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For the lower-qubit controlled rotation, the control is the upper qubit and only the subspace $|10⟩,|11⟩$ rotates. Write $u=\cos (\theta /2)$ and $v=\sin (\theta /2)$. Unitary conjugation of its diagonal populations gives $c{u}^2+d{v}^2$ and $c{v}^2+d{u}^2$, together with off-diagonal coherences. The ideal crush gradient stipulated in the source removes those coherences without changing populations. ThereforeThe other two populations remain $a,b$. This controlled-transfer gate is not unitary as a complete operation because it includes dephasing.

For an upper-qubit target, the lower qubit is the control: the mixed subspace is $|01⟩,|11⟩$. ThusThe population $a$ of $|00⟩$ is unaffected in either direction because its control bit is zero. The dephasing step also leaves this diagonal population unchanged, so any sequence of these controlled-transfer gates preserves $a$.

Use the high-temperature state consistently to first order in $ϵ$. Applying the lower-target controlled-transfer gate with $\cos {\theta }_1=1/3$ givesThe upper-target gate with $\cos {\theta }_2=0$ then equalizes the second and fourth populations. The result isThis is a pseudo-pure state; the identity part is unchanged by unitary gates and does not contribute to traceless NMR signals.

For $n$ equal, weakly polarized spins, the high-temperature expansion is $\rho \simeq 2^{-n}[I+(ϵ/2){\sum }_j{Z}_j]$. The all-zero population is $a_0=2^{-n}(1+nϵ/2)$. Preserving this largest population and averaging the other $2^n-1$ populations givesThus the signal-bearing effective purity falls approximately as $n/2^n$ at fixed small per-spin polarization. The linear expansion requires $n|ϵ|\ll 1$. More generally, if the exact single-spin polarization is $r=\tanh (ϵ/2)$, the same population-preserving averaging gives $p_n=[(1+r{)}^n-1]/(2^n-1)$, which still decreases exponentially for fixed $0<r<1$. This scaling concerns high-temperature ensemble preparation, not an impossibility theorem for every initialization method.