Angular Momentum Operators

Angular momentum operators are the self-adjoint generators of rotations on a quantum Hilbert space. The symbol $\mathbf J=(J_x,J_y,J_z)$ denotes the generator appropriate to the system being rotated. It may be orbital angular momentum, spin angular momentum, or a total angular momentum that contains both.

The point of this page is operational: to say that $\mathbf J$ is angular momentum means that rotations are implemented by unitary operators of the form

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

The algebraic consequences of this definition are developed in Angular Momentum Algebra. The concrete orbital realization $\mathbf L=\mathbf R\times\mathbf P$ belongs to Orbital Angular Momentum.

Generator Definition

For a rotation by angle $\theta$ about a unit vector $\hat{\mathbf n}$, the one-parameter unitary family is

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

Near the identity,

$$U_{\hat{\mathbf n}}(\theta) = I - \frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf J + O(\theta^2).$$

Thus the generator is the coefficient of the infinitesimal rotation:

$$\hat{\mathbf n}\cdot\mathbf J = i\hbar \left. \frac{dU_{\hat{\mathbf n}}(\theta)}{d\theta} \right|_{\theta=0}.$$

This is the rotation-specific version of the general generator idea. The parameter $\theta$ is dimensionless, so the generator has units of angular momentum.

Components and Axes

The three standard components generate rotations about the coordinate axes:

$$U_x(\theta) = \exp\left( -\frac{i}{\hbar}\theta J_x \right),$$ $$U_y(\theta) = \exp\left( -\frac{i}{\hbar}\theta J_y \right),$$

and

$$U_z(\theta) = \exp\left( -\frac{i}{\hbar}\theta J_z \right).$$

For a general axis,

$$J_{\hat{\mathbf n}} = \hat{\mathbf n}\cdot\mathbf J = n_xJ_x+n_yJ_y+n_zJ_z.$$

The notation $\mathbf J$ is not saying that the components commute like ordinary coordinates. It says that the three components transform as a vector under rotations and generate rotations about the corresponding axes. The noncommutativity of rotations appears as noncommutativity among $J_x,J_y,J_z$.

Vector Operators

An operator-valued vector $\mathbf V=(V_x,V_y,V_z)$ transforms like an ordinary vector if

$$U(\mathcal R)^\dagger V_i U(\mathcal R) = \sum_j \mathcal R_{ij}V_j.$$

For infinitesimal rotations, this is equivalent to the commutator condition

$$[J_i,V_j] = i\hbar\sum_k\epsilon_{ijk}V_k.$$

For example,

$$[J_z,V_x]=i\hbar V_y, \qquad [J_z,V_y]=-i\hbar V_x, \qquad [J_z,V_z]=0.$$

Using

$$U_z(\theta)^\dagger V_x U_z(\theta) = V_x + \frac{i\theta}{\hbar}[J_z,V_x] + O(\theta^2),$$

one obtains

$$V_x\mapsto V_x-\theta V_y, \qquad V_y\mapsto V_y+\theta V_x,$$

which is the infinitesimal form of an active rotation about the $z$ axis in the conventions of Rotations in Three Dimensions.

Scalar operators are different. A rotational scalar $A$ satisfies

$$U(\mathcal R)^\dagger A U(\mathcal R)=A,$$

and infinitesimally

$$[J_i,A]=0.$$

This distinction between scalar, vector, and higher-rank tensor operators is the entry point to Irreducible Spherical Tensors.

Rotational Invariance

A Hamiltonian is rotationally invariant when it is unchanged under the unitary representation of rotations:

$$U(\mathcal R)^\dagger H U(\mathcal R) = H$$

for every rotation in the symmetry group. For rotations about a fixed axis, this condition becomes

$$U_i(\theta)^\dagger H U_i(\theta)=H.$$

Differentiating at $\theta=0$ gives

$$[H,J_i]=0.$$

When $H$ has no explicit time dependence, $J_i$ is then conserved. If the Hamiltonian is invariant under all rotations, all three components generate symmetries, and $J^2$ and one chosen component are natural labels.

Orbital Realization

For a spinless particle in three-dimensional space, the rotation generator is the orbital angular momentum

$$\mathbf J=\mathbf L=\mathbf R\times\mathbf P.$$

Its components are

$$L_i = \sum_{j,k}\epsilon_{ijk}R_jP_k.$$

In position space, with $\mathbf P=-i\hbar\nabla$, these are differential operators that rotate the angular dependence of the wavefunction. The full position-space formulas and domain cautions are handled in Position-Space Representation.

For a scalar wavefunction,

$$(U(\mathcal R)\psi)(\mathbf r) = \psi(\mathcal R^{-1}\mathbf r).$$

The inverse argument and the generator $\mathbf L$ are two descriptions of the same rotation action: one finite and geometric, the other infinitesimal and operator-theoretic.

Spin and Total Angular Momentum

Spin is angular momentum because it also generates rotations, but it does not come from $\mathbf R\times\mathbf P$. For spin-$1/2$,

$$\mathbf S = \frac{\hbar}{2}\boldsymbol\sigma,$$

and

$$U(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf S \right) = \exp\left( -\frac{i}{2}\theta\,\hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

The spinor-specific meaning of this formula is developed in Spin Rotations and SO(3) and SU(2) Preview.

