# Wavefunction normalization vs 1/√(|x| )

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1. Sep 28, 2014

### 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.

2. Sep 28, 2014

### 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.

3. Sep 28, 2014

### 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.

4. Sep 28, 2014

### 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.)