The transmon Hamiltonian in the regime is

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Problem

The transmon Hamiltonian in the regime EJ≫ECE_J \gg E_C is H=4ECn2−EJcos⁡φH = 4E_Cn^2 - E_J\cos\varphi, with [φ,n]=i[\varphi,n]=i. Expanding cos⁡φ=1−φ2/2+φ4/24−⋯\cos\varphi = 1-\varphi^2/2+\varphi^4/24-\cdots leaves a harmonic term 12EJφ2\tfrac12 E_J\varphi^2 plus a quartic perturbation −124EJφ4-\tfrac{1}{24}E_J\varphi^4 (dropping the constant −EJ-E_J). The harmonic part is a mode of angular frequency ωp=8EJEC/ℏ\omega_p = \sqrt{8E_JE_C}/\hbar with zero-point phase spread φzpf=(2EC/EJ)1/4\varphi_{\rm zpf} = (2E_C/E_J)^{1/4}, so φ=φzpf(a+a†)\varphi = \varphi_{\rm zpf}(a+a^\dagger) in terms of the usual ladder operators and number states ∣n⟩|n\rangle. ((a)) Using ⟨n∣(a+a†)4∣n⟩=6n2+6n+3\langle n|(a+a^\dagger)^4|n\rangle = 6n^2+6n+3, find the first-order energy shift of level nn from the quartic term, and hence the level energies EnE_n up to an additive constant. ((b)) Show that the n→n+1n\to n+1 transition frequency is fn,n+1=ωp2π−(n+1)EChf_{n,n+1} = \dfrac{\omega_p}{2\pi} - (n+1)\dfrac{E_C}{h}, and hence that the anharmonicity α/2π≡f12−f01=−EC/h\alpha/2\pi \equiv f_{12}-f_{01} = -E_C/h. ((c)) Evaluate f01f_{01} and α/2π\alpha/2\pi for EJ/h=18.0 GHzE_J/h = 18.0\,\mathrm{GHz} and EC/h=0.360 GHzE_C/h = 0.360\,\mathrm{GHz}.

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