= Solution
With the principal arctangent in part (a), both coefficients must be $1/\sqrt2$. Hence
$$
\boxed{\theta(t)=\frac\pi4,\qquad \Omega_{12}(t)=\Omega_{23}(t)\ne0.}
$$
For the positive-envelope transfer this is the crossing at a positive coupling. If both couplings have the same negative sign, their ratio still gives the stated principal-angle vector; a continuous-angle representative can differ by an irrelevant overall minus sign. At this point the instantaneous dark state is $(|3\rangle-|1\rangle)/\sqrt2$. In the example sketch this occurs at $t=4T$, where the two effective couplings cross. Equality of the raw envelopes $A_{12}$ and $A_{23}$ is not enough if the dipole moments differ: the condition is $A_{12}d_{12}=A_{23}d_{23}$, with the same effective relative phase. More generally the equality may occur once or several times, depending on the pulse shapes. Simultaneous zero couplings do not specify this superposition because the whole <Quantum Hamiltonian> is then degenerate.
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