+  \frac{\sigma_z}{2} \otimes \sigma_0 \otimes \frac{\sigma_z}{2} \right), \end{aligned}$$ where $\omega_0$ corresponds to a Larmor frequency of the center spin and $J$ a coupling constant between the center spin and the two side spins. The signal from the center spin is given as [@Cory1634; @Gershenfeld350; @liquidNMRQC] $$\begin{aligned}
 S_k(t) &=& {\rm Tr}\left( 
\left((\sigma_x+i \sigma_y) \otimes \sigma_0 \otimes \sigma_0 \right) 
e^{-i H t}\rho_ke^{iHt}
\right).\end{aligned}$$ Therefore, the expected normalized signals are given as $$\begin{aligned}
 S_\bigcirc(t) 
%&=& \frac{1}{4}e^{-t/T_2}
%\left( 1+ \cos Jt \right)\cos(\omega_0 t + \phi)\\ 
 &=& \frac{1}{4}e^{-t/T_2}
\left( e^{-iJt} + 2 +  e^{iJt} \right)\cos(\omega_0 t + \theta), \label{Sl}\\
 S_\circ(t) 
%&=& \frac{1}{4}e^{-t/T_2}
%\left( 1+ \cos (Jt + 2\phi) \right)\cos \omega_0 t \\
&=&  \frac{1}{4}e^{-t/T_2}
\left( e^{-iJt-i2\theta} + 2 +  e^{iJt+i2\theta} \right)\cos \omega_0 t \label{Ss}, \end{aligned}$$ where $T_2$ is a phenomenological transverse relaxation time constant that is introduced in order to express that the signal does not last forever. Equation (\[Sl\]) corresponds to the signal when the field is applied on the center spin, while Eq. (\[Ss\]) is when on the side spins. Note the difference in the position of $\theta$ in $S_k(t)$. The three terms in the parentheses in $S_k(t)$ correspond to three peaks that are observed when $S_k(t)$’s are Fourier Transformed, see Fig. \[fig\_fid\_th\].

