# Computing the product $(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n$

I want to compute the product $$(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n,$$ for a natural number $$n$$. For $$n$$ equal to 0 or 1, the computation is very simple obviously, but for such a low number as 2 the brute force calculation begins to be rather cumbersome and I cannot see any pattern emerging. I tried to find some connection with the Rodrigues' formula for the Hermite polynomials but I could not.

These operators come up in the algebraic approach to the quantum harmonic oscillator.

Explicit Example

To avoid any misunderstanding, I am going to show explicitly the computation for the case $$n=1$$: $$(\frac{d}{dx}+x)(-\frac{d}{dx}+x)=-\frac{d^2}{dx^2}+1+x\frac{d}{dx}-x\frac{d}{dx}+x^2=-\frac{d^2}{dx^2}+x^2+1.$$

One can think of a function $$f$$ the operators are acting on. For example, $$(\frac{d}{dx}\circ x) f= (\frac{d}{dx}x)f+x\frac{d}{dx}f=(1+\frac{d}{dx})f,$$ then $$\frac{d}{dx}\circ x=1+x\frac{d}{dx}$$

• Since this is physics related, can I ask what the stand-alone differential expression $\mathrm{d}/\mathrm{d}x$ means for your context ? – Rebellos Feb 15 at 19:47
• That simplification doesn't work though as operators don't commute. It simplifies to $(-\frac{d^2}{dx^2}-x\frac{d}{dx}+\frac{d}{dx}x+x^2)^n$. – Nick Guerrero Feb 15 at 20:00
• @Rebellos, $d/dx$ is simply the derivative operator. In this context, it is supposed that the operators are acting on some function. For example, the commutator between $d/dx$ and $x$ is computed as $[d/dx,x]f=d/dx(xf)-x(d/dx f)=f+d/dx f-d/dx f=f$, then $[d/dx,x]=1$. – jobe Feb 15 at 20:15
• @NickGuerrero Inasmuch as these operators ($\mathscr{L}^+$ and $\mathscr{L}^-$) do not commute, how would you show, as you asserted, that $$\left(\mathscr{L}^+\right)^n \left(\mathscr{L}^-\right)^n=(\mathscr{L}^+\mathscr{L}^-)^n?$$ – Mark Viola Feb 15 at 20:32
• @jobe See THIS, which provides a discussion for $(A+B)^n$ where $[A,B]\ne 0$ (i.e., the operators do not commute). – Mark Viola Feb 15 at 20:47

For convenience let us rewrite $$x,\partial_x$$ with $$a,b$$ so $$[a,b]=x\partial_x-\partial_x x=-1$$ (in the operator sense on the Schwartz space).

Lemma. For all $$n\in\mathbb N$$ $$[(a+b)^n,a-b]=2n(a+b)^{n-1}$$

Proof. As darij pointed out, one has $$[a+b,a-b]=2$$ (i.e. the case $$n=1$$). The trick then is $$\begin{split} [(a+b)^{n+1},a-b]&=(a+b)[(a+b)^n,a-b]+[a+b,a-b](a+b)^n\\ &=(a+b)2n(a+b)^{n-1}+2(a+b)^{n}=2(n+1)(a+b)^{(n+1)-1} \end{split}$$ which concludes the proof via induction. $$\square$$

Proposition. For all $$n\in\mathbb N_0$$ $$(a+b)^n(a-b)^n=\prod_{j=1}^n (a^2-b^2+(2j-1))$$

Proof. ($$n=0$$ is obvious). Note that $$(a-b)(a+b)=a^2-b^2-1$$. Using the previous lemma $$\begin{split} (a+b)^{n+1}(a-b)^{n+1}&=[(a+b)^{n+1},a-b](a-b)^n+(a-b)(a+b)(a+b)^n(a-b)^n\\ &=2(n+1)(a+b)^n(a-b)^n +(a^2-b^2-1)(a+b)^n(a-b)^n\\ &=\big( a^2-b^2+2n+1)(a+b)^n(a-b)^n=\prod_{j=1}^{n+1} (a^2-b^2+(2j-1)) \end{split}$$ which again concludes the proof via induction. $$\square$$

