Born's statistical interpretation of the wave function says that $|\Psi (x,t)|^2$ is the probability density of finding the particle at point $x$ at time $t$, then
$$\int_{-\infty}^\infty |\Psi (x,t)|^2 dx = 1 \tag{1}$$
In other words, the particle must be somewhere in space at a certain time $t$. Is this equivalent to saying that the particle must be somewhere in time at a certain position $x$?
My Thoughts: Why I can't write,
$$\int_{-\infty}^\infty |\Psi (x,t)|^2 dt = 1 \tag{2}$$
Whenever I look for the position of the particle, $\Psi$ stops obeying the Schrödinger equation and discontinuously collapses to a spike around some position $x$. If I was able to focus my microscope at say $x = 2$ and look for the particle across all of time (in this case, time is the measurement at not position so setting my microscope at $x = 2$ doesn't disturb anything), then $\Psi$ would collapse to a spike around some value of $t$ and this method of normalizing the wave function would be appropriate. However, humans can only sample at instants in time and look over all space (equation $\textbf{(1)}$). We can't sample at an instant in position and look over all time (equation $\textbf{(2)}$). Scientists can't search for the particle in time. Therefore $\textbf{(2)}$ is not appropriate. However, even though we can't do this as humans (search through time at will instead of staying anchored to the present), is it wrong to say that 'nature' can't accomplish $\textbf{(2)}$? Or do we have laws like the 2nd Law of Thermodynamics that say nature is prohibited from doing so? Is this an example that shows how space and time are not on equal footing?
Best Answer
For a quantum system with one degree of freedom on the closed interval $I$, the Hilbert space is $L^2(I)$. In this case the $I$ is the range for the spatial coordinate $x$, so that normalisation applies with respect to the Lebesgue measure on $I$. Now suppose that you have a dynamics described by the Hamiltonian $H$ on such Hilbert space, and that $\psi$ is an eigenstate of $H$ with eigenvalue $E$. The time evolution of $\psi$ is
$$\psi\mapsto e^{iEt}\psi.$$
If we naively try to integrate this function, we get to
$$\int_{\mathbb R}|e^{iEt}\psi(x)|^2\text d t = \int_{\mathbb R}|\psi(x)|^2\text d t = |\psi(x)|^2\int_{\mathbb R}\text dt,$$
which is infinite for every $x\in I$ for which $\psi(x)\neq 0$ or zero otherwise. We then have a problem trying to attach the meaning of probability to such integral. You can interpret this result as saying that a particle will pass through $x$ infinitely often provided $\psi(x)\neq 0$ if you wait indefinitely, but this information is already deductible from $\psi$ itself, there is no need to do such an integral.