# Entanglement cycles [closed]

Absolute amateur here with a question that was probably answered eighty years ago somewhere. Suppose we have three particles of a type such that if two are entangled, then if either particle is observed to be in state 0, then the other must be in state 1. Now suppose that A and B become entangled, then separate. Later, B and C become entangled, so that C is now bound up with both A and B. Then if A is observed to be in state 0, then B must be in state 1, whence C must be in state 0, whence A must be in state 1: uh-oh.

I see four possibilities here:

1. Another amateur is ignorant of, or has failed to appreciate, subtleties. What are those?

2. Some mechanism prevents the entanglement. What mechanism is that?

3. The entanglement can happen, and the resulting object becomes in principle unobservable—dark matter.

4. Quantum theory is broken.

Can anyone enlighten me? Thanks in advance.

• Quantum theory is broken, of course. What else. – Norbert Schuch Dec 30 '20 at 23:40
• Alternatively: 1, and thus 2. – Norbert Schuch Dec 30 '20 at 23:41
• Why bring quantum mechanics into it? Imagine that I have three shoes, all of which are either black or white. A and B are opposite colors. B and C are opposite colors. A and C are opposite colors. But if A is black, then B is white, so C is black. Uh oh! What could possibly explain this? – WillO Dec 31 '20 at 4:03

Another way of stating the problem is to point out that I simply can't write down the final state which you try to create. A state with particle's A and B entangled, and not entangled with C, could be written as $$(|01\rangle+|10\rangle)|\psi_C\rangle$$. I can't there's no way to write down the state you describe because it's self-contradictory.