Example 16 Given that $\alpha, \beta$ are acute angles, and $\alpha-\beta=\frac{\pi}{3}$, try to find the value of $\sin ^{2} \alpha+\cos ^{2} \beta-\sqrt{3} \sin \alpha \cdot \cos \beta$.

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Notice that $\cos \beta=\sin \left(\frac{\pi}{2}-\beta\right)$, and $\alpha+\left(\frac{\pi}{2}-\beta\right)+\frac{\pi}{6}=\pi$. Therefore, we can construct $\triangle A B C$ with $\alpha, \frac{\pi}{2}-\beta, \frac{\pi}{6}$ as its interior angles, i.e., $\angle A=\alpha, \angle B=\frac{\pi}{2}-\beta, \angle C=\frac{\pi}{6}$. Let the circumradius of $\triangle A B C$ be $R$. By the Law of Sines, we get
$$
B C=2 R \cdot \sin \alpha, A C=2 R \cdot \sin \left(\frac{\pi}{2}-\beta\right)=2 R \cdot \cos \beta, A B=2 R \cdot \sin \frac{\pi}{6}=R .
$$

By the Law of Cosines, we have $B C^{2}+A C^{2}-2 B C \cdot A C \cdot \cos \angle C=A B^{2}$,
which gives $4 \sin ^{2} \alpha+4 \cos ^{2} \beta-4 \sqrt{3} \cdot \sin \alpha \cdot \cos \beta=1$,
thus $\sin ^{2} \alpha+\cos ^{2} \beta-\sqrt{3} \sin \alpha \cdot \cos \beta=\frac{1}{4}$.