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In the introductory pages of Griffith's book on Quantum Mechanics, he says:

But wait a minute! Suppose I have normalized the wavefunction at time $t=0$. How do I know that it will stay normalized as time goes on and $\Psi$ evolves?

He then goes on to show that

$$(1)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{d}{dt}\ (\int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx)=0$$

From $(1)$, he argues that if

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \int_{-\infty}^{\infty} |\Psi(x,t=0)|^2 dx=1$$ then, $$ \ \ \ \ \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx=1$$

So basically, in order to prove that $\Psi(x,t)$ is normalized he uses $(1)$ but in order to prove $(1)$ he constrains $\Psi(x,t)$ to be normalized by saying that,$$\frac{d}{dt}\ (\int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx)=\frac{i\hbar}{2m}\bigg|\bigg(\Psi^*\frac{\partial\Psi(x,t)}{\partial x}-\Psi\frac{\partial\Psi^*(x,t)}{\partial x}\bigg)\bigg|_{-\infty}^{\infty}=0 $$
For the above equality he argues:

But $\Psi(x,t)$ must go to zero as x goes to ($+$ or $-$) infinity-otherwise the wavefunction would not be normalizable.

Writing down the whole exact derivation is time taking so I have instead summarized the main concerns above and attached a clear, visible picture of his proof here below:

enter image description here

I have issues with the justification of [1.26].

To me, it seems that this is a circular proof since in order to force that $\Psi(x,t)$ (which is the wavefunction at any arbitrary time) must go to zero as $x$ goes to infinity we have to assume that $\Psi(x,t)$ must be normalized which is what we are trying to prove!

Is this proof loose or am I missing/misunderstanding something obvious?

Qmechanic
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1 Answers1

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He does not say that $\Psi(x, t)$ must be normalized; he says it must be normalizable, meaning that

$$\int_{-\infty}^{\infty} |\Psi(x, t)|^2\text{d}t < \infty.$$

It is easy to see why $\Psi(x, t)$ must be normalizable. The wavefunction is assumed to be normalized at $t = 0$, and, since the wavefunction evolves continuously by the Schrodinger equation, it follows that the wavefunction must have finite norm for all $t > 0$, which, by definition, means that the above equation holds. It is from normalizability that Griffiths argues that the wavefunction should vanish at infinity (although, as mentioned in the comments of your question, wavefunctions do not necessarily have to vanish at infinity to be normalizable).

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