Two Qubit Gates

In this example we will solve for a selection of two-qubit gates using a simple two-qubit system. We will use the UnitarySmoothPulseProblem template to solve for the optimal control fields.

Defining our Hamiltonian

In quantum optimal control we work with Hamiltonians of the form

\[H(t) = H_{\text{drift}} + \sum_{j} u^j(t) H_{\text{drive}}^j,\]

Specifically, for a simple two-qubit system in a rotating frame, we have

\[H = J_{12} \sigma_1^x \sigma_2^x + \sum_{i \in {1,2}} a_i^R(t) {\sigma^x_i \over 2} + a_i^I(t) {\sigma^y_i \over 2}.\]

where

\[\begin{align*} J_{12} &= 0.001 \text{ GHz}, \\ |a_i^R(t)| &\leq 0.1 \text{ GHz} \\ \end{align*}\]

And the duration of the gate will be capped at $400 \ \mu s$.

Let's now set this up using some of the convenience functions available in QuantumCollocation.jl.

using QuantumCollocation
using PiccoloQuantumObjects
using NamedTrajectories
using LinearAlgebra

using PiccoloPlots
using CairoMakie

⊗(a, b) = kron(a, b)

# Define our operators
σx = GATES[:X]
σy = GATES[:Y]
Id = GATES[:I]

# Lift the operators to the two-qubit Hilbert space
σx_1 = σx ⊗ Id
σx_2 = Id ⊗ σx

σy_1 = σy ⊗ Id
σy_2 = Id ⊗ σy

# Define the parameters of the Hamiltonian
J_12 = 0.001 # GHz
a_bound = 0.100 # GHz

# Define the drift (coupling) Hamiltonian
H_drift = J_12 * (σx ⊗ σx)

# Define the control Hamiltonians
H_drives = [σx_1 / 2, σy_1 / 2, σx_2 / 2, σy_2 / 2]

# Define control (and higher derivative) bounds
a_bound = 0.1
da_bound = 0.0005
dda_bound = 0.0025

# Scale the Hamiltonians by 2π
H_drift *= 2π
H_drives .*= 2π

# Define the time parameters
T = 100 # timesteps
duration = 100 # μs
Δt = duration / T
Δt_max = 400 / T

# Define the system
sys = QuantumSystem(H_drift, H_drives)

QuantumSystem: levels = 4, n_drives = 4

Now let's plot the pulse and the population trajectories for the first two columns of the unitary, i.e. initial state of $\ket{00}$ and $\ket{01}$. For this we provide the function plot_unitary_populations.

plot_unitary_populations(prob.trajectory, control_name=:a)

For fun, let's look at a minimum time pulse for this problem

min_time_prob = UnitaryMinimumTimeProblem(prob, U_goal; final_fidelity=.995)

DirectTrajOptProblem
   timesteps            = 100
   duration             = 375.53104362185144
   variable names       = (:Ũ⃗, :a, :da, :dda, :Δt)
   knot point dimension = 45

solve!(min_time_prob; max_iter=300)

fid_final_min_time = unitary_rollout_fidelity(min_time_prob.trajectory, sys, drive_name=:a)
println(fid_final_min_time)
@assert fid_final_min_time > 0.99

0.9949999904078358

And let's plot this solution

plot_unitary_populations(min_time_prob.trajectory, control_name=:a)

It looks like our pulse derivative bounds are holding back the solution, but regardless, the duration has decreased:

duration = get_duration(prob.trajectory)
min_time_duration = get_duration(min_time_prob.trajectory)
println(duration, " - ", min_time_duration, " = ", duration - min_time_duration)

375.53104362185144 - 355.48583992851627 = 20.045203693335168

Mølmer–Sørensen gate

Here we will solve for a Mølmer–Sørensen gate between two. The gate is generally described, for N qubits, by the unitary matrix

\[U_{\text{MS}}(\vec\theta) = \exp\left(i\sum_{j=1}^{N-1}\sum_{k=j+1}^{N}\theta_{jk}\sigma_j^x\sigma_k^x\right),\]

where $\sigma_j^x$ is the Pauli-X operator acting on the $j$-th qubit, and $\vec\theta$ is a vector of real parameters. The Mølmer–Sørensen gate is a two-qubit gate that is particularly well-suited for trapped-ion qubits, where the interaction between qubits is mediated.

Here we will focus on the simplest case of a Mølmer–Sørensen gate between two qubits. The gate is described by the unitary matrix

\[U_{\text{MS}}\left({\pi \over 4}\right) = \exp\left(i\frac{\pi}{4}\sigma_1^x\sigma_2^x\right).\]

Let's set up the problem.

# Define the goal operation
U_goal = exp(im * π/4 * σx_1 * σx_2)

# Set up and solve the problem

prob = UnitarySmoothPulseProblem(
    sys,
    U_goal,
    T,
    Δt;
    a_bound=a_bound,
    da_bound=da_bound,
    dda_bound=dda_bound,
    R_da=0.01,
    R_dda=0.01,
    Δt_max=Δt_max,
    piccolo_options=PiccoloOptions(bound_state=true),
)
fid_init = unitary_rollout_fidelity(prob.trajectory, sys, drive_name=:a)
println(fid_init)

    constructing UnitarySmoothPulseProblem...
	using integrator: typeof(UnitaryIntegrator)
	control derivative names: [:da, :dda]
	applying timesteps_all_equal constraint: Δt
0.12249963901581312

solve!(prob; max_iter=100)

# Let's take a look at the final fidelity
fid_final = unitary_rollout_fidelity(prob.trajectory, sys, drive_name=:a)
println(fid_final)
@assert fid_final > 0.999

0.9999999791551119

Again, looks good!

Now let's plot the pulse and the population trajectories for the first two columns of the unitary, i.e. initial state of $\ket{00}$ and $\ket{01}$.

plot_unitary_populations(prob.trajectory, control_name=:a)