Einstein and "spooky actions"

[Mr Spinkles]

They are entangled from your POV, as you are playing God who knows everything. But to Alice, or Bob, unless they are told of the set-up of this experiment before they take their measurement, they don't know.

Okay. I have always been assuming they are told of the setup. Or, they perform the preparation of the singlet themselves so they don't need to be told.

(Tangentially, I don't see why this matters. Alice and Bob could just as well be measuring devices, programmed or controlled in real-time by a single human experimenter instead of two. Does every measuring device in the lab need to be "told" the setup of the experiment before taking measurements? All that matters is that "someone" could hypothetically figure it all out from the information available--you call that God's POV, I call it a competent experimenter's POV.)

It's an interpretation I don't subscribe to, which I have said before.

Okay. So, at worst, what I said (and what Einstein said) about nonlocality in QM was not "wrong". It just wasn't an interpretation of QM you subscribe to.

 

[idav]

Okay. So, at worst, what I said (and what Einstein said) about nonlocality in QM was not "wrong". It just wasn't an interpretation of QM you subscribe to.

Seems to be that the other interpretation has some unknowns in it. Like with the particle with Alice and Bob. If the premise is that that particle A collapses the state of particle B then you have spooky action. If the premise is that particles A and B were always in that state to begin with then no spooky action necessary. The spooky action interpretation doesn't say how this is possible it is just assumed that is whats happening but as noted there's an alternative.

 

[PolyHedral]

(Tangentially, I don't see why this matters. Alice and Bob could just as well be measuring devices, programmed or controlled in real-time by a single human experimenter instead of two. Does every measuring device in the lab need to be "told" the setup of the experiment before taking measurements? All that matters is that "someone" could hypothetically figure it all out from the information available--you call that God's POV, I call it a competent experimenter's POV.)

I agree, although IMO, however you interpret QM, every object has to have some well-defined state regardless of who knows what about it. There has to be something for God to see, even if nobody in practice can measure it.

I don't care whether one's interpretation of QM is irreducibly statistical or a multiverse interpretation (or any relative state interpretation), everyone still has to deal with the fact that the "system" represented by the e.g., Schroedinger's wave function does not correspond to a physical system in any known way except through the use of another mathematical operation (not just an actual physical measurement)

What "physical system" are you trying to make it correspond to, and why?

 

[Mr Spinkles]

Seems to be that the other interpretation has some unknowns in it. Like with the particle with Alice and Bob. If the premise is that that particle A collapses the state of particle B then you have spooky action. If the premise is that particles A and B were always in that state to begin with then no spooky action necessary. The spooky action interpretation doesn't say how this is possible it is just assumed that is whats happening but as noted there's an alternative.

Right but there are holes in the alternatives too. For example, did the particles exist in any sort of state before they were measured (entangled or otherwise)? Zaybu's (most recent) interpretation is a tradeoff, which gets around action-at-a-distance, but at the cost of ambiguity on the existence of an objective reality. IMO the cost of that transaction outweighs the benefit. I would say the number of unknowns, as you put it, are therefore fewer in my interpretation.

 

[LegionOnomaMoi]

What "physical system" are you trying to make it correspond to, and why?

Everything subatomic, and because without doing so we have an irritating new formalism for linear algebra that is useless because every advantage of the bra and ket formalism disappears when we don't "make it correspond" to any physical system. The QM formalisms become a combination of annoying notational schemata and an unheard of way of deriving probability that we can't actually use. We can't use it in any experiment, because we can't do anything with it. It is a description of a state that of a system that never interacts with anything, is never observed, was never observed, and will never be observed or measured or prepared or known about in any other way other than writing down symbols.

We are back to classical mechanics, but with an irritating way of representing systems on paper. Unless, of course, we have a way to relate these systems to experiments. This is done through a combination of mathematical means (e.g., Hermitian operators) and physical observations/measurements, which have become incredibly sophisticated from ion traps to amazing imaging technologies and a plethora of methods to enable quantum coherence while obtaining information (in both the formal quantum information sense and the colloquial sense) about quantum systems.

