Angular Momentum Algebra

Angular momentum algebra is the operator algebra of rotation generators. It applies to orbital angular momentum, spin angular momentum, and total angular momentum whenever the components generate rotations.

The defining commutator is

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

This single relation controls noncommuting components, multiplet structure, ladder operators, and the allowed angular momentum labels.

Rotation Generators

A rotation by angle $\theta$ about a unit vector $\hat{\mathbf n}$ is represented by

$$U(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf J \right).$$

The three components $J_x,J_y,J_z$ generate rotations about the corresponding axes. Their noncommutativity reflects the noncommutativity of rotations in three dimensions.

Commutation Relations

The component relations are

$$[J_x,J_y]=i\hbar J_z, \qquad [J_y,J_z]=i\hbar J_x, \qquad [J_z,J_x]=i\hbar J_y.$$

Because the components do not commute, a state cannot generally have sharp values of $J_x$, $J_y$, and $J_z$ at the same time.

Representation Assumptions and Normalization

The familiar discrete labels follow for finite-dimensional unitary irreducible representations of $SU(2)$, or equivalently for self-adjoint generators on an invariant finite-dimensional subspace. More general nonunitary or infinite-dimensional representations of the complexified Lie algebra need not have this spectrum.

Physics convention uses self-adjoint $J_i$ with dimensions of angular momentum. A common mathematics convention uses dimensionless anti-Hermitian generators

$$T_i=-\frac{i}{\hbar}J_i, \qquad [T_i,T_j]=\epsilon_{ijk}T_k.$$

Both describe the same Lie algebra after a change of normalization. They must not be mixed within one calculation.

The Casimir Operator

Define

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

This operator commutes with each angular momentum component:

$$[J^2,J_i]=0.$$

For example,

$$\begin{aligned} [J^2,J_x] &= [J_y^2,J_x]+[J_z^2,J_x] \\ &= -i\hbar(J_yJ_z+J_zJ_y) +i\hbar(J_zJ_y+J_yJ_z) =0, \end{aligned}$$

and cyclic permutations give the other two commutators.

It is the Casimir operator of the angular momentum algebra. Physically, it measures the total angular momentum magnitude while one chosen component, conventionally $J_z$, labels orientation within a multiplet.

Simultaneous Eigenstates

Since $J^2$ and $J_z$ commute, one uses simultaneous eigenstates

$$\lvert j,m\rangle.$$

They are defined by

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

and

$$J_z\lvert j,m\rangle = \hbar m\lvert j,m\rangle.$$

For fixed $j$,

$$m=-j,-j+1,\ldots,j.$$

The number of states in the multiplet is $2j+1$.

Ladder Structure and Allowed Labels

Define

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

The algebra implies

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

and the operator identities

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

If $\lvert j,m\rangle$ is normalized, positivity of the squared norms gives

$$\lVert J_+\lvert j,m\rangle\rVert^2 = \hbar^2[j(j+1)-m(m+1)] \ge0,$$ $$\lVert J_-\lvert j,m\rangle\rVert^2 = \hbar^2[j(j+1)-m(m-1)] \ge0.$$

A finite multiplet therefore has a highest weight $m_{\max}=j$ and a lowest weight $m_{\min}=-j$. Since neighboring weights differ by one, $2j$ is a nonnegative integer:

$$j=0,\frac12,1,\frac32,\ldots.$$

Choosing phases consistently yields

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

Matrix Construction and Small Representations

Order the basis as $\lvert j,j\rangle,\lvert j,j-1\rangle,\ldots,\lvert j,-j\rangle$. Then $J_z$ is diagonal, $J_+$ has only one adjacent off-diagonal band fixed by the ladder coefficient, $J_-=J_+^\dagger$, and

$$J_x=\frac{J_++J_-}{2}, \qquad J_y=\frac{J_+-J_-}{2i}.$$

For $j=\tfrac12$,

$$J_i=\frac{\hbar}{2}\sigma_i,$$

where $\sigma_i$ are the Pauli matrices. For $j=1$,

$$J_z = \hbar \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&-1 \end{pmatrix}, \qquad J_+ = \hbar\sqrt2 \begin{pmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{pmatrix}.$$

These matrices are representations of the same abstract algebra, not spatial vectors whose entries are $x$, $y$, and $z$ components.

Reducible Representations and Multiplicity

A general unitary representation can decompose as

$$\mathcal H \cong \bigoplus_j \left( \mathbb C^{n_j}\otimes V_j \right),$$

where $V_j$ is the spin-$j$ irrep and $n_j$ is its multiplicity. The pair $(j,m)$ does not distinguish multiple copies; an additional label is required. Such multiplicities appear naturally when several angular momenta are coupled.

