and, in particular, the phase of the basis vectors. For the two operators, it remains a question, however, whether we can choose the basis in such a way that $\hat X_A$ and $\hat P_B$ are quadratures conjugate to each other acting on the same oscillator, or whether they belong to two independent atomic oscillators. To answer this question in general, we introduce the unnormalized vectors $$\begin{aligned}
  \label{eq:a-vector}
  {|{a}\rangle} &\equiv
  \big( \hat A^{(1)} - \langle A^{(1)} \rangle_0 \big) {|{\phi_0}\rangle},
  \\
  \label{eq:b-vector}
  {|{b}\rangle} &\equiv
  \big( \hat B^{(1)} - \langle B^{(1)} \rangle_0 \big) {|{\phi_0}\rangle},\end{aligned}$$ and analyze the following three cases.

#### Parallel case.

If ${|{a}\rangle}$ and ${|{b}\rangle}$ are parallel to each other (${|{b}\rangle} = \lambda e^{i\varphi} {|{a}\rangle}$), then $\hat X_A$ and $\hat P_B$ belong to the same atomic oscillator. Indeed, we can set ${|{\phi_1}\rangle} = {|{a}\rangle} /
\|a\|$, so that $$\begin{gathered}
  \label{eq:Xab-parallel}
  \hat X_A = \hat X_1,
  \qquad
  \hat P_B = \hat X_1 \cos\varphi + \hat P_1 \sin\varphi,\end{gathered}$$ where $\varphi = \arg {\langle{a}|{b}\rangle}$.

For example, in the vicinity of the fully polarized spin coherent state (${|{\phi_0}\rangle} = {|{F, m_F^x=F}\rangle}$), the transverse components of the total angular momentum operator, $\hat{\mathbf J} = \sum_{j=1}^N \hat{\mathbf F}^{(j)}$, define the quadratures of the same atomic oscillator. The 