Wouldn't 3 or more particle entanglement allow passing classical information?

Question

My (simple) understanding of entanglement is that by measuring the spin of one entangled particle, the other entangled particles' spin changes to the opposite of measured particle. This act of measurement doesn't allow any communication. (Obviously entanglement isn't just the spin but also other physical properties, but for simplicity of this question).

However, since more than two particles can be entangled, what if a group of 3 particles are entangled, with Alice having 2 and Bob having 1.

At predetermined time intervals (eg every 0.1ms), Bob measures his entangled particle if he wishes to communicate a 1 in binary times. This will cause Alice's two entangled particles (in the same entanglement group) to have the same, opposite spin direction of Bob's particle.

Bob does not measure his entangled particle if he wishes to communicate a 0. This keep all of the 3 entangled particles still in a superposition.

At every 0.1ms, Alice measures both of her entangled particles. If Bob has already measured the entangled particle, Alice will find that both of her entangled particles have the same spin. However, if Bob has not measured, then when Alice measures one of her entangled particles, the other entangled particle will change to the opposite spin state, and Alice will find her two entangled particles are different. This would mean that information would be passable.

What's the problem with this?

Answer 1

What's the problem with this?

The problem is that this:

My (simple) understanding of entanglement is that by measuring the spin of one entangled particle, the other entangled particles' spin changes to the opposite of measured particle.

is not true.

What is possible for example is to use 3 particles to make this state: $$\newcommand{\ketrm}[1]{\left|\mathrm{#1}\right\rangle} \newcommand{\ketThree}[3]{\ketrm{#1}\ketrm{#2}\ketrm{#3}} \frac{1}{\sqrt 2}(\ketThree{up}{down}{down} + \ketThree{down}{up}{up})$$

so that if we measure up for the first one, the two other will be down and vice versa, but it doesn't allow us to pass information.

EDIT: Ok so what you want to do is something like: $$\frac{1}{\sqrt 2}(\ketThree{up}{down}{up} + \ketThree{down}{up}{up})$$

So that if particle "a" is up, "b" and "c" are different, and if a is down, b and c are both up.

Whatever Bob does, Alice has 1/2 to measure b as down and c as up, and 1/2 to measure b and c as up.

Answer 2

Your understanding of entanglement is wrong. The probabilities of measurement results on Alice's particles are not affected at all by any measurements performed on Bob's particle. What happens when Bob's particle is measured? The measurement apparatus differentiates into two versions, each of which has recorded one of the possible outcomes. Alice's particles become correlated with Bob's when the measurement results are compared because the decoherent systems that carry the measurement outcomes also carry locally inaccessible information. Their observables depend on what Bob has measured but the expectation values of those observables do not depend on the measurement, see

http://xxx.lanl.gov/abs/quant-ph/9906007

http://arxiv.org/abs/1109.6223.