For a superconducting qubit with a frequency of 5 GHz, which is closest to the energy difference hf between |0⟩ and |1⟩ expressed as a temperature, hf/k_B? Use Planck's constant h = 6.63×10⁻³⁴ J·s and Boltzmann's constant k_B = 1.38×10⁻²³ J/K.

Answer: B. About 0.24 K

hf = 6.63×10⁻³⁴ × 5×10⁹ ≈ 3.3×10⁻²⁴ J, and dividing by k_B gives about 0.24 K. In thermal equilibrium the fraction thermally excited to |1⟩ is set by exp(−hf/k_B T), so at 20 mK, well below this, hf/k_B T ≈ 12 and thermal excitation is almost negligible. Conversely, at liquid helium temperature, 4 K, k_B T is about 17 times hf and the device cannot be used as a qubit.

• A. This uses the reduced Planck constant ħ, h divided by 2π, instead of h.
• B. Correct. 3.3×10⁻²⁴ J ÷ 1.38×10⁻²³ J/K ≈ 0.24 K.
• C. This multiplies h by the angular frequency 2πf instead of the frequency f.