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Simple ftl setup using Stern-Gerlach

  1. Oct 4, 2004 #1
    Hi,

    I've been reading David Albert's book "Quantum Mechanics and Experience," and it got me thinking about how one could use Stern-Gerlach sorts of measurements on entangled electron pairs (or electron-positron pairs) to transmit a statistical signal from one to the other.

    Let's say you have two entangled beams, one of electrons, and the other of positrons, and their spins are negatively correlated. The two beams travel in opposite directions at the same velocity through a vacuum. Bob is receiving the electrons, and Alice gets the positrons.

    To keep it simple, they're at rest with respect to each other, and each is at an equal but opposite distance, D, from the source.

    (1) Bob uses a Stern-Gerlach magnet to split his electron beam by spin along the Y axis at a distance D from the source. It is always on, and it creates two beams.

    (2) Alice has a Stern-Gerlach magnet oriented on the X axis at a distance D+delta from the source. She examines the next bit to be sent, and if it is a '1', she turns on her magnet and she splits her positrons by spin along the X axis for five seconds. If it's a zero, she turns off the magnet.

    (3) Bob has another two magnets around his two electron beams at a distance of D+2*delta from the source, once again aligned on the Y axis, and a screen that displays the beams.

    It seems to me that if Alice has not turned on her magnet, then Bob will see two spots on his detector, which is the normal outcome of a Stern-Gerlach experiment.

    However, if Alice has turned on her magnet, then the state vectors of her positrons (and thus Bob's electrons) will have moved to the X axis, so Bob's second detector will split the beam again, and Bob will see four spots. If this is true, and if D is great enough (> 6 light-seconds), then a bit will have been communicated faster than the speed of light, which is indicated by whether Bob sees two spots or four spots.

    This setup is different from most of the FTL ideas I've seen since an individual electron will only show a diffent spin on the Y axis half the time when Alice turns on her magnet. However, if you have enough particles for each bit, you can get the error to be arbitrarily small, so that's why I specified 5 seconds for Alice to power her magnet.

    I'm studying this stuff on my own, so I've probably overlooked some simple or subtle principle that would keep it from working, but I can't figure out what it is, so I'd appreciate it if someone could point out the flaw in this setup.
     
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  3. Oct 4, 2004 #2

    vanesch

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    No, because after Bob's first magnet, we know with certainty what is the state of the electron in each beam, so Bob will always see two spots (one up-up and one down-down).

    If the initial state is 1/sqrt(2) { |y+>|y-> - |y->|y+> } (first = electron, second is positron), then after the first magnet of Bob, an electron that goes into the upper beam has a state |y+>|y-> and an electron that goes in the lower beam has a state |y->|y+>. Whatever Alice measures in the X direction (or not) doesn't affect this ; she will measure the state |y-> = 1/sqrt(2) (|x+> + |x->) in the first case, and |y+>=1/sqrt(2) (|x+> - |x->) in the second case (or vice versa, I didn't do the calculation). But Bob will measure always a |y+> in the first case, and a |y-> in the second case.

    cheers,
    Patrick.
     
  4. Oct 4, 2004 #3
    Thanks for your reply, Patick. The equations make this easier to discuss.
    Note: I'm going to change the ascii notations a little for clarity, so that Ye+ means "the electron's spin on the Y axis is +" and eUP1 means the electron is deflected up by the first magnet).

    Well, since Bob hasn't actually measured the electron yet, its spin state has not yet assumed a deterministic value. So, it's still in the initial state [though Ye is now entangled with the position of the electron as deflected by the magnet, so the full state is 1/sqrt(2) (|Ye+>|eUP1>|Yp-> - |Ye->|eDOWN1>|Yp+). To keep things manageable, I'm going to assume that this will not affect the argument that follows].

    So, after the electron goes through the first Y magnet, Alice measures X on the positron, and macroscopically observes the result.

    Since Bob did not physically measure anything, Alice will measure 1/sqrt(2) (|Xe+>|Xp-> - |Xe->|Xp+).
    So, after she observes the result, the state will be either |Xe->|Xp+> or else |Xe+>|Xp->.

