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

$[l_{z},l_{x}^{2}]=i\hbar (l_{y}l_{x}+l_{x}l_{y})$ $[l_{z},l_{y}^{2}]=-i\hbar (l_{y}l_{x}+l_{x}l_{y})$ $[l_{z},l_{z}^{2}]=0$

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.

References

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

This entry is a derivative of the Public domain work [1].