# Connection between algebraic and analytic method of quantum harmonic oscillator

I am studying Quantum harmonic oscillator, There are 2 methods to solve Harmonic oscillator one is algebraic method and another is analytic method , Wave functions derived from 2 methods are equivalent,"Zettili" shows this equivalence as follows

$$\exp\left({-x^2\over2x_0^2}\right)\Biggl(x-x_0^2{d\over dx}\Biggr)\exp\left({x^2\over2x_0^2}\right)=x_0^2{-d\over dx}$$

An application of this opperator n times leads to

$$\exp\left({-x^2\over2x_0^2}\right){\Biggl(x-x_0^2{d\over dx}\Biggr)}^n\exp\left({x^2\over2x_0^2}\right)=(-1)^n({x_0^2})^n{-d^n\over dx^n},$$

I don't get this 2 steps.

Arranging this 2 equations gives

$${\Biggl(x-x_0^2{d\over dx}\Biggr)}^n\exp\left(-{x^2\over2x_0^2}\right)={x_0^n}\exp\left({\mathbf-x^2\over2x_0^2}\right)H_n\left({x\over x_0 }\right)$$

Where $$H_n$$ are Hermite polynomials.

Please clarify me to move further.

In first step just take a function $$f(x)$$ and operate it and you will get $$exp({\mathbf-x^2\over2x_0^2})\Biggl(x-x_0^2{d\over dx}\Biggr)exp({ x^2\over2x_0^2})f(x)=x_0^2{\mathbf-d\over dx}f(x)$$Applying the operator n times gives, $$exp({\mathbf-x^2\over2x_0^2})\Biggl(x-x_0^2{d\over dx}\Biggr)exp({x^2\over2x_0^2})exp({\mathbf-x^2\over2x_0^2})\Biggl(x-x_0^2{d\over dx}\Biggr)exp({x^2\over2x_0^2})exp({\mathbf-x^2\over2x_0^2})\Biggl(x-x_0^2{d\over dx}\Biggr)exp({x^2\over2x_0^2}).......exp({\mathbf-x^2\over2x_0^2})\Biggl(x-x_0^2{d\over dx}\Biggr)exp({x^2\over2x_0^2})=exp({\mathbf-x^2\over2x_0^2})\Biggl(x-x_0^2{d\over dx}\Biggr)^n exp({x^2\over2x_0^2}).$$ Because adjacent exponential cancels and become 1. Operating the right hand side of the equation $$n$$ times give $$(-1)^n({x_0^2})^n{d^n\over dx^n}$$ Edit I will make the proof of first step more elaborate. An operator means it should have to act on some function $$f$$ $$\Biggl(e^{-x^2 \over 2x_0^2}(x-x_0^2{d\over dx})e^{x^2 \over 2x_0^2}\Biggr)f=e^{-x^2 \over 2x_0^2}xe^{x^2 \over 2x_0^2} f-e^{-x^2 \over 2x_0^2}x_0^2{d\over dx}(e^{x^2 \over 2x_0^2} f)$$ Apply product rule to the second term and cancelling the exponentials we will get,
$$\Biggl(e^{-x^2 \over 2x_0^2}(x-x_0^2{d\over dx})e^{x^2 \over 2x_0^2}\Biggr)f=xf-(xf+x_0^2{d\over dx}f)=-x_0^2{d\over dx}f$$ Thus we get $$\Biggl(e^{-x^2 \over 2x_0^2}(x-x_0^2{d\over dx})e^{x^2 \over 2x_0^2}\Biggr)=-x_0^2{d\over dx}$$