Inquiry about operator algebra

Question

I've just began studying some quantum mechanics, and I'm not sure why certain rules in operator algebra are correct. For instance, in this book it is stated that

$$\left(\frac{d}{dx}+v(x)\right)\left(\frac{d}{dx}+w(x)\right)= \frac{d^2}{dx^2}+\frac{dw}{dx}+(v(x)+w(x))\frac{d}{dx}+vw.$$

when I try to work it out

$$\left(\frac{d}{dx}+v(x)\right)\left(\frac{d}{dx}+w(x)\right)= \frac{d}{dx}\left(\frac{d}{dx}+w(x)\right)+v(x)\left(\frac{d}{dx}+w(x)\right)$$

$$=\frac{d^2}{dx^2}+\frac{dw}{dx}+v(x)\frac{d}{dx}+v(x)w(x)$$

Why is the result cited in the book true but mine not?

Answer 1

We have $(\frac{d}{dx}w(x))f=\frac{d}{dx}(w\cdot f)=w\cdot f'+w'\cdot f=(w\frac{d}{dx}+w')f$. It is not true that multiplication by $w$ followed by differentiation is the same as multiplication by $w'$.

Answer 2

There is a hidden subtlety in here. The equation you have started above is true in the book, but only because that is how these operators work. A differential operator is really a linear map from a space of functions to functions, and so in order to see what is going on when you compose the maps as you have done above (what looks like multiplication is really just composition) you should look at how it acts on an arbitrary function. Explicitly:

$(\frac{d}{dx}+v(x))(\frac{d}{dx}+w(x))f=(\frac{d}{dx}+v(x))(\frac{df}{dx} +(w\cdot f)(x))$

Then applying the Leibnitz rule to the product, you will get the result stated in the book