= Solution
Write $a=\langle\psi|w\rangle=2^{-N/2}$, which is real and satisfies $0<a<1$ for $N\geq1$. The invariant two-dimensional span has orthonormal eigenvectors
$$
|+\rangle=\frac{|\psi\rangle+|w\rangle}{\sqrt{2(1+a)}},\qquad
|-\rangle=\frac{|\psi\rangle-|w\rangle}{\sqrt{2(1-a)}},
$$
with <eigenvalues> $\Delta(1+a)$ and $\Delta(1-a)$. In this basis,
$$
|\psi\rangle=\sqrt{\frac{1+a}{2}}|+\rangle+\sqrt{\frac{1-a}{2}}|-\rangle,
\quad
|w\rangle=\sqrt{\frac{1+a}{2}}|+\rangle-\sqrt{\frac{1-a}{2}}|-\rangle.
$$
Transfer therefore requires the two evolved eigenphases to differ by an odd multiple of $\pi$:
$$
e^{-i2\Delta at}=-1.
$$
The first positive time is
$$
\boxed{t_0=\frac\pi{2\Delta a}=\frac{\pi\sqrt{2^N}}{2\Delta}.}
$$
Equivalently, exponentiating on the invariant span gives the full evolution
$$
e^{-iH_Gt}|\psi\rangle=e^{-i\Delta t}
\bigl[\cos(\Delta at)|\psi\rangle-i\sin(\Delta at)|w\rangle\bigr].
$$
At $t_0$ this is $-ie^{-i\Delta t_0}|w\rangle$, so the requested <global phase> can be chosen as $\alpha=-\Delta t_0-\pi/2$ modulo $2\pi$. Since the two vectors are linearly independent, their relative coefficient cannot produce exact transfer at an earlier positive time. This is <continuous-time quantum search>, with time proportional to the square root of the search-space dimension for fixed $\Delta$. Restoring units multiplies the stated time by $\hbar$. The degenerate zero-qubit case would already have $\psi=w$, so there would be no smallest positive transfer time; the formula concerns the nontrivial $N\geq1$ search.
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