Talk:Solovay–Kitaev theorem

Latest comment: 3 years ago by Omnissiahs hierophant in topic What about U(2q)?

What about U(2q)?

edit

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(2q). It seems from reading the paper, they really do mean SU(d). I assume that it is considered to be known that U(1) is efficiently generated by a few phase shift gates, and that  , which in turn is extensible to U(2q) via random coupling gate (e.g. CNOT) or as shown in the extended SK[1]??? Can someone please explain this? · · · Omnissiahs hierophant (talk) 19:41, 31 March 2021 (UTC)Reply

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. Fawly (talk) 22:27, 8 April 2021 (UTC)Reply
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 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 explained here. · · · Omnissiahs hierophant (talk) 02:38, 11 April 2021 (UTC)Reply

References

  1. ^ Dawson, Christopher M.; Nielsen, Michael (2006-01-01). "The Solovay-Kitaev algorithm". Quantum Information & Computation. arXiv:quant-ph/0505030.