This answer depends on the timescale of your system. Often time, the problem can be rephrased as follows.
If you start with a state a $\vert\psi_0(0)\rangle$, evolving under a Hamiltonian $\hat H$ so that $U(t)$ is the resulting evolution, one can ask for what small time $\Delta t$ will be probability of finding the system in $\vert\psi_0(0)\rangle$ be greater than $1-\epsilon$, where $\epsilon$ is assumed small and will depend on $\Delta t$. In other words, one ask for what $\Delta t$ is
$$
\vert \langle \psi_0(\Delta t)\vert \psi_0(0)\rangle \vert^2<1-\epsilon. \tag{1}
$$
with $\vert\psi_0(t)\rangle =U(t)\vert\psi_0(0)\rangle$.
This defines "quickly" in the sense that, if you remeasure within the time interval $\Delta t$, there is only a small probability $\epsilon$ that your system will have evolved out of your initial state. Given your tolerance $\epsilon$ you can find the appropriate $\Delta t$ to stay within this tolerance.
Assuming $\hat H$ does not depend explicit on $t$ for simplicity, $U(t)=e^{-i\hat H t/\hbar}$ and for small times one can usually write
\begin{align}
\vert\psi_0(\Delta t)\rangle &\approx
\left(\hat 1-i\frac{\Delta t}{\hbar} \hat H
-\frac{(\Delta t)^2}{2\hbar^2}\hat H^2 \right)
\vert\psi_0(0)\rangle\\
\langle \psi_0(0)\vert\psi_0(\Delta t)\rangle&=
1 -i\frac{\Delta t}{\hbar}
\langle \psi_0(0)\vert \hat H\vert \psi_0(0)\rangle
-\frac{(\Delta t)^2}{2\hbar^2}
\langle\psi_0(0)\vert H^2\vert\psi_0(0)\rangle
\end{align}
from where you can complete the calculation of
$$
\vert \langle \psi_0(0)\vert\psi_0(\Delta t)\rangle \vert^2\le 1-\epsilon
$$
and find $\epsilon$ in terms of $\Delta t$ and the matrix elements of $\hat H$.
Thus for instance (using $\hbar=1$), let's say we take
$$
\hat H=\sigma_x+\sigma_z=\left(
\begin{array}{cc}
1 & 1 \\
1 & -1 \\
\end{array}
\right)
$$
and suppose $\vert\psi_0(0)\rangle=\left(\begin{array}{c}1\\0\end{array}\right)$. Then
$$
U(\Delta t)=
\left(
\begin{array}{cc}
\cos \left(\sqrt{2} \Delta t\right)
-\frac{i \sin \left(\sqrt{2} \Delta t\right)}{\sqrt{2}} & -\frac{i \sin \left(\sqrt{2} \Delta t \right)}{\sqrt{2}} \\
-\frac{i \sin \left(\sqrt{2} \Delta t\right)}{\sqrt{2}} & \cos \left(\sqrt{2}
\Delta t \right)+\frac{i \sin \left(\sqrt{2} \Delta t\right)}{\sqrt{2}} \\
\end{array}
\right)
$$
and
$$
\langle \psi_0(0)\vert U(\Delta t) \vert\psi_0(0)\rangle
=\cos \left(\sqrt{2} \Delta t \right)-\frac{i \sin \left(\sqrt{2} \Delta
t\right)}{\sqrt{2}}
$$
so that, expanding, one finds
$$
\vert\langle\psi_0(0)\vert U(\Delta t)\vert\psi_0(0)\rangle\vert^2
\approx 1-(\Delta t)^2
$$
so that, in this case, "quickly" means $(\Delta t)^2<\epsilon$.