I Entanglement: is there 'action at a distance' due to measurement?

[timmdeeg]

Two particles, A and B, are entangled. The measurement of A yields spin-up.

What happens to B in that instant of time?

a) is B still a quantum object and as such doesn't possess definite spin properties before measurement?

b) has B the definite property spin-down even prior to its spin measurement?

What means action on a distance in this context, if at all?

 

[PeroK]

Two particles, A and B, are entangled.

It's conceptually clearer to describe this as a system of two particles in an entangled state. See below.

The measurement of A yields spin-up.

Again, it's conceptually clearer to say that the system of two particles is measured and particle A is measured as spin-up about a given axis.

What happens to B in that instant of time?

The system of two particles is now in an unentangled state. The state of the non-local system has changed, as result of the measurement. The particles are now independent, although each is in an eigenstate of spin about a given axis.

a) is B still a quantum object and as such doesn't possess definite spin properties before measurement?

There's no problem with a particle being in an eigenstate of spin about a given axis. The measurement of the entangled system has effectively done that to both particles. The spin about other axes is indeterminate - as it would be with any measurement of spin about a given axis.

b) has B the definite property spin-down even prior to its spin measurement?

Again, this is no different from any measurement of a particle. After the measurement, the particle is in an eigenstate of spin about that axis. Repeated measurements of spin about the same axis must result in the same outcome.

It seems to have been a common theme recently that people asking about QM don't realise that systems in an eigenstate of a given observable will have a definite measurement outcome for that observable. This is not something where entanglement contradicts the rest of QM.

What means action on a distance in this context, if at all?

It has no meaning, since QM says nothing about any action or mechanism to enforce correlation between measurements. The simplest position is to treat the quantum state as a non-local, mathematical object that allows you to calulate the probabilities of measurement outcomes. And, sometimes, those probabilities can be 0 or 1.

Again, there has been a common misconception recently that a probability of 0 or 1 somehow contradicts the probabilistic nature of QM. It doesn't. Probabilities of 0 or 1 are precisely what you get in the case of an eigenstate of the measured observable.

 

[Nugatory]

What happens to B in that instant of time?

Quantum Mechanics only describes measurement results, and we haven’t made a measurement yet. As far as QM is concerned, all that we can say about B is that if and when we measure it on the same axis the result will be spin-down. Note that this is not the same thing as saying that B is spin-down.

a) is B still a quantum object and as such doesn't possess definite spin properties before measurement?

That depends on your choice of interpretation.

b) has B the definite property spin-down even prior to its spin measurement?

That also depends on your choice of interpretation.

What means action on a distance in this context, if at all?

It means that we’re speculating about some mechanism (which isn’t part of QM) that could explain (which is not something QM claims to do) why we get the measurement results that QM predicts, and we’re finding it hard to do this without introducing some sort of hypothetical faster-than-light effect in which measuring A changes something about B.

 

[FactChecker]

b) has B the definite property spin-down even prior to its spin measurement?

This seems like asking if there is a local hidden variable, determining the spin of both, that was set before the separation of the particles. That was a great question until Bell's Theorem was proposed and experimental results were obtained.

 

[timmdeeg]

Thanks to everybody for clarifying this question.

I think that's what I've been missing:

The system of two particles is now in an unentangled state. The state of the non-local system has changed, as result of the measurement. The particles are now independent, although each is in an eigenstate of spin about a given axis.

 

[pines-demon]

Two particles, A and B, are entangled. The measurement of A yields spin-up.

What happens to B in that instant of time?

Depends on the interpretation of quantum mechanics. Per relativity of simultaneity, which measurement is performed first depends on the reference frame (for one oberserver A measures first, for another B measures first) so whatever answer has to take this into account.

Some interpretations are:

• Some kind of instantaneous collapse or nonlocal interaction, some kind of influence would travel faster than light but it does not violate relativity because (somehow) it cannot be used for faster-than-light communication. This is the closest we have to action at a distance.
• Something alike many-worlds, where the universe branches from measurements. This creates two universes one where A measured up a B down and one where B measures down and A up. If this happens instantaneously or not, does not seem to matter much because the branching kind of handles the issue.
• Superdeterminism, theories were the measurements and the experiments are somehow part of a big script and the results are pre-determined since the Big Bang. No matter how careful the experimenters are and what the underlying mechanics truly is, the results are always correlated in such a way to reproduce quantum mechanics (conspiracy).
• Retrocausal theories: in the transactional interpretation some kind of signal travels into the past to make sure the results are compatible (shake hand mechanism). Arguably per relativity this is the same as the instantaneous/non-local version.

Note that in some theories like Bohmian mechanics, the spin is not part of the particle but of the wavefunction guiding the particle.

 

[PeterDonis]

What happens to B in that instant of time?

That depends on which interpretation of QM you adopt.

 

[PeterDonis]

Moderator's note: Thread moved to interpretations subforum.

