Why must $\Psi (x,t)$ go to zero faster than $\frac{1}{\sqrt{|x|}}$? 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.  
 A: 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.
A: 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.
