Hermite polynomials $H_{n}(y)=\frac{1}{\sqrt{2^n}}\left( y -\frac{d}{dy} \right)^n$ equivalent form

Question

I want to transform the following espression: $$H_{n}(y)=\frac{1}{\sqrt{2^n}}\left( y -\frac{d}{dy} \right)^n$$ in $$H_{n}(y)=(-1)^n e^{y^{2}}\frac{d^n}{dy^n}e^{-y^{2}}$$ Honestly, I have no idea how to go. Thank you so

Answer 1

First of all you got a typo in the question: it must be $\left(2y - \frac{d}{dy}\right)^n\cdot 1$ on the left hand side for the equality to be true (just check for $n=1$). Also, as has been pointed out above, to separate this from an operator identity one should explicitly point out that the left hand side of the identity is operation on the identity $1$.

With this fixed the identity is most easily shown using induction. Assume the two representations agree for a given $n$, i.e.

$$\left(2y - \frac{d}{dy}\right)^n\cdot 1 = (-1)^n e^{y^2}\frac{d^n}{dy^n}e^{-y^2}$$

Now apply $\left(2y - \frac{d}{dy}\right)$ to both sides of the equation to get

$$\left(2y - \frac{d}{dy}\right)^{n+1}\cdot 1 = \left(2y - \frac{d}{dy}\right)\left[(-1)^n e^{y^2}\frac{d^n}{dy^n}e^{-y^2}\right]$$

The left hand side above is the first representation for $n+1$. If you expand the left hand side (but write $\frac{d}{dy}\frac{d^n}{dy^n} = \frac{d^{n+1}}{dy^{n+1}}$, don't try to expand the $n$'th derivatives) you should be able to show it reduces to the right hand side for $n+1$.

This shows that if the identity holds for a given $n$ then it holds for $n+1$. Finally show that the equality holds for the base-case $n=1$, i.e. $2y = (-1)e^{y^2}\frac{d}{dy}e^{-y^2}$, completing the induction proof.