# Does the wavefunction probabilities have to sum to 1? [duplicate]

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 $$U(t)=2^t$$

• 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. Jan 6, 2019 at 19:25
• @ZeroTheHero but what then is the equivalent of Unitarity?
– user84158
Jan 6, 2019 at 19:26
• unitarity preserves the norm, whatever that norm is. Jan 6, 2019 at 19:29
• @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?
– user84158
Jan 6, 2019 at 19:30
• 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) Jan 6, 2019 at 19:41

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