(\[Y1property\]), each application of the control operator $Y_1={\cal P}_{|0\rangle}$ leaves the state $|0\rangle$ intact but induces a negative sign for all other two-qubit states. It is interesting to compare the control operator $Y_1$ with what can be intuitively expected from early single-qubit UDD results. Suppose that the two qubits are unrelated at all, then in order to suppress the spin flipping of the first qubit (second qubit), we need a control operator $\sigma_z^1$ ($\sigma_z^2$). As such, an intuitive single-qubit-based control Hamiltonian would be $$\begin{aligned}
  H_{c,\text{single}}=\frac{\pi}{2}\sum_{j=1}^{N}\delta(t-T_j)(\sigma_{z}^{1}+\sigma_{z}^{2}).
  \label{intu}
  \end{aligned}$$ This intuitive control Hamiltonian differs from Eq. (\[control1\]), hinting an importance difference between two-qubit and single-qubit cases. Indeed, here the qubit-qubit interaction or the system-environment coupling may directly cause a double-flipping error ${|\uparrow\uparrow\rangle} \rightarrow
{|\downarrow\downarrow\rangle}$, which cannot be suppressed by $H_{c,\text{single}}$. The second key step is to split the Hamiltonian $H$ into two parts $H_0$ and $H'$, with the former commuting with $Y_1$ and the latter anti-commuting with $Y_1$. Once these two steps are achieved, the remaining part of our proof becomes straightforward by exploiting Eq. (\[ourTn1\]). These understandings suggest that it should be equally possible to preserve the coherence associated with entangled two-qubit states.

### Preserving coherence associated with entangled states

Consider 