Murray Gell-Mann on Entanglement

[Demystifier]

on many interesting questions that can be checked experimentally? What would be an example?

What orientation of the Stern-Gerlach apparatus will the experimentalist freely choose in the next experimental run.

 

[atyy]

Sure, what I meant is that there is not the slightest evidence for a failure of the purely quantum theoretical description.

If one believes the moon is there when one is not looking, then one should believe the quantum description is incomplete (or that many-worlds is correct).

 

[vanhees71]

But to me, the fact that the two halves of a quantum experiment--the system being measured, and the system doing the measuring--have such completely different properties according to the quantum formalism suggests to me that the burden of proof should be on the other side. Prove that hamiltonian dynamics is sufficient to account for all phenomena, including measurement processes.

Many-worlds is an attempt to do that. A. Neumaier claims that it can be done without many-worlds (although I don't understand his argument). But it seems to me that some kind of derivation of measurement from hamiltonian dynamics is needed before you can say that hamiltonian dynamics applies to everything.

The problem for me is that the standard way that quantum mechanics is done postulates properties for measurement devices and measurement interactions which it does not postulate for single particles, or any combination of particles. If you have a single electron that is in the spin state $\frac{1}{\sqrt{2}} (|U\rangle + |D\rangle$, then it doesn't make any sense to say that it is 50% likely to be spin-up and 50% likely to be spin-down. It is in the definite state "spin-up in the x-direction". If you have an interaction between two electrons, it doesn't make any sense to say that one electron has a 50% chance of observing the other to be spin-up. With a small number of particles, probability doesn't come into play at all. Definite values for dynamic variables doesn't come into play at all. But if you scale up one of the interacting systems to be a measurement device designed to measure spin, then it becomes unproblematic to say that the measurement device interacting with the electron has a 50% chance of going into the "observed spin-up" state, and 50% chance of going into the "observed spin-down" state. How did this probabilistic description arise from microscopic interactions that are non-probabilistic?

Given the state $\hat{\rho}_{\sigma_x=1/2}=|\sigma_x=1/2 \rangle\langle \sigma_x=1/2 |$ with $|\sigma_x=1/2 \rangle=\frac{1}{\sqrt{2}} (|\sigma_z=1/2 \rangle+|\sigma_z=-1/2 \rangle)$ means, according to minimally interpreted QT, with regard to a measurment of $\sigma_z$ not more and not less that you'll find with 50% probality up and with 50% probability down when measuring $\sigma_z$. That's it. $\sigma_z$ doesn't have a definite value due to the preparation in a state where $\sigma_x$ has a definite value "up".

In many-body systems we are often content with a view "macroscopically relevant" parameters; often these are thermodynamic quantities like temperature, density of a gas, or the center of mass/momentum and its velocity of a macroscopic object etc. From a microscopic point of view very little is known of the system, but the accuracy at which you need to know the macroscopic coarse grained quantities to give a pretty accurate description of the macroscopic object usually is so coarse that the macroscopic quantity is derived as the average (often a time average over times macroscopically small and microscopically large) over many microscopic degrees of freedom. Quantum effects tend to be averaged out in the vast majority of macroscopic situations (decoherence). The collective quantum behavior of a macrosopicy body are rare in the sense that usually you have to carefully prepare such states like superfluid helium or superconducting metals (low temperature!) etc.

 

[vanhees71]

If one believes the moon is there when one is not looking, then one should believe the quantum description is incomplete (or that many-worlds is correct).

Which feature of QT contradicts the fact that the moon is there when one is not looking? To the contrary the conservation laws ensuring that the moon is not spontaneously puffing out of existence are also at the very beginning of Q(F)T model building ;-).

 

[Ken G]

Haven't we gotten beyond the idea yet, in all this history of physics that we have to look at, that there is any such thing as a "correct" theory-- other than simply the theory that is doing for us what we want that theory to do for us? Theories certainly have lessons that teach us about how things work that surprise us, and then new theories have new lessons that show the old lessons were only waypoints in the journey. If anything the history of science has shown us, it's that.

