“Know how to solve every problem that has been solved.” “What I cannot create, I do not understand.” — Richard Feynman

Two-Level Systems by Hand — Why qubits are operated at the sweet spot

Exercises

Introductory Quantum Mechanics › Unit 3 · Two-level systems & spin-1/2 › Two-Level Systems by Hand › all problems

Practice Problem 6 of 9

Why qubits are operated at the sweet spot

Problem

A qubit’s diagonal asymmetry

Δ

is swept by a control voltage while the coupling is fixed at

V / h = 150

MHz. Find the transition frequency at

Δ/ h = 200

MHz and at

Δ = 0

. Then explain, via

d E_{gap} / d Δ

, why

Δ = 0

is the preferred operating point when the control voltage is noisy.

Solution

Predict before reading on. The gap formula from the worked example, twice — then differentiate it. What is the slope of $Δ^{2} + V^{2}$ at $Δ = 0$ ?

E_{gap} = 2 Δ^{2} + V^{2} : 2 20 0^{2} + 15 0^{2} = 500 MHz, 2 V = 300 MHz at Δ = 0

\frac{d E _{gap}}{d Δ} = \frac{2Δ}{Δ ^{2} + V ^{2}} = 0 at Δ = 0

At the avoided-crossing minimum the transition frequency is first-order insensitive to the noisy knob — fluctuations in $Δ$ enter only quadratically. Superconducting-qubit people call it the sweet spot; it is the reason coherence times jump when you park there.

Result

500 MHz; 300 MHz; d E / d Δ = 0 at the crossing