# Is this proof of Griffith's watertight? [duplicate]

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:

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?

• Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. Note that your summary doesn't really help people who can't see the picture because you refer to [1.26] in your actual question and the only place that identifier appears is in the picture. Sep 3, 2021 at 18:19
• See physics.stackexchange.com/q/382324/50583 for discussions of the notion that "wave functions must vanish at infinity" Sep 3, 2021 at 18:21

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).

• From "the wavefunction vanishes at infinity" he argues that it is normalized, not normalizable. Sep 3, 2021 at 18:29
• Oh right! Mind-numbing overlook.
– Lost
Sep 3, 2021 at 18:30
• Continuity isn't enough to ensure that a normalizable wavefunction remains that way. Consider the putative wavefunction $\psi(x,t) \propto (1 + (\kappa x)^2)^{t/T}$. This is normalizable up until time $t = T/4$ but then ceases to be so. Griffiths's proof isn't entirely watertight. Sep 3, 2021 at 21:32
• Sorry, the exponent should have a minus sign. Sep 3, 2021 at 23:03
• @tparker It seems you object to the assumption in (1). My question was simply a stupid overlook of words which was satisfactorily answered here but now, from your comment, it turns out there is actually an unjustified assumption DJ uses. Thanks a lot for pointing it out.
– Lost
Sep 5, 2021 at 11:08