Wikipedia:Reference desk/Archives/Mathematics/2013 March 31

= March 31 =

Dx - xD = 1, why does this not commute?
This equation: Dx - xD = 1 from the differential operator article says its basic to quantum physics. Obviously there is no commutativity otherwise Dx=xD, but I don't understand why. Any explanation? I've taken calculus and I would like to know what specific things I need to learn in order to understand this equation. --Idontnodi (talk) 02:03, 31 March 2013 (UTC)


 * To help others who wish to respond, here's a link to the relevant section of that article: Differential_operator. StuRat (talk) 02:13, 31 March 2013 (UTC)


 * This is really nothing more than the product rule in action. Apply both sides to a function f(x) and then use the product rule; hey presto. Looie496 (talk) 03:19, 31 March 2013 (UTC)


 * How is that done? I do understand the product rule d(uv)=udv+vdu which is the notation I am accustom to.  But I'm not familiar with the operator notation being used. Can someone rewrite this equation using Leibniz notation perhaps? Doesn't (d(y)/d(x)) x = x (d(y)/d(x))? I'm confused. --Idontnodi (talk) 03:42, 31 March 2013 (UTC)
 * Using the notation in the article and apply to f(x). D(x f(x)) - x D(f(x))= (Dx) * f(x) + x D(f(x)) - x D(f(x)) = (Dx) * f(x) = 1 * f(x). Or in you notation d(x * f) - x*d(f) = d(x)*f + x*d(f) - x*d(f)=d(x)*f=1*f.--Salix (talk): 04:33, 31 March 2013 (UTC)


 * According to these equations: Dx = 1 and xD = 0. Correct? --Idontnodi (talk) 05:27, 31 March 2013 (UTC)


 * They are both operators and need to be applied to a function. $$Dx = \frac{d}{dx}(x*?)$$, $$xD=x*\frac{d}{dx}(?)$$ where ? is replaced by some function. They don't have any particular value until the function is specified. If you apply them to the constant function 1 $$Dx(1) = \frac{d}{dx}(x*1)=1$$, $$xD(1)=x*\frac{d}{dx}(1)=0$$. --Salix (talk): 05:46, 31 March 2013 (UTC)


 * Thanks. --Idontnodi (talk) 08:03, 31 March 2013 (UTC)


 * Here's amother couple of operators to show how something like that can work. With $$\operatorname T = (\times 2), \operatorname S = (+1)$$, you can see that $$(\times 2)(+1)f - (+1)(\times 2)f = 1$$. We could write that as $$\operatorname T \operatorname S - \operatorname S \operatorname T = \operatorname {Constant} 1$$. In the original the 1 was the identity operator rather than producing the constant 1 when applied to any function. Dmcq (talk) 11:19, 31 March 2013 (UTC)


 * Not having ever used operators such as these, I almost wasted my time with that one. Thanks again!! --Idontnodi (talk) 15:04, 31 March 2013 (UTC)

This is perhaps what Dmcq is trying to say.
 * With $$\operatorname T(x) = 2x, \operatorname S(x) = 1+x$$, you can see that $$(\operatorname T\operatorname S-\operatorname S\operatorname T)x = 2(1+ x) - (1+2x) = 1$$.

Bo Jacoby (talk) 15:21, 31 March 2013 (UTC).