A regular octahedron has a side length of $\frac{2}{2}$ . It is inscribed in a cube. The cube's vertices are $P_{1}, P_{2}, P_{3}, P_{4}, P_{1}^{'}, P_{2}^{'}, P_{3}^{'},$ and $P_{4}^{'}$ . Vertices $P_{2}, P_{3},$ and $P_{4}$ are adjacent to $P_{1}$ , and for $1 \leq i \leq 4$ , vertices $P_{i}$ and $P_{i}^{'}$ are opposite to each other. The octahedron's vertices are located on the segments $P_{1} P_{2}, P_{1} P_{3}, P_{1} P_{4}, P_{1}^{'} P_{2}^{'}, P_{1}^{'} P_{3}^{'},$ and $P_{1}^{'} P_{4}^{'}$ . What is the side length of the cube?