1
$\begingroup$

Generally, Hermite polynomials can be described using the Rodrigues formula:

$$ H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2} $$

And the first few polynomials (for $n = 0,1,2,3,4,5...$) are well known:

$$ H_0(x) = 1\\ H_1(x) = 2x\\ H_2(x) = 4x^2 - 2\\ H_3(x) = 8x^3 - 12x\\ H_4(x) = 16x^4 - 48x^2 + 12\\ H_5(x) = 32x^5 - 160x^3 + 120x\\ $$

I am asked to derive two "ladder operators", $L_{\pm}$, (like the ones used for the harmonic oscillator in quantum mechanics), such that

$$ L_{\pm}H_{n}(x) = H_{n\pm 1}(x) $$

It seems the $L_{+}$ would be something like "multiply $2x$ each time and do other stuff" -- or more eloquently, "multiply $\frac{d}{dx} e^{x^2}$ and ..." -- but I'm having trouble getting a full answer.

$\endgroup$

1 Answer 1

1
$\begingroup$

Note that,

$$H_{n}'(x) = \frac{d}{dx}\left[(-1)^n \exp(x^2) \left(\frac{d}{dx}\right)^n \exp(-x^2)\right] = 2x H_n - H_{n+1}(x),$$

which can be rearranged so that,

$$H_{n+1}(x) = \left(2x-\frac{d}{dx}\right)H_n(x),$$

from which we conclude that,

$$\boxed{L_+ = 2x-\frac{d}{dx}}.$$

For $L_{-}(x)$ note that for the polynomials you have listed $H'_n(x)=2nH_{n-1}(x)$ holds. This suggests an induction proof.

Suppose $H'_n(x)=2nH_{n-1}(x)$ for some $n$ and consider the following,

$$H'_{n+1} = (L_+H_n(x))' = (2xH_n(x)-\color{red}{H'_n(x)})'$$ $$= (2xH_n(x)-\color{red}{2nH_{n-1}(x)})'$$ $$ = 2H_n(x) + \color{blue}{2xH'(x)} - 2nH'_{n-1}(x)$$ $$ = 2H_n(x) + \color{blue}{2x(2n)H_{n-1}(x)} - 2nH'_{n-1}(x)$$ $$ = 2H_n(x) + 2n(\color{green}{2xH_{n-1}-H_{n-1}'(x)})$$ $$ = 2H_n(x)+2n\color{green}{L_+H_{n-1}(x)} $$ $$=2H_n+2n\color{green}{H_n(x)}$$ $$ = 2(n+1)H_n(x) $$

From this I would conclude that the best we can do for $L_-$ is,

$$\boxed{L_- = \frac{d}{dx} },$$

the result when applied to $H_n(x)$ will introduce a factor of $2n$ which is different than the identity you are looking for.

$\endgroup$
1
  • $\begingroup$ This is probably a bit late for your purposes, but I saw the unanswered question during a search and figured I would attempt to answer it. $\endgroup$
    – Spencer
    Aug 15, 2016 at 4:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .