# Regarding to the asymptotic solution of quantum harmonic oscillator

In quantum mechanics, the radial equation of the SHO takes the form \begin{align} \frac{d^2 u}{dx^2}+\left(\epsilon-x^2-\frac{l(l+1)}{x^2}\right)u=0, \end{align} where $$x=\sqrt{\frac{m\omega}{\hbar}}r$$, $$\epsilon=\frac{2mE}{\hbar^2}$$. In the far field ($$x\to\infty$$) limit, the equation becomes \begin{align} \frac{d^2 u}{dx^2}-x^2 u=0. \end{align} Every literature says that the asymptotic solution of this equation is $$u(x)\sim e^{\pm\frac{1}{2}x^2}$$ without detailed explanation. I suspect that the asymptotic solution is not unique. If we take the ansatz $$u=e^{-ax^2}$$, then \begin{align} (-2a+(4a^2-1)x^2)e^{-ax^2}=0, \end{align} and this is valid for all $$a>0$$, since we are considering asymptotic behavior for $$x\to\infty$$. Then why do we insist on $$a=1/2$$ for the asymptotic solution?

You need to be careful on defining the asymptotics. From the equation: $$u''+\left(\epsilon-x^2-\frac{l(l+1)}{x^2}\right)u = 0$$ you want to know the behaviour of $$u$$ at infinity. The issue is that you have an irregular singular point at infinity so you don't have the usual power asymptotics. There is no general method to estimate the asymptotics of the solution. A usual trick is to rather estimate $$\ln u := v$$. The equation becomes: $$v''+(v')^2+\left(\epsilon-x^2-\frac{l(l+1)}{x^2}\right) = 0$$ For $$x\to\infty$$, you can safely neglect $$\epsilon, \frac{l(l+1)}{x^2}$$ and the idea is that due to the rapid growth, $$v''\ll(v')^2$$. You might have seen such similar approximation in optics to recover the eikonal equation from diffraction or in mechanics to get Hamilton-Jacobi equation from Schrödinger's equation. You are left with the asymptotic equation: $$(v')^2-x^2 = 0$$ which you solve as: $$v \sim -\frac{x^2}2$$ from which you deduce: $$u \asymp e^{-x^2/2}$$ You need to be careful as it does not give you an asymptotic equivalent unlike what you wrote. Indeed, it only gives an asymptotic equivalent of $$v$$ , which diverges. You'll need a precision $$o(1)$$ in $$v$$ to get an asymptotic equivalent for $$u$$, but you only have a precision $$o(x^2)$$. Check out Advanced Mathematical Methods for Scientists and Engineers by Bender and Orszag for more on the method (section 3.4 in particular).