Why hidden variables don't work

[swansont]

This was discussed recently. Here's a really good video that just came out explaining the situation with entanglement and Bell's theorem.

 

 

This walks through the scenario described by a link Strange had posted in another thread — what the correlations need to be for a hidden variable system with the detectors at 0º, 120º and 240º, what they are in QM, and in experiment.

[Mordred]

Good post

[robinpike]

Swansont, thank you for that video, it is very easy to follow.

 

However, there is something that I am misunderstanding about the detectors that perhaps you can clarify for me.

 

To simplify to the basics, if an electron with known spin, for example with 'up' spin because it has just passed through a vertically aligned detector, goes through a second detector which is at some other angle to the vertical, wouldn't the direction of spin of the electron as measured by the second detector always be 'up'?

 

I am not asking about the direction of spin of the electron after it has been through the detector, just what the result of the detector gives?

[michel123456]

Question

From this page http://erc.europa.eu/succes-stories/how-entangle-two-electrons-%E2%80%93-and-do-it-again-and-again

Splitting pairs
One specific example of this is called a ‘Cooper pair’, naturally formed by two electrons in a superconductor – a material with zero, or nearly zero, electrical resistance. “Our objective is to split the electrons of a Cooper pair and get one electron to one lead or electrode and the other member of the pair to the other lead,” says Dr Csonka. To do this he uses a ‘Single-electron transistor’ (SET) at each end of a superconductor to ‘trap’ the entangled electrons. “This acts like a turnstile where people can only enter one by one,” he explains. The SET allows only one electron to be added at a time, leaving the second electron of the pair free in the superconductor. “But superconductors don’t like single electrons,” continues Dr Csonka, “so the free electron moves immediately to the other end, where it enters another SET, so we have one member of the entangled pair at each end.” With leads attached to the SETs, the apparatus becomes an ‘entangler device’ that can produce and control entangled pairs efficiently. “Before we started, the efficiency for such devices was around 3%,” says Dr Csonka, “but we have already achieved 20% efficiency and are now working to optimise the transistors and increase the efficiency further.”

 

What is the "20% efficiency" part meaning?

Does that mean that when the 2 electrons are entangled they show up & down spin and that when they don't show up & down spin the experimentalist decides that they were not entangled? And thus calculates the "efficiency"?

Edited January 14, 2015 by michel123456

[imatfaal]

Question

 

From this page http://erc.europa.eu/succes-stories/how-entangle-two-electrons-%E2%80%93-and-do-it-again-and-again

 

 

What is the "20% efficiency" part meaning?

Does that mean that when the 2 electrons are entangled they show up & down spin and that when they don't show up & down spin the experimentalist decides that they were not entangled? And thus calculates the "efficiency"?

 

From what I can gather it is the percent of tunnelling event in which one of each cooper pair ends up in each single electron transistor rather than both tunnelling to the same one. The efficiency is Amount with non-locality / (amount in sensing gate + amount in tuning gate). So they are getting an amazing efficiency - for you guidance the efficiency of the current photonic entanglement methodlogy - spontaneous parametric down conversion is around 10^-10 entangled photons per non-entangled

[swansont]

Swansont, thank you for that video, it is very easy to follow.

 

However, there is something that I am misunderstanding about the detectors that perhaps you can clarify for me.

 

To simplify to the basics, if an electron with known spin, for example with 'up' spin because it has just passed through a vertically aligned detector, goes through a second detector which is at some other angle to the vertical, wouldn't the direction of spin of the electron as measured by the second detector always be 'up'?

 

I am not asking about the direction of spin of the electron after it has been through the detector, just what the result of the detector gives?

 

 

No. If you measure spin (or polarization) in one direction, or otherwise know its value (as with entanglement), and then use another detector at some other orientation, you can get either value for the spin/polarization. The probability will depend on the angle of the detector (with a cos² dependence)

[michel123456]

 

From what I can gather it is the percent of tunnelling event in which one of each cooper pair ends up in each single electron transistor rather than both tunnelling to the same one. The efficiency is Amount with non-locality / (amount in sensing gate + amount in tuning gate). So they are getting an amazing efficiency - for you guidance the efficiency of the current photonic entanglement methodlogy - spontaneous parametric down conversion is around 10^-10 entangled photons per non-entangled

I don't understand how you recognize an entangled photon from a non-entangled.

[Strange]

I don't understand how you recognize an entangled photon from a non-entangled.

 

In the first case, there is only one photon in the latter there are two. At the detectors or, in the case of this Cooper pair thing, at the two single-electron transistors. (In some cases, you might only detect one of the entangled pair, but in that case, for all practiocal purposes you just had one photon/electron which might as well have been unentangled.)

