Combine the two components that are not measured into $\hat{J}_{+}=\hat{J}_{x}+i\hat{J}_{y}$ and $\hat{J}_{-}=\hat{J}_{x}-i\hat{J}_{y}$. Their commutators with $\hat{J}_{z}$ make them step its eigenvalue up and down by exactly $\hbar$, while leaving $\hat{J}^{2}$ alone. So the states of given $\hat{J}^{2}$ form a ladder of $\hat{J}_{z}$ values one $\hbar$ apart, and the ladder must stop at both ends, because a component cannot exceed the length of the vector.

On a dial with fixed length, the needle can point at different heights but can never point higher than its own length. The ladder operators move the height by one notch; the ends of the dial stop them.

The five states of $j=2$, one for each $m$ from $-2$ to $2$, in units of $\hbar$. $$ moves a state one rung up, one rung down, and is the same on every rung. On the top rung gives zero; on the bottom one does.

$\hat{J}_{\pm}=\hat{J}_{x}\pm i\hat{J}_{y}$ are adjoints of each other. They commute with and step the eigenvalue of by .

• If $\hat{J}_{z}\ket{m}=m\hbar\ket{m}$, then $$: raising adds to , unless it gives zero.

Compute the commutator from the cyclic relations, then move $\hat{J}_{z}$ past $\hat{J}_{\pm}$ in : it picks up the commutator, which adds or removes one .

The commutator is linear.

The cyclic relations.

Take out $\pm\hbar$.

Move $\hat{J}_{z}$ to the right, using the commutator.

The raising operator adds $\hbar$ to the $z$ component; the lowering operator removes one. The total $\hat{J}^{2}$ is untouched.

The raising and lowering operators are each other’s adjoints, since $\hat{J}_{x}$ and $\hat{J}_{y}$ are Hermitian.

The ladder operators commute with $\hat{J}^{2}$: every state of a ladder has the same total angular momentum.

The commutator of the raising and lowering operators is $2\hbar\hat{J}_{z}$.

Great work! You now know how to:

• define the raising and lowering operators of angular momentum
• prove that they step $\hat{J}_{z}$ by one unit of $\hbar$
• rewrite $\hat{J}^{2}$ through and