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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 $U(t)=2^t$

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    $\begingroup$ you can define the prob. as you do but why do it this way? You still have to compute $\int \vert \psi\vert^2$ so no savings. You might as well set it to one and be done with it. $\endgroup$
    – ZeroTheHero
    Commented Jan 6, 2019 at 19:25
  • $\begingroup$ @ZeroTheHero but what then is the equivalent of Unitarity? $\endgroup$
    – user84158
    Commented Jan 6, 2019 at 19:26
  • $\begingroup$ unitarity preserves the norm, whatever that norm is. $\endgroup$
    – ZeroTheHero
    Commented Jan 6, 2019 at 19:29
  • $\begingroup$ @ZeroTheHero So instead of $U^\dagger U=1$ you could have $U^\dagger U$ just has to be a real number. (is that right?) Is there a name for such a matrix? $\endgroup$
    – user84158
    Commented Jan 6, 2019 at 19:30
  • $\begingroup$ Related/possible duplicate: physics.stackexchange.com/q/156367/50583, physics.stackexchange.com/q/167099/50583 (for (un)normalized states), physics.stackexchange.com/q/169936/50583 (for why we need unitarity) $\endgroup$
    – ACuriousMind ♦
    Commented Jan 6, 2019 at 19:41

1 Answer 1

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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.

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