This result reproduces the cases (aside from $$n=0$$, obvious)

• $$(a+b)(a-b)=a^2-b^2+1$$

• Making use of $$[a^2,b^2]=-4ab-2$$ (similar techniques) one gets $$\begin{split} (a+b)^2(a-b)^2=(a^2-b^2+1)(a^2-b^2+3)&=a^4-a^2b^2-b^2a^2+4a^2-4b^2+b^4+3\\ &=a^4-2a^2b^2-4ab+4a^2-4b^2+b^4+1 \end{split}$$

etc... I feel like this formula is the best thing one can hope for in terms of structure.

Edit: Thanks darij for the +200 rep!

• Great job! Pretty sure that a factorization into degree-$2$ factors is an answer to the OP. – darij grinberg Feb 15 at 21:26
• Thank you! I like how we both independently considered a commutator of the form $[x^m,y]$ (my Lemma / your eq. (1)) to approach the problem... – Frederik vom Ende Feb 15 at 21:30
• Thank you for your answer! I am very happy my question has received so amazing answers, but I am having a hard time choosing which one to accept. Your answer seems to be the more explicit that is possible without becoming too complicated. I would like to have the derivative on the right side of each term, but probably that would made the result very complicated as the obtained by @Sangchul Lee. – jobe Feb 17 at 1:12
• Glad to hear so! Of course one may use further tricks to shift all the derivatives ($b$'s) to the right such as $[x^2,\partial_x^2]=-4x\partial_x-2$ etc. although I'm not sure how simple or applicable that'd end up being. Anyways, just accept whatever answer helps (or will likely end up helping) you the most! – Frederik vom Ende Feb 17 at 7:35

Just some Sage-generated data to play around with:

For $$n = 0$$, the result is $$1$$.

For $$n = 1$$, the result is $$-\frac{\partial^{2}}{\partial x^{2}} + x^{2} + 1$$.

For $$n = 2$$, the result is $$\frac{\partial^{4}}{\partial x^{4}} - 2 x^{2} \frac{\partial^{2}}{\partial x^{2}} - 4 \frac{\partial^{2}}{\partial x^{2}} - 4 x \frac{\partial}{\partial x} + x^{4} + 4 x^{2} + 1$$.

For $$n = 3$$, the result is $$-\frac{\partial^{6}}{\partial x^{6}} + 3 x^{2} \frac{\partial^{4}}{\partial x^{4}} + 9 \frac{\partial^{4}}{\partial x^{4}} + 12 x \frac{\partial^{3}}{\partial x^{3}} - 3 x^{4} \frac{\partial^{2}}{\partial x^{2}} - 18 x^{2} \frac{\partial^{2}}{\partial x^{2}} - 9 \frac{\partial^{2}}{\partial x^{2}} - 12 x^{3} \frac{\partial}{\partial x} - 36 x \frac{\partial}{\partial x} + x^{6} + 9 x^{4} + 9 x^{2} - 3$$.

For $$n = 4$$, the result is $$\frac{\partial^{8}}{\partial x^{8}} - 4 x^{2} \frac{\partial^{6}}{\partial x^{6}} - 16 \frac{\partial^{6}}{\partial x^{6}} - 24 x \frac{\partial^{5}}{\partial x^{5}} + 6 x^{4} \frac{\partial^{4}}{\partial x^{4}} + 48 x^{2} \frac{\partial^{4}}{\partial x^{4}} + 42 \frac{\partial^{4}}{\partial x^{4}} + 48 x^{3} \frac{\partial^{3}}{\partial x^{3}} + 192 x \frac{\partial^{3}}{\partial x^{3}} - 4 x^{6} \frac{\partial^{2}}{\partial x^{2}} - 48 x^{4} \frac{\partial^{2}}{\partial x^{2}} - 36 x^{2} \frac{\partial^{2}}{\partial x^{2}} + 48 \frac{\partial^{2}}{\partial x^{2}} - 24 x^{5} \frac{\partial}{\partial x} - 192 x^{3} \frac{\partial}{\partial x} - 216 x \frac{\partial}{\partial x} + x^{8} + 16 x^{6} + 42 x^{4} - 48 x^{2} - 39$$.

