= \frac {\hat J_y} {\sqrt{2(\Delta J_y)_0^2}},
  \quad
  \hat P_z = \frac {\hat J_z} {\sqrt{2(\Delta J_z)_0^2}}.\end{gathered}$$ Here $(\Delta J_y)_0^2 = (\Delta J_z)_0^2 = J/2$ only for the coherent spin state, but not for a generic reference state. As long as the reference state is a product state and there is no entanglement among the particles, the corresponding collective atomic oscillators are in the ground state. Finally, we note that the Heisenberg uncertainty relation implies $[\hat X_y, \hat P_z]/i \le 1$. The reason why $\hat X_y$ and $\hat P_z$ may have a non-canonical commutation relation is discussed in the next subsection.

Operators acting on different oscillators {#sec:overlap}
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Let us now consider two collective operators, $\hat A$ and $\hat B$, that can be linearized in the above manner. As we have shown, we can assign to them the oscillator quadrature operators $\hat X_A$ and $\hat P_B$, respectively, such that $$\begin{aligned}
  \label{eq:X_A-implicit-def}
  \hat A &= \langle A \rangle_0 +
  \sqrt{2 (\Delta A)^2_0} \hat X_A,
  \\
  \label{eq:X_B-implicit-def}
  \hat B &= \langle B \rangle_0 +
  \sqrt{2 (\Delta B)^2_0} \hat P_B.\end{aligned}$$ Note that the position or momentum quadratures of the oscillator modes are fully equivalent and the distinction depends on the choice of the single-particle basis 