The plane intersects each face of the tetrahedron in a midline of the face; by symmetry it follows that the intersection of the plane with the tetrahedron is a square of side length 1. The surface area of each piece is half the total surface area of the tetrahedron plus the area of the square, that is, $ \frac{1}{2}\cdot 4\cdot \frac{2^2 \sqrt{3}}{4}+1=\boxed{1+2\sqrt{3}} $.