PlanetPhysics/Commutation Relations of Angular Momentum

As an application of the commutator algebra rules

$$ [A,B] = -[B,A] $$

$$ [A,BC] = [A,B]C +B[A,C] $$

let us calculate the commutators of the components of the angular momentum of a particle $$\mathbf{l} \equiv \mathbf{r} \times \mathbf{p} $$

One has

$$[l_x,l_y] = [yp_z - zp_y, zp_x - xp_z]$$

$$[l_x,l_y] = [yp_z,zp_x] + [zp_y,xp_z]$$

$$[l_x,l_y] = y[p_z,z]p_x + p_y[z,p_z]x$$

$$[l_x,l_y] = i\hbar(xp_y-yp_x)$$

$$[l_x,l_y] = i \hbar l_z$$

The other two commutators are calculated by cyclic permutation. Thus

$$ [l_x,l_y] = i \hbar l_z \,\,\,\,\,\,\, [l_y,l_z]=i\hbar l_x \,\,\,\,\,\,\, [l_z,l_x]=i \hbar l_y $$

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

$$[l_z,l_x^2] = i\hbar(l_yl_x+l_xl_y)$$ $$[l_z,l_y^2] = -i\hbar(l_yl_x+l_xl_y)$$ $$[l_z,l_z^2] = 0$$

Adding term by term, we obtain

$$ [l_z, \mathbf{l}^2] = 0 $$

where the operator $$ \mathbf{l}^2 = l_x^2 + l_y^2 +l_z^2 $$

is the square of the length of the vector $$\mathbf{l}$$.

The operators $$\mathbf{l}^2$$ and $$l_z$$ commute: they can therefore be simultaneously defined with infinite preicision. The pairs $$(\mathbf{l}^2,l_x)$$ and $$(\mathbf{l}^2,l_y)$$ obviosly possess the same property.