PlanetPhysics/Commutation Relations of Angular Momentum

As an application of the [[../CommutatorAlgebra/|commutator algebra]] rules

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

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

let us calculate the [[../Commutator/|commutators]] of the components of the [[../MolecularOrbitals/|angular momentum]] of a [[../Particle/|particle]] $𝐥\equiv 𝐫\times 𝐩$

One has

$[l_x,l_y]=[y{p}_z-z{p}_y,z{p}_x-x{p}_z]$

$[l_x,l_y]=[y{p}_z,z{p}_x]+[z{p}_y,x{p}_z]$

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

$[l_x,l_y]=i\hslash (x{p}_y-y{p}_x)$

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

The other two commutators are calculated by cyclic permutation. Thus

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

The three components of the angular momentum do not [[../Commutator/|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

$[l_z,l_x^2]=i\hslash (l_y{l}_x+l_x{l}_y)$ $[l_z,l_y^2]=-i\hslash (l_y{l}_x+l_x{l}_y)$ $[l_z,l_z^2]=0$

Adding term by term, we obtain

$[l_z,{𝐥}^2]=0$

where the [[../QuantumOperatorAlgebra4/|operator]]

${𝐥}^2=l_x^2+l_y^2+l_z^2$

is the [[../PiecewiseLinear/|square]] of the length of the [[../Vectors/|vector]] $𝐥$.

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

References

[1] Messiah, Albert. "[[../QuantumParadox/|Quantum mechanics]]: [[../Volume/|volume]] I." Amsterdam, North-Holland Pub. Co.; New York, Interscience Publishers, 1961-62.

This entry is a derivative of the Public [[../Bijective/|domain]] [[../Work/|work]] [1].