# Simple check for operaters communting, think im missing something very obvious!

1. May 28, 2009

### hoch449

1. The problem statement, all variables and given/known data

I am supposed to find if the following commutes: [Lx,Ly]

2. Relevant equations

Lx= -i$$\hbar$$[y($$\partial/$$$$\partial$$z) - z($$\partial/$$$$\partial$$y)]

Ly= -i$$\hbar$$[z($$\partial/$$$$\partial$$x) - x($$\partial/$$$$\partial$$z)]

where [Lx,Ly]=LxLy-LyLx

If it commutes then [Lx,Ly]=0

3. The attempt at a solution

[Lx,Ly]= (i$$\hbar$$)2{[y($$\partial/$$$$\partial$$z) - z($$\partial/$$$$\partial$$y)[z($$\partial/$$$$\partial$$x) - x($$\partial/$$$$\partial$$z)]}

After expanding this I got a result of 0. So my solution concluded that they commute.

The answer however is [Lx,Ly]= i$$\hbar$$Lx

I clearly expanded it wrong. I was hoping if anyone could explain how they expanded the LxLy-LyLx part. In my calculations I cannot seem to figure out how the answer contains a few more parts in the expansion which results in a non-commutation..

Thanks!

2. May 28, 2009

### George Jones

Staff Emeritus
I think you mean [Lx,Ly]= i$$\hbar$$Lz.

Hint: Consider $\left( L_x L_y - L_y L_x \right)f$, where $f$ is an arbitrary function of the spatial coordinates. Take things one step at a time, don't forget about the product rule for differentiation, and remove $f$ at the end.

3. May 28, 2009

### xepma

Keep in mind that you're working with operators, so it's not just a matter of simple expansion. To calculate such a commutator, let the whole thing act on a function.

$$[L_x,L_y]f(x,y,z)$$

Since you have all sort of derivatives you will find that the product comes into a play a number of times. For instance, one term gives:
$$y\frac{\partial}{\partial z} \left( z\frac{\partial}{\partial x}\right) f(x,y,z) = y \frac{\partial}{\partial x} f(x,y,z) + yz \frac{\partial^2}{\partial x\partialz}f(x,y,z)$$

One of these terms will be canceled by some other term, while the other one survives.

EDIT: OK, George beat me to it ;)

4. May 28, 2009

### Cyosis

Another way to do it is to use the two following commutator identities:

\begin{align*} &[A+B,C]=[A,C]+[B,C] \\ &[A,BC]=[A,B]C+B[A,C] \end{align*}

5. May 28, 2009

### hoch449

Perfect it all makes sense now! You guys are great thanks!