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Aug 6, 2018 at 9:50 history undeleted
Aug 6, 2018 at 9:50 history edited
Aug 6, 2018 at 9:31 comment @kludg Why would $U=U^\dagger$ for real $U$?? This is completely wrong. Try e.g. [1 -1;1 1] (normalized) - a pi/2 rotation about Y - which doesn't square to the identity. (Indeed, a Y rotation by pi/2 prepares the same superpositions as H starting from the computational basis.)
Aug 6, 2018 at 7:29 comment @NtwaliB. Ok; but most people would not overcomplicate things and just use $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ because it looks simpler.
Aug 6, 2018 at 7:24 comment I would choose the one that makes the computer do the least amount of work. If a Hadamard gate is implemented in terms of rotations, I will use a rotation about X as it saves me the use of two additional gates. If Hadamard is a native gate, I will use Hadamard since most people are used to it. Note: Hadamard generates $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ while $R_x(\frac{\pi}{2})$ generates $\frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle)$
Aug 6, 2018 at 7:19 comment @NtwaliB. Suppose you need just a superposition with 50% chances being $|0\rangle$ and 50% chances being $|1\rangle$; both $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $\frac{1}{\sqrt{2}}(|0\rangle - i |1\rangle)$ are equally good for you, but what would you choose?
Aug 6, 2018 at 7:12 comment You certainly answer the question and thanks for that. Though would you mind elaborating why one would specifically need $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ over say $\frac{1}{\sqrt{2}}(|0\rangle - i |1\rangle)$? I believe elaborating on that will shed more light to aid my understanding.