Orthogonality of the wavefunctions of the quantum harmonic oscillator

The orthogonality of the wavefunctions of the quantum harmonic oscillator $\psi_n(y)=H_ne^{-\frac{y^2}{2}}$ can be proven using the Hermite differential equation.

Substituting eq13 in eq22, we have

$\frac{d^2H_n}{dy^2}-2y\frac{dH_n}{dy}+2nH_n=0\;\;\;\;\;\;\;\;23$

where $H_n$ are the Hermite polynomials.

Question

Show that eq23 can be written as

$e^{y^2}\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_n}{dy}\biggr$+2nH_n=0\;\;\;\;\;\;\;\;24$

Answer

Carrying out the derivative, the LHS of eq24 becomes

$e^{y^2}\biggr\[e^{-y^2}\frac{d^2H_n}{dy^2}+\frac{dH_n}{dy}(-2y)e^{-y^2}\biggr\]+2nH_n=\frac{d^2H_n}{dy^2}-2y\frac{dH_n}{dy}+2nH_n$

Multiplying eq24 (with a change of index from $n$ to $m$ ) by $H_n$ and subtracting the result from the product of $H_m$ and eq24, we have

${H_me^{y^2}\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_n}{dy}\biggr$+2nH_mH_n-\biggr\[H_ne^{y^2}\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_m}{dy}\biggr$+2mH_mH_n\biggr\]=0}$

$e^{-y^2}2(m-n)H_mH_n=H_m\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_n}{dy}\biggr$-H_n\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_m}{dy}\biggr$\;\;\;\;\;\;\;\;25$

Substituting ${H_m\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_n}{dy}\biggr$=\frac{d}{dy}\biggr$H_me^{-y^2}\frac{dH_n}{dy}\biggr$-e^{-y^2}\frac{dH_n}{dy}\frac{dH_m}{dy}}$ and ${H_n\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_m}{dy}\biggr$=\frac{d}{dy}\biggr$H_ne^{-y^2}\frac{dH_m}{dy}\biggr$-e^{-y^2}\frac{dH_m}{dy}\frac{dH_n}{dy}}$ in eq25 gives

$e^{-y^2}2(m-n)H_mH_n=\frac{d}{dy}\biggr\[e^{-y^2}\biggr$H_m\frac{dH_n}{dy}-H_n\frac{dH_m}{dy}\biggr$\biggr\]$

Integrating both sides of the above equation with respect to $y$ ,

$2(m-n)\int_{-\infty}^{\infty}e^{-y^2}H_mH_ndy=\int_{-\infty}^{\infty}\frac{d}{dy}\biggr\[e^{-y^2}\biggr$H_m\frac{dH_n}{dy}-H_n\frac{dH_m}{dy}\biggr$\biggr\]dy=0$