Component Fluctuations

In $\lvert j,m\rangle$,

$$\langle J_x\rangle = \langle J_y\rangle =0,$$

and rotational symmetry about the $z$ axis gives

$$(\Delta J_x)^2 =(\Delta J_y)^2 = \frac{\hbar^2}{2} \left[j(j+1)-m^2\right].$$

Thus a sharp value of $J_z$ does not imply vanishing transverse angular momentum or transverse uncertainty.

Why Only One Component Is Chosen

Because

$$[J_x,J_y]\ne0,$$

one cannot generally diagonalize all three components simultaneously. The standard convention chooses $J_z$ together with $J^2$. Another axis could be chosen, but one must choose only one component at a time.

This is not a limitation of notation. It is physical noncommutativity.

Units and Dimensions

The operators $J_i$ have units of angular momentum. The quantum numbers $j$ and $m$ are dimensionless. Keeping $\hbar$ explicit helps distinguish the operator eigenvalues from the labels:

$$J_z\lvert j,m\rangle = \hbar m\lvert j,m\rangle.$$

Orbital, Spin, and Total Angular Momentum

The same algebra can be realized in different ways:

• orbital angular momentum $\mathbf L=\mathbf R\times\mathbf P$,
• spin angular momentum $\mathbf S$ acting on internal spin states,
• total angular momentum $\mathbf J=\mathbf L+\mathbf S$ or sums over subsystems.

The algebra is shared, but the physical meaning of the Hilbert space differs.

Common Mistakes

• Confusing $j$ with $m$.
• Treating angular momentum states as ordinary spatial vectors.
• Forgetting that $J_x$, $J_y$, and $J_z$ cannot all be sharp in general.
• Dropping $\hbar$ from eigenvalues while keeping it elsewhere.
• Assuming orbital and spin angular momentum have identical physical origins because they obey the same algebra.

Cross-Links

• Rotations Preview
• Rotations in Three Dimensions
• SO(3) and SU(2) Preview
• Angular Momentum Operators
• Generators
• Ladder Operators
• Eigenvalues of J² and Jz
• Orbital Angular Momentum
• Spin as Intrinsic Angular Momentum
• Total Angular Momentum
• Spherical Harmonics
• Complete Sets of Commuting Observables
• Toolkit Angular Momentum Algebra
• SU(2)
• Angular Momentum Formula Card
• From Angular Momentum to Helicity

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• B. C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, 2nd ed., Springer, 2015.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.

Exercises

1. For $j=2$, list all possible $m$ values and count the number of states.

Solution

The values are

$$m=-2,-1,0,1,2.$$

There are $2j+1=5$ states.

2. Prove $[J^2,J_z]=0$ directly from the component commutators.

Solution

Using $[AB,C]=A[B,C]+[A,C]B$,

$$[J_x^2,J_z] = -i\hbar(J_xJ_y+J_yJ_x),$$

while

$$[J_y^2,J_z] = i\hbar(J_yJ_x+J_xJ_y).$$

The terms cancel, and $[J_z^2,J_z]=0$.

3. Starting from $J_\mp J_\pm=J^2-J_z^2\mp\hbar J_z$, derive the coefficient multiplying $\lvert j,m\pm1\rangle$.

Solution

Write $J_+\lvert j,m\rangle=c_+\lvert j,m+1\rangle$. Then

$$|c_+|^2 = \langle j,m\rvert J_-J_+\lvert j,m\rangle = \hbar^2[j(j+1)-m(m+1)].$$

The standard phase convention takes the positive square root. The lowering coefficient follows analogously.

4. Use termination of a finite ladder to show that $2j$ is an integer and the multiplet has $2j+1$ states.

Solution

The highest and lowest weights satisfy $m_{\max}=j$ and $m_{\min}=-j$. Repeated lowering changes $m$ by one, so $j-(-j)=2j$ must be a nonnegative integer. Counting both endpoints gives $2j+1$ weights.

5. Construct $J_x$ for $j=1$ from the displayed $J_+$ and verify $\langle1,0\rvert J_x^2\lvert1,0\rangle=\hbar^2$.

Solution

Because $J_-=J_+^\dagger$,

$$J_x = \frac{\hbar}{\sqrt2} \begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix}.$$

Acting twice on the middle basis vector gives $J_x^2\lvert1,0\rangle=\hbar^2\lvert1,0\rangle$, so the stated expectation value follows.

6. Explain why $J^2$ and $J_z$ can label the same state but $J_x$ and $J_z$ generally cannot.

Solution

$J^2$ commutes with $J_z$, so simultaneous eigenstates can be chosen. But $[J_x,J_z]=-i\hbar J_y$, which is generally nonzero, so $J_x$ and $J_z$ cannot generally be simultaneously diagonalized.