# Tag Info

0

Initially the pair is in state $(P\otimes M)+(M\otimes P)$. (I am writing "P" and "M" instead of "+" and "-" so as not to confuse the state "plus" with the addition operation in the state space.) You observe the first electron and happen to measure $P$. Now the pair is in the state $P\otimes M$. Note that this state is not entangled (i.e. it is a tensor ...

2

The flaw in your argument is that the claim "entanglement will instantly replicate the photons' paths and such the pattern onto Bob's screen [sic]" is incorrect. The statistics for the measurement outcomes of any experiment performed on one subsystem of a maximally entangled pair is independent of what goes on with the other subsystem. In your case, from ...

4

The problem with this sort of scheme is that Alice has no control over the results of her measurements, since those are random. This means that she can control which basis Bob's spin is projected on, but she cannot control which of the basis states gets chosen. Bob will then see a random mix of results which turns out to contain no trace of what Alice was ...

4

Your error seems to be the misconception that entanglement will magically make the results of any experiment of photon B exactly mimic those of a similar experiment done on its entangled partner. Entanglement is more subtle than that and must be treated carefully. In particular, there are many different types of entanglement. For example, photons may be ...

Top 50 recent answers are included