usingblock. As you note, you can't use measurement to do that in an adjointable operation; given that resetting is a special case of measurement, the
Resetoperation will also fail to preserve adjointability. The trick is that you need to coherently reset the qubit to |0⟩ without measurement. You can do that because you know exactly what state it's in at the end of the
using, and can unprepare it using only unitary operations. The
applyfeature of Q# or the
ApplyWithCAoperation will be very handy here.
By the way, @crazy4pi314 will be streaming with @bettinaheim about the Q# compiler today in a little under three hours if anyone is interested to join at https://twitch.tv/crazy4pi314. More details at https://twitter.com/crazy4pi314/status/1271837229221101569.
'Tand that returns an output of type
'Uis written as
'T => 'U; thus,
unitary : (Qubit = > Unit is Adj + Ctl)indicates that unitary takes a
Qubitinput and returns
is Adj + Ctltells you that
unitaryis adjointable and controllable (very important for some of the contest problems!). In this case, we can read the signature of
Solveas telling us that it takes an adjointable and controllable single-qubit operation, and returns an
Adjointfunctor to an adjointable operation doesn't change the type of that operation. For example,
Qubit => Unit is Adj + Ctl, such that
Adjoint Xalso has type
Qubit => Unit is Adj + Ctl. In the example from the contest problem,
Adjoint unitaryis an operation with type
(Qubit => Unit is Adj + Ctl). By contrast, for a controllable operation
Controlled opdoes modify the type to add the new control register. For more detail, you can checkout my book with @crazy4pi314 at bit.ly/qsharp-book, or the Microsoft Quantum docs at https://docs.microsoft.com/quantum/user-guide/using-qsharp/operations-functions#controlled-and-adjoint-operations.