Everything about angular momentum follows from three commutators: the commutator of two components is $i\hbar$ times the third, in cyclic order. They are the algebra of rotations — rotating about $x$ then $y$ differs from $y$ then $x$ by a rotation about $z$ — and they are all the rest of the chapter uses. Any three operators obeying them are called an angular momentum, whether or not they are $\hat{\mathbf{r}}\times\hat{\mathbf{p}}$: spin will be one that is not.

The rules of chess do not mention wood or ivory: anything moving by the rules is a chess piece. Any three operators with these commutators are an angular momentum, whatever they are made of.

The components of angular momentum obey cyclic commutation relations, and the square of the vector commutes with each component. From here on $\hat{\mathbf{J}}$ means any three operators that obey them.

• The order matters: $[\hat{J}_{y},\hat{J}_{x}]=-i\hbar\hat{J}_{z}$. The cyclic order is .

Expand the commutator into four terms. Two vanish because every operator in them belongs to a different direction. In each of the other two only one pair fails to commute, a coordinate and its own momentum, and the two surviving pieces make up $\hat{L}_{z}$.

Write out the two components.

$[\hat{y}\hat{p}_{z},\hat{x}\hat{p}_{z}]$ and vanish: all their factors commute.

Only $\hat{p}_{z}$ and $\hat{z}$ fail to commute.

Only $\hat{z}$ and $\hat{p}_{z}$ fail to commute.

Add the two.

The commutator of two components is $i\hbar$ times the third, taken in the cyclic order $x$, $y$, $z$. Reversing the order changes the sign.

Swapping the two operators in a commutator changes its sign.

The square of the angular momentum commutes with every component. The length of the vector and one component can be known together.

From now on, any three Hermitian operators with these commutators are called an angular momentum, whether or not they are built from $\hat{\mathbf{r}}$ and $\hat{\mathbf{p}}$.

Great work! You now know how to:

• derive the commutator of two components of $\hat{\mathbf{L}}$
• use the cyclic relations and their reverses
• explain why $\hat{L}^{2}$ and one component can be known together