A wave function that is normalized initially remains normalized

Question

Suppose that $\Psi(x,t)$ is normalized at time $t=0$. Show that this implies that $\Psi(x,t)$ is normalized at all other times.

I know that this makes intuitive sense, and we'd certainly want our models of reality to have this property - if the particle has a 100% chance to be found somewhere in space initially, it should ALWAYS have a 100% of being found. I just don't know how to prove this mathematically.

I can at least get started by saying that, since $\Psi(x,0)$ is normalized, we know $$\int_{-\infty}^{\infty}\Psi^*(x,0)\Psi(x,0) \mathrm{d}x=1$$

Can anyone give a hint about where to go from here?

Answer 1

I will elaborate on count_to_10's comment. Consider the following current

$$j=\frac{i}{2m}(\Psi\frac{d}{dx}\Psi^*-\Psi^*\frac{d}{dx}\Psi)$$

where $m$ is the mass of the particle and I am working in units where $\hbar=1$. This is the famous probability current of QM. It is related to the probability density $\rho(x,t)=\psi\psi^*$ via

$$\frac{d}{dx}j=\frac{d}{dt}\rho$$

it is easy to convince yourself of this. Just take the definition of $j$ and the Schrödiger equation and you will see that the equation above is satisfied. If you have taken an electromagnetism course you probably are guessing that we will have a charge associated with $\rho$ which is conserved. Let's prove it. Let's integrate

$$\int_{-\infty}^{\infty}dx\,\frac{d}{dx}j=\int_{-\infty}^{\infty}dx\,\frac{d}{dt}\rho$$

$$j(x\to\infty)-j(x\to{}-\infty)=\frac{d}{dt}\int_{-\infty}^{\infty}dx\,\rho$$

assuming that the wave function vanishes asymptotically

$$\frac{d}{dt}\int_{-\infty}^{\infty}dx\,\rho=0$$

and we have that $\int_{-\infty}^{\infty}dx\,\rho$ is time independent. But since you know that it is 1 at time 0 (since you assume the wave function is normalized) you know that it will remain normalized.