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Why must $\Psi (x,t)$ go to zero faster than $\frac{1}{\sqrt{|x|}}$ as $|x|$ goes to $\infty$?

According to Griffiths' Introduction to Quantum Mechanics, it must. I don't understand why, and this is in his footnote (while talking about normalizability), so there's no explanation as to why this must be so.

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    $\begingroup$ $\Psi(x,t)$ should be such that integration of its mode square on real line (which represents probability of finding the particle on real line) is 1. If the condition which Griffith mentions is not satisfied the integral would diverge. $\endgroup$
    – user10001
    Jul 14, 2012 at 11:34
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    $\begingroup$ Isn't this simply because the integral of $1/x$ diverges, whereas the integral for $1/x^{1+\delta}$ converges? As the probability must be finite, this sets the limit for $\Psi$. $\endgroup$
    – Alexander
    Jul 14, 2012 at 11:34
  • $\begingroup$ But the integral of $\frac{1}{\sqrt(|x|)}$ puts an $x^{1/2}$ term in the numerator, and the integral still blows up as $x$ goes to $\infty$. So, I assume he could have been more stringent if he had wanted to be. $\endgroup$
    – Joebevo
    Jul 14, 2012 at 12:12
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    $\begingroup$ You can have a square integrable wave function $\Psi (x)$ such that $\Psi (x) sqrt{|x|}$ doesn't converge for $|x| \rightarrow \infty$ : Choose $\Psi (x) = 1$ for $n \leq x \leq n+1/n^{2}, n \epsilon \mathbb{N}$ and $0$ otherwise. $\endgroup$
    – jjcale
    Jul 14, 2012 at 15:39

2 Answers 2

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Otherwise one could eventually bound the integral by $1/x$, which diverges to infinity as $\ln(x)$, and the function could not be normalized. Ofcourse, there is nothing special about $1/\sqrt{|x|}$, he could equally well have chosen $1/\sqrt{|x\ln(x)|}$. And in case you were wondering, there is no function, such that all eventually slower growing functions converge, and all faster growing functions diverge.

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  • $\begingroup$ Thanks. You read my mind with that last sentence. As an aside, I plugged your $1/\sqrt{|x\ln(x)|}$ function into wolfram alpha and it says "no result found in terms of standard mathematical functions" for the integral. How can you tell about its convergence properties? $\endgroup$
    – Joebevo
    Jul 14, 2012 at 12:40
  • $\begingroup$ I based it on the n'th prime beeing ~ nln(n), and their reciprocals diverge. $\endgroup$
    – TROLLHUNTER
    Jul 14, 2012 at 12:47
  • $\begingroup$ Griffiths' condition is the most general possible. $1/|xln|x||$ goes to zero somewhat faster than $1/|x|$ $\endgroup$
    – user10001
    Jul 14, 2012 at 13:08
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My understanding of why Griffiths picked $\frac{1}{\sqrt{ |x| }}$ as an upper bound for $\Psi$ is from a dimensional analysis perspective. For example $\Psi^*\Psi$ is a probability density. So $\Psi^*\Psi$ for 1 dimensions must have units of $\frac{1}{|distance|}$ in order for $\int^\infty_{-\infty}\Psi^*\Psi\,dx=1$ (for normalized $\Psi$). Thus $\Psi$ must have units of $\frac{1}{\sqrt{ |distance| }}$. But we know that the integral $\int^\infty_{-\infty}\frac{1}{\sqrt{ |x| }}\frac{1}{\sqrt{ |x|}}\,dx=\int^\infty_{-\infty}\frac{1}{|x|}\,dx$ will diverge. So a function for $\Psi$ must go to zero faster than $\frac{1}{\sqrt{ |x| }}$ in order to have any hope of $\int^\infty_{-\infty}\Psi^*\Psi\,dx$ converging.

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