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The Quantum Answer To The Thought Experiment

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Posted by Harvey on November 2, 2003 04:03:03 UTC

Bruce,

I received the answer to the thought experiment which I'll restate for everyone's benefit:

Okay, let's say we have Alice and Bob are going to be separated by some cosmological distance (say Alice is on earth and Bob is about to leave to M31). Prior to departure, Alice and Bob entangle a number of pairs of quantum particles. The entangled particle pairs are such that there are only two possible outcomes for each measurement of a single pair such that the probabilities of the outcomes add to unity. Let's say that the most likely measurement of a particular pair is to occur 49 out of every 50 measurements (state U is more likely). The least likely measurement occurs 1 out of every 50 measurements (state D is least likely). However, the particles are entangled in such a way that Alice and Bob cannot get the same measurement (i.e., one will get state U and the other state D). So, if Alice measures her particular particle (that is entangled with Bob's particle) she is 49/50th times more likely to see state U for that particle and Bob will see state D under most circumstances. Likewise, if Bob makes the measurement first of his particle that is of the entangled pair, then Bob is 49/50th times more likely to see state U, and therefore Alice will see state D if Bob gets state U.

Just before leaving Bob gets a little scared. Bob expresses concern to Alice that his spaceship might breakdown upon reaching M31, so he should be able to signal to Alice to send a spaceship, but decides he cannot wait for a signal to travel at the pokey speed of light. So, they make a few notes as to how Bob can signal Alice to send a spaceship if needed.

Are you still with me? Okay, Bob leaves for M31. Do da do da do da do da. Long trip..., but Bob finally arrives. He gets to planet Nine in M31, and the alarm goes off. Bob has to alert Alice if she should send a spaceship. Bob checks his instrument panel and sure enough, his flux capacitor is malfunctioning, so he needs the spaceship to be sent.

He reviews his notes and he sees that at 12:00PM he had the option to make a ton of measurements on his entangled pairs (stuffed neatly away in the carry-on compartment), or he has the option of not making the measurements. If he makes the measurements, he knows that for the vast majority of his measurements, he will get state U for his measurements since he knows that Alice has not made a measurement (he assumes she is not willing to screw him on this long distance trip), so he is confident that he will see mostly state U's on all of his measurements.

He looks at his notes again and sees that he is to make the measurements if he wants Alice to send the spaceship. Okay, he needs the spaceship, so Bob opens the carry-on compartment, pulls out all of his entangled pairs of particles, and measures each one. And, each one he happens to get state U in his measurements. Now, he crosses his fingers, and hope Alice will send the spaceship.

Back on earth, Alice waits a few days (just in case special relativity screwed up the simultaneity issue), and finally, as agreed before Bob left, she begins to make her measurements to see if she should send to Bob the spaceship that she promised she would do.

She makes her measurements, and she sees state D's in all the particles that she measures. She computes the probability of getting all state D's in her measurements, and it comes to be exceedingly small. Alice knows that Bob needs the spaceship. Being the polite scientist that she is, she calls NASA and a spaceship is sent. Bob and crew are saved to travel the star for another day.

-----------------------------------------

The answer to this issue, is that the following assumption is incorrect:

if Alice measures her particular particle (that is entangled with Bob's particle) she is 49/50th times more likely to see state U for that particle and Bob will see state D under most circumstances. Likewise, if Bob makes the measurement first of his particle that is of the entangled pair, then Bob is 49/50th times more likely to see state U, and therefore Alice will see state D if Bob gets state U.

It is false that whoever measures the particle first will have a higher probability of getting state U. Rather, when the two particles are entangled, it is no longer true that both particles have greater probability to measure state U. Rather, one particle will have a greater probability to measure state U (49/50 times), and the other entangled particle will have a greater probability to measure state D (49/50 times).

This is a result of entanglement. Entanglement requires that both entangled particle probabilities for state U or state D equal to unity. If one particle is 49/50 times more likely to have state U, then by the action of entangling another particle with that particle, the other particle must have probability 49/50 times more likely for state D.

Since Bob cannot influence the outcome of Alice's measurement by making his measurement, he cannot tell Alice to send a spaceship. Bob is screwed.

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