 

[PeterDonis]

The system of two particles is now in an unentangled state.

That's only true for interpretations where collapse is a real process, and which interpret the quantum state as describing individual systems. For interpretations like the MWI (where collapse is not a real process, evolution is always unitary) and statistical interpretations (where the quantum state doesn't describe individual systems, only ensembles), the statement quoted above is not true.

 

[timmdeeg]

Coming back to the term eigenstate,

The system of two particles is now in an unentangled state. The state of the non-local system has changed, as result of the measurement. The particles are now independent, although each is in an eigenstate of spin about a given axis.

There's no problem with a particle being in an eigenstate of spin about a given axis. The measurement of the entangled system has effectively done that to both particles. The spin about other axes is indeterminate - as it would be with any measurement of spin about a given axis.

could you please explain how it relates to the term eigenvalue.

Has a particle in an eigenstate inevitably also an eigenvalue?

 

[PeroK]

Has a particle in an eigenstate inevitably also an eigenvalue?

Yes, an eigenstate of a given observable will result in a definite measurement outcome. The measurement value is the eigenvalue associated with that eigenstate.

 

[timmdeeg]

Thanks!

 

[DrChinese]

There's no problem with a particle being in an eigenstate of spin about a given axis. The measurement of the entangled system has effectively done that to both particles. The spin about other axes is indeterminate - as it would be with any measurement of spin about a given axis.

…

Everything you say around these points is correct. I would emphasize the word “effectively”, because there is a quirk. You measure A and get polarization H, and let’s say that places B into an eigenstate certain to produce V on the same basis.

Q: Are initially polarization entangled PDC photons A and B in the same quantum state (HV?) as 2 PDC produced photons C and D entangled - but not on the polarization basis - which are known to be HV?

A: No! A and B can be used in entanglement swapping, but C and D cannot.

Apparently the “collapse” (or whatever you might imagine) defies even this mechanistic attempt as a description. Photon B is not exactly in the same state as D unless and until it is actually measured on that basis.

Go figure. :)

 

[martinbn]

Go figure. :)

Bohr would have said "Obviously".

 

[timmdeeg]

Quantum Mechanics only describes measurement results, and we haven’t made a measurement yet. As far as QM is concerned, all that we can say about B is that if and when we measure it on the same axis the result will be spin-down. Note that this is not the same thing as saying that B is spin-down.

Do you say that if we don't measure B then it isn't spin-down? It has an eigenvalue but not an eigenstate associated with it.

Is it wrong to say that if we measure A we effectively measure the system consisting of the entangled pair of particles A and B? But if true wouldn't it mean that the outcomes are A has spin-up und B has spin-down?

 

[Morbert]

The system of two particles is now in an unentangled state.

That's only true for interpretations where collapse is a real process, and which interpret the quantum state as describing individual systems. For interpretations like the MWI (where collapse is not a real process, evolution is always unitary) and statistical interpretations (where the quantum state doesn't describe individual systems, only ensembles), the statement quoted above is not true.

Some sharpening of statements is needed here. In MWI, the unitary evolution would apply to the biparticle-and-environment system rather than the biparticle system. The biparticle system (i.e. the system with environmental degrees of freedom traced out) would be in a correlated/anticorrelated state but not an entangled state.

 

[martinbn]

Do you say if we don't measure B then it isn't spin-down? And thus it hasn't an eigenvalue und an eigenstate associated with it.

What does it mean for B to have an eigenvalue?

 

[DrChinese]

1. Quantum Mechanics only describes measurement results, and we haven’t made a measurement yet. As far as QM is concerned, all that we can say about B is that if and when we measure it on the same axis the result will be spin-down. Note that this is not the same thing as saying that B is spin-down.

2. It means that we’re speculating about some mechanism (which isn’t part of QM) that could explain (which is not something QM claims to do) why we get the measurement results that QM predicts, and we’re finding it hard to do this without introducing some sort of hypothetical faster-than-light effect in which measuring A changes something about B.

1. Really well said, as I read this a second time. This seemingly minor detail is the essence of a lesson in QM that is sometimes missed: don’t speculate on the behavior of a quantum system outside of a measurement. (Peres: Unperformed measurements have no results.) This is demonstrated physically by the little quirk I mentioned yesterday.

2. Agree completely: Modern experiments make holding on to locality so difficult, I have finally tipped over the edge on this point. However, I lack any concept of an underlying mechanism - everything I’ve read as possible explanations appears to be ruled out by one experiment or another.

 

[PeterDonis]

In MWI, the unitary evolution would apply to the biparticle-and-environment system rather than the biparticle system.

And the measuring device. The entanglement first spreads to the measuring device when the measurement takes place, and then to the environment as decoherence takes place.

The biparticle system (i.e. the system with environmental degrees of freedom traced out) would be in a correlated/anticorrelated state but not an entangled state.

