Centers of adjacent faces of a unit cube are joined to form a regular octahedron.  What is the volume of this octahedron?

$ \textbf{(A) } \frac 18 \qquad \textbf{(B) } \frac 16 \qquad \textbf{(C) } \frac 14 \qquad \textbf{(D) } \frac 13 \qquad \textbf{(E) } \frac 12$
1. **Identify the problem**: We need to find the volume of a regular octahedron formed by joining the centers of adjacent faces of a unit cube.

2. **Visualize the octahedron**: The centers of the faces of a unit cube form a regular octahedron. Each face of the cube has a center, and these centers are equidistant from each other.

3. **Divide the cube**: Consider the unit cube. By joining the centers of adjacent faces, we effectively divide the cube into two congruent pyramids. Each pyramid has a square base (one of the faces of the cube) and a height equal to half the side length of the cube.

4. **Calculate the base area**: The base of each pyramid is a square with side length 1 (since it is a face of the unit cube). Therefore, the area of the base is:
   \[
   \text{Base Area} = 1 \times 1 = 1
   \]

5. **Calculate the height**: The height of each pyramid is the distance from the center of the cube to the center of one of its faces. This distance is half the side length of the cube, which is:
   \[
   \text{Height} = \frac{1}{2}
   \]

6. **Volume of one pyramid**: The volume \( V \) of a pyramid is given by:
   \[
   V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
   \]
   Substituting the values we have:
   \[
   V = \frac{1}{3} \times 1 \times \frac{1}{2} = \frac{1}{6}
   \]

7. **Volume of the octahedron**: Since the octahedron is composed of two such pyramids, the total volume of the octahedron is:
   \[
   \text{Volume of Octahedron} = 2 \times \frac{1}{6} = \frac{1}{3}
   \]

The final answer is \(\boxed{\frac{1}{3}}\).