> if I have a one-qubit state that's restricted to the set of states that are linear combinations of ±|0> ± |1>, then are only two such states up to global phas

> if I have a one-qubit state that's restricted to the set of states that are linear combinations of ±|0> ± |1>, then are only two such states up to global phase, and those states are orthogonal, so I can just measure without loss and treat it as a classical bit. Am I missing something? It's not a one-qubit state limited to ±|0> ± |1>, it's a two-qubit state limited to a|00> + b|01> + c|10> d|11> where a^2+b^2+c^2+d^2 = 1 and a,b,c,d are real. Also the state can be entangled with other qubits, as long the non-180-degree phase information is not between the various a's, b's, c's, and d's. For example, if you have the state (1/2 |000> - 1/sqrt(2) |010> + i/2 |101>)(|00> + |11>), then you can copy the first two qubits into the bell pair on one side then pull them out on the receiving side to end up in the state 1/2 |00000> - 1/sqrt(2) |01001> + i/sqrt(2) |10110>. E.g. see this circuit doing just that: http://i.imgur.com/wlcVAZG.png http://i.imgur.com/wlcVAZG.png Glad you liked the puzzle.