    Doesn't this mean that if Bob were to measure Xe in his two beams his results would negatively correlate with Alice's? (not that this would not give him any information, of course.)

    If so, it should follow that Ye has become indeterminate once again even if Bob does not measure the Xe's, and so this would yield four spots after the second set of magnets was applied to the beams.
     
    Last edited: Oct 4, 2004
  5. Oct 4, 2004 #4

    chroot

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    It looks to me like Bob has two detectors in two different places, one on each side of Alice. If I'm reading this right, then Bob is not really one person, but two.

    - Warren
     
  6. Oct 4, 2004 #5

    pervect

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  7. Oct 4, 2004 #6
    Sorry if I was unclear, but the idea is that Bob is distance D to the left of the source, and Alice is distance D to the right of the source.
     
  8. Oct 4, 2004 #7
    Thanks for your reply. While it is true that a cascade of two detectors on the same axis will always produce the same results if the state vector does not change between the two detectors, if an intermediate detector measures a property that is incompatible with the first, then this no longer holds.

    For example, if the first detector measures the spin on the Y axis, you get two beams, one moving upwards and one downwards.

    If you place magnets around each of those beams, aligned this time on the X axis, you get two more splits, two to the left and two to the right, yielding four in all. These second magnets have changed the state vector of all particles to an eigenstate of the X axis spin operator (although we won't really know which one until we let them illuminate the screen).

    If you then, for each of those four beams, again placed magnets measuring the spin on the Y axis, each of those beams would split again in the upwards and downwards directions, giving you eight dots on your screen.

    At least, that's how I understand it. This is because by measuring the spin on the X axis, you make the spin on the Y axis no longer determinate.

    In my FTL setup, Alice performs the X axis measurement on the entangled positron, and what I'm proposing is that this also measures the X axis spin of the electron, and thus makes both of their Y axis spins indeterminate.
     
  9. Oct 5, 2004 #8
    I think may have problems to understand what does a measurement on a quantum state.
    A local measurement, for example the spin of Alice, only changes the state local of Alice but not the one of bob.
    The only thing you can say is: the state of bob is |+> or |-> if the result of measurement of Alice is |-> or |+> that's all. You have no action on the Bob state as Bob’s state also includes particles where a measurement on alice would give the result |->.
     
  10. Oct 5, 2004 #9

    vanesch

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    I see what you mean. Let us consider standard measurement theory, using collapse and all that.
    Let us consider the initial state 1/sqrt(2) (|y+>|y->-|y->|y+>)

    Bob's first splitter: Y (because we do not undo this, we can track back afterwards through which branch the electron went, so this is a real measurement).

    A) Upper beam -> the state is now |y+>|y-> with probability 1/2.
    We rewrite this for Alice: |y+> ( 1/sqrt(2) [|x+>+|x->])

    A1) Alice finds X+ -> the state is now |y+>|x+> with probability 1/2.
    Bob measures Y again: he finds Y+ with certainty (eigenstate).

    A2) Alice finds X- -> the state is now |y+>|x-> with probability 1/2
    Bob measures Y again: he finds Y+ with certainty (eigenstate).

    B) Lower beam -> the state is now |y->|y+> with probability 1/2

    B1) Alice finds X+.... etc...
    Bob measures Y again: he finds Y- with certainty (eigenstate).

    ....

    The reason for this is, again, because the measurement occurs already in the first Y apparatus of Bob. If Bob would do a quantum erasure measurement (that is, letting the two separated beams interfere again, such that we will not be able to find out through which channel of the first Y apparatus the electron went), then this should not be considered as a measurement, and the superposition remains. If Alice then measures an |x+> state, say, then we could better have written the original superposition as 1/sqrt(2) [|x+>|x->-|x->|x+>] (which is the same as the y state I wrote earlier) ; now after Alice's measurement the state will be |-x>|+x>. But Bob's Y measurement doesn't exist anymore...

    cheers,
    Patrick.
     
  11. Oct 6, 2004 #10
    Patrick,

    Thanks for the insights. It is very interesting that splitting the beam creates a measurement, and that recombining the beam can erase its effect.

