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e solved exactly:
A(t + ∆t) = e −i(Tmn X t→t+∆t σ x+T mn Y t→t+∆t σ y+(T nn−T mm) Z t→t+∆t σ z)∆t/2 A(t) (3.48)
Since the transition matrix elements T mn are usually not known exactly and will be
calibrated away in the experiment, it is more useful to express the drive strengths
X, Y , and Z in terms of the frequency of rotation they cause. This equation then
becomes:
A(t + ∆t) = e −iπ∆t (X t→t+∆t σ x+Y t→t+∆t σ y+Z t→t+∆t σ z) A(t) (3.49)
Since the Z t→t+∆t σ z term corresponds to a rotation of the state around the Z-axis,
it can be used to emulate an off-resonant drive. For example, a drive causing a
rotation at a rate of 20 MHz around the Y-axis that is detuned from the qubit by
5 MHz is simulated as:
A(t + ∆t) = e −iπ∆t (20 MHz σ y+5 MHz σ z ) A(t) (3.50)
Effectively, this leads to a final rotation around an axis that is tilted out of
the X/Y-plane of the Bloch sphere. The vector therefore no longer traces out the
great-circle through the | 1 〉-state, but instead rotates faster with less amplitude
as shown in Figure 3.4c. This picture also visualizes nicely why a drive that is
far off resonance with a given transition can safely be ignored, as it does not
significantly move the state away from the pole.
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