The Jaynes-Cummings block of excitation number

PHYSICS · Multi-part · Fermi University Physics · Quantum Information & Technology · Physical realizations and quantum technologies · Physical realizations and quantum technologies

Quantum Information and Technology, Ch Superconducting qubits and circuit QED

Problem

The Jaynes-Cummings block of excitation number nn, in the basis {∣n+1,0⟩,∣n,1⟩}\{|n+1,0\rangle,|n,1\rangle\} (photon number, qubit state, ∣0⟩|0\rangle the qubit ground state), has exact eigenvalues (n+12)ℏωr±ℏ(n+1)g2+Δ2/4,(n+\tfrac12)\hbar\omega_r \pm \hbar\sqrt{(n+1)g^2+\Delta^2/4}, with Δ=ωq−ωr>0\Delta = \omega_q-\omega_r > 0 the qubit-resonator detuning and gg the coupling strength. ((a)) For gn+1≪Δg\sqrt{n+1} \ll \Delta, expand the square root to first order in (g/Δ)2(g/\Delta)^2 and find the energy shift, relative to its uncoupled value, of the dressed level that reduces to ∣n+1,0⟩|n+1,0\rangle as g→0g\to0, and of the dressed level that reduces to ∣n,1⟩|n,1\rangle. ((b)) Using the n=0n=0 block, find the shift of the level connected to ∣1,0⟩|1,0\rangle and hence the resonator frequency when the qubit is in ∣0⟩|0\rangle. ((c)) Using the n=0n=0 block (for the shift of ∣0,1⟩|0,1\rangle) and the n=1n=1 block (for the shift of ∣1,1⟩|1,1\rangle), find the resonator frequency when the qubit is in ∣1⟩|1\rangle, and hence show the dispersive shift is χ=g2/Δ\chi = g^2/\Delta. ((d)) Evaluate χ/2π\chi/2\pi for g/2π=45.0 MHzg/2\pi = 45.0\,\mathrm{MHz} and Δ/2π=900 MHz\Delta/2\pi = 900\,\mathrm{MHz}.

Browse the problem library