# square of differential operator

• Apr 5th 2010, 11:36 AM
cribby
square of differential operator
Can differential operators be squared algebraically in the usual way? What would this even mean? Is this just the differential operator applied twice consecutively?

For example, does it make sense to compute $(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y})^2 = (\frac{\partial}{\partial x})^2 -2i\frac{\partial}{\partial x}\frac{\partial}{\partial y}-(\frac{\partial}{\partial y})^2$? If so, what is to be done with, say, $(\frac{\partial}{\partial x})^2$? Is this just $\frac{\partial ^2}{\partial x^2}$?
• Apr 5th 2010, 12:04 PM
HallsofIvy
Quote:

Originally Posted by cribby
Can differential operators be squared algebraically in the usual way? What would this even mean? Is this just the differential operator applied twice consecutively?

For example, does it make sense to compute $(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y})^2 = (\frac{\partial}{\partial x})^2 -2i\frac{\partial}{\partial x}\frac{\partial}{\partial y}-(\frac{\partial}{\partial y})^2$? If so, what is to be done with, say, $(\frac{\partial}{\partial x})^2$? Is this just $\frac{\partial ^2}{\partial x^2}$?

Yes, in "operator" notation, the basic operation is composition so $\left(\frac{\partial }{\partial x}- i\frac{\partial }{\partial y}\right)^2$ $= \frac{\partial^2}{\partial x^2}- 2i\frac{\partial^2}{\partial x\partial y}- \frac{\partial^2}{\partial y^2}$