How can I simplify commutators with a useful trick?

[Fys]

\left[L_{x},L_{y}\right]=\left[yp_{z}-zp_{y},zp_{x}-xp_{z}\right]<br /> =\left[yp_{z},zp_{x}\right]-\left[zp_{y},zp_{x}\right]-\left[yp_{z},xp_{z}\right]+\left[zp_{y},xp_{z}\right]<br />

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

 

[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

 

[Fys]

Yes thanks
but how can i work this out?

 

[malawi_glenn]

Yes thanks
but how can i work this out?

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 ;)

 

[Fys]

oke thanks :P

 

[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].