Dork Scratchings

I'm still a bit obsessed with the GHZ experiment.  This is the one where you prepare a GHZ state $\lvert 000 \rangle + \lvert 111 \rangle$ and send one qubit to each of 3 protagonists: Alice, Bob, and Charlie.  (I will be ignoring shared normalizer constants throughout this post, as I find they don't add anything to the understanding.) If you do the maths it turns out that when all three choose to measure in the $\lvert + \rangle$, $\lvert - \rangle$ basis (shorthand for $\lvert 0 \rangle + \lvert 1\rangle$ and $\lvert 0 \rangle - \lvert 1\rangle$) then they are guarranteed to get a parity zero result.  On the other hand if only one measures in this basis and the other two measure in the $\lvert +i\rangle$, $\lvert -i\rangle$ basis (shorthand for $\lvert 0 \rangle + i\lvert 1\rangle$ and $\lvert 0 \rangle - i\lvert 1\rangle$) they are guarranteed to get a parity one result.  As I showed in an earlier post this appears to be incompatible with the outcomes being prede...

It has occurred to me that the cartoon I drew two days ago will appear to most people as absurd: Alice measures one half of an entangled pair and instantly splits into two versions of herself!  Bob measures the other half and also splits, but each Alice is paired with one of the Bobs in such a way that, should they meet later to discuss their results, everything remains consistent. Yet if you replace Alice and Bob with two qubits - to get a quantum circuit like the one below - no quantum computing expert would find this outcome absurd. The Einstein Podolsky Rosen experiment in circuit form with non-classical Alice and Bob Alice is qubit q0 and Bob is q3 .  Qubits q1 and q2 are entangled into the Bell state $(\lvert 01 \rangle + \lvert 10 \rangle)/\sqrt 2$ and then q1 is shared with Alice, and q2 with Bob.  In the standard (collapse-based, non-multiversal) description Alice and Bob measure q1 and q2 , collapse the q1q2 wavefunction, and find that one gets 0 while the...

I just read this article on phys.org which claimed to "resolve the paradox" that entanglement appears to imply faster than light communication.  I don't think it actually resolves the paradox, it just side steps it.  Here's how the argument goes (and I paraphrase): Alice and Bob share a pair of entangled qubits in a Bell state $\left( \lvert 01 \rangle + \lvert 10 \rangle \right)/\sqrt{2}$. When Alice measures hers it "collapses" the pair's state immediately even if Bob is a great distance away.  However, this does not imply faster than light communication because, although Alice now knows what Bob will measure, Bob does not know until he makes a measurement himself, or receives a message from Alice which can only travel at the speed of light. This argument misses the point somewhat.  If the state of both qubits changes instantly when Alice measures her qubit (e.g from $\left( \lvert 01 \rangle + \lvert 10 \rangle \right)/\sqrt{2}$ to $\lvert 01 \rangl...