Why does the commutator Dx - xD equal 1?

[tirnanog84]

We're doing differential operators in my Differential Equations class right now, and our professor assigned the following problem to us:

(D-x)(D+x)

Which inevitably gives us the following terms as part of the final answer: Dx-xD

The answer in the book tells me that Dx-xD = 1, and some preliminary research has told me that this is true. What I couldn't find was the why. Why, or how, does Dx-xD result in 1? And does -Dx+xD = -1?

 

[lurflurf]

We're doing differential operators in my Differential Equations class right now, and our professor assigned the following problem to us:

(D-x)(D+x)

Which inevitably gives us the following terms as part of the final answer: Dx-xD

The answer in the book tells me that Dx-xD = 1, and some preliminary research has told me that this is true. What I couldn't find was the why. Why, or how, does Dx-xD result in 1? And does -Dx+xD = -1?

product rule gives
Dx=1+xD
might be easier to see with a function
D(xy)=(Dx)y+x(Dy)=y+x(Dy)=(1+xD)y
so
Dx=1+xD

 

[balakrishnan_v]

That is because
$$\frac{d}{dx}\left(xf(x)\right)-x\frac{d}{dx}\left(f(x)\right)=xf'(x)+f(x)-xf'(x)=f(x)$$
$$(Dx-xD)f(x)=f(x)$$
Hence Dx-xD=1

 

[tirnanog84]

Thanks!