    As I understand it, if Bob split the electron beam on the y axis, the entanglement would no longer apply. But if Bob then recombined the two y beams and split the reconstituted beam on the x axis, he would restore the entanglement and observe a perfect -1 correlation with Alice's positrons' x axis measurements. This is pretty cool!

    Thanks again.

    Bruce
     
  12. Oct 7, 2004 #11

    vanesch

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    Hi again,

    That's the way I also understand it. It gives some people so much trouble that they prefer to go back to more classical models. One cannot deny that quantum theory has some trouble in defining what exactly is a measurement and what not, except by going to some extreme views.
    Indeed, the "realist" objection to what we just described is: yeah, but when the electron gets into the first magnet, IS IT OR ISN'T IT measured ?? If it is, and we apply the projection literally, how are we going to reconstitute the entanglement afterwards then ? If we don't project at that moment, what stops us from having Alice doing her projection, in the way you proposed, changing the pure Y state at Bob's into a pure X state. If Bob can then track back what must have happened at Y, we have contradictory measurements. So what gives ?

    I'll tell you mine. It is probably not universally accepted, but I think it works in all cases. I think that to use quantum theory consistently, there can be only one true measurement, in one place, at one moment, namely at the end of the experiment. All the rest are entanglements.
    In fact, if it weren't for this single final measurement, I'd be talking about many-worlds interpretations. But they have difficulties reproducing the right probabilities if never ever any measurement is done.

    Let's look at the situation you described, and let's say that it is Bob who measures, at the end of the entire experiment. He also measures Alice's state. I leave out all normalization factors like 1/sqrt(2).

    First, we have |y+>|y->-|y->|y+>
    Next, entanglement with first magnet at Bob's:

    |Y+>|y+>|y-> - |Y->|y->|y+>

    Next, entanglement with Alice's X magnet, and her piece of paper:

    |Y+>|y+>(|x+>|X+>+|x->|X->) - |Y->|y->(|x+>|X+>-|x->|X->)

    =|Y+>|y+>|x+>|X+>+|Y+>|y+>|x->|X-> - |Y->|y->|x+>|X+>
    + |Y->|y->|x->|X->

    Finally, entanglement with Bob's second Y magnet:

    |Y2+>|Y+>|y+>|x+>|X+>+|Y2+>|Y+>|y+>|x->|X-> - |Y2->|Y->|y->|x+>|X+>
    + |Y2->|Y->|y->|x->|X->

    And now Bob measures:

    1/4 a chance to have Y2+, Y+, X+
    1/4 a chance to have Y2+, Y+, X-
    1/4 a chance to have Y2-, Y-, X+
    and 1/4 a chance to have Y2-,Y-,X-.

    The measurement also implies the collapse of the state of Alice's paper with her measurement on. You see that no case Y2+ Y- occurs.
    It is when you consider multiple measurements at different times and places that strange things seem to occur. It is my way out of the "faster than light collapse without information transfer". The thing I sacrifice is the "universality of measurement": there can be only one, at one moment (the last one), and one place (where *I* am).

    Now let us treat the case of a quantum erasure. After Bob's first Y magnet, and Alice's X magnet we have the state:

    =|Y+>|y+>|x+>|X+>+|Y+>|y+>|x->|X-> - |Y->|y->|x+>|X+>
    + |Y->|y->|x->|X->

    When Bob now introduces an inverse Y-field that mixes perfectly again the |Y+> and |Y-> states back into |Ymix> (let us assume that no extra phase shifts are introduced in the procedure), we find:

    =|Ymix>|y+>|x+>|X+>+|Ymix>|y+>|x->|X-> - |Ymix>|y->|x+>|X+>
    + |Ymix>|y->|x->|X->
    = |Ymix> (|y+>|x+>|X+>+|y+>|x->|X->-|y->|x+>|X+>+|y->|x->|X->)

    and we can neglect this factor:
    |y+>|x+>|X+>+|y+>|x->|X->-|y->|x+>|X+>+|y->|x->|X->

