6. $\sqrt{1+\frac{\sqrt{6}}{3}}$.

Let $\angle B D M=\theta$, in $\triangle B D M$,
$$
\begin{array}{l}
B D=1, B M=M D=\frac{\sqrt{3}}{2} \\
\Rightarrow \cos \theta=\frac{\sqrt{3}}{3}, \sin \theta=\frac{\sqrt{6}}{3} .
\end{array}
$$

Rotate $\triangle B D M$ around $D M$ so that $\triangle B D M$ lies in the plane $A C D$, at this point, point $B$ is at $B^{\prime}$. The minimum value sought is the length of $A B^{\prime}$.
$$
\begin{array}{l} 
\text { It is easy to see that, } \angle A D B^{\prime}=\theta+30^{\circ} . \\
\text { Therefore, } A B^{2}=A D^{2}+D B^{\prime 2}-2 A D \cdot D B^{\prime} \cos \angle A D B^{\prime} \\
= 1+1-2 \cos \left(\theta+30^{\circ}\right)=1+\frac{\sqrt{6}}{3} \\
\Rightarrow A B^{\prime}=\sqrt{1+\frac{\sqrt{6}}{3}} .
\end{array}
$$
6. Given a regular tetrahedron $ABCD$ with edge length 1, $M$ is the midpoint of $AC$, and $P$ lies on the segment $DM$. Then the minimum value of $AP + BP$ is $\qquad$.