6. $P$ is a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a, b>0)$ in the first quadrant, $Q$ is the point symmetric to $P$ with respect to the origin $O$, $P H \perp x$-axis at point $H$, and the line $H Q$ intersects the hyperbola at point $M$ (different from $Q$). If the slope of the angle bisector of $\angle M P Q$ is 1, then the eccentricity of this hyperbola is $\qquad$
6. $\frac{\sqrt{6}}{2}$.

From the slope property of the hyperbola, we have
$$
k_{M P} k_{M Q}=\frac{b^{2}}{a^{2}} \text {. }
$$

From the given conditions, we get
$$
k_{M Q}=\frac{1}{2} k_{P Q}, k_{P M} k_{P Q}=1 \text {. }
$$

Thus, $\frac{b^{2}}{a^{2}}=\frac{1}{2}$, and it is easy to find that the eccentricity is $\frac{\sqrt{6}}{2}$.