Wavefunction normalization vs 1/√(|x| )

[Maharshi Roy]

In Griffith quantum mechanics, it is written that for a wavefunction to be normalizable, it is essential that the wavefunction approaches zero before 1/ √(|x|) as x tends to infinity...
Please explain from where this condition has been derived.

 

[Hypersphere]

The idea is that the wave function must be square integrable, so it must fall off to zero faster than functions which aren't.

Or put another way, integrate $\left( 1/\sqrt{|x|}\right)^2$ and see what you get.

 

[dextercioby]

There are square integrable functions which don't fall off to 0 when going to infinity. Griffiths' condition nicely singles out the so-called Schwartz test functions.

 

[Hypersphere]

You're right. I guess I should have said that the wave function must be square integrable and satisfy all other physical conditions of wave functions. (It certainly would be weird if the probability of finding the particle infinitely far away is non-zero.)