For a particle with both position and spin degrees of freedom, the total generator is

$$\mathbf J=\mathbf L+\mathbf S.$$

For several subsystems, total angular momentum is the sum of the generators acting on the tensor-product factors:

$$\mathbf J_{\rm tot} = \mathbf J_1+\mathbf J_2+\cdots.$$

The addition and change-of-basis machinery begins later in Coupled and Uncoupled Bases.

What J Is Not

The operator $\mathbf J$ is not always $\mathbf R\times\mathbf P$. That formula defines orbital angular momentum. Spin systems and internal multiplets can carry angular momentum without a literal spatial orbit.

The operator $\mathbf J$ is also not a classical vector with simultaneously sharp components. In a generic state, $J_x$, $J_y$, and $J_z$ cannot all have definite values. The standard strategy is to diagonalize $J^2$ and one component, usually $J_z$.

Finally, $J_z$ is not special in nature. It is special only because a quantization axis has been chosen. A different axis would be described by $J_{\hat{\mathbf n}}=\hat{\mathbf n}\cdot\mathbf J$.

Common Mistakes

• Defining all angular momentum as $\mathbf R\times\mathbf P$ and then trying to force spin into that formula.
• Forgetting the factor of $\hbar$ in the rotation exponential.
• Confusing the active rotation of vector operators with a passive coordinate change.
• Treating the three components of $\mathbf J$ as simultaneously measurable ordinary components.
• Using $\mathbf L$ as the rotation generator for a spinful particle when the correct total generator is $\mathbf J=\mathbf L+\mathbf S$.
• Assuming $[H,J_z]=0$ implies full rotational invariance; it only proves symmetry about the chosen axis.

Cross-Links

• Rotations in Three Dimensions
• SO(3) and SU(2) Preview
• Generators
• Angular Momentum Algebra
• Ladder Operators
• Eigenvalues of J² and Jz
• Orbital Angular Momentum
• Position-Space Representation
• Spin Rotations
• Pauli Matrices
• Irreducible Spherical Tensors
• SO(3)
• SU(2)
• Wigner D-Matrices
• Angular Momentum Formula Card

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.

Exercises

1. Starting from $U_{\hat{\mathbf n}}(\theta)=\exp[-i\theta\,\hat{\mathbf n}\cdot\mathbf J/\hbar]$, show that $\hat{\mathbf n}\cdot\mathbf J=i\hbar\,dU_{\hat{\mathbf n}}/d\theta|_{\theta=0}$.

Solution

Expand the exponential near $\theta=0$:

$$U_{\hat{\mathbf n}}(\theta) = I - \frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf J + O(\theta^2).$$

Therefore

$$\left. \frac{dU_{\hat{\mathbf n}}}{d\theta} \right|_{\theta=0} = - \frac{i}{\hbar}\hat{\mathbf n}\cdot\mathbf J.$$

Multiplying by $i\hbar$ gives

$$i\hbar \left. \frac{dU_{\hat{\mathbf n}}}{d\theta} \right|_{\theta=0} = \hat{\mathbf n}\cdot\mathbf J.$$

2. Use the vector-operator commutators with $J_z$ to find the infinitesimal rotation of $V_x$ and $V_y$.

Solution

For a small $z$ rotation,

$$U_z(\theta)^\dagger V_i U_z(\theta) = V_i + \frac{i\theta}{\hbar}[J_z,V_i] + O(\theta^2).$$

Using

$$[J_z,V_x]=i\hbar V_y, \qquad [J_z,V_y]=-i\hbar V_x,$$

gives

$$V_x\mapsto V_x-\theta V_y, \qquad V_y\mapsto V_y+\theta V_x,$$

to first order in $\theta$.

3. Show that rotational invariance about the $i$ axis implies $[H,J_i]=0$.

Solution

Rotational invariance about the axis means

$$U_i(\theta)^\dagger H U_i(\theta)=H.$$

Differentiate at $\theta=0$. Since

$$\left.\frac{dU_i}{d\theta}\right|_{\theta=0} = - \frac{i}{\hbar}J_i, \qquad \left.\frac{dU_i^\dagger}{d\theta}\right|_{\theta=0} = \frac{i}{\hbar}J_i,$$

the derivative gives

$$\frac{i}{\hbar}J_iH - \frac{i}{\hbar}HJ_i = 0.$$

Thus

$$[J_i,H]=0,$$

equivalently $[H,J_i]=0$.

4. A spin-$1/2$ particle moves in three-dimensional space. Which generator rotates only its spatial wavefunction, which generator rotates only its spinor, and which generator rotates the full physical state?

Solution

The orbital generator $\mathbf L$ rotates only the spatial dependence. The spin generator $\mathbf S$ rotates only the spinor index. The full rotation generator is

$$\mathbf J=\mathbf L+\mathbf S.$$

It rotates the complete state, including both position dependence and spin components.