= Solution
<Quantum error correction> encodes a <logical qubit> redundantly, obtains an <error syndrome> without measuring its unknown logical amplitudes, and applies a recovery that reverses the specified errors. A <decoherence-free subspace> instead uses a symmetry of a specified noise model: each noise operator acts as a scalar on the entire encoded space, so no syndrome-dependent recovery is required. This is passive protection and can also be viewed as a special case of <quantum error correction>.
For the <two-spin decoherence-free encoding>, choose
$$
|0_L\rangle=|01\rangle,\qquad |1_L\rangle=|10\rangle,\qquad
|\psi_L\rangle=\alpha|01\rangle+\beta|10\rangle,
\qquad |\alpha|^2+|\beta|^2=1.
$$
Simultaneous <Pauli Z gates> give
$$
(Z\otimes Z)|01\rangle=-|01\rangle,\qquad
(Z\otimes Z)|10\rangle=-|10\rangle,
$$
so
$$
\boxed{(Z\otimes Z)|\psi_L\rangle=-|\psi_L\rangle.}
$$
Both logical components acquire the same <global phase>; their <relative phase> and amplitudes survive. The <density matrix> is exactly unchanged, so even an unknown random choice between applying $I$ and $Z\otimes Z$ causes no logical <dephasing>.
More strongly, the encoding protects against arbitrary <collective dephasing>:
$$
[R_z(\phi)\otimes R_z(\phi)]|\psi_L\rangle
=e^{-i\phi(Z\otimes I+I\otimes Z)/2}|\psi_L\rangle
=|\psi_L\rangle,
$$
because both basis states have total $Z$ <eigenvalue> zero. At $\phi=\pi$, $R_z(\pi)\otimes R_z(\pi)=-Z\otimes Z$, consistent with the previous overall phase. The protection relies on correlated noise: independent $Z$ rotations generally change the logical relative phase, and this two-spin code does not protect a whole <logical qubit> against arbitrary collective rotations about all three axes or against arbitrary single-spin errors.
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