11. The volume of the regular tetrahedron $ABCD$ is $1, O$ is its center. The regular tetrahedron $A'B'C'D'$ is symmetric to the regular tetrahedron $ABCD$ with respect to point $O$. Then the volume of the common part of these two regular tetrahedrons is $\qquad$ .
11. ${ }_{2}^{1}$.

If we place the tetrahedron $A-BCD$ on a horizontal plane, it is easy to see that the distance from the center to point $A$ is $\frac{3}{4}$ of the distance from point $A$ to the base. Therefore, the reflective symmetry plane is a horizontal plane equidistant from point $A$ and the base. Thus, it cuts the small tetrahedron at point $A$ to be $\frac{1}{2}$ the size of the original tetrahedron. Similarly, the tetrahedra cut at points $B$, $C$, and $D$ are also $\frac{1}{2}$ the size of the original tetrahedron. When these four smaller tetrahedra are cut from the original tetrahedron, the remaining part is the intersection of two tetrahedra, with a volume of $1-4 \times \binom{1}{2}^{3}=\frac{1}{2}$.