addition and multiplication
For all a, b in F, both a + b and a · b are in F

Scalars are not in the set V, so scalar multiplication cannot be the operator ·

Things like the multiplicative inverse become easy. All we need is an arbitrary scalar that maps a vector to the coordinate set of an arbitrary point multiplied by -1

That gives you the additive, not multiplicative inverse.

Such operators on trig functions are different even though they are real-valued. Basic single-variable calculus requires specific rules for multiplication and division of functions. What you can't "possibly see" is high school mathematics or beginning college mathematics.

The addition of functions I had in mind was,

I don't know about you, but that looks the same whether or not f or g is a trig function. It's also the only sensible way I know of to define addition of functions. Would you disagree?

So prove it.

I'll get back to you on that. It's late.

Wrong.

They are not identical.
(If they were, the difference would be the additive identity.)

Completely wrong. In fact, it is so off target that you cannot use linear algebra, complex analysis, or anything that refers to dimensions and/or vectors if you think that statement is true.

Why? Two linearly indepedent n-dimensional vectors will span a 2D subspace of nD space, will be homeomorphic to R^2 if the original space was homeomorphic to R^n.

You are smart guy. Probably a lot smarter than I am.

But whether you had bad teachers or you are just out of practice, you have fundamentally misrepresented the basics.

You've misunderstood.

 

[DallasApple]

So Einstein said "spooky action" one time?Then someone on the internet said it?

Hmmmm....I'm not quite sure what to think about that.

 

[DallasApple]

xkcd: Duty Calls

 

[LegionOnomaMoi]

Let's get the easy stuff out of the way:

They are not identical.
(If they were, the difference would be the additive identity.)

Additive identity. Right.
"There are three basic kinds of operations that can be used to solve any system of linear equations. These operations are simple transformations of the original system of equations that preserve the information content about the solution.





The elementary operations are:

• Interchange two of the equations in the system.
• Multiply one of the equations through by a number other than zero.
• Add a multiple of one equation to another.

Clearly, these elementary transformations preserve the information content of the original system of equations for the simple reason of being reversible. In other words, after applying a sequence of these operations to the original system we'll end up with another system that may look very different from the initial one, but by applying the reversed sequence in reverse we can go back to the beginning. Notice that we undo the first kind of elementary transformation (switching two rows) by doing it again. That is, the reverse of switching is itself!" (source)

Note that contrary to whatever you think you were claiming, the transformations resulting from row operations preserve the original matrix. Otherwise, why the hell would we use them to solve systems? Matrices aren't filled with scalars but coefficients. How do you add or subtract the coefficients without any variables? You don't.

Of course, a matrix may not be a system of equations but a linear transformation Ax=b. But then you wouldn't be swapping rows, would you?

Scalars are not in the set V

There is no "the set V", although thinking of vector spaces as sets has some merit (there are important distinctions, however, that may be missed by equating a set with a space). More importantly, for the hundredth time, VECTOR SPACES are DEFINED by taking the linear combinations of vectors. This means that every single vector space other than the empty space is defined by some collection of vectors whose linear combinations span that space. Linear combinations include both vector addition and scalar multiplication. Ergo, YOU CANNOT HAVE a vector space V without scalars! I need only two vectors two span R2. The vector space, however, is not these two vectors. It is their linear combination, and thus the elements that make up the space include infinitely many scalars.

You will never understand the concept of vector spaces unless you have a firm grasp of what determines the dimensionality of a space and a vector because you will keep talking about vectors existing in spaces they can't.

Intuitively, a field is a set F

Intuitively" is not formally. Why might this be important? Because were it true, standard formulations of Boolean algebras or any algebraic structure which used set theoretic operators: union and intersection for addition and multiplication, respectively. Also, you'd lose the extensive use of group, ring, and field theories and notations that are used with abstract automata where the operations are defined on e.g., letters:

Automata, simply by virtue of being discrete and defined (usually) in terms of basic operations quite frequently have an algebraic structure. This includes, but is not limited to, fields. The formal definition doesn't require "addition" and "multiplication" but simply two binary operators defined on the set such that the Field axioms (if one is talking about a field, vs. a field over some set) are satisfied by the triple {F, +, +}. It does not require that these binary operations be defined as addition and multiplication, simply as operations that will hold true for the every a is an element of F.

