Double-sided double slit experiment

[Danijel Gorupec]

A double slit experiment is performed. In the middle of the experiment table, there is a particle source. The source emits entangled particle pairs, one by one. The two particles in each pair are, of course, emitted in opposite directions.

 

There are two identical double-slit experiment setups - the left one (A) and the right one (B). Each setup consists of one screen with two slits and one display screen. The fringe pattern is visible on both displays screens, A and B.

 

Is this true? Or is there any reason why there would be no fringes in the case of entangled particles?

 

If the experimenter at the side A (Alice) now makes a measurement effort to determine through which slot each particle actually traveled, the fringe pattern disappears. Because particles are entangled, fringes will disappear simultaneously on both, the A and the B side (even if the experimenter at the B side (Bob) made no similar measurements).

 

Do I understand this correctly? Alice can make immediate effects on (possibly very distant) Bob’s display.

[imatfaal]

the (delayed) quantum eraser experiment uses the fact that entanglement can provide path information seemingly without any interaction with the signal photon (the polarisation of the paired photon tells you which path the signal took) - this path information (even though obtain only through the entangled pair) is enough to destroy the interference. If in your set up the measurement of particle A was accurate enough to determine which path the paired particle took then there will be no interference.

 

as the (delayed) quantum eraser experiment is an actual reality - rather than a gedankan - you might want to have a look at it. The delayed bit in brackets describes that even if the paired photon is measured way after the signal has been measured (ie much much longer path) then the interference is still destroyed.

[swansont]

It's going to depend on what the entangled states are. If it's e.g. spin, this is independent of any interference pattern, so "which path" information is of no effect.

[Enthalpy]

I bet detection at one slit on side A does not destroy the interferences at side B.

 

One general idea is that acting one one particle does not influence the other - and much less, influence it at distance, which would transmit information faster than light. Entangled states only introduce correlations between the behaviour of particles, not the behaviour of each one. Please take with care.

 

-----

 

If the detector is after the slit at A, then the photon is still undetermined when passing the slits of A, and so is the entangled photon when passing the slits of B. I suppose a detector behind one slit at B would be correlated with the detector behind one slit at A, but not a detector at the fringes of B.

 

It could be something like: over enough photon pairs, you see fringes at B whatever you do at A. At B, you see photons only in bright areas of the fringe, and these areas are accessible by photons through either slit of B. When you detect a photon at B's fringe and you observed at A that the entangled photon passed through one slit, it only means that the photon passed the screen at B through the corresponding slit, fine.

 

-----

 

One other possibility, maybe better:

- Fringes at B whatever you do at A, if you observe B without condition on the measure A

- If you observe B only when seeing a photon at Ax, no fringes

- If you observe B only when seeing a photon at Ay, no fringes

- But if you observe B over all results of Ax and Ay, then the two uniform lightings at B interfere and give fringes.

 

This second possibiilty is not a transfer of information neither, because the detector at A does not influence the behaviour at B. When the photon pair "decides" to be seen at Ax - independently of the presence of the detector - then it behaves differently at B as well.

[Danijel Gorupec]

It's going to depend on what the entangled states are. If it's e.g. spin, this is independent of any interference pattern, so "which path" information is of no effect.

 

I think I understand. You tell me that entangelment does not need to contain all properties - only some properties can be in the entangled state... Does it mean that even non-identical particles (not even particle-antiparticle pairs) can have some properties entangled? I am a "Quantum Newbie".

the (delayed) quantum eraser experiment uses the fact that entanglement can provide path information seemingly without any interaction with the signal photon (the polarisation of the paired photon tells you which path the signal took) - this path information (even though obtain only through the entangled pair) is enough to destroy the interference. If in your set up the measurement of particle A was accurate enough to determine which path the paired particle took then there will be no interference.

 

as the (delayed) quantum eraser experiment is an actual reality - rather than a gedankan - you might want to have a look at it. The delayed bit in brackets describes that even if the paired photon is measured way after the signal has been measured (ie much much longer path) then the interference is still destroyed.

