Suppose we have the following differential equation using operator notation: $$(D-x)(D+x)y=0\tag{1}$$ where $$D=\frac{d}{dx}$$

Now I could rewrite $(1)$ as $$\begin{align}\require{enclose}(D-x)(D+x)y&=\left(\frac{d}{dx}-x\right)\left(y^{\prime}+xy\right)\\&=y^{\prime\prime}+\bbox[#AFA]{\left(xy\right)^{\prime}}-xy^{\prime}-x^2y\\&=y^{\prime\prime}+\bbox[#AFA]{y+\enclose{downdiagonalstrike}{xy^{\prime}}}-\enclose{downdiagonalstrike}{xy^{\prime}}-x^2y\\&\implies \fbox{$y^{\prime\prime}-x^2y + y = 0$}\tag{a}\end{align}$$

Or, by switching the order of the brackets; I could rewrite $(1)$ as $$\begin{align}\require{enclose}(D-x)(D\color{blue}{\textbf{ + }}x)y&=(D+x)(D\color{red}{\textbf{ - }}x)y\\&=\left(\frac{d}{dx}+x\right)\left(y^{\prime}-xy\right)\\&=y^{\prime\prime}\bbox[#FAA]{-\left(xy\right)^{\prime}}+xy^{\prime}-x^2y\\&=y^{\prime\prime}\bbox[#FAA]{-y-\enclose{downdiagonalstrike}{xy^{\prime}}}+\enclose{downdiagonalstrike}{xy^{\prime}}-x^2y\\&\implies \fbox{$y^{\prime\prime}-x^2y - y = 0$}\tag{b}\end{align}$$

Lastly, I could rewrite $(1)$ as $$\begin{align}\require{enclose}(D-x)(D+x)y&=(D^2-x^2)y\\&=\left(\frac{d^2}{dx^2}-x^2\right)y\\&=y^{\prime\prime}-x^2y\\&\implies\ \fbox{$y^{\prime\prime}-x^2y = 0$}\tag{c}\end{align}$$

There's no doubt there's most probably a simple explanation for it; but how can the same differential equation $(1)$ be written in three different ways: $(\mathrm{a})$, $(\mathrm{b})$, $(\mathrm{c})$?

Many thanks.

  • 1
    $\begingroup$ Now explain the downvote $\endgroup$ – BLAZE Jun 24 '16 at 7:30
  • 5
    $\begingroup$ Did not down vote, but the answer is obvious. Operators $D-x$ and $D+x$ do not commute. (You cannot switch the orders) $\endgroup$ – mastrok Jun 24 '16 at 7:32
  • 1
    $\begingroup$ @mastrok Thanks for that; it was not obvious to me. Do you have any sources/links that explain/prove why operators are not commutative? $\endgroup$ – BLAZE Jun 24 '16 at 7:34
  • 1
    $\begingroup$ @BLAZE: look e.g. at the quantum harmonic oscillator. Your observation derives from the elementary fact that $[D,x] =1$ with $[D,x]= Dx -xD$ the commutator. This statement is a statement about operators, thus it means that $[D,x]y =y$ or $(Dx-xD)y=y$. $\endgroup$ – Fabian Jun 24 '16 at 7:38
  • 2
    $\begingroup$ Since you already figured it out, why don't you answer the question yourself? :) $\endgroup$ – mastrok Jun 24 '16 at 7:43

$D+x$ and $D-x$ do not commute. Your (b) is a demonstration of this.

In (c) you are equating $(D+x)(D-x)$ and $D^2-x^2$, but that is wrong for a similar reason - in $D^2-x^2$ the first $D$ is no longer acting on the second $x$.

| cite | improve this answer | |
  • $\begingroup$ Many thanks for your answer; best regards. $\endgroup$ – BLAZE Jun 24 '16 at 10:31
  • $\begingroup$ Apart from the not getting any points; What is the difference between Community Wiki answers and those that are not? $\endgroup$ – BLAZE Jun 24 '16 at 10:46
  • $\begingroup$ You would have to check the help pages. I think you need less reputation to edit them. $\endgroup$ – almagest Jun 24 '16 at 10:47
  • $\begingroup$ Okay, thanks; I'll take a look. $\endgroup$ – BLAZE Jun 24 '16 at 10:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.