Depending upon the formalism one uses to represent some quantum system and the experimental design, the ways in which "observables" (i.e., the things that allow us to say anything at all about the atomic world we've been able to thanks to quantum physics) are used or described or whatever changes, but what never changes is the need for a way to say something about quantum systems. For that we require mathematical operations that act on wave functions. Otherwise, all we can say of the "state" of this thing out some n-dimensional complex vector space is, "look! I wrote down symbols on paper!".

 

[zaybu]

Okay. So, at worst, what I said (and what Einstein said) about nonlocality in QM was not "wrong". It just wasn't an interpretation of QM you subscribe to.

That interpretation, yours and Einstein, is based on the notion that the wavefunction, the one that is a solution to the Schroedinger's equation, is a real wave. I have demonstrated that it isn't a real wave. It's a mathematical function that yields, along with a whole formalism ( dual vectors, operators, Hilbert space, etc), nothing but probabilities. And therefore your interpretation should find its rightful place in the dustbin of history.

 

[LegionOnomaMoi]

What "physical system" are you trying to make it correspond to, and why?

Through a bit of searching for Mermin's quantum computing lecture notes (and the knowledge that the lecture notes existed in the first place because they contained a description of Dirac's notation I loved when I was trying to get used to it), I found them. And I found both the quote I remembered an apparently a second half I never knew existed: "Mathematicians tend to loathe Dirac notation, because it prevents them from making distinctions they consider important. Physicists love Dirac notation, because they are always forgetting that such distinctions exist and the notation liberates them from having to remember." (link to the lecture notes; the first chapter has the quote).

I looked for these not just for the quote but because they concern quantum computing, and for various reasons I thought you might find them more interesting and perhaps useful (I'm not sure how much of the material you are already so familiar with it's like teaching the Greek and Latin alphabets to a professor of classical languages). Also, I had hoped Mermin would provide what I was trying state in a clearer way (italics in original; emphases added):
"Quantum computers do an important part of their magic through reversible operations, which transform the initial state of the Qbits into its final form using only processes whose action can be inverted. There is only one irreversible part of the operation of a quantum computer. It is called measurement, and is the only way to extract useful information from the Qbits after their state has acquired its final form. Although measurement is a nontrivial and crucial part of the quantum computational process, in a classical computer the extraction of information from the state of Cbits is so straightforward a procedure that it is rarely even described as part of the computational process, though it is, of course, a nontrivial concern for those who design digital displays or printers." (part B of section 1.15).

So, why is it that measurement is the only irreversible part of quantum computing and what does it involve such that it is the "only way to extract useful information [of a Qbit's] state"? If we are trying to extract "useful information" about any quantum system, how might we do this? In particular, of what use is e.g., the Schrödinger equation is it does not correspond with any physical system? How might we use this or any other formal description of a quantum system to tell us anything useful about anything at all?

Also, as you have said in the past that the many-worlds interpretation which was built upon the work of Hugh Everett III, I'd like to quote from his dissertation On the foundations of Quantum Mechanics obtained through Proquest Dissertations and Theses, which differs, although I believe only in trivial ways (the Proquest version is a scanned version of the actual submitted dissertation, and thus I suspect that the underlined portions are not Everett's but Wheeler's or another member of the committee this thesis was submitted to), from the freely available version available e.g., here.

In it, Everett notes:
"A physical system is described completely by a state function Ψ, which is an element of a Hilbert space,and which furthermore gives information only to the extent of specifying the probabilities of the results of various observations which can be made on the system by external observers. There are two fundamentally different ways in which the state function can change:

Process 1: The discontinuous change brought about by the observation of a quantity with eigenstates φ1, φ2, ... in which case the state Ψ will be changed to the state φj with probability | (Ψ, φj) |^2

Process 2: The continuous, deterministic change of state of an isolated system with time according to a wave equation [I can't easily type the formalism but it is exactly the same as in the linked version except the U of the linked version is A in the original] where A is a linear operator...

[Here's the important bit]
"Not all conceivable situations fit into the framework of this mathematical formulation. Consider an isolated system consisting of an observer or measuring apparatus, plus an object system. Can the change with time of the state of the total system be described by Process 2? If so, then it would appear that no discontinuous probabilistic process like Process 1 can take place. If not, we are forced to admit that systems which contain observers are not subject to the same kind of quantum mechanical description we admit for all other physical systems....