 

I read about the Quantum Eraser experiment. The setup is very much like the one I am talking about... I am not sure I understand the 'delayed' part. Does it mean that the "faster-than-light" action is not strange enough; that there is also "time order is not important" action?

[swansont]

I think I understand. You tell me that entangelment does not need to contain all properties - only some properties can be in the entangled state... Does it mean that even non-identical particles (not even particle-antiparticle pairs) can have some properties entangled? I am a "Quantum Newbie".

AFAIK yes. A nuclear decay that emits one particle, e.g. a photon-less beta decay should have entanglement of spin and possibly momentum. The systems that are used in experiments are simpler, so you can prepare (and thus know) the original system's state and get the entanglement you want to investigate, and control e.g. the direction of the particles.

[imatfaal]

It's going to depend on what the entangled states are. If it's e.g. spin, this is independent of any interference pattern, so "which path" information is of no effect.

 

If you recreated the quantum eraser with some form of entanglement of particles and splitting of the paths of the the two paired particles such that one could tell that spin up went through one slit and spin down went through the other (as an exact analogue of the polarity in quantum eraser) - you could then measure the entangled particle (ie the idler) and thus know the spin state (of both particles), and thus the path information of the signal.

 

Would you then still get an interference pattern? I would say no - as which path information destroys the pattern. It is not the polarisation of the photons in the delayed quantum eraser that causes the interference - it is the standard explanation of young's.

[swansont]

 

If you recreated the quantum eraser with some form of entanglement of particles and splitting of the paths of the the two paired particles such that one could tell that spin up went through one slit and spin down went through the other (as an exact analogue of the polarity in quantum eraser) - you could then measure the entangled particle (ie the idler) and thus know the spin state (of both particles), and thus the path information of the signal.

 

Would you then still get an interference pattern? I would say no - as which path information destroys the pattern. It is not the polarisation of the photons in the delayed quantum eraser that causes the interference - it is the standard explanation of young's.

 

I agree. If the path and spin were correlated, I would expect that detecting one detects the other and destroys the interference.

[Danijel Gorupec]

Hmm.... I feel that words like "one could tell" and "detecting" are somewhat overused. In the Quantum Eraser experiment one should say: "filters are placed in front of slits so that no single photon can travel through both slits at once, and therefore the interference pattern disappers". No need to use mistical words that involve consciousness.

 

 

Anyway, the second part of my OP question was - is this the faster-than-light information propagation (Alice can send Morse code to Bob by destroying and establishing the interferrence pattern on Bob's display)? Can anyone comment? Thanks.

[swansont]

No, it can't be used for FTL communication.

[Enthalpy]

And I believe that the presence of a detector at A does not change the observation at B.

 

The observation at B may depend on what gets measured at A - and this result is decided by the particles, not by the observer nor the detector. So no information can be transmitted.

 

And if summing on all possible measures at A, one would get the standard observation at B, that is the fringes.

[Danijel Gorupec]

No, it can't be used for FTL communication.

 

I would like to know more because I don't see it. Obvously there is something I don't understand.

And I believe that the presence of a detector at A does not change the observation at B.

 

 

I only wish you are right . And I wish we are all victims of "Quantum Parading" [when physicist present world in a more bizare light than they should - just to earn themself higher chance to pick up a girl (or funds). You see, the world is all about probability ].

[Delta1212]

 

I would like to know more because I don't see it. Obvously there is something I don't understand.

 

 

I only wish you are right . And I wish we are all victims of "Quantum Parading" [when physicist present world in a more bizare light than they should - just to earn themself higher chance to pick up a girl (or funds). You see, the world is all about probability ].

From what I understand, and bearing in mind that quantum erasure is not something I'm well versed in, the measurement at B will look the same regardless of what happens at A (or, at least, the measurement at B when a filter is applied to A is indistinguishable from a measurement made when no filter was applied at A) until you have the results from A where a filter was applied, at which point a comparison will allow you to detect the difference in the results at B.

 

The results are created independently at a distance that would preclude sub-light communication between the detectors to explain the correlation, but the correlation is only observable when you compare the results with one another, and those must be communicated at sub-light speeds.

[swansont]

I would like to know more because I don't see it. Obvously there is something I don't understand.