For $$n = 5$$, the result is $$-\frac{\partial^{10}}{\partial x^{10}} + 5 x^{2} \frac{\partial^{8}}{\partial x^{8}} + 25 \frac{\partial^{8}}{\partial x^{8}} + 40 x \frac{\partial^{7}}{\partial x^{7}} - 10 x^{4} \frac{\partial^{6}}{\partial x^{6}} - 100 x^{2} \frac{\partial^{6}}{\partial x^{6}} - 130 \frac{\partial^{6}}{\partial x^{6}} - 120 x^{3} \frac{\partial^{5}}{\partial x^{5}} - 600 x \frac{\partial^{5}}{\partial x^{5}} + 10 x^{6} \frac{\partial^{4}}{\partial x^{4}} + 150 x^{4} \frac{\partial^{4}}{\partial x^{4}} + 150 x^{2} \frac{\partial^{4}}{\partial x^{4}} - 150 \frac{\partial^{4}}{\partial x^{4}} + 120 x^{5} \frac{\partial^{3}}{\partial x^{3}} + 1200 x^{3} \frac{\partial^{3}}{\partial x^{3}} + 1800 x \frac{\partial^{3}}{\partial x^{3}} - 5 x^{8} \frac{\partial^{2}}{\partial x^{2}} - 100 x^{6} \frac{\partial^{2}}{\partial x^{2}} - 150 x^{4} \frac{\partial^{2}}{\partial x^{2}} + 1500 x^{2} \frac{\partial^{2}}{\partial x^{2}} + 975 \frac{\partial^{2}}{\partial x^{2}} - 40 x^{7} \frac{\partial}{\partial x} - 600 x^{5} \frac{\partial}{\partial x} - 1800 x^{3} \frac{\partial}{\partial x} - 600 x \frac{\partial}{\partial x} + x^{10} + 25 x^{8} + 130 x^{6} - 150 x^{4} - 975 x^{2} - 255$$.

For $$n = 6$$, the result is $$\frac{\partial^{12}}{\partial x^{12}} - 6 x^{2} \frac{\partial^{10}}{\partial x^{10}} - 36 \frac{\partial^{10}}{\partial x^{10}} - 60 x \frac{\partial^{9}}{\partial x^{9}} + 15 x^{4} \frac{\partial^{8}}{\partial x^{8}} + 180 x^{2} \frac{\partial^{8}}{\partial x^{8}} + 315 \frac{\partial^{8}}{\partial x^{8}} + 240 x^{3} \frac{\partial^{7}}{\partial x^{7}} + 1440 x \frac{\partial^{7}}{\partial x^{7}} - 20 x^{6} \frac{\partial^{6}}{\partial x^{6}} - 360 x^{4} \frac{\partial^{6}}{\partial x^{6}} - 540 x^{2} \frac{\partial^{6}}{\partial x^{6}} + 120 \frac{\partial^{6}}{\partial x^{6}} - 360 x^{5} \frac{\partial^{5}}{\partial x^{5}} - 4320 x^{3} \frac{\partial^{5}}{\partial x^{5}} - 8280 x \frac{\partial^{5}}{\partial x^{5}} + 15 x^{8} \frac{\partial^{4}}{\partial x^{4}} + 360 x^{6} \frac{\partial^{4}}{\partial x^{4}} + 450 x^{4} \frac{\partial^{4}}{\partial x^{4}} - 9000 x^{2} \frac{\partial^{4}}{\partial x^{4}} - 6525 \frac{\partial^{4}}{\partial x^{4}} + 240 x^{7} \frac{\partial^{3}}{\partial x^{3}} + 4320 x^{5} \frac{\partial^{3}}{\partial x^{3}} + 15600 x^{3} \frac{\partial^{3}}{\partial x^{3}} + 7200 x \frac{\partial^{3}}{\partial x^{3}} - 6 x^{10} \frac{\partial^{2}}{\partial x^{2}} - 180 x^{8} \frac{\partial^{2}}{\partial x^{2}} - 540 x^{6} \frac{\partial^{2}}{\partial x^{2}} + 9000 x^{4} \frac{\partial^{2}}{\partial x^{2}} + 31050 x^{2} \frac{\partial^{2}}{\partial x^{2}} + 9180 \frac{\partial^{2}}{\partial x^{2}} - 60 x^{9} \frac{\partial}{\partial x} - 1440 x^{7} \frac{\partial}{\partial x} - 8280 x^{5} \frac{\partial}{\partial x} - 7200 x^{3} \frac{\partial}{\partial x} + 8100 x \frac{\partial}{\partial x} + x^{12} + 36 x^{10} + 315 x^{8} - 120 x^{6} - 6525 x^{4} - 9180 x^{2} - 855$$.

