Timeline for Having the derivative be an operator
Current License: CC BY-SA 3.0
4 events
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| May 3, 2012 at 13:24 | history | edited | FJRA | CC BY-SA 3.0 |
deleted 29 characters in body
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| May 3, 2012 at 13:05 | comment | added | FJRA | @celtschk you are totally right. I should not answer at nights... I will change it to a non-conmutative way. | |
| May 3, 2012 at 8:28 | comment | added | celtschk |
Your last example uses non-commuting operators (dx and x don't commute, nor do dy and y) and therefore violates the restriction you identified yourself (try da := D[#,a]&; applyOp[da a] to see a result which you surely didn't expect). However your suggestion has a more important flaw: You do not distinguish between the operator and the function it is applied to. Most importantly, the operator $\partial_x f(\hat x)$ is not the same as the operator $f'(\hat x)$, but equals the operator $(f'(\hat x) + f(\hat x)\partial_x)$.
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| May 3, 2012 at 8:08 | history | answered | FJRA | CC BY-SA 3.0 |