 

You haven't described an experiment in much detail, but it takes time for the particles to travel to the screen, and an interference pattern is comprised of many particles being detected, so being able to see that also takes time. It doesn't just blink on and off.

 

As Delta 1212 has mentioned, if you had an experiment where there actually was entanglement, you have to communicate the detection protocol, which is limited to c.

[Danijel Gorupec]

 

You haven't described an experiment in much detail, but it takes time for the particles to travel to the screen, and an interference pattern is comprised of many particles being detected, so being able to see that also takes time. It doesn't just blink on and off.

 

As Delta 1212 has mentioned, if you had an experiment where there actually was entanglement, you have to communicate the detection protocol, which is limited to c.

 

I dont' believe in FTL communication. But when I created this 'experiment' I was not able to find any reason why this would not be the FTL communcation. This puzzled me.... In my experiment the particle source is in the exact middle between two opposite displays. Therefore, I expect, both entangled particles will reach displays at the same time.

 

If now I imagine these two displays separated by a light year, there should be enough time for Alice to construct/destruct the pattern at Bob's side. I am searching for a reason why this should not be possible. Maybe, as the distance between displays increases, it becomes fundamentaly harder and harder to somehow 'aim' particles at slits - or there is any other reason why the time needed to confirm the pattern would also increase.

From what I understand, and bearing in mind that quantum erasure is not something I'm well versed in, the measurement at B will look the same regardless of what happens at A (or, at least, the measurement at B when a filter is applied to A is indistinguishable from a measurement made when no filter was applied at A) until you have the results from A where a filter was applied, at which point a comparison will allow you to detect the difference in the results at B.

 

The results are created independently at a distance that would preclude sub-light communication between the detectors to explain the correlation, but the correlation is only observable when you compare the results with one another, and those must be communicated at sub-light speeds.

 

In your first paragraph, do you reference the actual Quantum Eraser experiment, or the similar experiment I am proposing in OP?

[swansont]

I dont' believe in FTL communication. But when I created this 'experiment' I was not able to find any reason why this would not be the FTL communcation. This puzzled me.... In my experiment the particle source is in the exact middle between two opposite displays. Therefore, I expect, both entangled particles will reach displays at the same time.

 

If now I imagine these two displays separated by a light year, there should be enough time for Alice to construct/destruct the pattern at Bob's side. I am searching for a reason why this should not be possible. Maybe, as the distance between displays increases, it becomes fundamentaly harder and harder to somehow 'aim' particles at slits - or there is any other reason why the time needed to confirm the pattern would also increase.

 

 

You have not described an experiment in sufficient detail such that one can explain why FTL communication is impossible. You need to know how the which path information is obtained. All we know is that the particles are entangled and you are doing a double-slit experiment. That's not enough.

[Enthalpy]

 

[...] I expect both entangled particles will reach displays at the same time.

 

If now I imagine these two displays separated by a light year, there should be enough time for Alice to construct/destruct the pattern at Bob's side. I am searching for a reason why this should not be possible.

It is because the detector at A and the experimenter there do not influence the other particle at B, I maintain.

 

The particle pair decides its behaviour, not the detector. And you'll keep the same fringes at B - only the observation correlated wth the detection at A changes.

 

[ That is, fuzzy at B when the other photon is at Ax. Fuzzy at B when the other photon is at Ay. Fringes at B, resulting from the phased superposition of both fuzzy images at B, when the other photon is indifferently at Ax or Ay. ]

[ Meanwhile I doubt that last one ]

 

---------- There is a difficulty in the proposed setup ----------

 

The initial description said "Two entangled photons" and I challenge that.

 

Their side momentum, or angular direction if you prefer, is linked with a third particle which is the emitting atom or crystal. Through its recoil, the atom absorbs momentum and allows the photons to have directions not opposed.

 

Depending on the atom's recoil, even if the observer knows that one photons has passed through slit Ax, the other photon can still pass through Bx or By. A random recoil direction and intensity would fully blurr the experiment.

