F_{x+}^{(j)2} - \hat F_{x-}^{(j)2} \big)
  = \sum_{j=1}^N \big\{ \hat F_y^{(j)}, \hat F_z^{(j)} \big\}.\end{gathered}$$ Thus each atom is prepared in the same state $$\begin{gathered}
  \label{eq:twisted-state}
  {|{\phi_0^{\text{sq}}}\rangle}
  = \sum_{k=0}^F c_k {|{F, m_F^x = F - 2k}\rangle},\end{gathered}$$ with only even number of spin excitations. In such a reference state, the quadrature operators assigned to the twisting operator $\hat T$ and to the transverse angular momentum $\hat J_z$ belong to different atomic oscillators. Following the prescription above, we introduce the corresponding vectors $$\begin{aligned}
  \label{eq:t-vector}
  {|{t}\rangle} &\equiv
  \big(\hat T^{(1)} - \langle T^{(1)} \rangle_0 \big)
  {|{\phi_0^{\text{sq}}}\rangle}
  = \hat T^{(1)} {|{\phi_0^{\text{sq}}}\rangle},\\
  \label{eq:z-vector}
  {|{z}\rangle} &\equiv
  \big(\hat F_z^{(1)} - \langle F_z^{(1)} \rangle_0 \big)
  {|{\phi_0^{\text{sq}}}\rangle}
  = \hat F_z^{(1)} {|{\phi_0^{\text{sq}}}\rangle}.\end{aligned}$$ The expectation value $\langle T^{(1)} \rangle_0$ is zero, since it is zero in the initial $x$-polarized spin state, and $\hat T^{(1)}$ is a constant of motion in the process of internal squeezing. It is easy to see that $\langle F_z^{(1)} \rangle_0$ is also zero. Noting that $\hat T^{(1)}$ changes the azimuthal quantum number of the hyperfine sublevel by two, whereas $\hat F_z^{(1)}$ changes the same quantum number by only one, we immediately see that ${|{t}\rangle}$ is orthogonal to ${|{z}\rangle}$. Therefore, completely different atomic oscillators are accessed by the transverse components of 