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Only the two-qubit experiment is possible. Choose the spin singlet stateIt has eigenvalue $-1$ under each of $X\otimes X$, $Y\otimes Y$ and $Z\otimes Z$. Thus measuring either qubit in the same Pauli basis as the other always gives opposite eigenvalue labels. Each opposite-label pair occurs with probability $1/2$. More generally, collective-unitary covariance of the two-qubit singlet ensures opposite outcomes in any common orthonormal qubit basis.

For the sequential experiment, the first outcome is uniformly distributed and the Lüders rule prepares the corresponding eigenstate. Certain opposite outcomes in the computational basis force the fixed unitary matrix to have the formCertain opposite Hadamard basis outcomes require $⟨+|U|+⟩=(a+b)/2=0$, hence $b=-a$. Therefore $U=iaY$. But this operator preserves the two Pauli Y eigenstate rays: it gives the same Y-basis label on the second measurement, rather than the opposite one. This contradicts the third set of observations.

Equivalently, flipping all three bases would require $U^{†}XU=-X$, $U^{†}YU=-Y$, and $U^{†}ZU=-Z$. Conjugating $XY=iZ$ in these two ways gives both $iZ$ and $-iZ$, an impossibility. This three-axis obstruction to unitary qubit orthogonalization is a finite test of the obstruction behind the universal NOT gate.

The universal NOT gate is the isotropic qubit map with factor $-1$. It sendsFor $|\mathbf{s}|\le 1$, the output has the same two nonnegative eigenvalues as the input, exchanged, and still has trace one. Thus it maps every single-qubit density matrix to a density matrix. Its linear extension to all operators is $\mathcal{N}(A)=\mathrm{tr}(A)I-A$, not $I-A$ for arbitrary unnormalized $A$. Positivity on one qubit does not establish complete positivity when an ancilla is present.

The input is the Bell state $\rho =|{\Phi }^{+}⟩⟨{\Phi }^{+}|$, where $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. Acting on the first qubit reverses the signs of all three terms containing its Pauli matrices, so in computational basis order $00,01,10,11$ the output isIts eigenvalues are $\boxed{-1/2,1/2,1/2,1/2}$: in particular, its expectation value in $|{\Phi }^{+}⟩$ is $-1/2$. It therefore fails to be positive semidefinite, although its trace remains one. This single entangled state proves that the universal NOT gate is positive but not completely positive, so it cannot be a deterministic physical quantum channel on an arbitrary input.