1

The two forms of the Hermite polynomial $$H_{n}(y)=\left( 2y -\frac{d}{dy} \right)^n\cdot 1$$ $$H_{n}(y)=(-1)^n e^{y^{2}}\frac{d^n}{dy^n}e^{-y^{2}}$$ can be proven to be equivalent by induction (e.g see Hermite polynomials $H_{n}(y)=\frac{1}{\sqrt{2^n}}\left( y -\frac{d}{dy} \right)^n$ equivalent form)If I have some function $$G_{n}(y)=\frac{1}{\sqrt{2^n}}\left( y -\frac{d}{dy} \right)^n\cdot e^{-y^{2}}$$ which is the form you obtain when deriving the energy eigenstates of the quantum Harmonic oscillator, can we then express this in terms of the Hermite polynomial function? In textbooks, I've seen this interpreted as $$G_{n}(y)=e^{-y^{2}}H_{n}(y)$$ however I don't see why this would be the case, since we now have to consider the differentiation of the exponential. Am i misenterpreting the $\cdot 1$ here? Thanks for any help

daw
  • 54,637
  • 2
  • 44
  • 85
Adrien Amour
  • 275

1 Answers1

2

You have badly misdefined your Hermite functions $G_n$! The normalization of the fundamental Gaussian exponent matters...

Ignoring overall normalizations, which you may take care of later, you need to see $$ G_n(y) \propto \left (y-\partial_y \right )^n \cdot e^{-y^2/2} = e^{-y^2/2} \left (2y-\partial_y \right )^n \cdot 1 = e^{-y^2/2} H_n(y). $$

First, appreciate the fundamental operator braiding identity $$ \left (y-\partial_y \right ) \cdot e^{-y^2/2} = e^{-y^2/2} \left (2y-\partial_y \right ) \cdot 1 . $$

Repeat, recursively, to an arbitrary power of the raising operator in the parenthesis.

Cosmas Zachos
  • 6,223