    Let us rewrite the y-state of the electron in x basis:
    (|x+>-|x->)|x+>|X+> + (|x+>-|x->)|x->|X-> - (|x+>+|x->)|x+>|X+>
    +(|x+>+|x->)|x->|X->

    = (+++) - (-++) + (+--) - (---) - (+++) - (-++)
    +(+--) + (---)

    = -|x->|x+>|X+> + |x+>|x->|X-> (times 2 which we neglect)

    Now Bob passes the electron through his X magnet, and finds:

    -|X->|x->|x+>|X+> + |X+>|x+>|x->|X->

    He now MEASURES his result on the X magnet and Alice's paper:

    1/2 chance to find X- X+
    1/2 chance to find X+ X-

    Perfect anticorrelation!

    cheers,
    Patrick.
     
  13. Oct 7, 2004 #12

    vanesch

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    I would like to add something. Let us assume that it is Alice who measures. Of course, if we wait until Alice has all the information, she'll do exactly the same thing as Bob. But what happens if Alice considers 2 consecutive measurements: one is her X measurement, and the second one is Bob's paper which arrives in the mail ?

    I repeat the common part:

    Now Alice MEASURES:

    1/2 chance to find X+ and the state of Bob's part is then:

    (|Y+>|y+>-|Y->|y->)

    1/2 chance to find X- and the state of Bob's part is then:

    (|Y+>|y+>+|Y->|y->)

    We don't need to track down the state of the positron or Alice's magnet.

    If Bob now puts in his second Y magnet, in the first case (Alice X+),
    the state will be:
    |Y2+>|Y+>|y+> - |Y2->|Y->|y->
    and upon reception of the paper, Alice will find 1/2 probability that Bob had:
    Y2+ Y+
    and 1/2 probability that Bob had:
    Y2- Y-.

    Again, she'll not see any Y2+ Y- on Bob's paper.

    The case when Alice first measured X- is similar.

    If Bob erased his state, then in the case Alice had X+, Bob had:
    |Ymix>(|y+>+|y->) = |x->

    and if he now put in an X magnet, the state would be:
    |X->|x->

    So Alice would find with certainty that bob had X- (perfect anticorrelation).
    Similar for the case that Alice had X+.

    So I can relax a bit my condition: measurements can be applied successively, as long as they are made by one single conscious observer.

    Is this to be taken seriously ? Honestly I don't know. But as an algorithm, it solves all ambiguity of what is and what is not a measurement in standard quantum theory.

    cheers,
    Patrick.
     
  14. Oct 7, 2004 #13
    Here is an other interpretation of quantum measurement. Always separate the “quantum measurement” in 2 blocks:

    -- the local interaction on the Particles: what they become after their interaction with the apparatus: we can stay at the quantum level. We can describe this interaction with a hamiltonian or with projectors.

    -- the statistical event (the apparatus): what a human see or what gives a quantum computation.

    The description of the local interaction given by the apparatus allows getting a description of the system evolution that always stays at the quantum level while the statistical events are always calculated/viewed separately (like probability, it is always a conditional probability: I get that if I’ve measured that and that etc …, example “the last measurement” of Patrick).

    What does not explain this point of view, is why a measure gives only a single value. However, this is not very important, if we “accept” that quantum physics deals with (special) statistics. : )

    For example, the quantum eraser at bob side:

    The first magnet changes the spatial move of particles at bob side: the particles with spin + goes left while the particles with spin – goes right: the “magnet” has introduced in fact a change in the spatial part of the wave function of the particles at bob side (momentum update). We still keep spin correlations with the other particles (we have, in this case a unitary update of the global state):

    Initial state |psi>=|sp_alice>|sp_bob>(|y+>|y->-|y->|y+>) --->

    |psi>=|sp_alice>|right_bob> (|y+>|y->-|sp_alice>|left_bob>|y->|y+>)

    Now if we introduce a new apparatus that recombines the left (spin +) and right particles (-), The spatial extension state part of the spin + spin – particles of bob side (moving left and right) will be updated into a common spatial extension (the initial spatial extension): we get back the same state of spins:

    |psi>=|sp_alice>|right_bob> (|y+>|y->-|sp_alice>|left_bob>|y->|y+>) --->

    |psi>=|sp_alice>|sp_bob>(|y+>|y->-|y->|y+>)


    The main advantage is that we keep thinking on “classical” view of interactions: the magnet changes the spatial direction of the particle using the spin interaction (i.e. what does a stern gerlach apparatus).

