Einstein and "spooky actions"

[PolyHedral]

No in QM they only have to agree on the outcomes of measurements.

QM is a predictive theory. It tells us, ahead of time, what outcomes we should expect from certain measurements. But, apparently, it doesn't, if the thing we're measuring is part of an entangled pair.

There's no confusion about the meaning of a state. The disagreemernt is that you believe that having knowledge of a state will affect the state of a particle, and that's not what QM says.

And, as noted, this causes a paradox.

 

[Reptillian]

It seems to me that the core of what ya'll have been arguing about is this: Alice makes a measurement and instantly knows Andromeda Galaxy dwelling Bob's particle is spin down. Bob uses the best predictive powers he has available (quantum mechanics) to try and determine the outcome of his measurement and the best he can do is say that he has a 50% chance of measuring spin up, and a 50% chance of measuring spin down. Alice knows that Bob's particle has a well defined spin prior to his measurement. Even with the best of his available knowledge, Bob is powerless to predict that. Bob says, "No, the particle exists in a linear combination of up and down" So then we're left with the question "How do we know all particles don't have well defined properties regardless of measurement according to some other nonlocal observer, but we're just powerless to predict them?" I can see Dirac shaking his head and asking "Just exactly how many angels can sit on the head of a pin?"

 

[zaybu]

Pardon me, just trying to follow along. From what I gather, zaybu takes issue with your statement that particle 1 affects the state of the particle 2. When all that is happening is the first measurement tells you the second measurement by process of elimination, which is what zaybu agreed with.

Spinkles believes in voodoo magic. Nothing will convince him otherwise.

 

[zaybu]

Where did I claim "her knowledge" affects anything? If you stand by what you shouted in all caps, then you must agree that Alice's measurement changes the state of the second particle. That is, if she measures, then the second particle will have a state that is different from what its state was before she measured. Do you agree, or not?

No, her knowledge depends on Bob making a measurement. If Bob makes no measurement, his particle stays in a singlet state. And Alice can dwiddle her fingers. The predictive power Alice has is if Bob actually makes a measurement. So her knowledges does not in anyway affect the state of Bob's particle. But Bob's measurement will affect the state of his particle.

 

[zaybu]

It seems to me that the core of what ya'll have been arguing about is this: Alice makes a measurement and instantly knows Andromeda Galaxy dwelling Bob's particle is spin down. Bob uses the best predictive powers he has available (quantum mechanics) to try and determine the outcome of his measurement and the best he can do is say that he has a 50% chance of measuring spin up, and a 50% chance of measuring spin down. Alice knows that Bob's particle has a well defined spin prior to his measurement.

On the condition that Bob will make a measurement. If Bob decides otherwise, Alice is left with a useless knowledge. Recall that it's particle's state (singlet in this case) + act of measurement that yields the up spin for Bob's particle.

Even with the best of his available knowledge, Bob is powerless to predict that. Bob says, "No, the particle exists in a linear combination of up and down" So then we're left with the question "How do we know all particles don't have well defined properties regardless of measurement according to some other nonlocal observer, but we're just powerless to predict them?" I can see Dirac shaking his head and asking "Just exactly how many angels can sit on the head of a pin?"

True, we are left only with speculation, not with what Spinkles believes, that QM proves voodoo magic.

 

[Reptillian]

On the condition that Bob will make a measurement. If Bob decides otherwise, Alice is left with a useless knowledge. Recall that it's particle's state (singlet in this case) + act of measurement that yields the up spin for Bob's particle.



True, we are left only with speculation, not with what Spinkles believes, that QM proves voodoo magic.

Spinkles believes (quite justly) that particle states are a sort of "real" property that the particle has independent of measurement. So according to Spinkles, and pretty much every living practitioner of quantum physics, a particle can't be in two different states. It's either in a linear combination of up and down (Bob's viewpoint) or the wavefunction has collapsed and it's in the spin down state. (Alice's viewpoint) The alternative is to say that states are relative, or maybe that there's a kind of meta-quantum mechanical state describing the entire system. (essentially a non-local hidden variable theorem) The problem I see with that is that we're left with fleas upon fleas backs ad infinitum.

 

[PolyHedral]

Spinkles believes (quite justly) that particle states are a sort of "real" property that the particle has independent of measurement. So according to Spinkles, and pretty much every living practitioner of quantum physics, a particle can't be in two different states. It's either in a linear combination of up and down (Bob's viewpoint) or the wavefunction has collapsed and it's in the spin down state. (Alice's viewpoint) The alternative is to say that states are relative, or maybe that there's a kind of meta-quantum mechanical state describing the entire system. (essentially a non-local hidden variable theorem) The problem I see with that is that we're left with fleas upon fleas backs ad infinitum.

