I don't understand the no-communication theorem.

Okay, first I'll say the bit I do understand about it: if Alice and Bob both have two atoms, such that Alice's atom 1 is entangled with Bob's atom 1, and atom 2 is likewise entangled. Alice writes to hers by doing nothing to atom 1 and reading the Z spin from atom 2 for a 0, and doing the reverse for a 1. When Bob reads the Z spin from his pair, he has no way of knowing if he got the 0 or 1 that Alice meant to send him, or if he read it too early and got random noise.

Instead of that let's make atom 1 a pair, and atom 2 a pair. Alice writes a bit, and Bob tries to read. Now there's a 50% probability of them matching, and indicating that he read too early, and a 50% probability that he either got a bit or read to early (...I think).

So let's make each atom in A and B a set of 50 atoms, all entangled separately, but Alice and Bob have the matching sets. Alice writes to all of them simultaneously, Bob reads from them all simultaneously, now the probability of Bob having a false positive is about 2%, and the bit correction schemes used on the internet can deal with that. Now Bob basically knows if he got a bit, or if he's read too early.

So let's give them both infinite tapes, each "cell" on the tape is a 10x10 grid of atoms, as described above. Alice always writes to even numbered cells, and reads from odd numbered cells; Bob does the opposite. And both write to one cell every second. They know if they get noise they should slow down, and to be safe we'll have them read the next two at the same time every tenth read. So if the tenth read returns noise, then we're going at the right speed, and if it has a bit then we need to speed up. So we can eventually synchronize the two clocks.

I don't understand why this set up fails aside from "you can't have infinite tapes".

  • $\begingroup$ Why would anything Alice does affect the probability that the two pairs match? They'll have a 50% probability of matching in both cases, yes? $\endgroup$
    – BowlOfRed
    Jan 7, 2015 at 22:29

2 Answers 2


You seem to be thinking that if Alice reads the Z spin on one of the entangled pair that Bob would be able to detect that. Alice can't control what the value is that will be read. She might read an up or a down. Bob will read the complement, but since that is the complement of a "random" value, it appears random as well and cannot be differentiated.

  • 1
    $\begingroup$ Yes, and Bob will also have no way to detect whether Alice has actually performed a measurement on the entangled twin of one of his particles, or whether he is the first one to measure a member of the pair--the statistics for his particle alone will be exactly the same either way. $\endgroup$
    – Hypnosifl
    Jan 7, 2015 at 23:00

To communicate a particular bit sequence from Alice to Bob, there must be a fixed assignment of a logical 1 to a state, say, s1, and a logical 0 to s2. But when Alice measures her object, the outcome is always 50/50 s1 or s2, regardless of whether she wants to send a 1 or a 0. Thus there is no way to encode a particular bit onto a state. The bit sequence to send is known by Alice but the result of her measurement is always random. Any method Alice could use to force her measured state to be s1 or s2, as needed for the bit she wants to send, would break the entanglement with Bob’s object.


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