 

There is more. This recoil is observable, and the observation can be transmitted on time to both observers. Such an observation might even destroy completely the link between both photons; as an other possibility, observing the atom only reduces the entanglement to the photons, with a shift in their relation that depends on the atom's measured momentum. I have no opinion about that.

 

This situation differs from experiments that use photon polarization. If an electron falls 3s -> 2p -> 1s as an idealized example, the atom is isotropic in 3s and 1s so both photons are correlated only among themselves.

 

I still don't know if this is an experimental or theoretical difficulty. If you find something that emits two photons in opposite directions necessarily, fine. Even pairs of 511keV gammas depend on the electrons' momentum.

Edited July 4, 2013 by Enthalpy

[Danijel Gorupec]

It is because the detector at A and the experimenter there do not influence the other particle at B, I maintain.

 

The particle pair decides its behaviour, not the detector. And you'll keep the same fringes at B - only the observation correlated wth the detection at A changes.

 

[ That is, fuzzy at B when the other photon is at Ax. Fuzzy at B when the other photon is at Ay. Fringes at B, resulting from the phased superposition of both fuzzy images at B, when the other photon is indifferently at Ax or Ay. ]

 

Enthalpy, I think you might be correct – some fuzziness might happen at the B side and therefore Alice cannot influence Bob’s display if two displays are sufficiently distant. I also understand your point about ‘entangled particles not perfectly opposed due to emitter recoil’. But are you sure this is true, or you are just guessing? I thought that momentum-entangled particles are perfectly opposed. In addition, is this recoil behavior fundamentally unavoidable and always large enough? Anyone knows?

 

 

You have not described an experiment in sufficient detail such that one can explain why FTL communication is impossible. You need to know how the which path information is obtained. All we know is that the particles are entangled and you are doing a double-slit experiment. That's not enough.

 

I actually don't think the method makes difference. The setup is central-symetrical regarding to the particle source. If Alice knows through what slit the particle passed at her side, she also knows through what slit the entangled particle passed on Bob's side.... However, as Enthalpy said, this is questionable. Also, I can add, the particle source is hardly a point... Only I don't know if these reasons are fundamental or practical.

 

BTW, I don't know about how in practice they obtain the 'which path' information in real double slit experiments. In most double slit experiment descriptions, nobody cares enough to explain the dirty part.

[Enthalpy]

 

[...] your point about ‘entangled particles not perfectly opposed due to emitter recoil’. But are you sure this is true, or you are just guessing? I thought that momentum-entangled particles are perfectly opposed. In addition, is this recoil behavior fundamentally unavoidable and always large enough? Anyone knows?

It's much guessing, and I should have made it clearer, my fault... What I'm sure is that recoil would permit imperfectly opposed paths and give usable early information about the angle difference to 180°. What does literature say about "momentum-entangled particles"?

 

There is more - and again, much guessing here: the position of the fringes needs a well-positioned light source, which implies that the light momentum (=direction) is imprecise. With one single photon, that would be just a Heisenberg relation. Could it be that the momentum entanglement between both photons suffers the same uncertainty? That is, the more precisely opposed their direction, then the less precise the position of the emission, or the positionS of the emissionS?

 

I suppose it's not a matter of position-momentum undetermination of the emitter, because an atom is heavy so its own momentum would have little effect on the photon pair's one. It's more that one atom could emit two linked photons in directions little correlated, and absorb easily the resulting momentum.

 

That looks better: take an atom as a source. A 3s -> 2p -> 1s through the peanut-shaped 2p would link the second emission directions but only weakly, like a short antenna, that is like sin(angle) between the 2p peanut direction and the photon direction, while the 2p peanut direction defines the first emission direction.

 

To get a more precise direction for the second photon, it takes an emitter wider than a single atom, and then the position of the emission must be unknown enough. Or the positionS of the emissionS - but one uncertain position common to both looks enough.

 

Same uncertainty dx*dp here as with one single photon. Maybe you're on the path to applying uncertainty relations to the entanglement precision of several particles.

Here's a paper about the relation between the precision of momentum entanglement and position precision:

http://arxiv.org/ftp/arxiv/papers/0905/0905.4830.pdf

 

First, they use a crystal to produce the photons, so the emission position is imprecise.

