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simplification of all hermitian gates where symmetric doublets have been removed. A phase shifting gate P have det(P) = the phase it shifts with, and have no other effect beyond changing the phase along a specific basis axis. Multiplying a hermitian gate with phase shifting gates (one per basis vector?) results in a gate from the group U(2). Alternatively as it is
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Any gate U that is in SU(2) have det(U)=1. The determinant of two gates in series is det(U_1 * U_2) = det(U_1) * det(U_2). Most gates that are listed have determinants -1. So, flipping it back to 1 (so it's in SU(2)) is just multiplying it with whatever other gate. Maybe one can think of SU(2) as a
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The extended version of
Solovay–Kitaev theorem shows that this is possible to navigate SU(d) efficiently, but quantum gates on q qubits belong to U(2). It seems from reading the paper, they really do mean SU(d). I
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My guess is either what you said (it's straightforward to extend to U) or that all the gates in quantum circuits are considered to be phase-shifted so that all its gates are in SU. Not sure about this, though.
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that it is considered to be known that U(1) is efficiently generated by a few phase shift gates, and that
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Dawson, Christopher M.; Nielsen, Michael (2006-01-01).
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