As an application of the commutator algebra rules
![{\displaystyle [A,B]=-[B,A]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37b04a066443e96c0b27d4b14f7f5d51c7b2b0c3)
![{\displaystyle [A,BC]=[A,B]C+B[A,C]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c7c96785faa7d6e5ecc7d8ada53830b24338c60)
let us calculate the commutators of the components of the angular momentum of a particle
One has
The other two commutators are calculated by cyclic permutation. Thus
![{\displaystyle [l_{x},l_{y}]=i\hbar l_{z}\,\,\,\,\,\,\,[l_{y},l_{z}]=i\hbar l_{x}\,\,\,\,\,\,\,[l_{z},l_{x}]=i\hbar l_{y}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e0ad03ac0853bfc13c7161b6195c942327ec60d)
The three components of the angular momentum do not commute in pairs. There is no complete orthonormal set common to any two of them. In other words, two components of angular momentum cannot, in general, be defined simultaneously with infinite precision. Note that
Adding term by term, we obtain
![{\displaystyle [l_{z},\mathbf {l} ^{2}]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/404194f3d4a1e29f82f6106a14c4e736cc856f37)
where the operator

is the square of the length of the vector
.
The operators
and
commute: they can therefore be simultaneously defined with infinite preicision. The pairs
and
obviosly possess the same property.