Sage code:

A.<x> = DifferentialWeylAlgebra(QQ)
x, dx = A.gens()

def r(n):
return (dx + x) ** n * (-dx + x) ** n

for i in range(7):
print "For $$n = " + str(i) + "$$, the result is $$" + latex(r(i)) + "$$.\r\n"


Note that it is easily seen that $$\left[\dfrac{\partial}{\partial x} + x, - \dfrac{\partial}{\partial x} + x\right] = 2$$. Thus, the operators $$\dfrac{\partial}{\partial x} + x$$ and $$- \dfrac{\partial}{\partial x} + x$$ themselves generate an isomorphic copy of the Weyl algebra, except for a scalar factor of $$2$$.

In view of the relation $$\left[\dfrac{\partial}{\partial x} + x, - \dfrac{\partial}{\partial x} + x\right] = 2$$, perhaps the following copypasta from some of my old homework will come useful.

Let $$\mathbb{N} = \left\{0,1,2,\ldots\right\}$$.

Now we need an easy fact from quantum algebra:

Proposition 1. Let $$A$$ be a ring (not necessarily commutative). Let $$x\in A$$ and $$y\in A$$ be such that $$xy-yx=1$$. Then, \begin{align} \left( xy\right) ^{n}=\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}, \end{align} where the curly braces denote Stirling numbers of the second kind.

Proof of Proposition 1. First of all, it is easy to see that $$$$x^{m}y=mx^{m-1}+yx^{m}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N} \label{darij1.pf.xy-yx.1} \tag{1}$$$$ (this allows $$m=0$$ if $$0x^{0-1}$$ is interpreted as $$0$$). Indeed, the proof of \eqref{darij1.pf.xy-yx.1} proceeds by induction over $$m$$ and is straightforward enough to be left to the reader.

We will now prove Proposition 1 by induction over $$k$$. The induction base is obvious, so we step to the induction step:

Let $$n>0$$. Assuming that $$\left( xy\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}x^{k}$$, we need to show that $$\left( xy\right) ^{n}=\sum\limits_{k=0} ^{n} \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}$$.

