I know that U1 equivalent gate of Qiskit in Q# is R1, but I would like to implement U2 and U3 gate of Qiskit in Q#, what is the best way ?
Thank you.
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$\begingroup$ Do you need to implement U2 directly or do you have a specific unitary in mind? What I mean by that is that U2/U3 are more general forms for unitaries, but algorithms typically only require a restricted gate that doesn't need to be expressed as U2/U3 $\endgroup$– C. KangSep 8, 2020 at 0:26
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1 Answer
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The gates Rx, Ry and Rz in Q# have the same definition as RXGate, RYGate and RZGate in Qiskit, and the U2Gate and U3Gate documentation offers decomposition of these gates into Rx, Ry and Rz, so I would just use that decomposition:
operation U2(psi : Double, lambda : Double, q : Qubit) : Unit is Adj + Ctl {
Rz(lambda, q);
Ry(0.5 * PI(), q);
Rz(psi, q);
}
and a similar one for U3.
answered Sep 8, 2020 at 2:11
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$\begingroup$ Thank you Mariaa, this is what I have tested but I have seen something : If I simulate U2 or Rz/Ry/Rz gate with both Pi/4 for lambda and psi in Qiskit, then my state vectors are [ 0.707+0j, 0.5+0.5j ] Then if I simulate Rz/Ry/Rz with Q#, then I dumpmachine to get state vectors, I got : ∣0❭: 0,500000 + -0,500000 i == ********** [ 0,500000 ] \ [ -0,78540 rad ] ∣1❭: 0,707107 + 0,000000 i == *********** [ 0,500000 ] --- [ 0,00000 rad ] So there is a minus sign wich is different from Qiskit, is it normal ? $\endgroup$ Sep 8, 2020 at 7:23
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$\begingroup$ These states are the same up to a global phase (1-i)/√2 - multiply Qiskit state by this number and you'll get the expression Q# gives you. So yes, this is normal $\endgroup$ Sep 8, 2020 at 7:32
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2$\begingroup$ Thank you Mariia, answer clear and complete as usual ! $\endgroup$ Sep 8, 2020 at 8:20