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If we call your constant_ampl $\Gamma(t)$, the constant shape in the optimized_field $S(t)$ and the actual control t -> 1 $\epsilon(t)$, it seems to me your Hamiltonian is

$$ \hat{H} = \Gamma(t) \hat{H}_1 + S(t) \epsilon(t) \hat{H}_2 + \Gamma(t) S(t) \epsilon(t) \hat{H}_3 $$

The product $\Gamma(t) S(t)$ is just another shape function for a ShapedAmplitude. So this should be totally fine, as something like:

ϵ(t) = 1
S(t) =  QuantumControl.Shapes.flattop(t, T=T, t_rise=0.3, func=:blackman)
Γ(t) = t^2
optimized_field = QuantumControl.Controls.ShapedAmplitude(ϵ; shape=S)
constant_ampl = QuantumControl.Controls.LockedAmplitude(Γ)
cross_term = QuantumControl.Controls.ShapedAmplitude(ϵ; shape=(t…

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@will-atom
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@will-atom
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@goerz
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goerz Jul 3, 2025
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@mabuni1998
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@goerz
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goerz Jul 3, 2025
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Answer selected by mabuni1998
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Converted from issue

This discussion was converted from issue #80 on February 10, 2025 16:52.