So which multiple of two is equal to the multiplicative inverse?

The way you write this it sounds as if there is one multiplicative inverse for the entire set. I'm not sure if that's what you mean, but just in case every element in the set would have to have a unique multiplicative inverse. However, it doesn't have to be 1 and you don't even need numbers. I can't recall what my point as for using the example of multiples of 2 and I know I meant integers not natural numbers (you actually can construct a field rather easily out of the integers using prime numbers to any n collection you wish), but the point was to get at this:

You've used three different phrasings to describe this so far:

• "You do realize that a vector space is, by definition, a field over some set of vectors, right?"
• "a linear vector space over the field K"
• an assignment of a vector F(P) to each point P.

When I say "a field over some set" or any reference to the field K I'm using standard notation from k-theory (in algebraic topology) applied to field extension. Field extension is to field theory what analysis is to calculus. Field theory is the basic level foundations of the binary operations and unary operations of an arbitrary field F extended to create algebraic structures in various sets or spaces. Boolean algebras, if you recall, are fields. They are created by taking an arbitrary set S, defining union, intersections, and the unary operation complement, and the power set algebra on S (actually, there are more ways one can create fields than this, but as this method is pretty much the basis for computer science I figured it's the best example).

Anyway, you don't need multiples of 2 but you do need to have n greater than or equal to 2. The trick then is to take n finite integers creating a finite field by using the value of n. For both multiplication and addition, for any two integers the operations are defined as they usually are, but with an additional step: the output is the remainder of the multiplied or added integers after they are divided by n. If memory serves, this will always get you a commutative ring but not all n's will have an inverse. However, it's easy enough to check. You just use an operation table.

It asks if the set V has specific properties and functions.

What V? And do you mean a field over a vector space (in which case, once again, the answer is yes) or are you asking if a vector space is a field? That second question relates to this:

That gives you the additive, not multiplicative inverse.

You add vectors, not scalars. And no, it doesn't.

The addition of functions I had in mind was

It's equally meaningless, if that's what you mean. The above says that the addition of two functions that maps nothing to a variable x which maps nothing from nowhere to an unspecified set.

It's also the only sensible way I know of to define addition of functions.

This is how the second arrow is used. It defines a general function not in terms of values but in terms of spaces or sets. In other words, the above notation (as is) refers to all functions that takes elements from the set R2 and maps them to R. You can't add functions using that kind of notation because there are infinitely many such functions. The other arrow is for mappings between defined sets.

Why?

Think of the Cartesian plane as R2. What defines a vector here? an increment, or directed amount of change, that is associated with a point R x R. Every vector in R2 is associated to a point (x, y). If you have a vector with coordinate entries x, y, & x, then where in R2 can they go? Nowhere. Because they don't exist in R2. They exist in R3, where three coordinates don't just make sense, but are required.

 

[Mr Spinkles]

Hi Legion,

Thanks for the lively discussion. As you know, I respect your views and admire your passion for physics. But if you'll permit me, instead of addressing all of your points, let me briefly explain where I'm coming from here. I think our approach could be modified in a way that serves both of us better.

You may have noticed that I prefer to establish small, simple truths first, truths everyone can agree on, while tolerating (for the moment) perhaps some loss of rigor and neglecting (for the moment) everything nonessential. As I see it, we can let the bigger, more complex questions come later, possibly modifying/qualifying the smaller things we already established. This might seem frustrating to someone like you, who prefers to tackle everything at once, tolerating no loss of completeness or rigor. But I believe that in this case, the perfect is the enemy of the good. It seems you hastily reject statements that are not perfect, and make mistakes comparing many different quotes from many different contexts. Thus by attempting to reach 100% perfection all at once, it becomes impossible for us to achieve any incremental improvement, because we are constantly getting ahead of ourselves, and talking past each other.

With that in mind, the reason I wanted to focus on one small, manageable question is because you have put an awful lot on the table, in response to (what I felt) was a very uncontroversial and modest claim on my part. You've said things which I agree with too, of course, e.g. QM is conceptually very different from CM, and there is a less satisfying (or you would suggest, perhaps no satisfying) connection between the mathematical representation and the physical reality.