How to make a quantum description of a closed universe; of approximate measurements; and of a system that contains an observer? These three questions have one feature in common. They all inquire about the quantum mechanics that is internal to an isolated system."

How does Everett propose to solve this? He gives a number of key points but for my purposes the important point is why these are "key" in Everett's view (emphasis added): "For any interpretation it is necessary to put the mathematical model of the theory of correspondence with experience."

Granted, much of Everett's work was influenced by Wheeler, and although his work formed the basic, foundational structure to those who actually applied to his work the name "many-worlds" he never used the term, but neither Deutsch nor anyone else has developed a formalism which enables us to treat quantum systems as merely abstract systems in a mathematical space that somehow are relevant to anybody or any theory without dealing with the measurement and correspondence issues. Nor do such solutions simply deal only with abstract mathematical entities in Hilbert space without relating these to physical reality in ways that involve measurements/observations.

My question, then, is how you propose to treat something like Schroedinger's wavefunction as sufficient by itself and to treat quantum systems in terms of Hilbert space alone and yet make quantum physics anything useful for physicists or anything capable of empirical findings or anything for experiments in the physical sciences?

 

[LegionOnomaMoi]

I was at the APS conference on Thursday (the bad one- Association for Psychological Science) in the exhibition/poster presentation room. Among the many publishing vendors (Elsevier, Wiley, Sage, etc.) was Springer. And one of the books on display was on Is Science Compatible with Free Will?: Exploring Free Will and Consciousness in the Light of Quantum Physics and Neuroscience. The author of the 3rd chapter/paper is Nicolas Gisin. For those who don't know, when it comes to lists of names (or abbreviations for papers), there's EPR (or EPRB), Bell, Aspect, & Gisin. Guess what his paper is all about? Nature's nonlocality.

 

[Mr Spinkles]

That interpretation, yours and Einstein, is based on the notion that the wavefunction, the one that is a solution to the Schroedinger's equation, is a real wave. I have demonstrated that it isn't a real wave. It's a mathematical function that yields, along with a whole formalism ( dual vectors, operators, Hilbert space, etc), nothing but probabilities. And therefore your interpretation should find its rightful place in the dustbin of history.

Obviously a function is a function, the question is whether it accurately represents something real. The simplest explanation is that reality exists even when we aren't looking at it, and that reality is represented by the formalism of QM. Ironically, your argument about conservation of angular momentum supports rather than refutes this interpretation.

I note you continue to wisely dodge my challenge to show the calculation of angular momentum explicitly. Care to answer it now? Here it is again:

Please show the calculation explicitly. After Alice measures, (2) = Total spin = spin of Alice's particle + spin of Bob's particle = ...? Spin of Alice's particle = up. Spin of Bob's particle = ... ? (Does Bob's particle have a definite spin? At one point you were saying Bob's particle remains in the singlet state!)

 

[zaybu]

Obviously a function is a function, the question is whether it accurately represents something real. The simplest explanation is that reality exists even when we aren't looking at it, and that reality is represented by the formalism of QM. Ironically, your argument about conservation of angular momentum supports rather than refutes this interpretation.

(1) If you solve the Maxwell's equations, you get a wave equation. You can go and make real electromagnetic waves, like Hertz did in 1886. These waves have amplitudes, frequencies, wavelengths, velocities, all of which can actually be measured.

(2) But in QM, the wavefunction, which is a solution to the Schroedinger equation, doesn't represent anything of that reality. The only valuable quantity in that wavefunction is the square of its amplitude, which gives the probability of finding the particle in a certain state, for example P(&#8593 = (<&#968; |&#8593;> )^2 = 1/2, for spin; for other cases, like the hydrogen atom, you get the energy levels, P = (<&#968; |&#968;(n)> )^2 = -13.6 ev/(n^2), where n = 1,2,3...

Big differences between (1) real waves, and (2) a probability function, the connection to reality is P = (<&#968; |&#968;(a)> )^2, where a is a label for some state in &#968;.

