Normalize wave function with respect to time instead of space

Question

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?

Answer 1

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.

Answer 2

Time isn't a quantum-mechanical observable; it's a label. To understand the difference we must consider classical mechanics, in which canonical coordinates are functions of a time label. In particular, time doesn't have a conjugate momentum with which it has a canonical Poisson bracket.

Similarly, in field theory the action is a spacetime integral over a function of spacetime-dependent fields and their derivatives. These fields play the role of canonical fields and their arguments play the role of a time variable, so that even space isn't an observable in this context because we measure the field amplitude at a spacetime event, not a single particle's location.

The functional derivative $\frac{\delta\phi (x)}{\delta \phi (y)}=\delta (x-y)$ and Poisson bracket $\{\phi(x),\,\pi(y)\}=\delta(x-y)$ generalise analogous results from quantum mechanics, and show how the labels relate the quantities that become observables when we quantise this theory.

Answer 3

Given that $$\int_{-\infty}^\infty |\Psi (x,t)|^2 dx = 1,\tag1$$ implies that the particle must be somewhere in space at any time does not imply that it must be somewhere in time at a certain position $x$. To see this consider the simple example of a particle in a one dimensional box (infinite potential well). The wavefunction $\psi$ and the probability density $|\psi|^2$ of the first few eigenstates look like those in the figure bellow

As you can see, there are certain points (nodes) such that $|\psi(t)|^2=0$ meaning that the particle would never be found there.

You can also notice that Eq. (1) represent a sum of probabilities therefore it must be dimensionless. This implies that the dimension of $\psi$ (for one dimensional systems) is $[\mathrm{length}]^{-1/2}$. Now if we assume that (1) and its physical interpretation as a normalization of probabilities are correct then the dimension of $$\int_{-\infty}^\infty |\Psi (x,t)|^2 dt,$$ must be $[\mathrm{time}][\mathrm{length}]^{-1/2}$ which means the above integral cannot be equal to one. It cannot therefore represent a probability.