A calculation involving differential operators

[Adorno]

Homework Statement

I have the following expressions for angular momentum components: $L_1 = x_2\frac{\partial}{\partial x_3} - x_3\frac{\partial}{\partial x_2}$, $L_2 = x_3\frac{\partial}{\partial x_1} - x_1\frac{\partial}{\partial x_3}$, $L_3 = x_1\frac{\partial}{\partial x_2} - x_2\frac{\partial}{\partial x_1}$, and I simply need to work out $L^2 = L_1^2 + L_2^2 + L_3^2$.

Homework Equations

N/A

The Attempt at a Solution

Well, the way I expand it gives $$L_1^2 = (x_2\frac{\partial}{\partial x_3} - x_3\frac{\partial}{\partial x_2})(x_2\frac{\partial}{\partial x_3} - x_3\frac{\partial}{\partial x_2}) = x_2\frac{\partial}{\partial x_3}x_2\frac{\partial}{\partial x_3} - x_2\frac{\partial}{\partial x_3}x_3\frac{\partial}{\partial x_2} - x_3\frac{\partial}{\partial x_2}x_2\frac{\partial}{\partial x_3} + x_3\frac{\partial}{\partial x_2}x_3\frac{\partial}{\partial x_2} = -x_2\frac{\partial}{\partial x_2} - x_3\frac{\partial}{\partial x_3},$$ and similarly $$L_2^2 = -x_1\frac{\partial}{\partial x_1} - x_3\frac{\partial}{\partial x_3}$$ and $$L_3^2 = -x_1\frac{\partial}{\partial x_1} - x_2\frac{\partial}{\partial x_2},$$ so that $$L^2 = -2x_1\frac{\partial}{\partial x_1} - 2x_2\frac{\partial}{\partial x_2} - 2x_3\frac{\partial}{\partial x_3}$$ But this is not the expression for $L^2$ that I'm supposed to get! So I must be doing something wrong. Can anyone help?

 

[Dick]

How can the derivative expression for L_1^2 not have any second derivatives in it?

 

[Adorno]

That's a good question. I guess it should. When I simplify a term like $x_2\frac{\partial}{\partial x_3}x_2\frac{\partial}{\partial x_3}$, do I have to use the product rule? That would get the second derivatives.

 

[Dick]

That's a good question. I guess it should. When I simplify a term like $x_2\frac{\partial}{\partial x_3}x_2\frac{\partial}{\partial x_3}$, do I have to use the product rule? That would get the second derivatives.

Yes, you do! I see what you were doing. E.g. d/dx(x*d/dy)f=d/dx(x*df/dy). Sure, it's the derivative of a product. Not the same as (dx/dx)df/dy.

 

[lanedance]

also, hopefully not to distract from teh question at hand, but do you know about index notation with einstein summation convention? Definitely worth learning and wiould simplify this problem

it that notation you could write the problem as:
$$L_i = \epsilon_{ijk}x_j \frac{\partial}{\partial x_k}$$

$$L^2 = L_i-L_j \delta_{ij}= L_iL_i$$

 

[HallsofIvy]

How can the derivative expression for L_1^2 not have any second derivatives in it?

And that is why we always use the "comutator", LM- ML, which will NOT have second differential operators.