I note you continue to wisely dodge my challenge to show the calculation of angular momentum explicitly. Care to answer it now? Here it is again:

There's no point in answering as it doesn't make sense to calculate the total spin if the individual spins of both particles haven't been measured.

 

[idav]

Right but there are holes in the alternatives too. For example, did the particles exist in any sort of state before they were measured (entangled or otherwise)? Zaybu's (most recent) interpretation is a tradeoff, which gets around action-at-a-distance, but at the cost of ambiguity on the existence of an objective reality. IMO the cost of that transaction outweighs the benefit. I would say the number of unknowns, as you put it, are therefore fewer in my interpretation.

The realist interpretation doesnt assume more since it would involve classic and local factors. To me the other interpretation is one that relies on some hidden variable that can some how transcend physical limitations like space and time.

 

[PolyHedral]

My question, then, is how you propose to treat something like Schroedinger's wavefunction as sufficient by itself and to treat quantum systems in terms of Hilbert space alone and yet make quantum physics anything useful for physicists or anything capable of empirical findings or anything for experiments in the physical sciences?

OK, we're talking about a perfectly isolated system that doesn't interact with anything, and I'm good with claiming that the system is perfectly described by the wavefunction, with no "discontinuous change" involved. You're saying this is insufficient because I don't have any mechanism for producing observable quantities out of that interpretation.

...What's doing the observing? How is it posible to observe a perfectly isolated quantum system?

It's not. When you have a wavefunction that describes the entire contents of the universe, there are no observables, because there are no observers which are not also part of the wavefunction. (Which means that the "observations" they make don't collapse the system at all but instead superpostion them.)

 

[Mr Spinkles]

(1) If you solve the Maxwell's equations, you get a wave equation. You can go and make real electromagnetic waves, like Hertz did in 1886. These waves have amplitudes, frequencies, wavelengths, velocities, all of which can actually be measured.

(2) But in QM, the wavefunction, which is a solution to the Schroedinger equation, doesn't represent anything of that reality. The only valuable quantity in that wavefunction is the square of its amplitude, which gives the probability of finding the particle in a certain state, for example P(&#8593 = (<&#968; |&#8593;> )^2 = 1/2, for spin; for other cases, like the hydrogen atom, you get the energy levels, P = (<&#968; |&#968;(n)> )^2 = -13.6 ev/(n^2), where n = 1,2,3...

Big differences between (1) real waves, and (2) a probability function, the connection to reality is P = (<&#968; |&#968;(a)> )^2, where a is a label for some state in &#968;.

I acknowledge the difference between a classical electromagnetic wave and a quantum wavefunction (or state vector). I would say the probabilities that you calculate in QM are not "the connection to reality", but rather, they are the connection to the part of reality which can be measured. In classical mechanics, that distinction is unimportant because we can measure everything. In quantum mechanics we cannot, so the distinction is critical.

The alternative is to say that "reality" is like swiss cheese, full of holes when and where we can't (or choose not to!) measure. Does the Moon exist when no one is looking? To me, it is simplest to say that the Moon always exists in some quantum state or another, even though we cannot (by definition) confirm or measure what the Moon is doing when we aren't looking. This was the simplest assumption before QM, and it continues to be the simplest assumption about reality with QM, IMO. The less-simple alternative is to say that the Moon doesn't exist except when we can (or choose to) look. That would be an unnecessarily more complicated and spookier sort of "reality", IMO.

There's no point in answering as it doesn't make sense to calculate the total spin if the individual spins of both particles haven't been measured.

It makes perfect sense, mathematically and physically. Mathematically, there is one equation with one unknown (spin of Bob's particle). Solve for the unknown.

Known:
S = total spin = 0
A = spin of Alice's particle = up
B = spin of Bob's particle = ... ?

S = A + B

Solve for B. There is nothing about this calculation, mathematically, that doesn't make sense. What does the math say?

On a more rigorous level, apply the appropriate spin operator to the (singlet) state. The mathematical formalism of QM says that Alice CANNOT, and DOES NOT, just measure her particle. She is ALSO measuring the other particle, because they are entangled. She may read off only +hbar/2 or -hbar/2 from her detection equipment, but the resulting state MUST be |up>|down> or |down>|up>, mathematically. It cannot be just |up> or |down> by itself, because those are not eigenkets of the original (singlet) state. Both particles HAVE, in a legitimate sense, been measured by Alice already (hence the term nonlocality).

