# Trick for commutators problem

1. Jan 30, 2008

### Fys

$$\left[L_{x},L_{y}\right]=\left[yp_{z}-zp_{y},zp_{x}-xp_{z}\right] =\left[yp_{z},zp_{x}\right]-\left[zp_{y},zp_{x}\right]-\left[yp_{z},xp_{z}\right]+\left[zp_{y},xp_{z}\right]$$

How next?

My book is not of much help
I Tried
$$\left[A,BC\right]=\left[A,B\right]C+B\left[A,C\right]$$

But that is too much work
Does anybody knows a useful trick or something
not only for this case, but for al commutators

Thanks guys

2. Jan 30, 2008

### malawi_glenn

Sometimes you must do the hard work. It depends from case to case regarding what trick you can use, no gereneral rule.

But the general formula for angular momenta commutators is:
$$[J_i , J_j ] = i\hbar \epsilon _{ijk} J_k$$

3. Jan 30, 2008

### Fys

Yes thanks
but how can i work this out?

4. Jan 30, 2008

### malawi_glenn

by working out all partial commutators and using the commutator between position and momentumoperators. Or starting from group theory of rotations.. but I assume this is introductory QM course, so start with the first option ;)

$$\left[yp_{z},zp_{x}\right]-\left[zp_{y},zp_{x}\right]-\left[yp_{z},xp_{z}\right]+\left[zp_{y},xp_{z}\right]$$

Start with $$\left[A,BC\right]=\left[A,B\right]C+B\left[A,C\right]$$
a couple of times til you get things like [x,p_y] ; [y,z] etc

I got the same task in my intro QM course, took me a couple of hours ;)

5. Jan 30, 2008

### Fys

oke thanks :P

6. Jan 30, 2008

### Avodyne

Easier: first work out [L_i,x_j] and [L_i,p_j] using the explicit form of L_i. Then use the [A,BC] formula to get [L_i, x_j p_k].