Taking 2 **time-dependant** states $\vert s\_1 \rangle , \vert s\_2 \rangle$ , you can do the following calculation
\begin{align\*}
i \hbar \partial\_t \langle s\_1 \vert s\_2 \rangle
&=
\big\{i \hbar \partial\_t \langle s\_1 \vert \big\} \vert s\_2 \rangle +
 \langle s\_1 \vert \big\{ i \hbar \partial\_t \vert s\_2 \rangle \big\}
\\ &=
-\langle s\_1 \vert H \vert s\_2 \rangle +
 \langle s\_1 \vert H \vert s\_2 \rangle 
\\ &=
0\end{align\*}
Where
$
 i \hbar \partial\_t \vert s \rangle = H \vert s \rangle $ and $ -i \hbar \partial\_t \langle s \vert = \langle s \vert H
$
were used in the process.

If the two states are identical (let's say $\vert s \rangle$), this implies that the state stays normalized over time, meaning the probability of finding the system in any state (that $\vert s \rangle$ has in its linear combination) stays 1 .

But from the above you can follow more generally
$$
\vert \langle s\_1 \vert s\_2 \rangle \vert^2 ~~\text{is constant over time} 
$$
Which I would not have expected, based on some intuition a priori. And I still don't have any after it. So my question is: **Is there a physical interpretation of this result?** (like in the case of two identical states) And if yes; **Is there an intuitive reason why it should be true?**