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Cheating a faster than light signal.

  1. Feb 23, 2007 #1
    A spin zero particle decays into a pair of spin 1/2 particles, each particle in a superposition state of up and down. The pair of particles head off in two opposite directions. One of them enters an apparatus that will collapse its wave function and decide the life and death of a cat based on the result. When the collapse occurs, the other particle immediately changes state from superposition state to eigenstate. I measure the state of the particle and place a bet with some chump about the condition of the cat. Sometime later, the verification of the cat's state arrives after having traveled at the speed of light, and I win the bet.

    No real information has traveled faster than light, but that's small comfort to the chump.
     
  2. jcsd
  3. Feb 23, 2007 #2

    vanesch

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    :tongue: You've almost re-discovered the EPR Bell paradox !

    Almost. Because the start was good (with the singlet state and so on).
    But afterwards you don't use it.

    Consider this:

    Out of a bag containing white and black balls, one picks a ball and cut it in two halves. Each half is sent off in two different directions: one to you, and the other one to the cat. If half a black ball enters the device, the cat is killed. If it is half a white ball, not so.

    Upon receipt of your half of the ball, you know that when it is black, the cat is dead now, and if it is white, the cat lives.
     
  4. Feb 23, 2007 #3
    But in your scenario, the chump will take my money. His confederate saw the balls when they took off and sent a light signal to him. He got the information before the half-a-ball arrived. In my scenario, his confederate didn't know what would happen to the cat when the spin zero particle decayed.
     
    Last edited: Feb 23, 2007
  5. Feb 23, 2007 #4

    JesseM

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    But why do you allow the confederate to see which ball was picked in the classical scenario, but forbid the confederate to see the result of your spin measurement in the quantum scenario? That seems a little arbitrary, you could just as well say that in the classical scenario the choice of which ball to cut in half was made in a sealed room by an impartial judge, while in the quantum scenario a confederate was secretly observing your spin measurement and passed that information on to the would-be chump using a light-signal. Basically it all just comes down to which macroscopic event you allow the confederate to see and which you keep secret from him.
     
  6. Feb 23, 2007 #5

    vanesch

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    Ok then :grumpy: :tongue2:

    Consider this, in YOUR scenario: on the track of the particle that goes my way, the confederate has installed a Stern-Gerlach machine which measures the spin of that particle, and lets it continue my way. He sends the result of his measurement with a light signal to the chump.

    :wink:
     
  7. Feb 23, 2007 #6

    Doc Al

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    But you forgot that the chump was hip to your scam and had his confederate give the remote Stern-Gerlach magnet a spin, randomizing its orientation. Now you'll have to wait for the morning paper to find out what happened to that cat.
     
  8. Feb 23, 2007 #7

    vanesch

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    :rofl: That's good :rofl: :rofl:
     
  9. Feb 23, 2007 #8
    The confederate knows what color half-a-ball was sent the moment it was sent. He can send a light signal that goes faster than the half-a-ball does. The chump gets the money.

    However, the only thing he knows when the zero-spin particle decayed and my particle was sent is that it is in a superposition state. He doesn't know what will happen when the measurement is taken some time later. The particles went in opposite directions, so the cat is farther away from the chump and I than the decay location. He can't send a meaningful signal until the particle is measured. At that time he will have to be located where the cat is, not where the decay occured. Now he can send a light signal, but he is too far away and the particle will arrive at my end before the light signal does. I get the money.
     
  10. Feb 23, 2007 #9
    Not only will he make a chump of me, but he will be in a position to make chumps of the people at his location. Either way, a faster than light signal got cheated.

    By the way, in my scenario, I was measuring a particle in an eigenstate. It was in that state because when the particle at the cat end was measured, my particle went into an eigenstate immediately. Spooky action at a distance.

    In your scenario, the confederate will be measuring a particle in a superposition state. He will know what is going to happen to the cat before it happens.

    It's all a cheat because no one really knows what happened to the cat, only what very likely happened. It's subject to the reliability of the poison, etc.
     
  11. Feb 23, 2007 #10

    vanesch

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    That's the point. He can measure HIMSELF one of the particles, where ever he likes. (in the same way as he could get a glimpse of the color of the balls).
     
  12. Feb 23, 2007 #11

    vanesch

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    But that is exactly the same as when he got a glimpse of the color of the ball!
     
  13. Feb 23, 2007 #12

    vanesch

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    If I may add something:

    The reason why one can always come up with an equivalent classical scenario in the case you mentioned (except for Doc Al's smart remark, which contains the essence in fact), is that as long as we are going to measure spins along one single axis (on the cat side, and at my side, or in between, by some spy), there is no observational difference between a true singlet state:

    |+>|-> - |->|+>

    and a statistical mixture of 50% |->|+> and 50% |+>|->, and the last situation can always be replaced by classical equivalences, because we have product states (each individual particle has a well-defined state, no entanglement).

    It is only when you start to look at measurements along different axes that there will occur funny correlations in the true singlet state, and not so in the mixture case.
     
  14. Feb 23, 2007 #13
    My word, is it possible to send a signal faster than light after all?
    As I (half-? mis-?) understand it, particles that are in a superposition state have different measurement statistics than particles that are in eigenstates. I thought the import of the Bell experiment was to tell whether the daughter particles entered the eigenstate at the time of the decay or the time of the measurement. They proved it was at the time of the measurement. Have I erred in my understanding?

