Does the wavefunction probabilities have to sum to 1?

Question

In quantum mechanics we are often told that $\int |\psi(x,t)|^2 dx^3 =1$. i.e. the probabilities have to sum to 1. And that this implies the time evolution operator is unitary.

But can't we define the probability as:

$$P(x,t) = \frac{|\psi(x,t)|^2}{\int |\psi(x,t)|^2 dx^3}$$

Then the probabilities would still add up to 1.

But then we needn't have a unitary time evolution operator. In fact wouldn't any complex operator do?

What am I missing here?

e.g. could you have a time evolution operator like $U(t)=2^t$

Answer 1

The correct statement is all physical wave functions must normalizable (and hence once normalized integrate to 1). Note that is is almost always what is meant by "wave function". The set of all such functions is called the Hilbert space.

Time evolution: As per your question about having non-unitary time evolution. Such a time evolution would not be physical, as it would not preserve the norm of the state. Namely, if we had some time evolution operator such that

$$\langle \psi| V^\dagger V | \psi \rangle \neq \langle \psi | \psi \rangle$$

Then that would imply that if we at some point had a probability of 1 to find the particle somewhere, then at some time later there would be some nonzero probability that we wouldn't find the particle anywhere (i.e. it would just "vanish"). Such a situation is unphysical.