 

[Ken G]

Which feature of QT contradicts the fact that the moon is there when one is not looking? To the contrary the conservation laws ensuring that the moon is not spontaneously puffing out of existence are also at the very beginning of Q(F)T model building ;-).

Yes, I think to say that any of the interpretations of QT are saying that the Moon isn't there when we are not looking at it misses the point-- that's Einstein's straw man. Of course the Moon is there when we are not looking at it, the problem is, it is only our ability to look at it that allows us to say it is there when we aren't.

(If that was too cryptic, what I mean is, it is precisely our ready access to looking at the Moon that allows us to say it is always there, which simply begs the question about implications for systems that are not so readily accessible to our perceptions that we can make such sweeping claims about them.)

 

[atyy]

Which feature of QT contradicts the fact that the moon is there when one is not looking? To the contrary the conservation laws ensuring that the moon is not spontaneously puffing out of existence are also at the very beginning of Q(F)T model building ;-).

If the moon is always there, then the moon has a trajectory, so particles have trajectories.

 

[Ken G]

That's the point of view that says if we think up a concept, call it a "particle", then start seeing things we'd like to treat as particles, like moons and electrons, that somehow means these things actually are particles, so must behave the same way. We don't get to say that-- nature tells us how things behave, and we take that and cook up some useful notions like "trajectory" and "being there when we're not looking", and sometimes these notions work, and sometimes they don't.

 

[vanhees71]

If the moon is always there, then the moon has a trajectory, so particles have trajectories.

Yes, FAPP the moon has a trajectory, but just in principle, how implies the statement that "the moon is there when one is not looking" that "the moon has a trajectory"? If I say there is a container with walls at a certain temperature, it includes the black-body radiation of this temperature (no matter whether I look at it or not), it also hasn't a trajectory.

 

[atyy]

Yes, FAPP the moon has a trajectory, but just in principle, how implies the statement that "the moon is there when one is not looking" that "the moon has a trajectory"? If I say there is a container with walls at a certain temperature, it includes the black-body radiation of this temperature (no matter whether I look at it or not), it also hasn't a trajectory.

You mean the black body radiation has classical field values.

 

[vanhees71]

You mean the black body radiation has classical field values.

I mean I can measure its spectrum (in principle).

 

[atyy]

I mean I can measure its spectrum (in principle).

But does its spectrum exist when you are not measuring it?

 

[Demystifier]

Which feature of QT contradicts the fact that the moon is there when one is not looking? To the contrary the conservation laws ensuring that the moon is not spontaneously puffing out of existence are also at the very beginning of Q(F)T model building ;-).

Again, you are mixing individual properties and ensemble properties. The property of moon being or not being there is an individual property, on which standard Q(F)T does not say much. The conservation law of Q(F)T is an ensemble property.

To sharpen the problem, consider the following questions:
Is the photon momentum there when nobody looks?
If yes, isn't that an assumption of hidden variables?
If no, isn't that a violation of momentum conservation?
If the question has no answer within Q(F)T, then is there an answer to the same question when photon is replaced by Moon?

 

[vanhees71]

But does its spectrum exist when you are not measuring it?

This is a nonsense question, because I cannot check whether it exists without looking at it ;-).

 

[vanhees71]

Again, you are mixing individual properties and ensemble properties. The property of moon being or not being there is an individual property, on which standard Q(F)T does not say much. The conservation law of Q(F)T is an ensemble property.

To sharpen the problem, consider the following questions:
Is the photon momentum there when nobody looks?
If yes, isn't that an assumption of hidden-variables?
If no, isn't that a violation of momentum conservation?
If the question has no answer within Q(F)T, then is there an answer to the same question when photon is replaced by Moon?

If I have prepared a photon with some (pretty well defined) momentum, then it's there due to this preparation procedure and it has a (pretty well defined) momentum, no matter whether I detect it or not. Maybe I'm again to naive to understand (and I've never understood this argument), why this is a problem at all.

