# How are quantum gates encoded in voltage/current waves for superconducting qubits?

From what I understand, the voltage and current waves propagate down the input transmission line, hit the nonlinear oscillator, and then bounce back toward the generator. How are gates such as the X or the Hadamard encoded into these waves?

The Hamiltonian for an LC oscillator with capacitance $$C$$ and inductance $$L$$, and coupled through a "drive" capacitor $$C_d$$ to a voltage source with voltage $$V_d(t)$$ is $$H \approx \frac{Q^2}{2C} + \frac{\Phi^2}{2L} + V_d(t) Q \frac{C_g}{C}$$ where the approximation is $$C \gg C_g$$ and where $$\Phi$$ and $$Q$$ are quantum operators representing the magnetic flux in the inductor and the charge across the capacitor. The commutator is $$[\Phi, Q] = i\hbar$$.
It's much easier to understand this system with raising/lowering operators and in a rotating frame. Let's switch to raising/lowering operators first. Define $$a = \Phi \frac{1}{2 \Phi_\text{zpf}} + i Q \frac{1}{2 Q_\text{zpf}} \quad$$ where $$\Phi_\text{zpf}^2 = \hbar Z / 2$$ is the zero point fluctuation of the flux, $$Q_\text{zpf} = \hbar / 2Z$$ is the zero point fluctuation in charge, and $$Z \equiv \sqrt{L/C}$$ is the characteristic impedance of the resonator. The commutator is $$[a,a^\dagger] = 1$$. In terms of this new operator, the Hamiltonian is $$H = \hbar \omega_r (a^\dagger a + 1/2) - i V_d(t) \frac{C_g}{C}Q_\text{zpf}(a - a^\dagger) \, .$$ Now we go to the rotating frame as described in this post, with rotation operator $$R(t) = \exp(i \omega_f t n)$$ where $$n = a^\dagger a$$. Doing so, we find $$H = \hbar \Delta a^\dagger a + i V_d(t)Q_\text{zpf}(- a e^{-i \omega_f t} + a^\dagger e^{i \omega_f t})$$ where $$\Delta = \omega_r - \omega_f$$ is the "detuning" of the resonator from the rotating frame and we dropped a constant term coming from the $$1/2$$ in the original Hamiltonian. Now suppose the drive voltage has a sinusoidal dependence $$V_d(t) = V_d \cos(\omega_d t + \phi) = \frac{V_d}{2} \left( e^{i (\omega_d t + \phi)} + e^{-i (\omega_d t + \phi)} \right)\, .$$ Then the Hamiltonian is $$H=\hbar \Delta a^\dagger a + i \frac{V_d}{2}Q_\text{zpf} \left( -a e^{i((\omega_d - \omega_f)t + \phi)} -a e^{-i((\omega_f + \omega_d)t + \phi)} + a^\dagger e^{i((\omega_f + \omega_d)t + \phi)} + a^\dagger e^{-i((\omega_d - \omega_f)t + \phi)} \right) \, .$$ Now suppose we choose $$\omega_f = \omega_d$$, i.e. the rotating frame rotates as the same frequency as the drive's oscillation frequency. In that case we get $$H=\hbar \Delta a^\dagger a + i \frac{V_d}{2} Q_\text{zpf} \left( -a e^{i \phi} -a e^{-i(2\omega_d t + \phi)} + a^\dagger e^{i(2 \omega_d t + \phi)} + a^\dagger e^{-i \phi} \right) \, .$$ The first and last terms have no time dependence while the other two terms oscillate. In the so-called "rotating wave approximation" those oscillating terms are dropped. Whether or not this is a good approximation depends on some subtle details that we won't go into here; it's an ok approximation for many applications. With those terms dropped, we get $$H=\hbar \Delta a^\dagger a + i \frac{V_d}{2} Q_\text{zpf} \left( -a e^{i \phi} + a^\dagger e^{-i \phi} \right) \, .$$ If we pretend that this oscillator has only two levels, then we can truncate the $$a$$ and $$a^\dagger$$ operators as $$a = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \quad a^\dagger = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$ so \begin{align} H &= \begin{pmatrix} 0 & -i (V_d Q_\text{zpf}/2) e^{i \phi} \\ i (V_d Q_\text{zpf} / 2) e^{-i \phi} & \hbar \Delta \end{pmatrix} \\ &= \begin{pmatrix} 0 & (V_d Q_\text{zpf}/2) (-i \cos\phi + \sin\phi) \\ (V_d Q_\text{zpf} / 2) (i \cos\phi + \sin\phi) & \hbar \Delta \end{pmatrix} \\ (\Delta = 0) \quad &= \frac{V_d Q_\text{zpf}}{2} \left( \sin \phi \, \sigma_x + \cos \phi \sigma_y \right) \, . \end{align} In the last line we took $$\Delta = 0$$ meaning that $$\omega_f = \omega_r = \omega_d$$. You can analyze the problem without that assumption but I did it here to illustrate a simple point. The drive just creates Hamiltonian terms proportional to $$\sigma_x$$ and $$\sigma_y$$. Therefore, the drive creates rotation around the X and Y axes of the Bloch sphere. The speed of rotations is proportional to $$V_d Q_\text{zpf}/2$$ and the angle on the Bloch sphere is determined by the phase $$\phi$$ of the drive.
In the case $$\Delta \neq 0$$, then we just have additional rotation about the Z axis of the Bloch sphere.