    Designate a sending station and a receiving station. From the midpoint of the two locations, send a stream of particles to both stations, all particles in superposition state. For each fixed interval of time at the sending station either do, or do not collapse the states of the particles. At the receiving station, for each fixed interval of time, gather from the measurement statistics whether the sender measured or didn't measure. You could send a Mozart symphony that way. This is pretty simple stuff, someone surely has already considered and rejected this idea. What's wrong.
     
  15. Feb 23, 2007 #14

    Doc Al

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    Nothing you do (or don't do) at remote location A can be detected simply by examining the data collected at remote location B. Only by comparing the two sets of measurement results do you deduce that the outcomes are correlated. (Which is not to say that something interesting isn't happening, but it's subtler than you think.)
     
  16. Feb 23, 2007 #15

    vanesch

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    No, according to QM, this is not possible, be reassured.

    That depends ! If they come in statistical mixtures of eigenstates in the right proportion, then you cannot make the difference. What I mean is the following:

    if the superposition is a |+> + b |->, then you cannot distinguish this, when measuring + or -, from a mixture of eigenstates, randomly presented, with a fraction |a|^2 of the time, the eigenstate |+> and the fraction |b|^2 of the time, the eigenstate |->.

    And this is exactly what happens in such an EPR-Bell situation. So, locally, you will never know whether you still have a superposition, or whether you get eigenstates (but different ones, according to the statistical outcomes on the other side). In other words, what concerns local statistical distributions, EPR situations have nothing particular.

    However, the *correlations* between the outcomes are peculiar.


    That's what doesn't work, exactly because of the reason I said above. You can consider that you receive OR a superposition, OR a mixture of eigenstates, but *locally* they will always give you the same statistics, no matter what you measure.
     
  17. Feb 23, 2007 #16
    Thanks. Actually, it's reasuring to know that I'm not going to get a Nobel prize for half-baked musings. This thread started as very light hearted, but has ended up quite educational. Let me summarize my new (half-? mis-?) understanding:

    If I do a statistical study of the measurements of a bunch of particles, I can't tell whether my counterpart measured his or not, even though his measurement has put my particles into eigenstates.

    The reason is that I can't distinguish between the following two scenarios:

    A bunch of particles all in superposition states of a certain proportion of up and down.
    A bunch of particles all in eigenstates, some up others down, and mixed in such proportion as to mimic the constitution of the superposition states.

    Can Bell's experiment be explained to me in this kind of simple-minded language?
     
    Last edited: Feb 23, 2007
  18. Feb 23, 2007 #17
    This is quite a coincidence. I have been reading the book Photons and Atoms by Claude Cohen-Tannoudji et.al. and had stopped in the middle of page 215. Now I picked it up for today's reading and get to the following near the bottom of the page (slightly edited):

    For example, a single-photon state which is a statistical mixture of eigenstates |1_1, 0_2> and |0_1, 1_2> with the weights |a|^2 and |b|^2 does not give fringes in w_1 (single counting rate) whereas these are observable with the superposition state a|1_1, 0_2> +b|0_1, 1_2>

    Can we look for interference fringes in the wave-like properties of the incoming particles?
     
  19. Feb 24, 2007 #18

    vanesch

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    In order to distinguish a statistical mixture and a superposition, you have to work in *another basis*. In the explanation I gave below, I was assuming that the two analysers are lined up (at Alice and Bob). If they aren't, things are slightly more complicated, but the result is the same.

    Indeed, consider this. The singlet state can be written as a superposition in any eigenbasis, it will always take on the same form:

    |z+> |z-> - |z-> |z+> = |n+>|n-> - |n->|n+>

    So if Alice put her analyzer along Z, we get, half of the time, that she obtained a + and half of the time that she obtained a -.
    In other words, we get 50% of |z-> at our side and 50% of |z+>.

    If, on the other hand, Alice put her analyser along the n axis, we get 50% of the time an |n+> and 50% an |n->.

    Now, imagine that we (bob) measure along the m axis, which makes theta_m degrees with the z-axis.
    In the first case, (Alice along z), in 50% of the cases, we get cos^2(theta_m) chance to find m+ and sin^2(theta_m) chance to find m-
    (that's the Born probability for spin 1 particles, when we have a state |z+>, and we do a measurement along |m+> and |m->).
    In 50% of the cases (we received a z-) we have sin^2(theta_m) chance to find m+ and cos^2(theta_m) to find m-.

    Overall, we find 0.5 (cos^2(theta_m) + sin^2(theta_m)) = 0.5 chance to find m+ and 0.5 chance to find m-. I think you see that we also get this result when Alice measured along n (change theta_m into theta_(mn), the angle between the n axis and the m axis).

    If Alice didn't measure anything, then we receive the original singlet state, and there too, no matter along which axis, we find 50% up and 50% down.

    This is not a coincidence of this particular setup. It can be shown in all generality (long ago I posted the proof here as an attachment, but I can't find it again): any measurement on a particular part of an entangled system gives exactly the same statistics, independent of whether, and which, measurement is performed on the other side.

    What is peculiar, are the *correlations* between the outcomes at both sides. But the projected statistics of the measurements on each side are independent of what is the measurement performed on the other side.
     
    Last edited: Feb 24, 2007
  20. Feb 24, 2007 #19

    vanesch

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  21. Feb 24, 2007 #20

    DrChinese

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    Vanesch's paper says it very well.

    And if I recall correctly, we already determined that entangled particles do not display interference characterics when they enter a double slit apparatus.
     
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