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In this paper about Google superconducting qubit system, I want to understand the origin of equation (1), and the origin of the $\nu/2 Z$ term in equation (2). Regarding the first point, I want to know how one can derive to obtain this equation. Regarding the second point, I understand that the equation describes the interaction frame transformation, but I don't see how this specific term arises from the transformation, as it is not involved in the original Hamiltonian in equation (1). It would be great if anyone could provide explanations on these points.

karavan
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1 Answers1

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In general, for any basis $| \psi_U(t) \rangle = U(t) | \psi(t) \rangle$, one can define the "effective" Hamiltonian that governs the dynamics via $$\partial_t |\psi_U(t)\rangle = (\partial_t U(t)) |\psi(t)\rangle + U(t) \partial_t |\psi(t)\rangle = (\partial_t U(t)) U^\dagger(t) U(t) |\psi(t)\rangle -i (U(t) H U^\dagger(t)) U(t) |\psi(t)\rangle =: -i H_U |\psi_U(t)\rangle, $$ with $H_U = i (\partial_t U(t)) U^\dagger(t) + U(t) H U^\dagger(t)$. Starting with the RWA-Hamiltonian $$H_{RWA}(t) = −\frac{1}{2} \omega_{01}(t) Z + (\Omega (t) e^{−i\nu t} \sigma_- + h.c.),$$ we can go into the interaction picture by considering $| \psi_I(t) \rangle = e^{-i \nu t Z/2} |\psi(t)\rangle$. As $\partial_t |\psi_I(t)\rangle = -i \nu t Z/2 e^{-i \nu t Z/2} |\psi(t)\rangle + e^{-i \nu t Z/2} \partial_t|\psi(t)\rangle$, the Schrödinger equation in this basis is governed by the Hamiltonian $$ H = e^{-i \nu t Z/2} H_{RWA} e^{i \nu t Z/2} + \frac{1}{2} \nu Z, $$ which gives you Eq. (2).

Refik Mansuroglu
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