What does $D_x-xD=1$ mean in mathematics and how can we prove it?

Question

Let $D$ be a differential operator, I came across this physics formula $D_x-xD=1$ which is one of the basic of quantum mechanics. I'm interested about its mathematical meaning. Is there any proof about it, or is it just because of experimental results that should believe it?

Answer 1

The formula is probably $Dx-xD=1$. If you have a differentiable function $f(x)$, then $$ D(xf(x))=f(x)+xf'(x)=f(x)+xD(f(x)), $$ where $D$ denotes the differentiation operator, that is, $$ D(xf(x))-xD(f(x))=f(x). $$ If, in more abstract terms, $x$ denotes the operator “multiply by the identity function”, $$ D(xf)-xD(f)=f $$ so the operator $Dx-xD$ is the identity operator $1$. In this context, composition is often denoted by simple juxtaposition and the identity operator with $1$ (not to be confused with the identity function).

Answer 2

Here's a more philosophical answer to go with @egreg 's mathematical one.

Solutions to this differential equation describe how the quantum mechanical world behaves. That world is described in mathematical terms. This equation is essentially an assumption in that mathematical model. Experiments have shown pretty conclusively that the model accurately predicts what happens in the real physical world.

Several caveats:

• The equation can be derived in the model as a consequence of other
  somewhat more basic assumptions, but it's still part of the model.
  There's no way to prove mathematically that the model is the right
  one for the real world.

• The model may not tell the whole story when you try to incorporate
  the force of gravity with the forces that we mostly understand in
  quantum mechanics. That's an active area of research in physics.