You also brought up density operators, limits of integration, is a measurement represented by an operator or not, have I described the most common formulation or not. I think you are absolutely mistaken about all of these points. I believe I answered your objections about density operators (to my satisfaction although perhaps not to yours), and your objection that the ket describes merely our knowledge, and your objection that what I was saying was not the "orthodox" or most-common formulation of QM. I spent a good deal of time reading quotations you kindly provided from numerous sources. But at the end of the day, those sources didn't contradict anything I had said. So I am sort of left scratching my head, it seems like you are throwing objections out to see what will stick.

Each time I answer one of your objections, it seems to me, you jump to a new objection without skipping a beat, without acknowledging that the previous objection has been addressed. The result is a never-ending battery of objections leading us far astray from the simple, modest proposal I originally put forward: Both [QM and CM] say in principle that any given physical system has a one-to-one correspondence with its mathematical representation. In classical mechanics the mathematical representation is a list of all the positions and momenta of all the particles, i.e. a point in a 6N-dimensional phase space; in QM it's a vector in Hilbert space.

That's a pretty unassailable statement. Whether it's ultimately right or wrong, satisfying or unsatisfying, it is true that QM does postulate that. May I remind you that I am a physicist? And, in terms of grades, scholarships, etc., I am evidently not an incompetent one? I don't say that out of hubris; I say it because it would be false modesty to pretend otherwise. I know, from studies and experience, that the above would be an uncontroversial thing to say in front of a room full of physicists. I know it because I've been in a building full of physicists every working day for many years now. And yet, to my great surprise, you objected to it! I really believed we must be talking past each other, so I clarified what I was saying with this question:

Is it wrong to say that QM postulates the existence of a ket which represents the complete state of a physical system, not just our knowledge of it, and this is the standard, most common way of formulating QM?

More recently in post #653 you said something I was also surprised by:

If it is a physical system, then complete knowledge of the state entails the ability (in principle) to completely predict the systems evolution. We frequently say complex systems are indeterministic, but we mean they are deterministic in principle, but in practice we are unable to determine their dynamics.

If we can't, even in principle, describe the evolution of a physical system, then either our knowledge of the system's state is incomplete, QM is incomplete, or (the most widely accepted view) the state of the system doesn't correspond to a physical system at all, merely a mathematical device to tell us given preparation a and measurement b, we'll get outcome x with y probability.

Emphases added. The first sentence in bold is clearly a non-sequitur. Otherwise you've simply defined out of existence a physically real, indeterminate system! That begs the question. I posited by assumption the existence of a physical system which does not obey deterministic laws of evolution. When we entertain this possibility, rather than rejecting it by assumption, there is a fourth option in your second paragraph: the physical system does not have a predestined evolution, and hence, naturally, our complete mathematical representation also does not have one. Or, as Griffiths put it fancifully somewhere, it's not that we don't know--God doesn't know. (BTW you switched from "predict" in the 1st para. to "describe" in the 2nd, I'm going to assume we can stick with "predict", since that is what actually captures the difference between a deterministic vs. non-deterministic evolution.)

Now, I am capable answering your other objections about expectation values and ensembles, and whether measurements are represented by operators or not .... and trust me, you are mistaken on these points .... but instead, we agreed to focus on what ought to be an even simpler, more self-evident question: Does Griffiths agree that QM postulates the existence of a ket which represents the complete state of a physical system, not just our knowledge of it, and this is the standard, most common way of formulating QM?

And yet, somehow it still seems like you are throwing a lot of hasty comments out and we are still talking past each other. For example, you now say:

Griffiths [is] quite clear that the operator in no way corresponds to a mathematical representation [of] physical measurement ...

First: if you don't think measurements are represented by operators acting on kets, you are simply wrong. It's stated unambiguously in Shankar for example, a standard graduate-level text. Perhaps it's a bit less obvious in undergrad-level texts (like Griffiths). Second: in Eq. [3.18] Griffiths says "Observables are represented by hermitian operators". You quoted this yourself. You can also read the generalized statistical interpretation Griffiths explains in Sect. 3.4. Clearly, for starters, operators correspond to measurement in some way, not in no way as you claimed.