 

Page 2: "two-photon position separation Δ(x1-x2) and the two-photon momentum sum Δ(px,1+px,2)"

people are aware of uncertainties in the correlation.

 

Page 6: "Assuming a perfect spatial correlation with x1 = x2, ∆ , , (is limited by the

emission area in the nonlinear crystal with radius w1,2"

Edited July 4, 2013 by Enthalpy

[swansont]

I actually don't think the method makes difference. The setup is central-symetrical regarding to the particle source. If Alice knows through what slit the particle passed at her side, she also knows through what slit the entangled particle passed on Bob's side.... However, as Enthalpy said, this is questionable. Also, I can add, the particle source is hardly a point... Only I don't know if these reasons are fundamental or practical.

 

In order for me to have any chance of explaining it, yes, knowing the method makes a difference. The experiment you describe is not the one with which I am familiar.

 

BTW, I don't know about how in practice they obtain the 'which path' information in real double slit experiments. In most double slit experiment descriptions, nobody cares enough to explain the dirty part.

http://grad.physics.sunysb.edu/~amarch/

[Danijel Gorupec]

Thanks guys. After reading all your comments, I must conclude that the space-momentum entanglement is not what I was originally thinking it is.

For some reasons (that are fundamental but unknown to me) there exists some uncertainty between entanglement particle trajectories – it seems that the angle between their paths cannot be guaranteed. I see no other explanation.

Therefore, in my original post, Alice will have difficulties to influence Bob’s display. Greater the distance between displays, larger the difficulties.

Obviously, now I have the new problem – understanding what exactly the space-momentum entanglement is.

[Iwonderaboutthings]

It's going to depend on what the entangled states are. If it's e.g. spin, this is independent of any interference pattern, so "which path" information is of no effect.

Why is the spin independent???

Doesn't QM deal with waves meaning circles??

 

 

Could this be because spin as a "completion" meaning " a completed circle as a " translation" " larger" is always still = 0 ?

 

For example:

x^2 = -1

 

 

In other words no matter how large a circle becomes with translation it shall always have the same perimeter, right?

 

I assume this is where radians comes in for the " diameter " as a linear x method of distance, but that is just a concept for now with pi ratio coming along to the picture.

 

The Pythagorean Theorem comes to mind as well which deals with squared units " again " in the outer perimeters..

 

 

Pythagorean Theorem

http://en.wikipedia.org/wiki/Pythagorean_theorem

Gravity, The Speed of Light and Pi ratio, h and the h bar are all used with this independent spin???

 

 

Hymmmm

Edited August 24, 2013 by Iwonderaboutthings

[swansont]

Why is the spin independent???

In the scenario, the which-path information did not depend on the spin. Thus spin entanglement and measurement would not yield which path information and affect the interference.

 

Doesn't QM deal with waves meaning circles??

Waves are not the same thing as circles. They have things in common, but they are not identical.

 

Could this be because spin as a "completion" meaning " a completed circle as a " translation" " larger" is always still = 0 ?

 

For example:

x^2 = -1

 

 

In other words no matter how large a circle becomes with translation it shall always have the same perimeter, right?

 

I assume this is where radians comes in for the " diameter " as a linear x method of distance, but that is just a concept for now with pi ratio coming along to the picture.

 

The Pythagorean Theorem comes to mind as well which deals with squared units " again " in the outer perimeters..

 

 

Pythagorean Theorem

http://en.wikipedia.org/wiki/Pythagorean_theorem

 

Gravity, The Speed of Light and Pi ratio, h and the h bar are all used with this independent spin???

 

 

Hymmmm:unsure:

Spin is independent of gravity. There is a limited relation to c, at least in the fact that spin is not physical spinning, because c is a speed limit for physical motion.

[Iwonderaboutthings]

In the scenario, the which-path information did not depend on the spin. Thus spin entanglement and measurement would not yield which path information and affect the interference.

 

 

Waves are not the same thing as circles. They have things in common, but they are not identical.

 

 

Spin is independent of gravity. There is a limited relation to c, at least in the fact that spin is not physical spinning, because c is a speed limit for physical motion.

Very well said and thanks