While solving the quantum harmonic oscillator differential equation in 3 dimensions:
$$ E \psi = -\frac{\hbar^2} {2\mu}\nabla^2 \psi + \frac{1}{2}\mu\omega^2r^2 \psi $$ for the radial equation, we get:
$$ \frac{d^2 R}{d \rho^2} +\frac{2}{\rho}\frac{dR}{d \rho}+\left (\frac{2E}{\hbar \omega} - \rho^2 - \frac{l(l+1)}{\rho^2}\right)R =0 $$ My doubt is: when we perform asymptotic analysis on the equation to guess the form of the solution we have: $$ \frac{d^2 R}{d \rho^2}- \rho^2R= 0 $$ for large values of $\rho$. And in all of my text books and notes, approximates the solution to a gaussian: $$ R(\rho) \approx e^{-\rho^2/2} $$ But, furthermore, they use $-\rho^2/2$ in the exponent, instead of just $-\rho^2$. Could someone explain and prove, why this approximation is valid for large values of $\rho$?
The factor of one-half is needed to make the equation come out right. The derivatives are
$ \frac{dR}{d\rho}=-\rho\,e^{-\rho^2/2} $
and
$ \frac{d^2R}{d\rho^2}=-e^{-\rho^2/2}+\rho^2\,e^{-\rho^2/2} $
But in the limit $\rho\gg1$, the second term on the right-hand side is much larger than the first. Hence
$ \frac{d^2R}{d\rho^2}\sim\rho^2\,e^{-\rho^2/2} $
and
$ \frac{d^2R}{d\rho^2}-\rho^2R\sim\rho^2\,e^{-\rho^2/2}-\rho^2\,e^{-\rho^2/2} = 0 $
as desired. Hope this helps!