Commutation relation of angular momentum

[p.p]

Just wondering if any1 would by any chance know how to calculate the commutation relations when L = R x P is the angular momentum operator of a system?
[Li, P.R]=?

P

 

[MathematicalPhysicist]

Is it the dot product?
You can write it down as:
[Li,PiRi] (using Einstein summation notation).
and Li=Rj Pk ejki, where ejik is levi-civita tensor.
and using the basic relation for [R,P]=1 (units of hbar).

 

[White]

$$[L_i,p_ir_i]=p_i[L_i,r_i]+[L_i,p_i]r_i$$

Inserting:
$$L_i=\epsilon_{ijk}r_j p_k$$
Gives:

$$[L_i,p_ir_i]=\epsilon_{ijk}p_i[r_j p_k,r_i]+\epsilon_{ijk}[r_j p_k,p_i]r_i$$
$$=\epsilon_{ijk}p_i r_j[p_k,r_i]+\epsilon_{ijk}[r_j,p_i]r_i p_k$$
$$=-i\hbar\epsilon_{ijk}p_i r_j\delta_{ik}+i\hbar\epsilon_{ijk}\delta_{ij} r_i p_k$$
$$=0$$

 

[dextercioby]

The fact that the commutator of the angular momentum and the scalar product of p and r is zero follows naturally. The definition of a vector operator applies for both p and r individually, thus their scalar product should be a scalar operator, therefore the desired commutator is zero.