How do IBM, D-Wave, etc. change phase physically? In real hardware?


1 Answer 1


If by 'change phase' you mean the relative phase of the $|0\rangle$ and $|1\rangle$ state, then the short answer is: IBM doesn't, D-Wave doesn't really care.

For IBM:

Whenever a phase gate (that is, a rotation around the $Z$-axis in the Bloch sphere picture) is applied in the circuit that needs to be run on the hardware, the compiler keeps track of this phase change but does not actually physically implement anything. Rather, it changes the orientation of the $X$- and $Y$-axis, so that any subsequent rotation around these new axes is as if they were the original axes preceded by the $Z$ rotation. In essence, the entire Bloch sphere is rotated 'through software' along the $Z$-axis.

This only works because of two things:

  • the $X$- and $Y$-axis are a relatively arbitrary choice (they of course still need to be orthogonal). We like to think about the $X$-axis as "that axis for which the angle along the $Z$-axis is $0$", but we can just as equal think of it the other way around: Setting the $X$-axis (completely arbitrary) then actually defines not only the $Y$-axis but also the (angle of) the $Z$-axis.
  • The non-trivial rotations (that is, the rotations along axes in the $X-Y$ plane) that can be physically implemented are not limited to any two axes in that plane; a continuous range of axes can be implemented. The angle of the rotation is determined by the phase of the incident microwave$^{1}$; any phase can then be used to implement any angle. Therefore, the physical operations that performs the $X$- and $Y$-rotations are the same x-rays, only differing in phase.

For D-Wave

D-Wave has an entirely different type of quantum computer: the adiabatic/annealing-model. I am far from an expert on this, but I do believe that this computational model can always be used in such a way that $Z$-rotations are irrelevant. Please correct me if I am wrong.

Further reference

A good introduction on transmons, the type of qubits that IBM uses, can be found here:

With D-Wave I'm much less familiar but I know that the 'standard' or 'beginning' Hamiltonian of the system is slowly transformed to the target Hamiltonian by applying (transverse (?)) magnetic fields.


  1. This is a gross oversimplification and should not be perpetuated:)
  • $\begingroup$ Concerning D-Wave, it is not gate-based quantum computer, so question how it implements Rz gate is irrelevant. $\endgroup$ May 27, 2020 at 6:28
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    $\begingroup$ In D-Wave system's, the effective Hamiltonian of the entire system is changed so that the answer of the question is encoded into the ground state of that Hamiltonian. Applying a gate in any gate-based quantum computer is also effectively changing the Hamiltonian, invoking a rotation around an axis or a coupling between certain states. The same goes for the adiabatic quantum computer - there are no gates, per se, but the phases of qubits (i.e. parts of the Hamiltonian) still might need to be changed to arrive at the desired full-system Hamiltonian. $\endgroup$
    – JSdJ
    May 27, 2020 at 8:10
  • $\begingroup$ Do you know for any video that explain all this, because I have hard time to understand it. Any video how IBM QC or D-Wave works at hardware level? Not math, but physics. Thanks. $\endgroup$
    – guest
    May 28, 2020 at 1:59
  • $\begingroup$ I've added a couple of links to some videos from QuTech academy $\endgroup$
    – JSdJ
    May 28, 2020 at 8:02
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    $\begingroup$ @JSdJ: I meant that you do not operate with gates directly as on e.g. IBM Q. However, thanks for the comment. $\endgroup$ May 28, 2020 at 8:25

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