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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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Lost
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    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. – ACuriousMind Sep 03 '21 at 18:19
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    See https://physics.stackexchange.com/q/382324/50583 for discussions of the notion that "wave functions must vanish at infinity" – ACuriousMind Sep 03 '21 at 18:21

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).

Solarflare0
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  • From "the wavefunction vanishes at infinity" he argues that it is normalized, not normalizable. – Solarflare0 Sep 03 '21 at 18:29
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    Oh right! Mind-numbing overlook. – Lost Sep 03 '21 at 18:30
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    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. – tparker Sep 03 '21 at 21:32
  • Sorry, the exponent should have a minus sign. – tparker Sep 03 '21 at 23:03
  • @tparker There are two things here as far as I can see: (1) Normalizabilty: D..J. starts with a normalizable wavefunction and assumes it to be so at all later times. ..(2) Normalized or not?: D.J. starts with a normalized wavefunction and using (1) shows that it remains normalized for all later times. – Lost Sep 05 '21 at 11:05
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    @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 05 '21 at 11:08
  • @tparker I would guess that generally while doing QM we rarely encounter wavefunctions of the sort you mentioned(?) but does the formalism of QM allow such wavefunctions? – Lost Sep 05 '21 at 11:26
  • @tparker If we consider Born's statistical interpretation of the wavefunction then, the wavefunction being normalizable is necessary. – Lost Sep 05 '21 at 11:28
  • @Lost Yes, as you say, Griffiths proves that if the wavefunction remains normalizable over time, then it also remains normalized. But he doesn't actually prove that it remains normalizable over time. – tparker Sep 06 '21 at 17:40
  • @Lost But it turns out that this is always the case, so the wavefunction indeed always remains normalized. The technical reason for this, which unfortunately might be a bit too advanced for someone just starting out in QM, is the fact that the Hamiltonian is a Hermitian operator. This fact and the Schrodinger equation together imply that the time-evolution operator is a unitary, which means that the states remain normalized over time. You can actually prove this solely using the fact that the Hamiltonian is a Hermitian operator, which assuming any details about its particular form. But this – tparker Sep 06 '21 at 17:42
  • requires somewhat abstract concepts like the state-vector formalism of QM. It's isn't too hard be requires a bit of setup, which you'll probably soon get to if you haven't already. – tparker Sep 06 '21 at 17:43
  • @tparker Thank you. I read about the derivation you mentioned and it is clear now how the state remains normalized because of SE and Hermiticity of the hamiltonian. One thing that I wanted clear up was: What about the normalizability though now? And the wavefunction you mentioned...is it an allowed wavefunction in QM? It clearly stops being a part of $L_2$ after a certain time as you mentioned. – Lost Sep 07 '21 at 10:48