A transmon qubit is fundamentally in LC circuit. How are gates implemented in a transmon qubit? How do we know what voltage corresponds to the $\sigma_x$ gate for example?


1 Answer 1


I recommend checking out this reference if you haven't already, specifically page 28. Basically, starting from the classical Hamiltonian of the circuit described in figure 12, we see that the classical external voltage drive $V_d(t)$ ends up scaling the circuit charge variable $Q$, which, when quantized, can be expressed as the operator $Q \rightarrow \hat{Q} \propto -i (a - a^\dagger)$. This operator has the matrix representation $$ \begin{align} -i(a-a^\dagger) = -i\begin{bmatrix} 0 & 1 & \dots \\ -1 & 0 & \\ \vdots & & \ddots \end{bmatrix} \end{align} $$ which, after truncating to a two-dimensional subspace, becomes the qubit $\sigma_y$ operator. If the coupling had involved the flux variable $\phi$ instead, then we would have obtained the $\sigma_x$ operator, though the two are equivalent up to a $\pi/2$ rotation about the $z$-axis. Provided the ability to physically rotate about any single axis in the equator of the Bloch sphere, we can choose an appropriate rotating frame in which we effectively rotate about any desired axis in the equator (this is why in Eq. 92 we can control whether we get a $\sigma_x$ or a $\sigma_y$ rotation).

If by $\sigma_x$ gate you mean the bit flip, this corresponds to computing the resulting unitary evolution under the above interaction Hamiltonian in the rotating frame, and choosing an envelope function whose integral over the gate time is $\pi$, which means that the net action of the evolution is a rotation of $\pi$ radians about the $x$-axis (Eq. 93).

P.S. remember that truncating to a two-level system is only valid due to the energy spectrum of the transmon, that the energy of $|2\rangle$ is somewhat well-separated from the computational subspace.


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