[michel123456]

 

In the first case, there is only one photon in the latter there are two. At the detectors or, in the case of this Cooper pair thing, at the two single-electron transistors. (In some cases, you might only detect one of the entangled pair, but in that case, for all practiocal purposes you just had one photon/electron which might as well have been unentangled.)

Sorry I read and re-read the experiment description & your explanations and I understand nothing.

 

This is what I understand:

You have 2 detectors at some distance from each other (the SET).

A pair of electrons is emitted. At one SET an electron is received, spin up (for sake of simplicity). Simultaniteously, another electron is received at the other SET, spin down. Where is the efficiency?

Is that when there is no ectron at the 2nd SET?

Or is that when no pair is emitted at all?

Or is that when the 2nd electron has spin up and thus must be discarded because it is unentangled?

Edited January 16, 2015 by michel123456

[Strange]

Sorry I read and re-read the experiment description & your explanations and I understand nothing.

 

This is what I understand:

You have 2 detectors at some distance from each other (the SET).

A pair of electrons is emitted. At one SET an electron is received, spin up (for sake of simplicity). Simultaniteously, another electron is received at the other SET, spin down. Where is the efficiency?

Is that when there is no ectron at the 2nd SET?

Or is that when no pair is emitted at all?

Or is that when the 2nd electron has spin up and thus must be discarded because it is unentangled?

 

It isn't clear how they define efficiency from that news story. My guess is that it is the number of times you get a pair rather than just one of the electrons.

[imatfaal]

 

It isn't clear how they define efficiency from that news story. My guess is that it is the number of times you get a pair rather than just one of the electrons.

 

 

I read in one of the papers published - it is a measure of the amount of times they get a non-localised response divided by the sum of responses from only the signal quantum dot plus only the training quantum dot.

[Strange]

 

 

I read in one of the papers published - it is a measure of the amount of times they get a non-localised response divided by the sum of responses from only the signal quantum dot plus only the training quantum dot.

 

Er, ... that's what I meant ...

[imatfaal]

I don't understand how you recognize an entangled photon from a non-entangled.

 

Additionally - in parametric down conversion in order to conserve energy then two entangled photons must be half the energy of the driving photon.

 

As this all happens in a prism of sorts, and amount of bending is dependant on wavelength then I would think they fly off at a different angle than unaffected driver photons would

Er, ... that's what I meant ...

 

My knowledge is now completely exhausted as well I read the paper three times and couldn't work out how these damn measurements were taking place. The two sides (as explained above) are labelled signal and training by the way - the efficiency was G_nonlocal / (G_signal + G_training). The only thing I could see was that G_s was given as I_s/V_s which would be conductance - which makes sense but I wasn't confident enough to put it up here till now.

 

ps - I just checked to see if I mean conductance or conductivity or if they are the same - and I did mean conductance and the symbol is capital G.

 

So it seems that the presence of one or two electrons in the trap changes the conductance to a measurable extent. For your guidance this is all done at a fraction of one Kelvin - so those sorts of levels of accuracy are possibly attainable

[michel123456]

 

Additionally - in parametric down conversion in order to conserve energy then two entangled photons must be half the energy of the driving photon.

 

As this all happens in a prism of sorts, and amount of bending is dependant on wavelength then I would think they fly off at a different angle than unaffected driver photons would

 

My knowledge is now completely exhausted as well I read the paper three times and couldn't work out how these damn measurements were taking place. The two sides (as explained above) are labelled signal and training by the way - the efficiency was G_nonlocal / (G_signal + G_training). The only thing I could see was that G_s was given as I_s/V_s which would be conductance - which makes sense but I wasn't confident enough to put it up here till now.

 

ps - I just checked to see if I mean conductance or conductivity or if they are the same - and I did mean conductance and the symbol is capital G.

 

So it seems that the presence of one or two electrons in the trap changes the conductance to a measurable extent. For your guidance this is all done at a fraction of one Kelvin - so those sorts of levels of accuracy are possibly attainable

I guess it is a simple "effect" of the efficiency thing, if one can say so. With 100% of efficiency the factor should be unity. But maybe my guess is wrong. It goes far above my head.

I wanted to be sure that the entangled electrons are detected as such and after that the spin is measured, and not the other way round.

[imatfaal]

I wanted to be sure that the entangled electrons are detected as such and after that the spin is measured, and not the other way round.

 

This was what I was desperately trying to ascertain - and not succeeding. The reason I think it is the former is that the latter is pretty useless; ie the ideal situation is that a percentage of the time you can look at your readings and say "we have a pair electrons which are at present entangled spread over our two transistors"

[Sensei]

To simplify to the basics, if an electron with known spin, for example with 'up' spin because it has just passed through a vertically aligned detector, goes through a second detector which is at some other angle to the vertical, wouldn't the direction of spin of the electron as measured by the second detector always be 'up'?

 

Checking this for photons is easier. Don't require to have expensive equipment.