    Seratend.
     
  15. Oct 7, 2004 #14

    vanesch

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    But that's exactly what I said ; replace |right_bob> with my |Y+> and we're done. Indeed, this |Y+> represents the position wavefunction with a blob in the upper path. I didn't mean to say that it was a global measurement apparatus state, if that is what was understood.

    I admit having been sloppy by not introducing "initial states" and explicit evolutions, but I think we're talking about the same things.

    The point I was trying to make is "when do we apply the Born rule of probabilities and the projection of the states".

    cheers,
    Patrick.
     
  16. Oct 7, 2004 #15
    :rofl:

    Sorry, I should have taken more time reading your post :biggrin:
    (I confused your “y” with a simple label and thus do not make the association to the position, I assumed implicit in the state itself:tongue2:).

    So let's go on a call for other propositions in the quantum measurement interpretation.

    Seratend.
     
  17. Oct 11, 2004 #16
    Another way to put it would be to say that when Bob sends the electron through the first magnet he "marks" the electron as having a specific spin, even though he doesn't "measure" it yet. (I borrowed this term from the delayed choice interferometry experiments, where they "mark" the idler photon with which-path information.) To "mark" a particle just means to entangle its state with a macroscopically measurable property, but it might be easier for people working at an introductory level to understand.

    This is a subtle point, but actually I think this distinction between "marking" the particle's state and "measuring" it lies very close to the heart of these matters.

    Bruce
     
  18. Oct 11, 2004 #17
    What comes to mind here is E-P-R ?

    http://math.ucr.edu/home/baez/physics/Quantum/bells_inequality.html

    One has to determine all probable outcomes, from an experiment that you cannot know TOTALLY!

    There are a number of factors needed to be addressed, one question is that what is to be determined as an Observation?..and does the initial setup interfere with the outcome/results?..are the spacial events of the Atom-Detector-Observer intermingled in a Time dependant manner?

    For instance, one can ask this:Does the Electron Know of the Experimental set-up prior to its initial state, if so then the Electron can just as well be classed as Observer Dependant. Observers measure end results?..as pointed out earlier, then Electron "observation" has a forward projection of its oncoming events that lay directly in it path!

    Electrons were given a special status by Feynman,(aftera conversation with Wienberg I believe?), in that they can go forwards and backwards in Time(not to be confused with faster than light transportations), this can be an assertion that Human observation will never know enough about an observed system, because we are dynamically locked within Spacetime,(Time dependant) locations at all times, even if we try and construct experiments wherby we assume a prior "knowledge" of determination.

    I should state that David Alberts book:Q M and Experience, touches upon some deep philosophical implications, and if you continue with the book I am sure you will benefit the authors amazing angles he gives in the later chapters, areally amazing read.
     
    Last edited: Oct 11, 2004
  19. Oct 12, 2004 #18
    Can you develop your view concerning the marking?
    More precisely what do you intend by the "entanglement with a macroscopically measurable property"?

    I understand your marking, formally, as the result of a measurement on the particle giving a particular result - the eigenvaue of a particular observable -(and after, I may proceed after with conditional probabilities for further measurement results).

    Seratend.
     
  20. Oct 13, 2004 #19
    Well, the important point is that marking does not collapse the wave function -- it puts the particle into a superposition of eigenstates. Later measurement will reveal, when the wavefunction collapses, what the state was when the particle was marked.
    This assumes, of course, that the wave function actually does eventually collapse (though Albert is not so sure that this ever happens...).

    Bruce
     
  21. Oct 13, 2004 #20

    selfAdjoint

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    This is not possible. If you interact with the wave function, with the extended entangled state, then you collapse it. This is what inhibits many schemes for turning entanglement into FTL.
     
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