In any event you did finally quote Griffiths describing the statistical interpretation, and said:

If you want to interpret a physical system as being nothing more than this: "All I am asserting is that there must be something- if only a list of possible outcomes of every possible experiment- associated with the system prior to measurement", then Griffiths posits we have complete knowledge of a physical system. It's not physical in any sense that we are familiar with nor with any known correspondence with anything physical, and the only thing we have are probability functions to describe the physical states of this physical system. However, if you think that what Griffiths states above (the wave function is a probability function), means that the physical system is a mathematical function that yields outcomes, then yes we do have complete knowledge of a "physical system".

What I'm hearing is that you concede Griffiths does indeed agree with the statement in question. (In the part you quoted, plain English demands it.) But Legion has objections to that interpretation, namely it's not physical, etc. And Griffiths acknowledges a number of things (namely the statistical interpretation) which you feel support your objection. Is that fair?

If so, that's all fine and well. Perhaps we can discuss your objections about it seeming physical/unphysical next. It's widely appreciated, and I happily acknowledge, that while the ket as a complete representation of an electron (say) is not wrong, as far as it goes, it is also not satisfying to everyone. That's a distinction which I have often found useful in physics.

 

[PolyHedral]

The way you write this it sounds as if there is one multiplicative inverse for the entire set.

Oops, I meant the multiplicative identity.:cover:

 

[LegionOnomaMoi]

Each time I answer one of your objections

I can understand how it seems that way, but every time you address something I write, even about Griffiths' view, you tell me what he doesn't say but what you interpret him as saying. You refer to complete knowledge of a physical system, and never does. It seems to me that you have conflated 2 things:

1) That QM postulates a mathematical representation of the system which contains everything there is to know about that system.
2) The QM postulates a mathematical representation of outcomes of measurements of a physical system which is described in terms of a mathematical "state" with no known correspondence between this state and the physical system but full knowledge of measurement outcomes.


"The wave function provides only a formal description and does not by itself make contact with the properties of the system. Using only the wave function and its evolution, we cannot make predictions about the typical systems in which we are interested, such as the electrons in an atom, the conduction electrons of a metal, and photons of light. The connection of the wave function to any physical properties is made through the rules of measurement. Because the physical properties in quantum theory are defined through measurement, or observation, they are referred to as ‘observables’. The quantum formalism represents the observables by Hermitian operators on the system Hilbert space." (Bell's Theorem and Quantum Realism; SpringBriefs in physics; 2012).

You have provided me only with your interpretations of what you have not read, or what you have read but the author doesn't ever say.

Yet the only framework for your interpretation you've stated is this:

This was the common understanding I was taught by 3 different professors in 3 different QM courses, using 2 different textbooks. If I made this statement at one of the physics seminars I don't think anyone would find it controversial ... but maybe I'm wrong ...

and this:

I'm just reiterating what Shankar and Griffiths say in their QM books.

The problem is the number of times I've read something like "The textbooks and most articles devoted to this subject focus on the initial and final states of the system S and the apparatus A, without describing in detail the coupled dynamics of S and A during the measurement process" or how the correspondence I'm talking about is said in some preface to be glossed over in textbooks or that the mathematical relations are described only in terms of measurement (which isn't usually described in depth nor is the language concerning it clear).

You are, if I understand correctly, a graduate biophysicist and I have no doubts about your competency, but after two years or so of reading graduate textbooks on quantum physics, research papers in quantum physics, monograph and volume series on quantum physics, I wonder if I am aware of how Griffiths statements relate to a vastly larger body of work on QM interpretations and the measurement problem than you are, because I honestly do not know how much you've worked with e.g., MZIs or optic fibers used with some avalanche photodiode or any number of ways (including those I have used) you are familiar with a technology like NMR instruments but (like me) have not used these in experimental quantum mechanics.

In classical mechanics the mathematical representation is a list of all the positions and momenta of all the particles, i.e. a point in a 6N-dimensional phase space; in QM it's a vector in Hilbert space.