We have \begin{align*} \left( xy\right) ^{n} & =\left( xy\right) ^{n-1}xy=\left( \sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}x^{k}\right) xy\ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( xy\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}x^{k}\right) \\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}\underbrace{x^{k}x}_{=x^{k+1}}y=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}\underbrace{x^{k+1}y}_{\substack{=\left( k+1\right) x^{k} +yx^{k+1}\\\text{(by \eqref{darij1.pf.xy-yx.1}, applied to }m=k+1\text{)}}}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}\left( \left( k+1\right) x^{k}+yx^{k+1}\right) \\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \underbrace{y^{k}y}_{=y^{k+1}}x^{k+1}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k+1}x^{k+1}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=1}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }k-1\text{ for }k\text{ in the second sum}\right) \\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array}[c]{c} \text{here, we extended both sums by zero terms, using the fact}\\ \text{that } \genfrac{\{}{\}}{0pt}{0}{n}{n+1} = \genfrac{\{}{\}}{0pt}{0}{n}{0} =0\text{ whenever }n>0 \end{array} \right) \\ & =\sum\limits_{k=0}^{n}\underbrace{\left( \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) + \genfrac{\{}{\}}{0pt}{0}{n}{k} \right) }_{\substack{= \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} \\\text{(by the recursion formula for Stirling numbers}\\\text{of the second kind)}}}y^{k}x^{k}\\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}. \end{align*} This completes the induction step, and thus the inductive proof of Proposition 1. $$\blacksquare$$

Proposition 2. Let $$A$$ be a ring (not necessarily commutative). Let $$x\in A$$ and $$y\in A$$ be such that $$xy-yx=1$$. Then, \begin{align} \left( yx\right) ^{n}=\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}, \end{align} where the curly braces denote Stirling numbers of the second kind.

Proof of Proposition 2. Just as in the proof of Proposition 1, we show that \eqref{darij1.pf.xy-yx.1} holds.

We will now prove Proposition 2 by induction over $$k$$. The induction base is obvious, so we step to the induction step:

Let $$n>0$$. Assuming that $$\left( yx\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}x^{k}$$, we need to show that $$\left( yx\right) ^{n}=\sum\limits_{k=0} ^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}$$.

We have \begin{align*} \left( yx\right) ^{n} & =\left( yx\right) ^{n-1}yx=\left( \sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}x^{k}\right) yx\ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( yx\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}x^{k}\right) \\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}\underbrace{x^{k}y}_{\substack{=kx^{k-1}+yx^{k}\\\text{(by \eqref{darij1.pf.xy-yx.1}, applied to }m=k\text{)}}}x\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}\left( kx^{k-1}+yx^{k}\right) x\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}\underbrace{x^{k-1}x}_{=x^{k}}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} k\underbrace{y^{k}y}_{=y^{k+1}}\underbrace{x^{k}x}_{=x^{k+1}}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k+1}x^{k+1}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\sum\limits_{k=1}^{n} \genfrac{\{}{\}}{0pt}{0}{n-1}{k-1} ky^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }k-1\text{ for }k\text{ in the second sum}\right) \\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n-1}{k-1} ky^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array}[c]{c} \text{here, we extended both sums by zero terms, using the fact}\\ \text{that } \genfrac{\{}{\}}{0pt}{0}{n-1}{n} = \genfrac{\{}{\}}{0pt}{0}{n-1}{-1} =0\text{ whenever }n>0 \end{array} \right) \\ & =\sum\limits_{k=0}^{n}\underbrace{\left( \genfrac{\{}{\}}{0pt}{0}{n-1}{k} k+ \genfrac{\{}{\}}{0pt}{0}{n-1}{k-1} \right) }_{\substack{= \genfrac{\{}{\}}{0pt}{0}{n}{k} \\\text{(by the recursion formula for Stirling numbers}\\\text{of the second kind)}}}y^{k}x^{k}\\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}. \end{align*} This completes the induction step, and thus the inductive proof of Proposition 2. $$\blacksquare$$

• Thank you very much for this code! I really needed something like this! The only point is that I am desperately looking for a closed formula for arbitrary $n$. Do you have any thoughts about that? – jobe Feb 15 at 20:36
• @jobe: My output doesn't give me much hope for a closed formula -- but of course, the Weyl algebra has several bases in which these elements could be expanded, and maybe one of them looks better. – darij grinberg Feb 15 at 20:39
• Thank you for the reference , but I could not open it. – jobe Feb 15 at 20:45