:yes:

 

[Reptillian]

Now zaybu, I could understand your position from an ensemble interpretation perspective. I myself was a huge ensemble interpretation fan back in the day. In which case, the formalism of quantum mechanics doesn't apply to individual particles (our Alice and Bob setup with the individual electron pair) but only to systems of particles. (say we have 100 particles with the Alice and Bob setup, all in the state psi as a linear combination of up and down) In that case, the wavefunction remains unchanged after Alice's measurement because the wavefunction doesn't apply to any individual particle...only the collection. Our original setup would be like in statistical mechanics where there is a possibility of me opening my bedroom window and all the air rushes out the window. Sure there is a probability of that happening in statistical mechanics, but it's almost infinitessimal because there are very few "states" with that configuration. The theory was never developed to deal with those sorts of scenarios. The problem with the ensemble interpretation is precisely that which the EPR paradox addresses...in the ensemble interpretation individual particles have well defined states, but quantum mechanics can't predict them...so it's incomplete. Bell seems to have dealt some damage to the ensemble interpretation.

 

[zaybu]

Spinkles believes (quite justly) that particle states are a sort of "real" property that the particle has independent of measurement. So according to Spinkles, and pretty much every living practitioner of quantum physics, a particle can't be in two different states. It's either in a linear combination of up and down (Bob's viewpoint) or the wavefunction has collapsed and it's in the spin down state. (Alice's viewpoint)

However, these two states are not equivalent. It's like comparing apples with oranges. Alice made a measurement. Bob didn't. Now in the old language, which I have avoided, people believed that the wavefunction was real. I don't. That's why I call it a state vector, rather than the wavefunction. And when you make a measurement (wave collapse in the old language), people believe that states could affect each other. I see the state vectors not as real entities but as a mathematical tool to calculate probabilities. That's what QM is to me, a probability theory. In this case, Alice is faced with a probability - 50% up, %50 down. She makes her measurement, and now it's 100% up. She can calculate Bob's measurement only because of the conservation of angular momentum. Otherwise, she would have to guess what Bob will measure. There's no wave to collapse as this wave is a mathematical entity. And so her knowledge cannot affect Bob's measurement. The real surprise would be for Bob to measure anything else but a down spin. It would mean that the law of conservation of angular momentum is false. Now that would hit the headlines.

 

[Mr Spinkles]

QM is a predictive theory. It tells us, ahead of time, what outcomes we should expect from certain measurements. But, apparently, it doesn't, if the thing we're measuring is part of an entangled pair.

No I believe you're mistaken. In QM theory the observable outcomes you measure are eigenvalues with various probabilities. The states of particles are vectors in Hilbert space. Some wiggle room is allowed in the latter as long as everyone agrees on the former, and with entangled pairs it works out.

 

[Mr Spinkles]

It seems to me that the core of what ya'll have been arguing about is this: Alice makes a measurement and instantly knows Andromeda Galaxy dwelling Bob's particle is spin down. Bob uses the best predictive powers he has available (quantum mechanics) to try and determine the outcome of his measurement and the best he can do is say that he has a 50% chance of measuring spin up, and a 50% chance of measuring spin down. Alice knows that Bob's particle has a well defined spin prior to his measurement. Even with the best of his available knowledge, Bob is powerless to predict that. Bob says, "No, the particle exists in a linear combination of up and down"

Yes but if after the experiment Alice and Bob compare notes, Bob will agree that every particle he measured was already in a definite up or down state, not a singlet state. Such is the predictive power of QM that Bob still noticed no contradiction between QM and his experiment given the knowledge he had at the time. But Alice was right, Bob's particles were not in the singlet state in this scenario.

This is elementary QM and there is no need to even consider Bob. Let's just say Alice is the only one doing measurements. Singlet state = (1/sqrt(2)) x [(up/down) + (down/up)]. A measurement of the spin of particle 1 which obtains +hbar/2, according to the rules of QM, collapses the state into (up/down). The same measurement which obtains the other possibility, -hbar/2, collapses the state into (down/up). In either case, both particles are no longer in the singlet state. The state (up/down), for example, is a simple product of single-particle spin eigenstates. When particles are in a state that can be written that way, they are no longer entangled. Each particle is now in its own spin eigenstate, not a singlet state. Namely particle 1 is in the (up) state, particle 2 is in the (down) state. This was not the case in the singlet state. If there is any doubt about this, Alice can easily confirm her suspicions about the states of the particles by doing additional measurements on them.

Again, I repeat emphatically this is what QM says. You can believe it, disbelieve it, love it, hate it. That's what the theory says. Let's not have any more confusion about this: if you want to argue differently, please make sure you distinguish your own personal convictions from this theory.

 

[Mr Spinkles]

If Bob makes no measurement, his particle stays in a singlet state.

So you do not stand by what you shrieked in all caps, then? Yes or no? Because what I wrote (and you shouted agreement with) clearly says that Bob's particle is not in a singlet state after Alice's measurement, whether Bob does anything or not.

