deletions | additions       diff --git a/Linear regime.tex b/Linear regime.tex  index b19ae31..1458d41 100644  --- a/Linear regime.tex  +++ b/Linear regime.tex 
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Since $\v{y}(t) = \begin{matrix}(0 & 0 & 0 & F)\end{matrix} \v{x}(t)$, we can find a closed-form solution for $\v{y}(t)$:  \[ \v{y}(t) = \begin{matrix}(M_0^3 F\begin{matrix}(M_0^3  t^3/6 & M_0^2 t^2/2 & M_0 t & 1)\end{matrix} \v{x}(0) \] Expanding $\v{x}(0)$ in terms of its perturbation components, the above equation simplifies:  \[ \v{y}(t) = M_0^3 F [M_0^3  t^3 \rho_0(0)/6 + M_0^2 t^2 \rho_1(0)/2 + M_0 t \rho_2(0) + \rho_3(0) \rho_3(0)]  \] Thus, for a particular $E(t)$, the output polarization maps to a cubic function of time (after the third pulse). System identification techniques to identify $M_0$ and $\{\rho_i(0)|1\le i\le 3\}$ could prove useful. (Here, $\rho_0$ is a constant, so it should still represent the ground state of the molecule, as it did before $E(t)$ was applied.)