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The eigenvalue equation for the quantum harmonic oscillator is $$\langle y | E\rangle '' +(2\epsilon-y^2)\langle y| E \rangle=0$$ where $\epsilon = \frac{E}{\hbar\omega}$ and $y=\sqrt{\frac{\hbar}{m\omega}}x$. In Shankar's book, he starts to solve this by taking the limit at infinity, making the equation $$\langle y | E\rangle '' -y^2\langle y| E \rangle=0$$ Apparently, the solution to this equation in the same limit is $\langle y | E\rangle = Ay^me^{\pm\frac{1}{2}y^2}$. I know how to work with kind of equation normally, but how do I solve this equation in the limit $ y \to \infty$?

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Define $\langle y | E\rangle = \psi_(y)$ and study the asymptotics of your ODE, $$ (\partial_y^2-y^2)~\psi (y)=0. $$ That means the leading behavior of ψ for large y; so , for example, if you had a polynomial in y, you just keep the highest order thereof: If you have an order-m polynomial, you'd just keep the $y^m$ monomial, since it dominates all lower powers in y at large y.

So, does $Ay^me^{\pm\frac{1}{2}y^2}$ satisfy your differential equation to leading order in y? Does $$\partial_y^2 (y^me^{\pm\frac{1}{2}y^2}) = (y^{m+2}+ O(y^{m+1} )~ ) e^{\pm\frac{1}{2}y^2} ~~?$$ Let's see: $$ \partial_y^2 (y^me^{\pm\frac{1}{2}y^2}) = \partial_y \Bigl ((m/y\pm y) y^me^{\pm\frac{1}{2}y^2}\Bigr )\\ = ((m/y\pm y)^2-m/y^2\pm 1)~y^m e^{\pm\frac{1}{2}y^2}\\ = (y^2 \pm 2m + m^2/y^2 -m/y^2 \pm 1 )~~y^m e^{\pm\frac{1}{2}y^2}, $$ So, indeed, the leading behavior of the parenthesis on the r.h.s. is $y^2$, your desideratum.


When you solve the full equation at all y, beyond asymptotics, $$ (\partial_y^2- y^2 + 2m +1)\psi_m(y)=0, $$ and you dismiss the solutions blowing up for large y, you find a celebrated solution, the Hermite functions, $$ \psi_m(y)\propto (-)^m e^{y^2/2} ~\partial_y^m e^{-y^2} = e^{-y^2/2} \Bigl (2y -\partial_y \Bigr)^m \cdot 1 , $$ which, presumably, your text is building up motivation for. They are the eigenfunctions of the Fourier transform.

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