JuliaQuantumControl
:TDSE or :LvN? #14
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When I solve a Liouville equation with dissipation, I use the convention |
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Replies: 1 comment 4 replies
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That's because the propagators calculate See also https://juliaquantumcontrol.github.io/QuantumPropagators.jl/stable/overview/#Propagation-methods |
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I'm sorry, I didn't realize I had posted in the wrong place. Let's say I want to use the LvN convention instead of TDSE. My problem is like: dρ/dt = -i[H,ρ] + AρA† - 1/2A†Aρ - 1/2ρA†A, so by the form of the equation, the LvN convention should work. Is that correct? If it is not correct, what should I do to use the LvN convention? My questioning is because I want to understand and not just get the result I need. As an example I took your example "Optimization of Dissipative Qubit Reset" (in Python) and rewrote it in a slightly modified version for Julia, where I want to get a control to go from the state ρ_th = ρ_q_th ⊗ ρ_t_th to the state ρ_tgt = |00⟩⟨00|, with maximum fidelity (of the qubit+TLS system). The code only works if I use TDSE. What do I need to do to make the code work with LvN? using LinearAlgebra, QuantumControl, QuantumControlBase, PyPlot, Dates
if !@isdefined(σz)
const ⊗ = kron
const 𝕚 = im
const 𝟙 = ComplexF64[1 0; 0 1]
const ket0 = ComplexF64[1, 0]
const ket1 = ComplexF64[0, 1]
const σx = ComplexF64[0 1; 1 0]
const σz = ComplexF64[1 0; 0 -1]
end
function tsince(tini)
# time in hhmmss elapsed from tini to now()
tg = round(Int64, (now()-tini).value/1e3)
tgh = tg÷3600 ≠ 0 ? "$(tg÷3600)h" : ""
tgm = tg%3600÷60 ≠ 0 ? "$(tg%3600÷60)m" : ""
(tg, "$tgh$tgm$(tg%3600%60)s")
end
function S(t, T)
# Shape function for the field update
QuantumControlBase.Shapes.flattop(t; T=T, t_rise=T/20)
end
function q_TLS_liouvillian(ε1, ωq, ωt, J, κ, β)
# Liouvillian for the coupled system of qubit and TLS
# drift qubit Hamiltonian
H0_q = -0.5 * ωq * σz
# drive qubit Hamiltonian
H1_q = -0.5 * σz
# drift TLS Hamiltonian
H0_T = -0.5 * ωt * σz
# qubit-TLS interaction
H_int = J * σx ⊗ σx
# Lift Hamiltonians to joint system operators
H0 = H0_q ⊗ 𝟙 + 𝟙 ⊗ H0_T + H_int
H1 = H1_q ⊗ 𝟙
# Define Lindblad operators
Nth = 1.0 / (exp(β*ωt) - 1.0)
# Cooling on TLS
L1 = √(κ * (Nth + 1)) * 𝟙 ⊗ (ket0*ket1')
# Heating on TLS
L2 = √(κ * Nth) * 𝟙 ⊗ (ket1*ket0')
H = (H0, (H1, ε1))
# return
liouvillian(H, [L1, L2]; convention=:TDSE)
end
function q_TLS_optimize()
ωq = 1.0 # qubit level splitting
ωt = 3.0 # TLS level splitting
J = 0.1 # qubit-TLS coupling
κ = 0.04 # TLS decay rate
β = 1.0 # inverse bath temperature
T = 25.0 # final time
nt = 2500 # number of time steps
tlist = collect(range(0, T, nt))
ε1 = t -> S(t, T)
L = q_TLS_liouvillian(ε1, ωq, ωt, J, κ, β)
# Define the optimization target
x_q = ωq * β / 2.0
ρ_q_th = diagm(ComplexF64[exp(x_q), exp(-x_q)]) / 2.0 / cosh(x_q)
x_t = ωt * β / 2.0
ρ_t_th = diagm(ComplexF64[exp(x_t), exp(-x_t)]) / 2.0 / cosh(x_t)
# initial state of the joint system {qubit and TLS}
ρ_th = ρ_q_th ⊗ ρ_t_th
# target state of the system {qubit and TLS}
ρ_tgt = (ket0⊗ket0) * (ket0⊗ket0)'
objectives = [
Objective(
initial_state=reshape(ρ_th, :),
generator=L,
target_state=reshape(ρ_tgt, :),
)
]
problem = ControlProblem(
objectives=objectives,
prop_method=:newton,
use_threads=true,
pulse_options=IdDict(
ε1 => Dict(
:lambda_a => 1.0,
:update_shape => t -> S(t, T),
),
),
tlist=tlist,
iter_stop = length(ARGS) ≥ 1 ? Int(parse(Float64, ARGS[1])) : 1,
J_T=QuantumControlBase.Functionals.J_T_ss,
check_convergence=res -> begin
((res.J_T < 1e-1) && (res.converged = true) && (res.message = "J_T < 10⁻¹"))
end
)
optimize(problem, method=:krotov)
end
function main()
tini = now()
opt = q_TLS_optimize()
ts = tsince(tini)[2]
fidelity = 1.0 - opt.J_T
plot(opt.tlist, opt.optimized_controls[1])
xlabel("t")
ylabel("Optimized control")
title("Fidelity: $fidelity, ts=$ts")
savefig("q_TLS.png")
println("F = $fidelity")
println("total time = $ts")
end
main()The following result is with 3000 iterations. I did it with fewer iterations in order to post the result.
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Not at all! I only created the Discussions forum after you posted the question.
What do you mean by "work"? For optimal control to work, the left hand side equation of motion has to be
Yes, it's correct, but it's not the equation of motion that our implementation of Krotov's method expects
There short answer is, you can't. The long answer is, if you really really know what you're doing, you could implement a custom propagator that evaluates
You seem to think that there is some fundamental difference between the :LvN and the :TDSE convention. There really isn't. It's the same equation of motion. It's only a question of what you call "L" (the Liouvillian). Maybe if you can formulate what exactly makes you uncomfortable about Don't be discouraged just because I'm telling you you can't use
Yes, that's expected: you're effectively putting an factor of
Nothing, really, within the limits of reasonable effort. It's just not going to work with |
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I meant that I wasn't getting the results I expected, but I knew it was because I was using :LvN in the wrong way. Sometimes I didn't express myself correctly in English. Now I don't feel uncomfortable anymore. Goerz, thank you very much for the explanation. |
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No worries! Don't hesitate to ask if you have any other questions! I'll be happy to help as best as I can. |
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That's because the propagators calculate
ρ(t+dt) = exp[-i L dt] ρ(t), which is the solution toi ρ̇(t) = L ρ(t), which has the same form as the time-dependent Schrödinger equation (convention=:TDSE). However, most textbooks define the LiouvillianLso that the equation of motion isρ̇(t) = L ρ(t)(convention=LvN). So the point is that if you useconvention=:TDSE, your propagator "just works" for both a Hilbert space state vector and a density matrix.See also https://juliaquantumcontrol.github.io/QuantumPropagators.jl/stable/overview/#Propagation-methods