Einstein and "spooky actions"

[PolyHedral]

Following post #336, I think I have found a solution to the Mach-Zehnder interferometer. I will work out the detail and post it on my blog. Then I'll let you know when that is done. What's surprising is that the solution is very simple if you use vectors.

It is rather. I've drawn in the relative phases of each of the beams-in-transit. As you can see, the beam towards A is cancelled out entirely.

I'm not sure. I'm betting it looks weird to say the least....I'm trying to make it easier to imagine by thinking of half the star being blocked out with a massive sunspot to sort of simulate night on the surface of the sphere, but it isn't helping...the best I can come up with is a kind of murky twilight. IDK

That's why I'm skeptical of the claims that Alice can send a shadow to Bob from outside his lightcone. Sure, Alice can send a shadow so that she thinks Bob can see a FTL shadow, but the actual mechanics of Relativity might say otherwise.

 

[idav]

Read the link. All of it. Carefully.

The realist interpretation being that Bobs particle was already at said state which couldnt be known until measurement. No spooky action at a distance.

 

[PolyHedral]

The realist interpretation being that Bobs particle was already at said state which couldnt be known until measurement. No spooky action at a distance.

Try squaring the realist interpretation with the diagram above. Which path did the photon "really" take?

 

[idav]

Try squaring the realist interpretation with the diagram above. Which path did the photon "really" take?

Not sure of an explanation for the photon interfering with itself but that would be the extent of the photon "knowing" a path to take.

 

[Mr Spinkles]

The realist interpretation being that Bobs particle was already at said state which couldnt be known until measurement. No spooky action at a distance.

Without nonlocality, that interpretation implies Bell's theorem. But Bell's theorem is in conflict with experiment. Ergo, to maintain that interpretation, you must accept nonlocality (or abandon the realism).

Again, I encourage you to read the link I posted and educate yourself on this topic if it interests you, before jumping to conclusions.

 

[idav]

Without nonlocality, that interpretation implies Bell's theorem. But Bell's theorem is in conflict with experiment. Ergo, to maintain that interpretation, you must accept nonlocality (or abandon the realism).

Again, I encourage you to read the link I posted and educate yourself on this topic if it interests you, before jumping to conclusions.

You only have to accept non locality when interpreting QM as spooky action. I'm not seeing a violation. Those experiments are certainly interesting and tricky but it isn't magic.

 

[zaybu]

It is rather. I've drawn in the relative phases of each of the beams-in-transit. As you can see, the beam towards A is cancelled out entirely.

That's why I'm skeptical of the claims that Alice can send a shadow to Bob from outside his lightcone. Sure, Alice can send a shadow so that she thinks Bob can see a FTL shadow, but the actual mechanics of Relativity might say otherwise.

I will be using that diagram on my blog. I know you took it from wikipedia. I'm just letting you know so you don't think I'm copying you.

 

[zaybu]

Try squaring the realist interpretation with the diagram above. Which path did the photon "really" take?

You've hit it on the nail. Language fails us when we are trying to describe a QM system. However, I think the math can compensate if it is done properly without going into the deep end à la Spinkles. I'm still working on putting the solution on my blog. I'm terrible at writing HTML.

BTW, I will show in my proof that whether we have particles or waves, all the photons must go to detector B (100%), and 0% at A.

 

[Mr Spinkles]

You only have to accept non locality when interpreting QM as spooky action. I'm not seeing a violation. Those experiments are certainly interesting and tricky but it isn't magic.

Since you can't be convinced, I rest my case. I can assure you, however, that I am not claiming physics is magic.

 

[Mr Spinkles]

Fudge! Kludge! Inelegance! Relativistic classical mechanics says nothing propogates faster than light ...

PolyHedral, I'm sorry but you are wrong. It will be impossible for you to see why you are wrong if you won't learn. And it will be impossible for you to learn if you think you have mastered everything and refuse to be taught.

 

[Mr Spinkles]

The concept of a singlet is part of a theory, it's not a fact.

Yes, it's part of the theory of quantum mechanics, or what you called "quantum logic, that is, use vectors in a Hilbert space". That was before you realized you needed to change your argument to avoid nonlocality. Keep wriggling ...

This is worth repeating:

zaybu, I don't usually say this, but your posts demonstrate that you really don't know what you are talking about. You keep confusing different concepts and you don't even know what an entangled quantum state is. In other words, your understanding of the uncontroversial stuff is not even to the point where you can make a sensible argument (right or wrong) about the controversial stuff.

There's no shame in that as far as it goes. But what is really a conversation-stopper is that you refuse to be educated.

Since continuing this discussion is pointless, I rest my case. We'll just have to agree to disagree. It is a fact, however, and not an opinion, that I have represented in this thread what quantum mechanics says, and the orthodox interpretation of quantum mechanics accepted by most physicists. FWIW.