Wrong. The biparticle system becomes entangled with the measuring device and the environment. It never goes to a non-entangled state. The entanglement between the two particles themselves is no longer maximal, because the entanglement spreads to more and more degrees of freedom, but the particles are never in a non-entangled state.

 

[timmdeeg]

What does it mean for B to have an eigenvalue?

It means, that in the case of B's measurement B has an eigenstate associated with that eigenvalue.

 

[Morbert]

And the measuring device. The entanglement first spreads to the measuring device when the measurement takes place, and then to the environment as decoherence takes place.

I include the measuring device in the environment but we can demarcate the two if needed later.

Wrong. The biparticle system becomes entangled with the measuring device and the environment. It never goes to a non-entangled state. The entanglement between the two particles themselves is no longer maximal, because the entanglement spreads to more and more degrees of freedom, but the particles are never in a non-entangled state.

Consider a biparticle-and-environment system initially in the state $\frac{1}{\sqrt{2}}(\ket{00}+\ket{11})\ket{\epsilon_\Omega}$. After measurement the system is in the state $\frac{1}{\sqrt{2}}(\ket{00}\ket{\epsilon_0}+\ket{11}\ket{\epsilon_1})$. The biparticle system is in the state$$\mathrm{tr}_\epsilon \rho = \frac{1}{2}(\ket{00}\bra{00} + \ket{11}\bra{11})$$which is an unentangled state.

 

[PeterDonis]

It means, that in the case of B's measurement B has an eigenstate associated with that eigenvalue.

But you were talking about a case where B has not been measured.

 

[timmdeeg]

But you were talking about a case where B has not been measured.

In more detail, supposed A has been measured, but B not yet. In this case B has an eigenvalue, but depending on the interpretation not an eigenstate. In this case if B is measured thereafter B will have an eigenstate associated with said eigenvalue.

 

[PeterDonis]

In more detail, supposed A has been measured, but B not yet. In this case B has an eigenvalue

No, it doesn't. Until B is measured it has no eigenvalue. Remember that until B is measured, you can't dictate in what direction its spin is being measured--which you would have to do to assign it an eigenvalue.

 

[Fra]

2. Agree completely: Modern experiments make holding on to locality so difficult

For me they rather make holding onto objective reality (realism) difficult if not impossible

lack any concept of an underlying mechanism - everything I’ve read as possible explanations appears to be ruled out by one experiment or another.

Especially if the "mechanism of interactions between parts" one seeks must necessarily be formulated in terms of functions of objective beables; wether deterministically or probabilistically , then I agree a mechanism seems impossible.

/Fredrik

 

[DrChinese]

For me they rather make holding onto objective reality (realism) difficult if not impossible

I agree with this too. I was formerly of the view that Local Realism is untenable (due to Bell's Theorem). But with more and more experiments: Both Realism (basically contradicted by the HUP anyway) and Einsteinian Locality (denying remote effects of any kind) seem untenable.

(And every hypothetical mechanism for nonlocality seems equally untenable LOL.)

 

[Fra]

I agree with this too. I was formerly of the view that Local Realism is untenable (due to Bell's Theorem). But with more and more experiments: Both Realism (basically contradicted by the HUP anyway) and Einsteinian Locality (denying remote effects of any kind) seem untenable.

(And every hypothetical mechanism for nonlocality seems equally untenable LOL.)

I have not yet seen any compelling reason for abandoning Einsteinian Locality(*).

I see hope in understanding thing while keeping a notion of locality, in terms of subjective reality or "subjective beables" where "mechanism of interactions between parts" takes on other forms that presumed in "local realism", because unlike objective beables, they do not commute, so you can't make a conceptually sound partition, or divisions that Barandes speaks of conditional probablities in terms of a mixed of incompatible beables.

Sometime there seems to be a funny relation between "objective" vs "global" and "subjective" vs "local", as it is what defines the relations between "parts", either by some "position" in a space, or "position" in information space.

(*) Except the obvious that at some point when we get into QG, spacetime itself maybe reformulated, then notions defined in terms of them such as locality needs be reformulated too)

/Fredrik

 

[Fra]

(And every hypothetical mechanism for nonlocality seems equally untenable LOL.)

Here we 100% agree as well!

/Fredrik

 

[timmdeeg]

No, it doesn't. Until B is measured it has no eigenvalue. Remember that until B is measured, you can't dictate in what direction its spin is being measured--which you would have to do to assign it an eigenvalue.

Ah, yes, thanks. I have confused eigenstate and eigenvalue. So, until B is measured it has an eigenstate of spin. Correct?

 

[pines-demon]

Ah, yes, thanks. I have confused eigenstate and eigenvalue. So, until B is measured it has an eigenstate of spin. Correct?

There is no eigenstate of spin in the sense that spin is not an observable by itself. What you can observe are projections of spin in given directions of space. So if Alice measures spin in some axis, the particle of B is going to be in an eigenstate of the projection of spin only in that same axis.