Now consider it physically. Physics says in this case the spin is conserved. So: Before Alice measures: spin = 0. After Alice measures, Alice's particle is spin up. Therefore, to conserve spin, after Alice measures Bob's particle must be down. The alternative, where Bob's spin is (????) does not make physical sense, because it implies the total spin is now (???). How can total spin go from known to (???)? And how can we even say it's conserved, in that case? You are doing a lot of wriggling and creating a lot of unnecessary complexity in order to avoid the "spooky action", when it would be simplest just to accept it.

You realize that "the spin" carried by a particle is different from what you measure, right? For example, suppose Bob's particle is in the spin eigenstate "up". You measure it and you get +hbar/2 (mathematically, an eigenvalue). You measure again and you get +hbar/2. You measure a third time and you still get +hbar/2. You measured 3 x hbar/2 but that is not the spin of the particle. You did not "create" an extra +hbar/2 worth of spin in the universe each time you measured. "The spin" of the particle was "up" or +hbar/2 before, during, and after all these measurements. That's what an eigenstate is--it's analogous to a classical state, it had a definite value even before you measured and it's value was not affected by your measurement. Your measurement of an eigenstate only reports the spin that the particle already definitely had. That's why eigenstates are not like generic quantum states (which would generally be in superposition with indeterminate spin until you measure).

You realize eigenstates are special in this regard, right?

 

[zaybu]

I acknowledge the difference between a classical electromagnetic wave and a quantum wavefunction (or state vector). I would say the probabilities that you calculate in QM are not "the connection to reality", but rather, they are the connection to the part of reality which can be measured. In classical mechanics, that distinction is unimportant because we can measure everything. In quantum mechanics we cannot, so the distinction is critical.

The alternative is to say that "reality" is like swiss cheese, full of holes when and where we can't (or choose not to!) measure. Does the Moon exist when no one is looking? To me, it is simplest to say that the Moon always exists in some quantum state or another, even though we cannot (by definition) confirm or measure what the Moon is doing when we aren't looking. This was the simplest assumption before QM, and it continues to be the simplest assumption about reality with QM, IMO. The less-simple alternative is to say that the Moon doesn't exist except when we can (or choose to) look. That would be an unnecessarily more complicated and spookier sort of "reality", IMO.

A simple explanation is that the moon is always being looked upon. We don't need humans to look at it, a detector can do the job, and so would my cat. In reality, the universe is always looking at the moon. The moon doesn't need us, petty creatures crawling on a piece of rock, to exist.

It makes perfect sense, mathematically and physically. Mathematically, there is one equation with one unknown (spin of Bob's particle). Solve for the unknown.

Known:
S = total spin = 0
A = spin of Alice's particle = up
B = spin of Bob's particle = ... ?

S = A + B

Solve for B. There is nothing about this calculation, mathematically, that doesn't make sense. What does the math say?

Theoretically, yes,one can calculate the second spin, but that does mean a spooky action at a distance is actually forcing the second particle into a quantum state. That's why I proposed the second scenario to make you realize that one's knowledge of a theoretical value does not bring forth such instantaneous force.

Of course, if you believe like you do that the wavefunction is real, then you will come to the conclusion there is a spooky action. The point is, the waves we use to calculate these outcomes don't represent real waves.

I have spent many years studying QFT, both in flat spacetime and curved spacetime, and what you don't seem to realize is that the wavefunction, the same object that is the solution to the Schroedinger equation, is no longer a state vector but becomes a field operator acting on the vacuum. You end up spending a lot of times studying such thing as <0I &#966;(x) &#966;(y)I0>, called a two-point correlation function, where &#966; is the field operator. The concept of a wave is off the radar in QFT. And people like you who've never study QFT are still stuck on the wave concept, when that is the least preoccupation. And fortunately, the people in the 40's and 50's ignored Einstein and went on to develop QFT, the most successful theory we have so far.

