= Solution
The two displayed amplitudes have modulus $1/\sqrt2$. Absorb their phases into Bob's basis rather than use complex <Schmidt coefficients>. Take $|u_0\rangle=|0\rangle$, $|u_1\rangle=|1\rangle$, and
$$
|v_0\rangle=e^{-i\pi/4}|1\rangle,\qquad
|v_1\rangle=e^{i\pi/4}|0\rangle.
$$
Both local bases are orthonormal. A <Schmidt decomposition> is therefore
$$
\boxed{|\psi\rangle_{AB}=\frac1{\sqrt2}|u_0\rangle|v_0\rangle+\frac1{\sqrt2}|u_1\rangle|v_1\rangle}.
$$
The two Schmidt coefficients are $1/\sqrt2$, the Schmidt number is two, and both reduced density operators are $I/2$. Their equal eigenvalues make this decomposition nonunique.
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