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All time-dependent Hamiltonian operators are multiples of the same fixed operator, so they commute and their time-ordered exponential reduces to an ordinary exponential. Let $P_1=(I-Z_S)/2$. With $\hslash =1$ and the given unit coupling integral,Since $X_D^2=I$,ThusIt is exactly the CNOT gate with $S$ controlling $D$, with no residual relative phase. Acting on $(c_0|0⟩+c_1|1⟩{)}_S|0{⟩}_D$ correlates the system basis label with the apparatus label, producing $c_0|00⟩+c_1|11⟩$ as required.

The Hadamard basis changes the Pauli matrices by $Z_S={\tilde{X}}_S$ and $X_D={\tilde{Z}}_D$. Thus, keeping tensor factors ordered as $S,D$, the Hamiltonian operator isIf $D$ has new-basis label plus, the second factor vanishes. If $D$ has label minus, the integrated generator on $S$ is $\pi (I-{\tilde{X}}_S)/2$, whose exponential is ${\tilde{X}}_S$ by part (a). The generator therefore has the same controlled form with $D$ as control in the complementary bases.

To use $S$ as the apparatus, prepare it in the ready state $|+{⟩}_S$ and later read it in the Hadamard basis. For an arbitrary input $\alpha |+{⟩}_D+\beta |-{⟩}_D$, part (b) givesThe two apparatus states are orthogonal, so its readout measures the corresponding projectors on $D$ and preserves each input eigenstate. The observable iswith outcomes $+1$ and $-1$. This is the measurement direction of a CNOT interaction in the complementary basis.

The ready state and readout basis matter. Keeping $S$ in the original $|0{⟩}_S$ would make the old-basis controlled gate act as the identity on $D$, producing no record. Thus the same interaction can serve either measurement direction, with different apparatus preparation and pointer readout; the Hamiltonian operator alone does not assign an intrinsic system/apparatus role.