 

[LegionOnomaMoi]

OK, we're talking about a perfectly isolated system that doesn't interact with anything, and I'm good with claiming that the system is perfectly described by the wavefunction, with no "discontinuous change" involved. You're saying this is insufficient because I don't have any mechanism for producing observable quantities out of that interpretation.

...What's doing the observing? How is it posible to observe a perfectly isolated quantum system?

It isn't. And this doesn't change if:

you have a wavefunction that describes the entire contents of the universe

because it is still not true that

there are no observables

The only way there are no "observables" is if no experiment can ever be done such that we have any reason to thing the wavefunction means anything at all.

because there are no observers which are not also part of the wavefunction. (Which means that the "observations" they make don't collapse the system at all but instead superpostion them.)

Which means that we should never see quantum effects, and should never have discovered it at all. The universe is encapsulated in this wavefunction, so why do we quantum physics? Or physics? What is it that experiments do? Why are our molecules not flying apart into branching universes and how is it that we can see quantum processes cohere? Why is it that there is no multiverse theory that has dispensed with observables, and where is this universal wavefunction described?

 

[idav]

I acknowledge the difference between a classical electromagnetic wave and a quantum wavefunction (or state vector). I would say the probabilities that you calculate in QM are not "the connection to reality", but rather, they are the connection to the part of reality which can be measured. In classical mechanics, that distinction is unimportant because we can measure everything. In quantum mechanics we cannot, so the distinction is critical.

The alternative is to say that "reality" is like swiss cheese, full of holes when and where we can't (or choose not to!) measure. Does the Moon exist when no one is looking? To me, it is simplest to say that the Moon always exists in some quantum state or another, even though we cannot (by definition) confirm or measure what the Moon is doing when we aren't looking. This was the simplest assumption before QM, and it continues to be the simplest assumption about reality with QM, IMO. The less-simple alternative is to say that the Moon doesn't exist except when we can (or choose to) look. That would be an unnecessarily more complicated and spookier sort of "reality", IMO.

Physics in the classical sense keeps the moon there regardless of qm. We cant really see at the atomic level but assuming the higgs boson isnt popping in and out of realkty then we should be able to be assured reality is here.

 

[PolyHedral]

Which means that we should never see quantum effects, and should never have discovered it at all.

I don't follow. Why does the observer becoming superpositioned mean that we should never see any quantum effects at all?

 

[LegionOnomaMoi]

I don't follow. Why does the observer becoming superpositioned mean that we should never see any quantum effects at all?

We have these things we call wave functions (among other things), that describe the states of quantum systems. You say the universe is one giant wavefunction. There is no "collapse" of quantum systems because the "observer" becomes superpositioned. How can the "observer" see any quantum processes when every time such an observation is made the "observer" exists in two different states? Do these states magically come back to make one magically unified observer who now has not observed anything quantum but actually been in a state unique to quantum physics?

We use wave functions to describe quantum systems. We perform experiments where "it is possible to freely and a posteriori decide which type of mutually exclusive correlations two already earlier measured particles have." (Ma, X. S., Zotter, S., Kofler, J., Ursin, R., Jennewein, T., Brukner, &#268;., & Zeilinger, A. (2012).
We have observed macroscopic molecules in two different "states" at the same time, we have been able to freely and consistently have a single photon act wave-like or particle-like over the course of one experiment by tuning measurement instruments back and forth.

What, exactly, does saying observers are in superposition states explain? How do you incorporate your universal wavefunction into physics such that it is of any use to anybody?

 

[zaybu]

Physics in the classical sense keeps the moon there regardless of qm. We cant really see at the atomic level but assuming the higgs boson isnt popping in and out of realkty then we should be able to be assured reality is here.

But even from the POV of QM, one can say that the moon is always being watched by humans, cats, detectors, IOW, the universe. Classical and quatum physics are on the same page on that.

 

[idav]

But even from the POV of QM, one can say that the moon is always being watched by humans, cats, detectors, IOW, the universe. Classical and quatum physics are on the same page on that.

QM would have it that the ”observer” collapses the wave. In the case of classical mechanisms higgs and gravity are the observers.

Edit: For spooky actions it has to be a vacuum. The moon has too much mass.