On Fri, Jan 29, 2016 at 12:09 PM, Henry Baker <hbaker1@pipeline.com> wrote:

Bose-Einstein statistics apply to bosons, correct?

If I have a number N of these bosons, *all in the "same" state*, then these are all maximally entangled, correct?

No; in fact they're completely disentangled: you've just said that each particle is in the same state, so we can write the total state as (tensor_{i=1}^{n} |psi_i>) // S_n The superpositions involved only involve the order of the particles, not their spin. Read the wiki article on Fock space.

If I "measure" one of these bosons (is this even possible -- to measure just one?),

Yes, that's possible.

this measurement necessarily puts *all* of my bosons into the same state ?

They're already all in the same state, by assumption.

Now, I'd like to have N entangled qubits, so that measuring any one will guarantee the same measurement on *all* of the other qubits.

It sounds like you want a GHZ state. The measurement still only occurs on one of the qubits, but the measuring device gets entangled with them, so you get |Device 0>(|000> + |111>)/sqrt(2) | XOR the state of the first GHZ qubit into the state of the device | V (|Device 0>|000> + |Device 1>|111>)/sqrt(2) So if YOU are the measurement device and you do subsequent measurements on the second and third GHZ qubits, you'll either see the qubits all being 0 or all being 1. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com