 

[zaybu]

Where did I claim "her knowledge" affects anything? If you stand by what you shouted in all caps, then you must agree that Alice's measurement changes the state of the second particle. That is, if she measures, then the second particle will have a state that is different from what its state was before she measured. Do you agree, or not?

What you wrote then was:

After Alice's measurement: ... 50% of the time ... What is the spin of the second particle? Down. ... The other 50% of the time ... What is the spin of the second particle? Up.

Now, you wrote that post with an assumption in your mind that Alice's measurement causes instantaneously the spin on Bob's end to be up. That's where we disagree.

I'm making no such assumption. It's spin up because the law of conservation of angular momentum demands it.

Note: we don't disagree on the result but we do disagree on the interpretation of that result.

My interpretation that it's spin up is founded on the law of conservation of angular momentum: yours, on voodoo magic.

 

[PolyHedral]

No I believe you're mistaken. In QM theory the observable outcomes you measure are eigenvalues with various probabilities. The states of particles are vectors in Hilbert space. Some wiggle room is allowed in the latter as long as everyone agrees on the former, and with entangled pairs it works out.

But it only works out later, and if we say that Bob's particle is "really" an eigenstate (even though Bob doesn't know it yet) that means a state change propogated faster than light!

 

[zaybu]

But it only works out later, and if we say that Bob's particle is "really" an eigenstate (even though Bob doesn't know it yet) that means a state change propogated faster than light!

:yes:

It's a problem for people who believe that the wavefunction is a real wave, when in reality, it is a mathematical function that allows the calculation of probabilities.

 

[PolyHedral]

:yes:

It's a problem for people who believe that the wavefunction is a real wave, when in reality, it is a mathematical function that allows the calculation of probabilities.

That sounds like you're saying there's some "reality" underlying the wavefunction which allows the WF to be relative with POV.

So what is it?

 

[Mr Spinkles]

Now, you wrote that post with an assumption in your mind that Alice's measurement causes instantaneously the spin on Bob's end to be up. That's where we disagree.

I wouldn't worry about assumptions in my mind, I would focus on the plain meaning of words on the page. The words on the page I wrote said Alice's measurement changes the state of both particles, from singlet to something else, without even considering additional measurements by Bob. You angrily shouted agreement. Were you wrong? Yes or no?

You do know that a singlet state is, by definition, an entangled two-particle state, right? A single particle cannot be in a singlet state.

Additionally, if a particle has a 100% chance of being measured spin-down, it cannot be in a "singlet" state. The only quantum state that has a 100% chance of being measured spin-down, is the spin-down eigenstate. This is just math.

So these are two good reasons why, when you say Bob's particle continues to be in a "singlet state" even after Alice's particle is not, you are talking nonsense.

I'm making no such assumption. It's spin up because the law of conservation of angular momentum demands it.

... A law which would be violated if you were correct, and after Alice's measurement Bob's particle remained in a "singlet" state all by itself (somehow) and carried with it no definite spin up or down.

According to what you have said:

Before Alice measures: Total spin = (total spin of singlet state) = 0
After Alice measures: Total spin = (total spin of Alice's particle) + (total spin of Bob's particle) = (spin-up) + (total spin of singlet state) = spin-up + 0 = spin-up

... which violates conservation of (spin) angular momentum. As I said above, a single particle "singlet" state makes no sense to begin with, but if we are generous we could also try to interpret your argument this way:

After Alice measures: Total spin = (total spin of Alice's particle) + (total spin of Bob's particle) = (spin-up) + (spin of one particle in a singlet state) = spin-up + (50% chance of spin-up) = 2 x spin-up

... which again violates conservation of angular momentum 50% of the time.

This is the third reason you are talking nonsense about the singlet state, in addition to the two reasons above. I am not surprised: you said from the beginning the way I wrote down the singlet state was "wrong". You were wrong about that. Little can be expected from such confused beginnings, particularly if the confused party refuses to admit it.

Note: we don't disagree on the result but we do disagree on the interpretation of that result.

I disagree, but let's entertain this as the most generous possible take on your argument. In that case, when you said the way I wrote the singlet state was "wrong", and the way I described quantum nonlocality was "wrong", you were wrong. It's just not your preferred interpretation, according to what you say now.

 

[PolyHedral]

You do know that a singlet state is, by definition, an entangled two-particle state, right? A single particle cannot be in a singlet state.

Well, a single spin-0 wavicle could be, but since the only one of those we know about is the Higg's...
(Pedantry? Where?)

 

[Mr Spinkles]

But it only works out later, and if we say that Bob's particle is "really" an eigenstate (even though Bob doesn't know it yet) that means a state change propogated faster than light!

Right. That is what is meant by "nonlocality" in quantum mechanics.

 

[Mr Spinkles]

Well, a single spin-0 wavicle could be, but since the only one of those we know about is the Higg's...
(Pedantry? Where?)

No I don't think so. A singlet state is, by mathematical definition, a two particle state of fermions with zero total spin. Any state with zero total spin is not a singlet state.