 

[Demystifier]

If I have prepared a photon with some (pretty well defined) momentum, then it's there due to this preparation procedure and it has a (pretty well defined) momentum, no matter whether I detect it or not. Maybe I'm again to naive to understand (and I've never understood this argument), why this is a problem at all.

The question, of course, refers to the case in which you have prepared the photon in a state with not pretty well defined momentum.

 

[vanhees71]

Then, of course, it doesn't have a (pretty well defined) momentum. So what?

 

[atyy]

This is a nonsense question, because I cannot check whether it exists without looking at it ;-).

So it is also a nonsense question whether the moon is there when you are not looking at it.

 

[vanhees71]

Yes, that's what I was saying the whole time. Also Bell ridiculed it in asking, whether you need a conscious being to provide the collapse to be sure that the moon is there. Would it be enough to have an amoeba or do you need a more complicated creature to take notice of the moon to make it come into existence? It's pretty much nonsense to claim that the moon isn't there if "nobody" is looking.

 

[Demystifier]

This is a nonsense question, because I cannot check whether it exists without looking at it ;-).

But then it is an equal nonsense to claim that Moon exists without looking at it.

 

[Demystifier]

Yes, that's what I was saying the whole time.

If you were saying that it is nonsense to say that the Moon is there when nobody looks, then why did you use the conservation law to argue that it is there when nobody looks?

 

[vanhees71]

If you were saying that it is nonsense to say that the Moon is there when nobody looks, then why did you use the conservation law to argue that it is there when nobody looks?

No I said that the claim the moon is not there is nonsense to begin with. It has been observed in the past. Then there are pretty well established conservation laws telling you that it won't puff out of existence only because nobody is looking at her.

 

[Demystifier]

Then, of course, it doesn't have a (pretty well defined) momentum. So what?

I have no further comments on this, because you explained your position consistently in #267.

 

[atyy]

No I said that the claim the moon is not there is nonsense to begin with. It has been observed in the past. Then there are pretty well established conservation laws telling you that it won't puff out of existence only because nobody is looking at her.

So if the moon has a trajectory, there are hidden variables.

 

[vanhees71]

Well, of course the photon is much more fragile than the moon. To be sure to have one photon after having created it, hasn't been absorbed by some material around. After all photon number is not conserved (as are various quantum numbers of the matter making up the moon).

It's by the way highly non-trivial to prepare exactly one photon; that's standard only for a few years with parametric downconversion where you can make a photon pair and measure one of the photons (absorbing it) to have the other photon as the prepared one-photon state

 

[vanhees71]

So if the moon has a trajectory, there are hidden variables.

Again, it's not necessary for the moon to have a trajectory only for "being there". Strictly speaking the moon has no trajectory since nothing has an exact trajectory, because this contradicts the position-momentum uncertainty relation. As a macroscopic object in the sense of the classical approximation, of course, its center of mass has a trajectory (not easy to calculate, as already made Kepler crazy ;-)).

I don't believe in hidden variables, but of course, I cannot disprove their existence. Maybe after all nature is deterministic with non-local interactions, but we are not clever enough (yet?) to find an adequate theory of such a possibility and also no experiment to observe the hidden variables (yet?).

 

[atyy]

Again, it's not necessary for the moon to have a trajectory only for "being there". Strictly speaking the moon has no trajectory since nothing has an exact trajectory, because this contradicts the position-momentum uncertainty relation. As a macroscopic object in the sense of the classical approximation, of course, its center of mass has a trajectory (not easy to calculate, as already made Kepler crazy ;-)).

I don't believe in hidden variables, but of course, I cannot disprove their existence. Maybe after all nature is deterministic with non-local interactions, but we are not clever enough (yet?) to find an adequate theory of such a possibility and also no experiment to observe the hidden variables (yet?).

The moon having a trajectory does not contradict the position-momentum uncertainty. It just means the x(t) exists, where x is the position of the particle.

If the moon being "there" does not mean it has a position, then what do you mean by "there"?

 

[vanhees71]

But $x(t)$ doesn't exist in the classical sense, because it's uncertain anyway. Given you start with a pretty well located particle. Then it has a pretty unsharp momentum, and thus with time, also the position uncertainty grows. So there are only trajectories in a coarse-grained sense, i.e., not in the sense of accurate values $x(t)$ as in classical physics!