 

For photons, there is needed array of polarization filters with the same alignment. The more filters in chain, the more "up" photons, but less quantity (the rest is reflected).

 

 

I showed how polarization filters works with polarized photons in this thread

http://www.scienceforums.net/topic/80366-particle-location/page-3#entry783255

So you can see it on your own eyes (photos).

Edited January 16, 2015 by Sensei

[david345]

Swansont, thank you for that video, it is very easy to follow.

 

However, there is something that I am misunderstanding about the detectors that perhaps you can clarify for me.

 

To simplify to the basics, if an electron with known spin, for example with 'up' spin because it has just passed through a vertically aligned detector, goes through a second detector which is at some other angle to the vertical, wouldn't the direction of spin of the electron as measured by the second detector always be 'up'?

 

I am not asking about the direction of spin of the electron after it has been through the detector, just what the result of the detector gives?

This can be explained with photons going through two polarizers. Polarizer 1 can give the photons a definite polarization. All that pass through polarizer 1 will have the same polarization. In this example we will say they are all vertically polarized. If there is a 30 degree angle between polarizer 1 and 2 then 75% of the photons will pass through polarizer 2. 75% is not a definite yes or no. If 4 photons pass through polarizer 1 then 3 will pass through polarizer 2 and one will be blocked. All four had the same polarization when they passed through the first polarizer. The angle between the two polarizers was the same for all four photons. Why did three pass through while one was blocked? The law used to determine the probability is Malus law. It was originally written to determine the intensity of the light waves. Quantum mechanics now uses the intensity of light waves to determine the probability of detecting the photon. The only two possibilities where the results are definite are when both polarizers are vertically polarized and when one is vertically polarized and the other is horizontally polarized. In the first case the probability of pass through is 100% in the second case the probability is 0%. In all other cases it is uncertain whether the photon will pass through the second polarizer. Edited January 20, 2015 by david345

[robinpike]

This can be explained with photons going through two polarizers. Polarizer 1 can give the photons a definite polarization. All that pass through polarizer 1 will have the same polarization. In this example we will say they are all vertically polarized. If there is a 30 degree angle between polarizer 1 and 2 then 75% of the photons will pass through polarizer 2. 75% is not a definite yes or no. If 4 photons pass through polarizer 1 then 3 will pass through polarizer 2 and one will be blocked. All four had the same polarization when they passed through the first polarizer. The angle between the two polarizers was the same for all four photons. Why did three pass through while one was blocked? The law used to determine the probability is Malus law. It was originally written to determine the intensity of the light waves. Quantum mechanics now uses the intensity of light waves to determine the probability of detecting the photon. The only two possibilities where the results are definite are when both polarizers are vertically polarized and when one is vertically polarized and the other is horizontally polarized. In the first case the probability of pass through is 100% in the second case the probability is 0%. In all other cases it is uncertain whether the photon will pass through the second polarizer.

 

Does anybody know the results for when a successively thinner and thinner polarizing filter is used for the second filter?

 

Using the above example, i.e. passing the light through one (thick) polarizing filter to produce light that is 100% polarized, and then passing that 100% polarized light through a very thin polarizing filter at a 30 degree angle. Is there a thickness (thinness) at which more than 75% of the light makes it through the second filter? And if so, is there variation in the angle of polarization of the light?

[swansont]

 

Does anybody know the results for when a successively thinner and thinner polarizing filter is used for the second filter?

 

Using the above example, i.e. passing the light through one (thick) polarizing filter to produce light that is 100% polarized, and then passing that 100% polarized light through a very thin polarizing filter at a 30 degree angle. Is there a thickness (thinness) at which more than 75% of the light makes it through the second filter? And if so, is there variation in the angle of polarization of the light?

 

Why would the thickness matter?

 

There would be no thickness for which a different amount made it through. If a different amount made it, then it's not polarizing the light. These polarizers work in situations where entanglement is not an issue, and it's only the polarizing ability that's tested (i.e. purely classical behavior)

[robinpike]

 

Why would the thickness matter?

 

There would be no thickness for which a different amount made it through. If a different amount made it, then it's not polarizing the light. These polarizers work in situations where entanglement is not an issue, and it's only the polarizing ability that's tested (i.e. purely classical behavior)

 

I was wondering if in the case of a very thin polarizing filter, whether there was a thickness that allowed more light through - presumably yes, and more interestingly, if when this happened, if the angle of polarization started to 'spread'?

[swansont]

 

I was wondering if in the case of a very thin polarizing filter, whether there was a thickness that allowed more light through - presumably yes, and more interestingly, if when this happened, if the angle of polarization started to 'spread'?

 

Yes, I suppose you could make a crappy polarizer. It probably wouldn't spread the polarization angle, as such, it would reduce the extinction of the orthogonal component. But crappy equipment isn't related to the issue of hidden variables.