The solution to the Shordinger's eq.

[rbwang1225]

When I read the book of quantum mechanics of Griffiths, he said that the wave function must go to zero faster than 1/(|x|^-(1/2)), as |x| ->infinity. I wonder why?
Any help would be appreciated.

 

[haushofer]

Because the wavefunction has the statistical interpretation that

<br /> \int_{-\infty}^{+\infty} | \psi(x,t) |^2 d^3 x = 1<br />

So psi has to be square integrable to make sense as a probability density.

 

[unusualname]

It depends on the potential, for a decreasing potential the wave function has plane wave type solution as |x| -> infinity, for increasing potential the wave function decreases at a rate depending on how fast the potential is increasing.

The general proof is quite subtle, eg see 'The Schrödinger Equation' - Berezin & Shubin , Kluwer Academic Publishers 1991.

I assume Griffiths had a specific potential function in mind.

(Square Integrable doesn't imply decreasing at infinity, although the counter examples are rather artificial mathematical constructs, and probably not physically interesting)