$\xi_0$ and $\xi_1$, and the single-particle basis vector ${|{\phi_1}\rangle}$ such that $$\begin{gathered}
  \label{eq:phi1}
  \hat{\mathcal O}^{(1)} {|{\phi_0}\rangle}
  = \xi_0 {|{\phi_0}\rangle}  + \xi_1  {|{\phi_1}\rangle}.\end{gathered}$$ Normalization implies that $\xi_0$ and $\xi_1$ are the mean and variance of the single-particle operator, respectively, $$\begin{aligned}
  \label{eq:xi0}
  \xi_0 &= {\langle{\phi_0}|} \hat{\mathcal O}^{(1)} {|{\phi_0}\rangle}
  \equiv \langle {\mathcal O} ^{(1)} \rangle_0,
  \\
  \label{eq:xi1}
  \xi_1^2 &= {\langle{\phi_0}|} \hat{\mathcal O} ^{(1)}
  {}^2 {|{\phi_0}\rangle} - \xi_0^2
  \equiv \big( \Delta {\mathcal O} ^{(1)} \big)^2_0 .\end{aligned}$$ Eq.  now simplifies to $\hat{\mathcal{O}} \approx N \xi_0 + \sqrt {2N} \, \xi_1 \hat X_1$. We can assign to $\hat{\mathcal O}$ a single oscillator quadrature variable $$\begin{gathered}
  \label{eq:X-from-O}
  \hat X_1 = \frac{\hat{\mathcal O} - \langle \mathcal O \rangle _0}
  {\sqrt{2 (\Delta \mathcal O)^2_0}},\end{gathered}$$ and in the reference product state of the system, by construction, this collective atomic oscillator is in the ground state and $(\Delta X_1)^2_0 = \tfrac12$.

As an example, we mention that the natural normalization factor for the quadrature operators , that are assigned to the transverse angular momentum components, is not the spin quantum number $J$ nor the macroscopic expectation value of $\hat J_x$. Rather, it is related to the variance of $\hat J_y$ and $\hat J_z$ in the reference state, respectively, $$\begin{gathered}
  \hat X_y 