 

[Demystifier]

I don't believe in hidden variables, but of course, I cannot disprove their existence.

A hypothetic question: If you lived in time when Boltzmann had a theory that thermodynamics is a consequence of motion of atoms, while Mach argued against existence of atoms because there was no any direct experimental evidence for atoms, on whose side would you be at that time (given the evidence at that time)?

 

[atyy]

But $x(t)$ doesn't exist in the classical sense, because it's uncertain anyway. Given you start with a pretty well located particle. Then it has a pretty unsharp momentum, and thus with time, also the position uncertainty grows. So there are only trajectories in a coarse-grained sense, i.e., not in the sense of accurate values $x(t)$ as in classical physics!

But what are you coarse graining?

 

[A. Neumaier]

If one believes the moon is there when one is not looking, then one should believe the quantum description is incomplete

This is a statement without any logical support.

It is enough to believe that the mass density (an expectation value computable in principle from the density operator of the solar system) is positive in the volume occupied by the moon. Since this a macroscopic observation, one can work to a very good approximation in the limit where Planck's constant is set to zero and the classical description is therefore valid. No discrepancy at all with quantum physics!

It only conflicts with the ridiculous view that the highly idealized axioms introduced in an introductory quantum mechanics course or textbook define quantum mechanics.

 

[vanhees71]

A hypothetic question: If you lived in time when Boltzmann had a theory that thermodynamics is a consequence of motion of atoms, while Mach argued against existence of atoms because there was no any direct experimental evidence for atoms, on whose side would you be at that time (given the evidence at that time)?

That's difficult to say. As a physicist probably I'd have taken the side of Mach, although if I'd have been be a bit more open minded and taking the vast evidence of "atomism" from chemistry at the time, maybe I'd have taken Boltzmann's side since his model at least didn't contradict any evidence (and it was appealing mathematically).

 

[A. Neumaier]

If the moon is always there, then the moon has a trajectory, so particles have trajectories.

The moon need only to have a mean trajectory, given by the expectation of the center of mass of the position operators of its atoms. Its standard deviation is far below the radius of the moon and hence negligible.

 

[stevendaryl]

Given the state $\hat{\rho}_{\sigma_x=1/2}=|\sigma_x=1/2 \rangle\langle \sigma_x=1/2 |$ with $|\sigma_x=1/2 \rangle=\frac{1}{\sqrt{2}} (|\sigma_z=1/2 \rangle+|\sigma_z=-1/2 \rangle)$ means, according to minimally interpreted QT, with regard to a measurment of $\sigma_z$ not more and not less that you'll find with 50% probality up and with 50% probability down when measuring $\sigma_z$. That's it.

Under what circumstances does an electron measure its own spin? Never, right? So it doesn't make any sense at all to say that an isolated electron has a 50% probability of being spin-up in the z-direction. What about a pair of electrons? When does one electron measure the spin of another electron? Never, right? So for a pair of electrons, probability doesn't make any sense.

Probability only makes sense for an interaction in which one of the subsystems is a macroscopic measuring device.

 

[stevendaryl]

If I have prepared a photon with some (pretty well defined) momentum, then it's there due to this preparation procedure and it has a (pretty well defined) momentum, no matter whether I detect it or not. Maybe I'm again to naive to understand (and I've never understood this argument), why this is a problem at all.

To me, if you and your equipment are all described by the same physics as electrons and photons, etc., then to say that "I prepared things in such-and-such a way" means "Me and my equipment were put into such and such a macroscopic state". So there is a notion of "state" for macroscopic objects that does not depend on yet another system to prepare them in that state. They can put themselves into a particular state. But you're saying that for an electron, or a photon, or any microscopic system, the only notion of state is a preparation procedure by a macroscopic system. That seems incoherent to me. At best, it's a heuristic, but it can't possibly be an accurate description of what's going on. If macroscopic systems have properties without being observed, then why can't microscopic systems?