Angular Momentum as a Lie Algebra

Angular momentum uses the Lie algebra of rotations to organize a vector space into finite multiplets. The two labels $j$ and $m$ identify, respectively, the multiplet and a vector within it. This primer connects the representation language to the formulas; Angular Momentum Algebra gives the full spectrum construction and physical applications.

Helpful background. Commutators describe the algebraic operation, while representations explain invariant subspaces.

The Hermitian generators $J_i$ of rotations satisfy

$$[J_i,J_j]=i\hbar\epsilon_{ijk}J_k.$$

Mathematics texts often use dimensionless anti-Hermitian generators $T_i=-iJ_i/\hbar$, for which $[T_i,T_j]=\epsilon_{ijk}T_k$. Mixing these normalizations is a common source of missing factors of $i$ and $\hbar$.

Casimir operators and angular-momentum labels

The quadratic Casimir

$$J^2=J_x^2+J_y^2+J_z^2$$

commutes with every generator. In a finite-dimensional unitary irreducible representation of $\mathfrak{su}(2)$, one may choose

$$J^2\lvert j,m\rangle = \hbar^2j(j+1)\lvert j,m\rangle, \qquad J_z\lvert j,m\rangle = \hbar m\lvert j,m\rangle,$$

with $j=0,\tfrac12,1,\ldots$ and $m=-j,-j+1,\ldots,j$. Half-integer representations are representations of $SU(2)$ but do not descend to single-valued representations of $SO(3)$.

Ladder Identities

Define $J_\pm=J_x\pm iJ_y$. Then

$$[J_z,J_\pm]=\pm\hbar J_\pm, \qquad [J_+,J_-]=2\hbar J_z,$$

and

$$J_\mp J_\pm = J^2-J_z^2\mp\hbar J_z.$$

With the standard phase convention,

$$J_\pm\lvert j,m\rangle = \hbar\sqrt{j(j+1)-m(m\pm1)} \lvert j,m\pm1\rangle.$$

These identities are the minimal algebraic input for the physical ladder construction and for tensor-product coupling.

Reading the labels

An irreducible multiplet is an invariant subspace with no smaller nonzero invariant subspace. The operator $J^2$ has the same eigenvalue on every vector in that multiplet. Choosing the $z$ axis then selects a basis in which $J_z$ is diagonal. Thus $j$ describes the whole rotation representation, while $m$ describes a component relative to the chosen axis.

For $j=1/2$, the representation has two basis vectors. In the order $m=1/2,-1/2$,

$$J_z=\frac{\hbar}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad J_+=\hbar\begin{pmatrix}0&1\\0&0\end{pmatrix}.$$

The raising operator maps the lower basis vector to the upper one and annihilates the upper one. Rotating a state generally produces a superposition of the $m$ basis vectors; it does not change $j$ within this irreducible space.

Exercise

For $j=1$, use the ladder formula to find the nonzero entries of $J_+$ in the ordered basis $m=1,0,-1$. Why is $J_+$ not Hermitian?

Solution

The two allowed upward steps have coefficient $\hbar\sqrt2$, while the step from $m=1$ vanishes. Hence

$$J_+=\hbar\begin{pmatrix}0&\sqrt2&0\\0&0&\sqrt2\\0&0&0\end{pmatrix}.$$

Its adjoint is the lowering operator $J_-$, which has the entries below the diagonal. The Hermitian components are $J_x=(J_++J_-)/2$ and $J_y=(J_+-J_-)/(2i)$.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957 — angular-momentum basis and ladder conventions.
• B. C. Hall, Lie Groups, Lie Algebras, and Representations, 2nd ed., Springer, 2015 — finite-dimensional representations of $\mathfrak{su}(2)$.