two vectors in Eqs  and , $$\begin{aligned}
  \label{eq:y-CSS}
  {|{y}\rangle} &\equiv
  \hat F_y^{(1)} {|{\phi_0}\rangle}
  = \sqrt{F/2} \; {|{F, m_F^x=F-1}\rangle},
  \\
  \label{eq:z-CSS}
  {|{z}\rangle} &\equiv
  F_z^{(1)} {|{\phi_0}\rangle}
  = i {|{y}\rangle}\end{aligned}$$ are parallel to each other with $\varphi = \pi/2$, so $\hat X_y = \hat X_1$ and $\hat P_z = \hat P_1$, and $$\begin{gathered}
  \label{eq:Jy-Jz-CSS}
  \hat J_y \approx \sqrt{NF} \hat X_1,
  \qquad
  \hat J_z \approx \sqrt{NF} \hat P_1.\end{gathered}$$

In Appendix \[app:two-oscillators\], we show in general that $\hat X_A$ and $\hat P_B$ defined in Eqs  and  are conjugate quadratures of the same atomic oscillator if and only if ${|{\phi_0}\rangle}$ is a minimum uncertainty state with respect to $\hat A^{(1)}$ and $\hat B^{(1)}$.

#### Orthogonal case.

If ${|{a}\rangle}$ and ${|{b}\rangle}$ are orthogonal to each other (${\langle{a}|{b}\rangle}=0$), then $\hat X_A$ and $\hat P_B$ belong to completely different atomic oscillators. Choosing ${|{\phi_1}\rangle} = {|{a}\rangle} / \|a\|$ and ${|{\phi_2}\rangle} = -i {|{b}\rangle} / \|b\|$ yields $$\begin{gathered}
  \label{eq:Xab-orthogonal}
  \hat X_A = \hat X_1,
  \qquad
  \hat P_B = \hat P_2.\end{gathered}$$

In recent experiments [@PhysRevLett.101.073601; @PhysRevLett.99.163002], the uncertainty in $\hat J_z$ was significantly reduced by coherently squeezing the spin of each individual spin-4 cesium atom via two-axis counter-twisting [@PhysRevA.47.5138] described by the Hamiltonian, $$\begin{gathered}
  \label{eq:twist}
  \hat T \equiv \sum_{j=1}^N  \frac1{2i}
  \big( \hat 