6. Given a quadrilateral pyramid $P-ABCD$ whose four lateral faces are isosceles right triangles with leg lengths of 1, where $\angle APB = \angle APD = \angle PBC = \angle PDC = 90^{\circ}$, then the height of this quadrilateral pyramid is $\qquad$.
6. $\frac{\sqrt{5}-1}{2}$.

As shown in Figure 2, connect $A C, B D$ intersecting at point $O$. From the problem, we know that points $P, A, C$ are all on the perpendicular bisector plane of segment $B D$.

Construct $P H \perp$ plane $A B C D$, then the foot of the perpendicular $H$ must be on $A C$, $O P^{2}+O B^{2}=1=O C^{2}+O B^{2}$, thus $O P=O C$, so $\angle P O A=2 \angle P C O$.

Since $\sin \angle P C O=\frac{P H}{P C}=\frac{P H}{\sqrt{2}}$, then $\cos \angle P O A=$ $1-2\left(\frac{P H}{\sqrt{2}}\right)^{2}=1-P H^{2}$.

Also, $P A \perp P B, P A \perp P D$, so $P A \perp$ plane $P B D$, then $P A \perp P O$, so $\cos \angle P O A=$ $\sin \angle P A O=\frac{P H}{P A}=P H$.

From $1-P H=P H^{2}$, we get $P H=\frac{\sqrt{5}-1}{2}$.