 

[zaybu]

Yes, it's part of the theory of quantum mechanics, or what you called "quantum logic, that is, use vectors in a Hilbert space". That was before you realized you needed to change your argument to avoid nonlocality. Keep wriggling ...

The only thing I did is to change the initial scenario only very slightly to show that your arguments don't hold.

Since continuing this discussion is pointless, I rest my case. We'll just have to agree to disagree. It is a fact, however, and not an opinion, that I have represented in this thread what quantum mechanics says, and the orthodox interpretation of quantum mechanics accepted by most physicists. FWIW.

I disagree that your opinion prevails among physicists, perhaps among the old folks who have learned their QM back some 30 years ago. But the interesting thing is that eventually they will all be dead. For the young generation and the next ones coming up, that old language of the dinosaur age will disappear, and the sooner the better.

 

[zaybu]

It is rather. I've drawn in the relative phases of each of the beams-in-transit. As you can see, the beam towards A is cancelled out entirely.

As promised, I have posted my solution here:

strings of ideas: Mach–Zehnder interferometer Particle or Wave?

 

[PolyHedral]

PolyHedral, I'm sorry but you are wrong. It will be impossible for you to see why you are wrong if you won't learn. And it will be impossible for you to learn if you think you have mastered everything and refuse to be taught.

The only forces which exist in Relativistic classic mechanics are elctromagnetism and gravity, both of which are limited to c or less. The only possible contention is what shadows look like, but like I said, please do the math for that one.

 

[PolyHedral]

As promised, I have posted my solution here:

strings of ideas: Mach–Zehnder interferometer Particle or Wave?

What on earth does -> represent in that algebra? It's not "implies," and it's not "equality," so what is it?

Also, your result is wrong: you imply the the outgoing photon should be the same phase as the incoming one, when it is actually phase-inverted.

 

[zaybu]

What on earth does -> represent in that algebra? It's not "implies," and it's not "equality," so what is it?

It's normal use for an -> when the state vector is not normalized

You can always write

(1)| I > → |E> + |S> ,
as
(2)| I > = a|E> + b|S> , where a and b are constant which will be needed if you are going to do the calculation and the state vector has to be normalized. This is standard practice in QM. If the argument doesn't need normalization than (1) is good enough

Also, you're result is wrong: you imply the the outgoing photon should be the same phase as the incoming one, when it is actually phase-inverted.

A phase factor brings in an additional correctional factor of e^(ik), where k is a constant, the phase factor. When you normalize, you will get the complex conjugate e^(-ik) in the dual vector, which will yield 1. In this case, we are trying to show that all the photons (100%) will go to detector B, the phase factor plays no role.

 

[PolyHedral]

It's normal use for an -> when the state vector is not normalized

You can always write

(1)| I > → |E> + |S> ,
as
(2)| I > = a|E> + b|S> , where a and b are constant which will be needed if you are going to do the calculation and the state vector has to be normalized. This is standard practice in QM. If the argument doesn't need normalization than (1) is good enough

I'm still not following why the southern vector disappears. Your analysis doesn't even seem to consider it.

A phase factor brings in an additional correctional factor of e^(ik), where k is a constant, the phase factor. When you normalize, you will get the complex conjugate e^(-ik) in the dual vector, which will yield 1. In this case, we are trying to show that all the photons (100%) will go to detector B, the phase factor plays no role.

I'm fairly sure normalization only involves changes to length, not phase, considering the probability depends on the mod-squared, which a phase change has no effect on.

 

[idav]

Since you can't be convinced, I rest my case. I can assure you, however, that I am not claiming physics is magic.

The debate is on the interpretation of the experiment, assuming we are looking at the same experiment. Non-locality at faster than speeds of light is like saying it's magic. Admittedly the superposition stuff points to a form of non-locality but nothing that goes beyond measurable parameters, in other words we see casual influences just as you would running the test with water waves. At the end of your link even says there are other interpretations and non-locality hinges on that interpretation.

I've also read the article on the Alice and Bob experiment. I came to the same conclusion and there answer was needing better detectors. After all we have been able to actually observe simultaneous states without collapsing the wavefunction. Only time will tell who is right but I leave that to their respective fields.

 

[zaybu]

I'm still not following why the southern vector disappears. Your analysis doesn't even seem to consider it.

How does it disappear? Can you point out the equations that seem to trouble you?

I'm fairly sure normalization only involves changes to length, not phase, considering the probability depends on the mod-squared, which a phase change has no effect on.

Correct.

 

[PolyHedral]

How does it disappear? Can you point out the equations that seem to trouble you?

In your analysis it doesn't. You just assume a priori that there will be no output towards A, which is begging the question. (Specifically, you seem to say the second beam splitter will not actually split the beam travelling right, but instead always let it pass straight through.)

Correct.

You're not making any sense. Does the output towards B match the phase of the input, or not?