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 operational definition is that $\mathbf J$ generates the corresponding unitary rotations:

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

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

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

For finite spin matrices, or on vectors in the square-generator domain for an unbounded orbital generator, the expansion near the identity is

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

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 determines the measured vector components after actively rotating the state 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

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

For a time-independent $J_i$, this gives conservation. The Hamiltonian itself may depend on time if the same rotation symmetry holds at every instant and the propagator preserves the relevant domains. 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 $J_z$.

The notation $z$ is a coordinate choice. A physical external field can, however, select a preferred axis and make only the component along that axis conserved. A different axis would be described by $J_{\hat{\mathbf n}}=\hat{\mathbf n}\cdot\mathbf J$.

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.

References

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