Ironically, another Griffiths (Robert) is quite clear here, but rather than quote, it is enough to say that he compares classical vs. quantum as follows: "To summarize, two states of a classical particle have the same physical interpretation if and only if they have the same mathematical description. The case of a quantum particle is not nearly so simple"

That's a pretty unassailable statement.

It is not at all unassailable for one reason: the use of the word physical. As Heisenberg's uncertainty principle tells us, we cannot ever have complete knowledge of a quantum physical system (which you are well are of). So how can we mathematically represent our complete knowledge of the state of a system that we claim it is impossible to know the exact state of?

May I remind you that I am a physicist?

I recognize that, and you are being too modest. The problem is that while I have actually neglected my own field more and more (especially after having to move), I have spent a vast amount of time solely on quantum physics, whether experiments, reviews, proceedings, etc. So I have a problem: on the one hand, you are clearly better equipped to deal with physics than I, as I am not a physicist, yet I am capable of reading physics literature this:

I know, from studies and experience, that the above would be an uncontroversial thing to say in front of a room full of physicists.

is not what I have encountered.

I know it because I've been in a building full of physicists every working day for many years now.

Most of what I have read on biophysics (which isn't a whole lot) concerns a great deal of the kind of studies used in neuroscience, but without restriction to merely the brain and nervous system. On the other hand, I've read an even smaller amount that differs radically, e.g., Detection of Liquid Explosives and Flammable Agents in Connection with Terrorism (NATO Science for Peace and Security Series B: Physics and Biophysics). I worked in an interdisciplinary field myself (and hope to return ASAP), and I know how broad the spectrum of specialties within such fields can be. Hence my question about your experience in experimental quantum physics at the level beyond the use of quantum dots, NMR, FRET, SNOM, etc. I can say that I used NMR technology to measure the spins orientations in neural regions using hydrogen transfer from hemodynamic fluctuations as a proxy, but this doesn't mean I have any clue what preparing a 3-photon GHZ state entails.

You clearly have vastly more experience with physics than I. But this is not about physics, but a very particular topic in physics.

The first sentence in bold is clearly a non-sequitur. Otherwise you've simply defined out of existence a physically real, indeterminate system!

I simply used the term as it has been even in statistical mechanics. A physical system is deterministic in principle. I have no qualms with asserting that physical systems can be really, truly indeterministic, and that

the physical system does not have a predestined evolution, and hence, naturally, our complete mathematical representation also does not have one

But what do we have complete knowledge of? It might be the physical system, as you posited, or not. But until you can demonstrate any evidence anywhere that QM posits complete knowledge of a physical system as the standard interpretation and that this physical system has a one-to-one correspondence with the mathematical representations, then statements like this:

and trust me, you are mistaken on these points

give me nothing.

First: if you don't think measurements are represented by operators acting on kets

I know that they are. That has nothing to do with my original issue which was the correspondence that does not exist in QM as it does in classical physics because we are describing a state that we derived from theory and preparation and postulating that the observables (represented by operators acting on kets) are the only things we can say regarding any physical properties of the system.

It's stated unambiguously in Shankar for example

You mean on page 124, in the discussion of postulates, where he says that "the system had a well-defined state-vector

before measurement, though we did not know

, and has a well-defined state-vector

after measurement, although all we know is that it lies within a subspace

" or perhaps you referring to the next page when he states "Notice how in quantum theory, measurement, instead of telling us what the system was doing before the measurement, tells us what it is doing just after the measurement"? And perhaps you can take me through 4.1 where the postulates are compared.

a standard graduate-level text

Surely first year, though? After all, the author does say it can be used as an undergrad text.

operators correspond to measurement in some way, not in no way as you claimed.

Allow me to clarify: they don't correspond to measurements in the classical sense. See the Shankar quote above (or any number of other quotes I supplied on the radical separation between bijective formalisms in classical physics and the unknown relation in quantum.

In any event you did finally quote Griffiths describing the statistical interpretation, and said:

What I'm hearing is that you concede Griffiths does indeed agree with the statement in question

I honestly don't know, because I don't know if he would describe the clearly probabilistic approach and statistical interpretation he begins and ends with as relating to your postulated indeterminate system which appears nowhere in his text.