Commutation relation of angular momentum

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The discussion focuses on calculating the commutation relations for the angular momentum operator defined as L = R x P. It explores the expression [Li, PiRi] using Einstein summation notation and the Levi-Civita tensor. The basic relation [R, P] = 1 is applied, leading to a detailed calculation that ultimately shows the commutator [L_i, p_ir_i] equals zero. This result is attributed to the scalar nature of the product of p and r, confirming that the commutator of angular momentum with this scalar product is zero. The conclusion emphasizes that both p and r are vector operators, reinforcing the scalar operator property of their product.
p.p
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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
 
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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).
 
[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
 
Last edited:
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.
 

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