Let $$X, Y$$ be operators which are central, meaning that $$[X,Y] = XY-YX = c$$ for some scalar $$c$$. Then

1. As a form of binomial theorem, we have

$$(X+Y)^n = \sum_{\substack{a,b,m \geq 0 \\ a+b+2m=n}} \frac{n!}{a!b!m!2^m} c^m Y^b X^a = \sum_{a=0}^{n} \binom{n}{a} P_{n-a}(c,Y)X^a,$$

where $$P_n$$ is defined by the following sum

$$P_n(c, x) = \sum_{m=0}^{\lfloor n/2\rfloor} \frac{n!}{(n-2m)!m!2^m} c^m x^{n-2m} = \left( c\frac{d}{dx} + x \right)^n \mathbf{1}.$$

In particular, if $$c = -1$$ then $$P_n(-1, x) = \operatorname{He}_n(x)$$, where $$\operatorname{He}_n$$ is the probabilists' Hermite polynomial. Similarly, if $$c = 1$$, then $$P_n(1, x) = i^{-n} \operatorname{He}_n(ix)$$.

2. Under the same condition, for any polynomials $$f, g$$ we have

$$f(X)g(Y) = \sum_{m \geq 0} \frac{c^m}{m!} g^{(m)}(Y)f^{(m)}(X).$$

In our case, $$[\frac{d}{dx}, x] = 1$$, and so, we can use both formulas to give a complicated, but still explicit expression for the product of $$(\frac{d}{dx}+x)^n$$ and $$(-\frac{d}{dx}+x)^n$$. Combining altogether,

\begin{align*} &\left( \frac{d}{dx} + x \right)^n \left( -\frac{d}{dx} + x \right)^n \\ &= \sum_{p \geq 0} \frac{1}{p!} \Bigg( \sum_{\substack{a_i, b_i, m_i \geq 0 \\ a_i+b_i+2m_i = n-p}} \frac{(-1)^{a_2+m_2} (n!)^2}{a_1!a_2!b_1!b_2!m_1!m_2!2^{m_1+m_2}} x^{b_1+b_2} \left( \frac{d}{dx} \right)^{a_1+a_2} \Bigg). \end{align*}

The following is a sample Mathematica code, comparing this formula with the actual answer for the case $$n = 3$$.

Coef[n_, c_, l_] := n!/(l[[1]]! l[[2]]! l[[3]]! 2^l[[3]]) c^l[[3]];
T[n_] := FrobeniusSolve[{1, 1, 2}, n];
(* Compute (x+d/dx) (x-d/dx)^n f(x) *)
n = 3;
Nest[Expand[x # + D[#, x]] &, Nest[Expand[x # - D[#, x]] &, f[x], n], n]
Sum[1/p! Sum[ Sum[ Coef[n, 1, l1] Coef[n, -1, l2]
(-1)^l2[[2]] x^(l1[[1]] + l2[[1]]) D[f[x],
{x, l1[[2]] + l2[[2]]}], {l1, T[n - p]}], {l2, T[n - p]}], {p, 0, n}] // Expand
Clear[n];


• Impressive! Your answer gives an explicit expression for the product as I have asked for, but it is indeed complicated! I am still trying to figure out how using it effectively in the problems I have in mind, but that is my fault. Thank you! – jobe Feb 17 at 1:19
• @jobe, Glad it helped! This formula may be still far from being qualified as generally useful, but a form of combinatoric interpretation of the coefficients in the expansion of $(X+Y)^n$ is available, which may possible be useful for some applications: $$\frac{n!}{(n-2m)!m!2^m}=[\text{# of m-matchings in \{1,\cdots,n\}}].$$ So $P_n(c,x)$ computes the sum of weights over all possible matchings on $\{1,\cdots,n\}$, where each matching is weighted by the factor $c^m x^{n-2m}$, i.e. each matched pair receives the weight $c$ and each unmatched vertice receives the weight $x$. – Sangchul Lee Feb 17 at 1:32