3. In a square $ABCD$ with side length 2, $E$ is the midpoint of $AB$. Now, fold the figure along segments $EC$ and $ED$ so that segments $EA$ and $EB$ coincide (point $B$ overlaps with $A$), forming a tetrahedron $CDEA$. Then the volume of this tetrahedron is $\qquad$
3. $\frac{\sqrt{3}}{3}$.

As shown in Figure 2, it is clear that $\triangle A C D$ is an equilateral triangle with a side length of 2. Taking the midpoint $F$ of $C D$,
we have $A F=\sqrt{3}, C D \perp$ plane $A E F$, $E F=2, A E=1$.
By the Pythagorean theorem, we know that $\triangle A E F$ is a right triangle.
Thus, $S_{\triangle A E F}=\frac{1}{2} A E \cdot A F=\frac{\sqrt{3}}{2}$,
$$
V=\frac{1}{3} S_{\triangle A E F} \cdot C D=\frac{\sqrt{3}}{3} .
$$