Can the chance of finding a particle diminish over time? Let's assume we have a wave function described by a wave equation and it is a function of space and time $\psi : \mathbb{R}^4 \rightarrow \mathbb{C}$.
This function needs to be normalized, so if I understood the bra-ket notation well:
$$\langle \psi | \psi \rangle = {\iint}_{-\infty}^{\infty} \psi^*(x,t) \psi(x,t)dx dt = 1$$
(asterisk means conjugate.)
But I see some problems here:


*

*Wavefunctions are time reversible this means half of it at the past, so that would mean there is 50% chance that the particle is never found at all.

*In order to get finite value for a full domain integral, the value of the function must approach zero as we go towards the infinite. That would mean that finding the particle diminish over time. That's interesting because if there are nothing that would measure it, then it's gone for good.


Am I missing something or is this an integral over space only and not time?
 A: To reply to the title, which doesn't specify the domain of integration, the answer is yes unless the whole configuration space is considered.
For concreteness, let $\Gamma$ be the configuration space of your system, and let $D$ be any (Borel) subset of $\Gamma$. Given that a particle is described by a time-dependent wave-function $\psi:\mathbb R\to L^2(\Gamma)$, the probability of finding it in the domain $D$ at the time $t$ is
$$\text{Pr}_\psi(t,D) = \int_D|\psi(t,x)|^2\text d\Omega(x),$$
where $\Omega$ is a regular probability measure on $\Gamma$ (usually the Lebesgue measure when $\Gamma = \mathbb R^n$). This probability can diminish over time, but since the total probability of finding the particle in $\Gamma$ is 1, this means that the probability of finding the particle is increasing in the complement $\Gamma\smallsetminus D$. A physical interpretation of this behaviour is that the particle is, on average, drifting away from the region $D$ towards other regions of the configuration space.
A: Where did you get that formula? The correct normalization does not involve the time integral.
Denoting with $| \psi(t) \rangle$ the state at time $t$, the normalization condition reads
$$ \tag{A} \langle \psi(t) | \psi(t) \rangle = \int d^3 x | \psi(x,t) |^2 = 1, \forall t. $$
What this is telling you is that at any time $t$ the particle must be somewhere.
Moreover, the unitarity of the time-evolution tells you that at any time $t$ the space integral in (A) is 1, so if you also integrate with respect to time you get a trivially infinite result. Vice versa, if the integral you report if finite, the evolution cannot be unitary.
