# 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 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 ? Feb 15, 2019 at 19:47
• That simplification doesn't work though as operators don't commute. It simplifies to $(-\frac{d^2}{dx^2}-x\frac{d}{dx}+\frac{d}{dx}x+x^2)^n$. Feb 15, 2019 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 $[d/dx,x]=1$. Feb 15, 2019 at 20:15
• @NickGuerrero Inasmuch as these operators ($\mathscr{L}^+$ and $\mathscr{L}^-$) do not commute, how would you show, as you asserted, that $$\left(\mathscr{L}^+\right)^n \left(\mathscr{L}^-\right)^n=(\mathscr{L}^+\mathscr{L}^-)^n?$$ Feb 15, 2019 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). Feb 15, 2019 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. Feb 15, 2019 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... Feb 15, 2019 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. Feb 17, 2019 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! Feb 17, 2019 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, for each integer $$n \geq 0$$, we have \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 $$n$$. 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, for each integer $$n \geq 0$$, we have \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 $$n$$. 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? Feb 15, 2019 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. Feb 15, 2019 at 20:39
• Thank you for the reference , but I could not open it. Feb 15, 2019 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! Feb 17, 2019 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 $x$. Feb 17, 2019 at 1:32