When an RF pulse is interacting on resonance with a transmon qubit, it leads to rotation of the qubit around an axis in the XY plane (in a reference frame rotating in the transition frequency of the qubit). What is the exact relation between the pulse amplitude and phase to the rotation of the qubit? Can you give a proof?


1 Answer 1


Kirchhoff to Lagrangian

Let's approximate the transmon as a parallel LC resonant circuit. Suppose we connect a voltage source through a coupling capacitor $C_d$ (d for "drive") to a transmon qubit. If the voltage of the source is $V_d(t)$, then Kirchhoff's equations for the circuit are $$\frac{1}{C/C_d} \dot{V}_d(t) = \ddot{\Phi}(t) + \frac{\omega_0^2}{1 + C_d / C} \Phi(t)$$ where $C$ is the capacitance of the transmon, $\Phi(t)$ is the flux of the transmon, and $\omega_0 \equiv 1 / \sqrt{LC}$. This differential equation is reproduced by the following Lagrangian \begin{align} \mathcal{L} &= \frac{1}{2}C \dot{\Phi}^2 - \frac{1}{2L}\Phi^2 + \frac{1}{2} C_d (\dot{\Phi} - V_d(t))^2 \\ &= \underbrace{\frac{1}{2}(C + C_d) \dot{\Phi}^2 - \frac{1}{2L}\Phi^2}_{\mathcal{L}_0} + \underbrace{\frac{1}{2} C_d V_d(t)^2 - C_d V_d(t) \dot{\Phi}}_{\mathcal{L}_d} \end{align} as you can check yourself.

Lagrangian to Hamiltonian

The first part of $\mathcal{L}$ corresponds to the free oscillation of the qubit and we call it $\mathcal{L}_0$. Analyzing $\mathcal{L}_0$, we find that the canonical position and momentum are $$\text{position} = \Phi \qquad \text{and} \qquad \text{momentum} \equiv \frac{\partial \mathcal{L}_0}{\partial \dot{\Phi}} = (C + C_d)\dot{\Phi} \equiv Q\, .$$ Denoting the canonical momentum $\partial \mathcal{L}_0 / \partial \dot{\Phi}$ as $Q$ is justified by the following reasoning. When $C_d \rightarrow 0$, we have $\partial \mathcal{L}_0 / \partial \dot{\Phi} = C \dot{\Phi}$. But $\dot{\Phi}$ is just the voltage on the qubit, so using the definition of capacitance $C = Q / V$ we see that $\partial \mathcal{L}_0 / \partial \dot{\Phi} = Q$ where $Q$ is the charge on the qubit. It's then clear that adding the drive capacitor just shifts the capacitance associated with the canonical momentum, which makes sense because the drive capacitor simply adds to the qubit's own capacitance since they're connected in parallel. Of course, in practical systems $C_d \ll C$.

Now let's look at $\mathcal{L}_d$. The first term in $\mathcal{L}_d$ doesn't involve the dynamical variables at all, so we can ignore it. If we convert the second term to a Hamiltonian through the usual procedure, we find that the drive Hamiltonian $H_d$ is $$H_d = C_d \dot{\Phi}V_d = \frac{1}{1 + C/C_d} Q V_d \approx \frac{C_d}{C} Q V_d \, .$$ Using the theory of the quantum harmonic oscillator, the $Q$ operator can be written in terms of raising and lowering operators as $$Q = -i Q_\text{zpf} (a - a^\dagger) $$ where $Q_\text{zpf} = \sqrt{\hbar / 2 Z_{LC}}$ is the zero point fluctuation of the circuit's charge, and $Z_{LC} \equiv \sqrt{L/C}$ is the characteristic impedance of the oscillator. If we restrict to the lowest two levels, then $-i(a - a^\dagger) = \sigma_y$ so we get $$H_d(t) = \frac{C_d}{C} Q_\text{zpf} V_d(t) \, \sigma_y \, .$$

Interaction picture (a.k.a. rotating frame)

Of course, there's still the free oscillation part of the Hamiltonian $H_0$, which in the two-level approximation is $$H_0 = -\hbar \frac{\omega_0}{2} \sigma_z \, .$$ Because $H_0$ and $H_d$ don't commute, in order to make progress we have to go into the interaction picture. When you do that, if we assume that the drive voltage has the form $V_d(t) = V_0 \cos(\omega_0 t + \phi)$, then we get [2] $$H_d^\text{interaction} = \frac{1}{2} \frac{C_d}{C} Q_\text{zpf} V_0 \left( \cos(\phi) \sigma_x - \sin(\phi) \sigma_y \right) \, .$$

Final result

Without loss of generality, we can pick $\phi = 0$ and compute the unitary time evolution operator $$U(t) = \exp \left( - \frac{i}{\hbar} \int_0^t H_d^\text{interaction}(t') dt' \right) = - \frac{i}{\hbar} \frac{1}{2} \frac{C_d}{C} Q_\text{zpf} V_0 \sigma_x t \, .$$ When you have an exponent of a Pauli matrix like $\exp(-i \theta \sigma_x)$, the angle of rotation on the Bloch sphere is $2 \theta$ [1]. Therefore, in our case the Bloch sphere rotation angle (in radians) is $$\text{rotation angle} = \frac{C_d}{C} \frac{Q_\text{zpf}}{\hbar} V_0 t \, .$$ Now that we found the Bloch sphere rotation angle when $\phi=0$, it's clear from symmetry and the form of $H_d^\text{interaction}$ that changing $\phi$ keeps the same rotation angle on the Bloch sphere but changes the azimuth.

A form that I particularly like can be found by for the case of a pi-pulse where the Bloch sphere rotation angle is $\pi$. In that case, we can rewrite the result we found above as $$(\text{condition for pi-pulse)} \quad 1 = \frac{1}{\pi \sqrt{2}} \left( \frac{C_d}{C} \right) \sqrt{\frac{R_K / 8 \pi}{Z_{LC}}} \frac{V_0 t}{\Phi_0 / 2\pi}$$ where here $R_K \equiv h / e^2$ is the resistance quantum and $\Phi_0 \equiv h / 2e$ is the flux quantum.


[1] You can check this by Taylor expanding the exponential function and grouping even and odd terms.

[2] Here we are using the so-called rotating wave approximation.


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